### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Low Noise Elec Sys Dsgn ECE 6416

GPA 3.64

### View Full Document

## 6

## 0

## Popular in Course

## Popular in ELECTRICAL AND COMPUTER ENGINEERING

This 0 page Class Notes was uploaded by Cassidy Effertz on Monday November 2, 2015. The Class Notes belongs to ECE 6416 at Georgia Institute of Technology - Main Campus taught by William Leach in Fall. Since its upload, it has received 6 views. For similar materials see /class/233862/ece-6416-georgia-institute-of-technology-main-campus in ELECTRICAL AND COMPUTER ENGINEERING at Georgia Institute of Technology - Main Campus.

## Reviews for Low Noise Elec Sys Dsgn

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 11/02/15

Chapter 6 Noise Speci cations 61 Signalto Noise Ratio The signaletoenoise ratio is usually measured at the output of an ampli er where the signal and noise voltages are larger and easier to measure It is given by NR Ugo113w where 1130 is the meanesquare signal output voltage and 112 is the meanesquare noise output voltage It is usually speci ed in dB by the relation 10 log 11301150 When calculating the SNR with the W 7 In ampli er noise model it is convenient to make the calculation at the ampli er input When the source is modeled by a Th venin source the SNR is given by SNR 113115 where 113 is the meanesquare source voltage and 112 is the meanesquare equivalent input noise voltage When it is modeled by a Norton source it is given by SNR where is the meanesquare source current and is the meanesquare equivalent input noise current Expressions are derived below for the SNR for both cases The source impedance and admittance which maximizes the SNR are also derived 6 1 1 Th venin Source When the source is modeled by a Th venin equivalent circuit as in Fig 61 the signaletoenoise ratio is given by SNR 113115 When Eq 43 is used for 115 it follows that the SNR is given by 3 U3 vi 4kTR5Af U 2w Re 72 2393 lZSlZ 61 where ZS R5 jXS and y YT It is expressed in dB by the relation 1010g It is maximized by minimizing 112 The source impedance which minimizes 11 can be obtained by setting avid8R5 0 and BuiiBXS 0 and solving for R5 and X5 The solution for R5 is negative Because this is not realizable R5 0 is the realizable solution for the least noise The source impedance which 2 is given by minimizes Um u ZSRSXS0Wquot 62 TL 115 116 Copyright 1999 7 2009 by W Marshall Leach Jr Vn 1i Amplifier Z0 Figure 61 Ampli er With Th venin source Because minimum noise occurs for R5 0 it can be concluded that a resistor should never be connected in series With a source at an ampli er input if noise performance is a design criterion If a series resistor is required eg for stability it should be much smaller than Rs Although the output impedance of a source is usually xed the SNR can be improved by adding a reactance X1 in series With the source Which makes the total series reactance equal to the imaginary part of ZS in Eq 62 Its value is X1 71mZ5 7 Won2 When this is the case 112 is given by U5 4kTR5A f u 1 7 7 QVTRSW39H iiRi 63 612 Norton Source When the source is modeled by a Norton equivalent circuit as in Fig 62 the signal7to7noise ratio is given by SNR When Eq 424 is used for 2 it follows that the SNR is given by 392 392 25 i 2 4kTG5Afule5l2 S N R 5 I I 64 2on2 Re 7Y5 2 Where Y5 G5 jBS and 39y 7 It is expressed in dB by the relation 10log It is maximized by minimizing The source admittance Which minimizes can be obtained by setting BiiiBGS 0 and BiiiBBS 0 and solving for G5 and B5 The solution for G5 is negative Because this is not realizable G5 0 is the realizable solution for the least noise The source admittance Which minimizes is given by Y5GsBs0 YiU n n Because minimum noise occurs for G5 0 it can be concluded that a resistor should never be connected in parallel With a source at an ampli er input if noise performance is a design criterion If a parallel resistor is required eg as part of a bias network it should be much larger than 1G5 Although the output admittance of a source is usually xed the SNR can be improved by adding a susceptance Bl in parallel With the source Which makes the total parallel susceptance equal to the imaginary part of Y5 in Eq 65 Its value is Bl 71m Y5 Winu When this is the case 2392 is given by 7 TH 4kTG5Af Ugag may if 1 7 72 66 62 NOISE FACTOR AND NOISE FIGURE 117 Is Figure 62 Ampli er With Norton source Vn 1i Amplifier 20 62 Noise Factor and Noise Figure The noise factor F of an ampli er is de ned as the ratio of its actual SNR and the SNR if the ampli er is noiseless Where the temperature is taken to be the standard temperature To When it is expressed in dB it is called noise gure and is given by NF 10log In this section the noise factor is derived for an ampli er driven by a Th venin source and by a Norton source Often it is convenient to express F in terms of the ampli er noise resistance Rquot and noise conductance G de ned in Eqs 218 and 219 and the correlation impedance Z and correlation admittance YA de ned in Eqs 312 and 313 These are related to 112 2 and 39y by U2 TL 6 Rquot 4kT0Af 7 G i 2 6 g quot 4kT0Af 39 39U U ZRYJX vi w wi 69 Y7G7B39y7 11 77quot37139 63910 TL TL Note that Rquot and G respectively represent normalized values of 112 and ii Where the normalization factor is 4kT0Af 62 1 Th venin Source Consider the ampli er model in Fig 61 If the ampli er is noiseless the signalitoi noise ratio given by SNR iii11 Where 113 is the meanisquare source voltage and 1125 is the meanisquare thermal noise voltage generated by the source impedance The noise factor F is obtained by dividing the noiseless ampli er SNR by Eq 61 to obtain F US115 1 115 Qunin Re 7Z9 lZsl2 611 115 uts 4kT0R5Af It follows from this expression that a noiseless ampli er has the noise factor F 1 An alternate expression for F is obtained When the ampli er noise parameters are 118 Copyright 1999 7 2009 by W Marshall Leach Jr expressed in terms of Rn7 G and Z7 It is Rquot 2G ReZAZ on lZSlZ 1 T R S F 612 Optimum Source Impedance The value of ZS which minimizes F is called the optimum source impedance and is denoted by Zapt It is obtained by setting BFBRS 0 and BFBXS 0 and solving for R5 and X5 The impedance is given by Zapt Raptanpt u Rn 177127339YiinG 7X273X7 613 TL TL Note that the imaginary part of Zap is equal to the imaginary part of ZS in Eq 62 which maximizes the signal7to7noise ratio The corresponding minimum value of the noise factor is called the optimum noise factor and is given by u 239 Fm 7 1 w 1 7 v 7 1 2G Ry Ram 614 Relating F to Fmquot and the Source Impedance We next wish to express F in terms of me and Zapt It follows from Eqs 612 and 614 that the difference F 7 me is given by m 2G EYES XVXS an 12512 F7 RS 726quot Ry Ram m 7 2G RmRs Xme an 12512 615 S where X7 7X0pt has been used The square in the numerator of this expres7 sion can be completed by adding and subtracting the term G REP Xgpt G lZaptlZ This leads to the equation Rquot Gquot 115 7 Ram X5 7 X01712 712017112 1 RS F 7 me 616 From Eq 6137 we have lZaptlZ It follows that F can be written F 7 F Gquot R R 2 X X 2 i 5 017057 apt 7 Ga 2 7 me R 1Z5 7 Z0171 617 5 62 NOISE FACTOR AND NOISE FIGURE 119 U in and 39y as Functions of me and Zap If G me and Zap Rap jX0pt are given for an ampli er7 it follows from Eqs 68 and 613 that 2393 4kTGnAf 618 LOW 619 1 RomX01002 X 2 u 239 620 7139 QkT A YT O f 7 1 7 1 7 unzn where sgn Xapt Xopt lXUPtl39 F a Function of Re ection Coe icients The noise factor can also be expressed as a function of the re ection coef cients of the source and the ampli er input Imagine a zero length transmission line having a characteristic impedance ZC connecting the source to the ampli er In Eq 6177 let Z5 21 622 gm 623 R Z5 2 624 Where the re ection coef cients T5 and TOP are given by S l 625 Top 626 When the expressions for ZS7 Zapt and R5 are substituted into Eq 6177 the expression for F reduces to 4anc r5 7 ram 1 7 ram 1 7 W Example 1 Calculate F and NF for the ampli er in Example 1 in Chapter 4 for which RS 759 and um 569 nV Assume Af 1 Hz and T T0 290K F me 627 Solution The meanesquare thermal noise voltage of the source is 1125 4kTR5 120 X 10 18V2 Thus the noise factor and noise gure are u 569 X 10quot2 7 F 7 1125 12 X 10718 270 NF 1010g270 143dB 120 Copyright 1999 7 2009 by W Marshall Leach Jr 622 Norton Source Consider the ampli er model in Fig 62 If the ampli er is noiseless7 the signalitoi noise ratio given by SNR 232225 Where is the meanisquare source current and ii is the meanisquare thermal noise current generated by the source impedance The noise factor F is obtained by dividing the noiseless ampli er SNR by Eq 64 to obtain 23922392 2392 2y22u Re Y 392 F sis s nn 39Ys2n 225 4kT0G5Af 628 An alternate expression for F is obtained when the ampli er noise parameters are expressed in terms of Rn7 Gquot7 and Y7 It is B m 2H Re ij5 G F1 G5 629 Optimum Source Admittance The value of Y5 that minimizes F is called the optimum source admittance and is denoted by Yam It is obtained by setting BFBGS 0 and BFBBS 0 and solving for G5 and BS The admittance is given by Yap Gapt JBapt V1 72 171 11 En BA 13 630 TL Note that this is the reciprocal of the optimum source impedance7 ie Yap 1Z0pt Also7 the imaginary part of Yap is equal to the imaginary part of Y5 in Eq 65 which maximizes the signalitoinoise ratio The corresponding minimum value of the noise factor is given by u 239 Fm 1 02 7 1 7 7 1 2H G7 G 631 Relating F to Fmquot and the Source Admittance We next Wish to express F in terms of me and Yam It follows from Eqs 629 and 631 that the difference F 7 Fm is given by B m 2H GG 313 G F 7 me 721G7 G R m e 2R Gupta BB G GS where B Bap has been used The square in the numerator of this expression can be completed by adding and subtracting the term Rn G310 ngt Rn lYaptlZ This leads to the equation R G i G2 B a B2 a m G F 7 F G 633 S 632 62 NOISE FACTOR AND NOISE FIGURE 121 From Eq 630 we have lYaptlZ It follows that F can be written Rn F me E G5 7 Gopt2 7 B5 7 301002 S me g m 7 30le 634 S u in and 39y as Functions of Fmquot and Yap If Rquot me and Yap Gap jBapt are given for an ampli er it follows from Eqs 67 and 630 that u 4kTRnAf 635 Sgn BM 636 1 l GaptBapt2 2 Bum U 637 7139 2kT A 7r 0 f FrLin 7 1 7 63938 unzn Where sgn Bum Bapt lBoptl39 F as a Function of Re ection Coe icients The noise factor can also be expressed as a function of the re ection coef cients of the source and the ampli er input Imagine a zero length transmission line having a characteristic admittance YE connecting the source to the ampli er In Eq 634 let Y5 Yci 639 Yap Yci 640 a g y m 16 3113 641 Where the re ection coef cients F5 and TOP are given by S 642 Top 643 When the expressions for Y5 Yam and G5 are substituted into Eq 634 the expression for F reduces to 46m m 7 PM 2 2 1F0pt 1411 644 122 Copyright 1999 7 2009 by W Marshall Leach Jr 623 The Noise Factor Fallacy The noise factor can be a misleading speci cation If an attempt is made to minimize F by adding resistors either in series or in parallel with the source at the input of an ampli er the SNR is always decreased This is referred to as the noise factor fallacy or the noise gure fallacy Potential confusion can be avoided if lowinoise ampli ers are designed to maximize the SNR This is accomplished by minimizing the equivalent noise input voltage If a series resistor must be included at the ampli er input its value should be much smaller than the source impedance If a parallel resistor must be included at the ampli er input its value should be much larger than the source impedance The following example illustrates the noise factor fallacy Example 2 An ampli er has an input resistance R 150 9 For Af 1Hz its noise parameters are u 2nV in 10 pA and p 01 It is driven from a source having an output resistance RS 50 Q a Calculate 115 F and NF 17 A resistance is added in series with the source impedance to minimize F Calculate the new 115 F and NF c Calculate the changes in the noise gure and the signalitoinoise ratio Solution a 52 4kTR5Af 53 2puninR5 iiRi 525 X 1018v2 u F Tb l 4kTR5A f NF 10log F 817 dB 656 Rm 7 quot 200 9 2n 55 4kTRmA f 53 205 an 231 12 X 1017v2 u F 3 5 4kTR0ptAf 7 NF 10log F 574 dB c The decrease in the noise gure is 8171 7 5740 243 dB The dB decrease in the SNR is 10log 12 X 1017525 X 10518 359 dB This example illustrates how the noise gure appears to be decreased by adding resistance in series with the ampli er input However the signalitoinoise ratio is lowered The fallacy comes from treating the added resistance as part of the source rather than part of the ampli er In reality the ampli er noise is increased by the added resistor but the source noise remains constant The correct way to calculate the noise factor with the added resistor is F viiAllsTRsAf which gives F 15 and NF 1176 dB In this case the noise gure decreases by 63 NOISE MATCHING NETWORK EXAMPLES 123 1176 7 817 359 dB This is the same as the dB decrease in the signalitoinoise ratio Figure 63 shows the calculated noise gure versus R5 for the example Curve a shows the correct noise gure which is a monotonically increasing function of R5 Curve b shows the incorrect noise gure that is calculated with the added resistance considered to be part of the source and not the ampli er It shows a minimum at a source resistance of 200 9 This example illustrates how the noise factor is degraded with the addition of a resistor in series with the source A resistor in parallel with the source has the same effect However if a lossless matching network is used between the source and the ampli er input the noise gure can be minimized while simultaneously minimizing the noise This is illustrated in the following sections 40 Noise Figure N O 102 10 Source Resistance Figure 63 a Correct plot of NF versus R5 b lncorrect plot of NF versus Rs 63 Noise Matching Network Examples To obtain a minimum noise gure without degrading the SNR a noiseless matching network can be used between the source and the ampli er input This is illustrated in the following examples in each case it is assumed that the source impedance is real If it is not a reactive element can be used in series or in parallel with the source to cancel the reactive component Alternately the reactive component in the matching network that is closest to the source can be adjusted to cancel the source reactance The examples assume that the source is modeled by a Thevenin equivalent circuit It is straightforward to apply the networks to the case where the source is modeled by a Norton equivalent circuit 631 Transmission Line Matching Example Example 3 With a 50 9 transmission line test xture the re ection coe icient seen looking out of the input terminals of an ampli er that minimizes its noise factor F at 900 MHZ is determined to be Tap 0712370 Determine the lengths 61 and 124 Copyright 1999 7 2009 by W Marshall Leach Jr 62 of the transmission lines in the network of Fig 64 which cause the source ime pedance seen by the ampli er to be the optimum source impedance Zapt The source impedance is BS 50 Q The characteristic impedance of the two transmission lines in the network is ch ch 759 Amplifier 20 V Zi AVi ZL Figure 64 Transmission line noise matching network for Example 3 I V Solution With ZC 50 Q ie the line impedance of the text xture used to measure Tupi the optimum source impedance is given by 1 Top Z Z 645 opt c 1 7 Tom It follows that 1 07cos 237O j07sin 2370 2 50 1226 13529 quot1 1 7 07cos237 730751112370 3 The re ection coef cient at the ampli er end of T1 is given by Z0 t 7 ch I 1 646 11 Zopt l ch which yields 1226 j1352 7 75 04230 03539 1226j1352 75 3 T11 This has the magnitude lT lll 05988 It follows from Eq 5129 that the reactance of T2 is given by 1 7 ME 12 647 X2 iZc1 If the positive solution is used7 this gives 1 7 059332 4253 9 059332 1 7550 7 1 7 7550 63 NOISE MATCHING NETWORK EXAMPLES 125 By Eq 5130 the electrical length of T2 is given by 662 tan 1 648 c Thus we have 42 58 662 tanil 2958 By Eq 5127 the re ection coef cient at the source end of T1 is given by RslleZ Z61 F 649 12 Rslle2 Zc1 which yields 7 50Hj4258 7 75 7 1 12 7 50j4258 75 7 704654 j03767 It follows from Eq 5133 that the electrical length of T1 is given by 1 T12 6 650 61 2mm lt gt which gives 704654 103767 04830 103539 As a check of the impedance seen looking into T1 we have Z Z 1 391le2 ch1tan 552 C Z61 j Rslle2tan W2 2102 j1220 5 1226 13529 7 4296 j2728 3 361 arglt 5239O which is the desired value 632 Lumped Element Matching Examples Example 4 For the ampli er of Example 5 determine X1 and X2 in the noise matching network shown in Fig 65 Solution To transform RS into Zapt we set R5 jX171jX271 Rap jX0pt71 to obtain I I I Rs 7 3X1 7 i 7 Ropt onpt R X12 X1 7 RaptXo2pt By separating this equation into its real and imaginary parts it is possible to solve for X1 and X2 to obtain X1 1125 651 126 Copyright 1999 7 2009 by W Marshall Leach Jr l R pg JP Ampli er 20 Vs jXZ Vi zi AVi ZL V0 Figure 65 Lumped element noise matching network for Example 4 R3 X12 RonptRopt 7 X1 The solution for X1 yields X1 i105 9 For X1 1053 9 X2 has the value X2 727099 For X1 71053 9 X2 has the value X2 846SQ X2 652 Example 5 At a frequency of 900 MHZ it is determined that the optimum source impedance for an ampli er is Zap 30 7160 Q If R5 50 9 determine the values of X1 and X2 in the noise matching network shown in Fig 66 R jX2 J Amplifier 20 Vs 2X1 Vi Zi AVi ZL Figure 66 Lumped element noise matching network for Example 5 lt a 71 Solution To transform R5 into Zapt we set R1 jX171 jX2 RaptjX0pt to obtain R1 ijl Egg ng jX2 Ropt anpt By separating this equation into its real and imaginary parts7 it is possible to solve for X1 and X2 to obtain 11 X1 5 653 Rs3m 7 1 R X2 X0pt 7 1 7 X1 654 R5 The solution for X1 yields X1 i6129 For X1 61297 X2 has the value X2 78459 For X1 7612 9 X2 has the value X2 73559 64 NOISE TEMPERATURE 127 64 Noise Temperature The internal noise generated by an ampli er can be expressed as an equivalent inputitermination noise temperature When the source is represented by a Th venin equivalent circuit the noise temperature Tquot is the temperature of the source resis tance that generates a thermal noise voltage equal to the internal noise generated in the ampli er when referred to its input For the Th venin source the noise temperature is de ned by 4anR5Af 53 2w Re 72 if 25 655 where RS Re Z5 It follows that the noise temperature is given by 53 2w Re 72 if lZSlZ Tquot 4kR5Af 656 When the source is represented by a Norton equivalent circuit the noise temper ature is the temperature of the source conductance that generates a thermal noise current equal to the internal noise generated in the ampli er when referred to its input It is given by 52 lel2 2w Re 7Y5 if Tquot 4kG5A f 657 where G5 Re The noise temperature is related to the noise factor by Tn F 7 1T0 658 This holds for either the Th venin or the Norton source Example 6 Calculate the noise temperature of the ampli er of Example Qfor which F 656 Solution Tn 656 7 1 X 290 1612 K 65 Noise Factor of a Multistage Ampli er 651 F as a Function of Equivalent Noise Input Voltages The circuit model for a multistage ampli er is shown in Fig 67 It is shown in Section 43 that the equivalent noise input voltage is given by Vm 2 Gmlzal Vm Vm1 Vm N 659 Gm1Z01Gm2Z02 39 39 39 GmN71Z0N71 Where Vrtil Vts Vrtl 1n1Z57 Vrtij an Iananil for 2 S j lt N7 N is the number of stages and Zaj is the output impedance of the jth stage The 128 Copyright 1999 e 2009 by W Marshall Leach Jr transconductance ij is the ratio of the output current from the jth stage to its Th venin or openecircuit input voltage It is given by In 39 gm 39 G J 1 660 m Vmoc Z0671 t Zij Where gmj IajWj ande is the input voltage across Zl39j It follows that the noise factor of the ampli er is given by 2 F Uni 4kT Re Z5 Af 1 2 11239 U quot 12 4kT Re Z5 Af quot11 igmlz 2 UMN 2 l 661 le1Z01Gm2Z02quot39GmN71Z0N71l ZS Vm391 Amplifier 1 sz Amplifier 2 I I ll l Vs Zi1j4gtlai zo1 212239 ltgtIozljzoz Zl V0 Figure 67 Multiestage ampli er 652 F as a Function of the F of Each Stage We Wish to express F as a function of the noise factor of each stage The noise factors are given by U2 F1 m1 4kTReZ5R5Af F 7 4kT Re Z01 Af 11512 7 1 11512 2 4kTReZ01Af 4kTReZ01Af F 4kT Re Z0N1 Af vim N 4kTRe Z0N1 Af 11ng quoti 662 1 l 4kT Re gown Af Note that 115 includes the thermal noise generated by Z5 For 2 2 112 does not M include the thermal noise generated by ZOO1 because its thermal noise is included in 1151071 Thus the thermal noise generated by ZOO1 must be added to 11239 m in 65 NOISE FACTOR OF A MULTISTAGE AMPLIFIER 129 the expression for Fj This may seem contradictory However by de nition the noise factor of an ampli er requires that the thermal noise of the source be included in its equivalent noise input voltage It follows from Eq 661 and 662 that F can be written F2 7 1 Re Z01 leZallZReZs FN 7 1 Re Z0N1 l 2 le1Z01Gm2Z0239quotGmN71ZaN71l Re Z5 FF1 663 This is the desired expression 653 F as a Function of Avaliable Power Gains In the analysis of rf ampli ers it is common to express F as a function of the available power gains of the stages For the rst stage the current through Z is I Vg Z5 Z The power delivered by the source to Z is given by 113 Re Zn 113R 6 64 lZsZi1l2 RSR 12X5X 12 P1 1Rez The maximum value of PM is called the available input power Pail It is solved for by setting 8P18R1 0 and 8PM 8X 1 0 and solving for R and X The solutions are R R5 and X 7X5 ie Zn Z It follows that Pail is given by p 7 U 3 6 65 a 4ReZ5 39 For the second stage the current through Z12 is Z1 1 I quot 666 12 01 Z01 Z12 where 01 Gles The power delivered to Z12 is the power output from the rst stage It is given by 2 2 M Re 22 667 2392 lzll2 p1 Z39ZR201 Rez a 12 l l lZ01Z 2l2 lZal Z12 l2 Following the derivation for Pail this is a maximum when Z12 Z31 It follows that Paul is given by lelzal l2 U P 668 quot1 4Re 201 l l The available power gain Gal of the rst stage is given by P G Z 2R Z Gal 101 l m1 all e s Pail Re Z01 130 Copyright 1999 7 2009 by W Marshall Leach Jr Similarly the available power gain Gaj of the jth stage is given by G Paaj lejZOjl2 Re Zaj11 6 70 a Paij Re Zaj I With these de nitions it follows that Eq 663 for F can be written FF1 F271 FN71 671 G61 Ga1Ga2 39 39 39GaN71 which is the desired result If Gal can be made large enough the above equation implies that F 2 F1 However increasing Gal may not make F2 7 1 Ga1 approach zero For example consider the case where Z01 R01 jO In this case Gal is given by lelRallZRe Z5 G a1 R01 Gmll2R01Rezs 672 This equation shows that Gal can be made arbitrarily large by increasing R01 The contribution to F by the secondestage noise is given by 211 in Re 7 R01 F2 7 1 1 112 392 Gal 4kT0 Re Z5 Afog1 1731 l 2quot 63973 The latter term in this equation is independent of R01 It follows that this term cannot be made arbitrarily small by making R01 arbitrarily large Let Tquot be the noise temperature of the overall ampli er and TM the noise teme perature of the jth stage Eq 658 can be used to express the noise factors in Eq 671 in terms of the noise temperatures It follows that the noise temperature of the multistage ampli er is given by T 2 TnN T quot n quot1 G61 Ga1Ga2 39 39 39GaN71 674 Example 7 Use Eqs 671 and 658 to calculate the noise factor and noise temperature of the twoistage ampli er in Example 5 of Chapter4 for which 11 5nVin1 2pA 39y 0 Af 1Hz RS 1kQ Z01 R01 20kQGm1 35 1 S and Gmg 22571 S Solution 2 2 2 2 u v u i R F ml 1 quot quot 5 281 1 4kTR5A f 4kTR5A f 11239 112 i2R2 F quot12 1 quot quot 5 608 2 4kTR01A f 4kTR01A f F F1 4 M 281 lelzoll Rs Tn F7 1T0 526K For this example it follows that the noise factor is determined by the rst stage 66 EFFECT OF A MATCHING NETWORK ON NOISE 131 66 Effect of a Matching Network on Noise 66 1 Th venin Source Figure 68 shows a lossless matching network between the input to an ampli er having an input impedance Zi Bi le39 and a signal source having an output impedance ZS RS jX5 Denote the input impedance of the matching network by Zl39m Rim inm and its output impedance by Zam Ram anm We wish to develop a relation which can be used to re ect the noise sources V and In to the input side of the matching network Lossless Mafclhing H Zs Vts Nefwork Vn Amplifier 20 1 ON E N 3 E I 5 E N s in 5 LiJ 5 Figure 68 Ampli er with a Th venin source and an input matching network The total power delivered to the matching network by the source can be written V39erst 2 u 4kTR5Af Z5 l Re Zv R 675 l lZsZiml2 1m Looking into the output terminals of the matching network7 the Th venin equivalent circuit consists of an openicircuit voltage in series with the impedance Zam Let the openicircuit voltage be written VIM 5 Vamts where VIM 5 is the component due to V5 and Wmts is the component due to Vts The output power from the matching network is the power delivered to Zi by V and V25 This can be written Vams Vamts 2 4kTRamAf 2 Reg Um P R39 676 am Zam Zl39 Zam Zi 2 l Because the matching network is lossless7 we have Pl39m Pam This leads to the relation 2 R39 Z Z v vim 4kTRamAf u 4kmA f 677 The thermal noise component of this equation is 7 l Zam 2 4kTRamAf 7 Bi 4kTR5Af 678 which leads to 2 Ram Rim Zi Zam 679 132 Copyright 1999 7 2009 by W Marshall Leach Jr When this is used in Eq 677 we obtain 62 4kTR0mA f R u 4kTRSA f 680 ams R 5 To obtain the total meanesquare openecircuit voltage at the ampli er input the contributions of V and In must be added to Eq 680 The result is 111206 1 ng Ugmts U72 Qunin Re 39Yzjm lzoml2 13m 63 4kTR5A f 62 5 26 Re 723m if 20m R 0m U vii 681 S It follows from this expression that the meanesquare equivalent noise input voltage in series with V is 62 4kTR5Af 15 63 2w Re yzgm if 20m 682 am The noise factor is given by F U72 1 UTZL T Re T lZOTLl2 4kT0R5Af 4kT0R0mAf 39 This is the same as the noise factor calculated at the output ofthe matching network The basic reason that the noise factors at the source and at the output of the matching network are equal is because a lossless matching network cannot add noise Thus it follows that the signaletoenoise ratio is also the same at the input to the matching network as it is at the input to the ampli er However these conclusions do not hold for a lossy matching network Eq 683 can be used to predict the noise factor for any arbitrary matching network For example Zam might be chosen to be the optimum source impedance to minimize F Alternately it can be chosen for a conjugate impedance match to maximize the power gain Such calculations are illustrated in Example 8 For a conjugate match the condition Zam Z must hold In this case Ram R and the expression for 112 in Eq 682 reduces to 62 4kTR5Af 63 2w Re 72 if 212 684 l The corresponding noise factor is F 112 1 112 2unin Re 7Z z Z 2 685 4kTR5A f 4kTRA f Because of the dependence of 112 and F on Z it is dif cult to predict from these equations how changes in Z affect the noise This is because Vquot Iquot and 39y in the 66 EFFECT OF A MATCHING NETWORK ON NOISE 133 noise model are in general related to Z1 For example V Iquot 39y and Zi may all be functions of the bias current in the ampli er input stage A change in the bias current to vary Z can cause a change in V In and 7 Thus the effects cannot be examined in detail unless the relations between the variables are known Example 8 An ampli er is driven from a source with a resistive output impedance RS 50 9 At the operating frequency f 10 MHZ the ampli er has a resistive input impedance Bi 259 and the noise parameters vnm 0447nVxE 31 pA E and 39y 012 7 j044 a Calculate the noise gure with a conjugate impedance matching network between the source and the ampli er 1 Calculate the noise gure if the matching network is designed so that the ampli er sees its optimum source impedance c Calculate the decrease in power gain with the second matching network Solution a For a conjugate impedance match the noise gure is u 2m Re 72 if lZ Z NF 101 1 0g 4kTRAf 506 dB b The optimum source impedance is given by Eq 613 It is 20m M1 7 ml 3 quot 13j6359 TL Thus the minimum noise gure is u 2w Re 72317 if Zml NF 39 10l 1 Wquot 0g 4kT Re 20m A f 441 dB This is 0648 dB lower than for part a c By Eq 680 vgms for part a is R 25 v 0m 2 i v2 05 07715 R Us 7 5 5 For part b it is R 13 vgms mvg Ev 026vg 5 The signal power delivered to the ampli er input is U2 p 23 For part a we have 0 5 2 Us 25 0005u3 p 39 al 50 25 2 134 Copyright 1999 7 2009 by W Marshall Leach Jr For part b 611 2 225 00043911S 13 3635 25 It follows that the ampli er power gain drops by the factor 0004390005 with the optimum source impedance This is a decrease of 122 or 0567 dB PM 662 Norton Source Figure 686 shows an ampli er with a Norton source at its input The solutions for the meanesquared noise current in parallel with IS and the noise factor follow the derivations of Eqs 684 and 685 for the ampli er with a Th venin source The meanesquare input noise current is given by 4kTG5Af u mm 2w Re 7Y0 239 686 G5 Gum where G5 Re and Yam Gum jBam The noise factor is given by F 1 U lyamlg 2Wquot ReWYom ii 687 4kTGsAf 4kTG0mA f For a conjugate match7 Yam In this case and F are given by 392 G5 2 2 2 i 2 2m 4kTGSAf a v m l 2w Re m 2 688 l 39 2 2 F 23 1 u l 2on2 Re 2 689 4kTGsAf 4kTGlA f Lossless Matching Network Vn Amplifier 20 l l 18 ileIts Yum Yam In Vi liYi Alli ZL V0 Figure 69 Ampli er with a Norton source and an input matching network 67 Noise Circles ln rf design7 the contours of constant F on the Smith chart for the re ection coe e icient seen looking out of an ampli er input are important These contours are circles Thus they are called noise circles These circles are developed below for both the impedance and admittance Smith charts 6 7 NOISE CIRCLES 135 67 1 Th venin Source Fig 610 shows an ampli er with a Th venin source and a lossless input matching network Let the impedance seen looking into the output of the matching network be Zam Ram anm Imagine a zero length transmission line of characteristic impedance ZC connected between the matching network and the ampli er input Let Lam be the re ection coe icient seen looking into the output of the matching network It is given by Z Z 0m C Tam Zam Z6 690 By Eq 6277 the noise factor is given by 4Gan lram 7 TaptlZ F Fmquot 2 2 691 1 7 ram 1 7 mm This equation can be rearranged into the form r 7 r 2 lam m Z 692 1 1 0m where 2 is given by F 7 l1 7 Taptlg Z 693 4GnZC Lossless Matching Z3 Nefwork Ii Amplifier Zo Vi 2i AVi ZL V0 Figure 610 Ampli er with a Th venin source and a lossless input matching net work Let Lam p jq and TOP pap jq0pt With these de nitions7 Eq 692 can be written p2 1 z 223023 q2 1 z 2qmq z 7 mm 694 After dividing both sides of this equation by 1 2 and completing the squares7 the equation can be reduced to Pepi 2 gapt 2 7 Z 1 lraptl2 p 12 q 12 712 12 I or equivalently 2 Z Z Z 1 7 0p 696 1 2 1 2 136 Copyright 1999 7 2009 by W Marshall Leach Jr This equation represents a circle on the Smith chart with center at the point Top 1 b 69 1 12 7 2 C Z 1 7 lroptl 1 z 1 2 For 2 a constant Eq 696 represents a circle of radius 0 that is centered at the point 1 Fapt 1 z on the Smith chart On this circle the noise factor is constant and is given by and a radius given by 12 698 42GnZC on F 1 7 ram Wquot Ram F me 120 7 ml 699 Because F is constant for z a constant it follows that the contours of constant F on the Smith chart are the circles de ned by Eq 696 For Zam apt the circles degenerate into a point located at 1 Tupi Because ba lm1 0pt Re1 0pt is independent of Zam it follows that the centers of the noise circles lie on a straight line passing through the origin and the point Tupi on the Smith chart Figure 611a shows an example impedance Smith chart with the point Tupi 0517135 0 and three surrounding noise circles labeled 1 1 1 2 and F3 corresponding to z 03 2 05 and z 07 respectively Figure 611 Smith charts showing the optimum re ection coefficient 1 0 and three noise circles a lmpedance chart b Admittance chart Example 9 An ampli er is driven from a Theuenin source with an output imi pedance ZS R5 60 Q The ampli er has an input spot noise current inxA 6 7 NOISE CIRCLES 137 20pA The optimum noise gure is Nme 15dB when the source imi pedance is Zap 30 7120 Q a If a zeroilength transmission line having a char acteristic impedance ZC 509 is connected between the source and the ampli er calculate the reflection coe icient F5 seen looking out of the ampli er input 1 Calculate the optimum reflection coe icient Tupi c Use Eq 699 to calculate the noise gure when the ampli er is driven from the source d Calculate the center coordinates and the radius of the noise circle that Tam lies on e Calculate the equivalent spot noise input voltage Solution a Looking out of the ampli er input7 the re ection coef cient is Z5 7 ZC 1 5 Z5Zc 0091 b For ZS Zapt the optimum re ection coef cient is Zopt 7 Zc T 70176 7 0294 7 20 Zc C G 7 7 0 025 S quot 4kT0Af 39 me 101VFm10 1413 G F me Fquot RS 7 201742 1954 5 NF 1010g F 291 dB F me 1 Twig 0159 Z 4GnZC R r a M 70152 1 z b Im PM 70254 1 2 12 0351 c L 17 12 12 U V quotl F X 40035 137nV Hz Af 138 Copyright 1999 7 2009 by W Marshall Leach Jr 672 Norton Source Fig 612 shows an ampli er with a Norton source and a lossless input matching network Let the admittance seen looking into the output of the matching network be Yam Gum jBam Imagine a zero length transmission line of characteristic admittance YE connected between the matching network and the ampli er input Let Tam be the re ection coe icient seen looking into the output of the matching network It is given by 7 Yb 7 me 7 Y Yam By Eq 6447 the noise factor is given by 4Rnyc lram 7 Faptl2 l1 F0117tl21 7 lramlg ram 6100 F me 6101 This equation can be rearranged into the form lram 7 TaptlZ 7 y 6102 1 7 mm where y is given by F 7 l1 Faptl2 6103 y 4Rnyc Lossless Mulching Nefwork i Amplifier Z0 Isa Y3l Yim gt lt Yom Vi n An ZL V0 Figure 612 Ampli er with a Norton source and a lossless input matching network Let Tam p jq and TOP pap jqapt With these de nitions7 Eq 6102 can be written 132 1 y 2pm 12 1 y Qqapzq y 7 llsz 6104 After dividing both sides of this equation by 1 y and completing the squares7 the equation can be reduced to 7 papt 2 7 gapt 27 y 17 lraptl2 p 1y q 1y 71y 1y 39 or equivalently 2 L lt1 7 M 6106 Tom 7 1y 1y 6 8 GAIN CIRCLES 139 This equation represents a circle on the Smith chart with a center at the point rapt F b 6107 a J 1 y C y lilraptl2 1y 1y For y a constant Eq 6106 represents a circle of radius 0 that is centered at the point P Tapt 1 y on the Smith chart On this circle the noise factor is constant and is given by and a radius given by 12 6108 4anYc F F v mm 1F0pt2 l Rn lyam 7 Yaptl2 am Because F is constant for y a constant it follows that the contours of constant F on the Smith chart are the circles de ned by Eq 6106 For Yam am the circles degenerate into a point located at T Tupi Because ba 1m Tupi Re Tupi is independent of Yam it follows that the centers of the noise circles lie on a straight line passing through the origin and the point Tupi on the Smith chart Figure 611b shows an example admittance Smith chart with the point Tupi 0517135 0 and three surrounding noise circles labeled F1 F2 and F3 corresponding to y 0568 y 0983 and y 133 respectively These values are chosen to give the same radii of the corresponding circles on the two charts in the gure It follows from Eqs 699 and 6109 that the value of F on corresponding circles is not the same unless the following relation holds an Ram lzam 7 0ptl2 Rn lYam aptlZ am 68 Gain Circles Compromises are often made in rf ampli er design between lowest noise and highest gain In Sec 67 the contours on the Smith chart of constant noise factor are derived In this section the contours of constant gain are derived Like the noise factor contours the constant gain contours are circles Thus they are called gain circles Given the Smith chart with both the noise circles and the gain circles plotted for a particular ampli er the effect of the input matching network on both noise and gain can be easily visualized 68 1 Th venin Source Consider the ampli er model of Fig 610 where the source is represented by a Th venin equivalent The signal power B delivered to Zi is the power output Pam 140 Copyright 1999 7 2009 by W Marshall Leach Jr of the matching network with the thermal noise of the source zeroed By Eq 676 and 6807 B is given by 2 R39 R P Pam 211 12 u 6111 lzam l Zil lzam l Zil Rs where Ram Re Zam and R5 Re Z5The maximum value of Pi occurs when Z0m Z and is given by 2 v P 5 6112 Kmart 4R5 Thus the relative ef ciency 7 of the input matching network can be written 13139 4Ri Ram 6113 Pima lZl39 Zamlg where 0 S 7 S 1 Let us next express Zi Bi Zam and Ram in Eq 6113 as functions of re ection coe icients Imagine a zero length transmission line of characteristic impedance ZC connected between the matching network and the ampli er input Let Tam be the re ection coe icient looking into the output of the matching network and let Ti be the re ection coe icient looking into the ampli er input We can write 1 12 Z39 Z 6114 1 cl in lt gt Viz 11 1r 7 1711112 R 2 17111171 Zcuimg 6115 1 F0m Z Z 6116 0m c1 iram 261Fam 11 1711077112 B quotm 6117 quotm 2 171m 171 m 6114 2 l 1 Ti 1 ram 1 Firam Z39 Z Z 2Z 6118 0 C17ril171 0m C171 171 0m With the use of these relations7 Eq 6113 can be written 1 71112 1 71112 7 6119 l1 7 Firale With the substitutions Tam p jq and Ti pi y the above equation becomes r 711 2 r q 1 7 1 7611112 1 1 7 7 1quot 2 quot l l 2 6120 1 i 1 i 7I 1131 71 1 7 1 711th 232 68 GAIN CIRCLES 141 After completing the squares7 this equation can be reduced to 2 i 1 7 Fi z 2 17n 1717n m2 6121 7117 F quotm 14km This represents the equation of a circle on the Smith chart with a center at the point 7117 Fajb 17077011112 6122 and a radius given by 017 l1 il2 q 6123 1 i 1 7 7I W2 Because ba lm Re is independent of r it follows that the centers of the gain circles lie on a straight line passing through the origin and the point 1 on the Smith chart Figure 613a shows an example impedance Smith chart with the point 1 03145O labeled no and three surrounding gain circles labeled n1 712 and n3 corre sponding to gains lower than 710 by 1 dB7 2dB7 and 3dB7 respectively Figure 613 a lmpedance Smith chart showing example gain circles corresponding to 07 71 72 and 73 dB b Corresponding admittance Smith chart Example 10 The ampli er of Example 2 in Chapter 5 is driven from a Th venin source with the output impedance Z RS 50 Q The amplifier has the input impedance Zi Bi le39 316 7 j127 2 and a gain of 398dB a Calculate the value of r if no matching network is used at the ampli er input 1 Calculate a b and c for the gain circle on which 7 lies c Calculate the new overall gain of 142 Copyright 1999 7 2009 by W Marshall Leach Jr the ampli er if a conjugate matching network is added between the source and the ampli er Assume zeroelength transmission lines with the characteristic impedance ZC 50 Q for the Smith chart calculations Solution a Because there is no input matching network7 Zam ZS RS 50 Q W 0277 or 7 557dB lztztm12 0 Z Z 7 i7 c 7 7 7 Zi ZC 70642 30557 R 1 a quot 61 2 0373 1707701111 WHO 12 0324 1707701111 17 1quot2 c l ll 7 0493 2 1 171 77lril c The addition of a conjugate matching network at the input would increase the dB gain by 10log 171 to 398 557 454 dB 682 Norton Source Consider the ampli er model of Fig 6127 where the source is represented by a Norton equivalent Let Y5 G5 jB57 Yam Gum jBam7 and Yi Gi jBl39 Eq 6111 can be transformed into the equivalent equation for the Norton source with the substitutions Z 119 20m 1mm Ram Gum114ml R5 Gs 11512 1112 lYilg and u me After these substitutions are made7 it follows from Eq 6111 that the meanesquare signal current through Yi is given by 2 Yi Gum 2 s 2 239 i 6124 1 Yt Yam Gs The power delivered to Y is given by 1 G39 392 G39G P if Re if g 6125 K W M Yaml Gs The maximum value of Pi occurs when Yam and is given by 2392 13imam 5 Thus the relative ef ciency 7 of the input matching network can be written 7 7 4CTViCTan l Yoml2 6127 6 8 GAIN CIRCLES 143 With this value of 7 the equations for the gain circles for the Norton source are the same as they are for the Thevenin source given by Eqs 6121 through 6123 This follows because the gain circle equations involve only the power ratio 7 and the input re ection coef cient Fl Fig 613b shows the gain circles on the admittance Smith chart corresponding to those in Fig 613a on the impedance Smith chart Example 11 An ampli er designed to operate at the frequency f 19GHz with a Norton source having an output impedance ZS R5 50 9 At 19 GHZ the ampli er has the specifications optimum source re ection coe icient for minimum noise F017 05216880 minimum noise gure Nme 139 dB noise resistance Rquot 204 9 input re ection coe icient F 06817860 The re ection coe icients are measured with a test xture having the characteristic impedance ZC 50 Q a A noise matching network is to be used between the source and the ampli er On an admittance Smith chart plot the point representing F010 and the noise circle for which the noise is 025 dB higher than its minimum value b A conjugate impedance matching network is to be used between the source and the ampli er On the same chart plot the point representing the re ection coe icient seen looking out of the ampli er input and the gain circle for which the gain is 1 dB lower than its maximum value Solution a The point representing F010 is shown on the chart in Fig 614 At this point the noise factor is Fmquot 101VFm10 1377 For 025 dB higher noise the noise gure is NF1 Nme 025 164 dB The corresponding noise factor is F1 101VF110 1459 Eq 6103 can be used to calculate y to obtain F1 me l1 Faptlg 0082 9 man where YE 1ZC 002 S The coordinates a b of the center of the 7025 dB noise circle and its radius 0 are calculated from Eqs 6107 and 6108 as follows a w 0174 1 21 12 Im PM 044s 1 21 2 12 c L 17 0239 1y 1y The noise circle is shown labeled F1 in Fig 614 144 Copyright 1999 7 2009 by W Marshall Leach Jr Figure 614 Admittance Smith chart for Example 11 b For a conjugate impedance matching network7 the re ection coef cient seen looking out of the ampli er input is Fam Ff 068186 0047j0678 This re ection coef cient maximizes the normalized gain of the matching network and is labeled no in Fig 614 On the gain circle for which the normalized gain of the matching network is 71 dB7 the value of 7 in Eq 6127 is 711 107110 0794 The coordinates 0717 of the center of the 71 dB gain circle and its radius 0 are calculated from Eqs 6122 and 6123 as follows aLFi20o42 1 1 Wlril I 1 L z jgg 1 1 Wlril V2 c 1 M 2 17710269 117nl1 l The gain circle is shown labeled 711 in Fig 614 If the source admittance lies in the intersection of the noise circle and the gain circle7 the noise factor is within 025 dB of its minimum value and the relative power gain of the input matching network is within 1 dB of its maximum value 69 MEASURING THE NOISE FACTOR 145 69 Measuring the Noise Factor 691 Method 1 This method is the most general one because it does not require knowledge of either the ampli er gain or its noise bandwidth Consider the noise model of an ampli er given in Fig 615 Consider the source to be a white noise source having the spectral density Sq Vng Af ugAf and output resistance R5 The total noise voltage at the output can be written ZL V A V V V I R Z39 a Z0ZL515 RSZinSH AZ Z L 1 Vs Vts VT 135 6128 Z0 ZL R5 Z1 The meanisquare value is given by Zi AZL Z 2 S B 4kTRB 2 ZOZL Rszi l f quot1 5 quot1 quot 2w39n Re 7R5 ii 111512 2 SE f B F X 4kTR5Bn 6129 7 AZL Z T Z0ZL RSZ where B is the ampli er noise bandwidth and F is the noise factor given by Eq 611 ts Vn I Amplifier Z 1 0 I Figure 615 Ampli er driven by a white noise source Let 1151 be the value of 115 with the noise source at the input set to zero7 ie Sq 0 Now7 let Sq be increased until the rms output voltage increases by a factor F ie 110 r1101 It follows by taking the ratio of the two meanisquare voltages that S f B S f 2 7 11 n 11 T 1 F X 4kTR5Bn 1 F X 4kT0R5 6130 The above equation can be solved for F to obtain S F f 6131 r2 7 1 X 4kT0R5 146 Copyright 1999 7 2009 by W Marshall Leach Jr In making measurements a commonly used value for r is r In this case the output noise voltage increases by 3dB when the source is activated Note that the expression for F is independent of B A and Z 692 Method 2 This method replaces the white noise source in Fig 615 with a sinusoidal source The frequency should be chosen for maximum gain The noise bandwidth B of the ampli er must be known An often used alternative is to estimate B with the equation B 133 6132 where B3 is the 73 dB bandwidth This expression is exact if the ampli er has a firsteorder lowepass response or a secondeorder bandepass response The meanesquare output voltage is given by 2 AZL Z 0 U 7 ZOZL RSZ 2 63 4151353 62 26 Re 7R5 13 11112 6133 where u is the meanesquare openecircuit source voltage With V 0 the mean square noise output voltage is measured Denote this by 1151 Increase VS until the rms output voltage increases by the factor T It follows by taking the ratio of the two meanesquare voltages that 2 T2 7 1 l U 2 4kTR5Bn 112 2on2 Re 7R5 2 lel 2 u 1 5 6134 T F X 4kT0BnR5 Solution for F yields U2 F 5 r2 7 1 X 4kT0BnR5 6135 This method requires knowledge of the noise bandwidth of the amplifier 693 Method 3 Unlike the above methods this method requires knowledge of both B and the ampli er gain The gain is measured with a sineewave source having an open circuit output voltage V5 and an output resistance R5 where R5 is the value of the source resistance for which the noise factor is to be measured With the source connected to the ampli er input adjust the voltage to obtain a convenient voltage at the ampli er output Denote this by VA The frequency should be chosen for maximum gain Next disconnect the source from the ampli er input and measure 610 CALCULATION OF NOISE PARAMETERS 147 its openicircuit output voltage Denote this by V51 Let A0 be the magnitude of the gain at the test frequency With reference to the model in Fig 6157 it is given by V01 AZL Z1 V221 Z0 ZL R5 Z1 This is the gain including the loading effects at the input and at the output The next step is to measure the noise bandwidth B An often used alternative is to estimate B with Eq 6132 The source is then replaced with a resistor of value RS and the amplifier noise A0 6136 output voltage is measured The meanisquare value is given by 2 AZL z 2 2 7 2 m 7 Z0 ZL R5 Zi u 4kT0R5BnFA0 6137 U TL This equation can be solved for F to obtain 2 1 F 6138 4kT0R5BnA l 610 Calculation of Noise ParaIneters Let the noise factor F be measured for N values of source admittance7 where N 2 4 Denote the noise factor values by E and the source admittance values by Y5 G jBl397 where 1 S 239 S N By Eq 6297 we can write G2 Bv2 B39 G F 1 RnQ 2BLG7 2RnB7 quot 6139 where Y7 G7 jBA is the correlation admittance given by Eq 610 The object is to use the measured values of F to determine the noise resistance Rn7 the noise conductance Gquot7 and the correlation admittance Y7 De ne the meanisquare error function N G2Bv2 2 7 7 7 e 7 1 B mac 3 G 2 72 B U T 6140 myg Q lt gt The values of Rn7 G G7 and BA which minimize 62 represent a best estimate of the noise parameters These values can be obtained by simultaneous solution of the set of equations BegBRquot 07 BegBOquot 07 8628 QRHGV 07 and 862 8 QRHBV 07 where QRHGA and QRHBA are considered independent variables This procedure leads to the following solution m gamin 2ampG 7 2min HIRE A 1 2 E 7 1 6141 26min 1 148 Copyright 1999 7 2009 by W Marshall Leach Jr where the matrix A is given by 02w 2 02 0232 0232 0232 B 1 E f N Z 6 Z a A 012B2 B B12 B1 31 1021 3 Z 20 GZB2 1 B 1 Z Z a Z Z 0 l The matrix A is singular if the ratios of the corresponding elements in any two rows or in any two columns is a constant With a 4 X 4 matrix there are 6 6142 combinations of two rows and two columns Because the matrix is symmetrical only the rows need be considered It follows that the matrix is singular if the values of G and Bi lie on one of the contours de ned by G2 132 kf 6143 G 7 1222 32 kg 6144 G B 7 132 kg 6145 G 1343 6146 G k5 6147 B k6 6148 where k1 through k6 are constants Fig 616 shows example plots of these equations on the G B plane The k s are chosen so that the curves intersect at two common points The curves are labeled a through f corresponding in order to Eqs 6143 through 6148 Two curves are labeled c d and f corresponding to positive and negative values of k3 k4 and kg 3 CE A b C d 0 K Figure 616 Example contours on which the matrix A is singular Chapter 9 Noise in MOSFETs Whereas the JFET has a diode junction between the gate and the channel7 the metaleoxide semiconductor FET or MOSFET di ers primarily in that it has an oxide insulating layer separating the gate and the channel The circuit symbols are shown in Fig 91 Each device has gate Gr7 drain D7 and source S terminals Four of the symbols show an additional terminal called the body B which is not normally used as an input or an output It connects to the drainesource channel through a diode junction ln discrete MOSFET S7 the body lead is connected internally to the source When this is the case7 it is omitted on the symbol as shown in four of the MOSFET symbols ln integratedecircuit MOSFET S7 the body usually connects to a dc power supply rail which reverse biases the bodyechannel junction In the latter case7 the soecalled body e ect must be accounted for when analyzing the circuit Channel Depletion MOSFET Enhancement MOSFET D D D D N GAE GAET lDB lt34le G El1DB s s 1 G 1 lt34er G c a P 117 ID lWD lWD D D D D Figure 91 MOSFET symbols The principle noise sources in the MOSFET are thermal noise and icker noise generated in the channel Flicker noise in a MOSFET is usually larger than in a JFET because the MOSFET is a surface device in which the uctuating occupancy of traps in the oxide modulate the conducting surface channel all along the chane nel The relations between the icker noise and the MOSFET geometry and bias 211 212 Copyright 1999 7 2009 by W Marshall Leach Jr conditions depend on the fabrication process In most cases the icker noise when referred to the input is independent of the bias voltage and current and is inversely proportional to the product of the active gate area and the gate oxide capacitance per unit area 9 1 Device Equations The discussion here applies to the nechannel MOSFET The equations apply to the pechannel device if the subscripts for the voltage between any two of the device terminals are reversed eg 1105 becomes 1150 The nechannel MOSFET is biased in the active mode or saturation region for 11135 2 1105 7 11TH where 11TH is the threshold voltage This voltage is negative for the depletionemode device and pos itive for the enhancementemode device It is a function of the bodyesource voltage UTH VTo 7 V lt15 39UBS 91 where VTO is the value of 11TH with 1135 0 39y is the body threshold parameter b is the surface potential and 1133 is the bodyesource voltage In the saturation and is given by region the drain current is given by k W 2D 3 1 Mos 1105 UTH2 for 1105 2 VTo 0 for 1105 lt VTO where W is the channel width L is the channel length A is the channelelength modulation parameter and k is given by e kl M00000 Mot 0 In this equation 0 is the average carrier mobility Cm is the gate oxide capacitance per unity area cm is the permittivity of the oxide layer and tax is its thickness It is convenient to de ne a transconductance coefficient K given by K K0 1 AUDs where K0 is the zeroebias value of K given by 7 k W7 000 E K 7 7 94 0 2 L 2 L l With this de nition the drain current can be written iD K 1105 7 UTH2 Note that K plays the same role in the MOSFET drain current equation that 8 plays in the JFET drain current equation given in Eq 81 In some texts the K in Eq 95 is replaced with K2 In this case the factor 12 in Eq 94 is omitted 91 DEVICE EQUATIONS 213 Figure 92 shows the typical variation of drain current with gateitoisource volt age for a constant drainitoisource voltage and zero bodyitoisource voltage In this case7 the threshold voltage is a constant7 ie 11TH VTO For 1105 S VTO7 the drain current is zero For 1105 gt VTO7 the drain current increases approximately as the square of the gatetoisource voltage The slope of the curve represents the smallisignal transconductance gm7 which is de ned in the next section Fig 93 shows the typical variation of drain current with drainitoisource voltage for a sevi eral values of gateitoisource voltage 1105 and zero bodyitoisource voltage 1135 The dashed line divides the triode region from the saturation or active region In the sat uration region7 the slope of the curves represents the reciprocal of the smallisignal drainisource resistance To which is de ned in the next section Drain Current I 0 VTO Gate to Source Voltage Figure 92 Drain current iD versus gatetoisource voltage 1105 for constant draini toisource voltage UDs Triode Saturation Increasing T as Drain Current Drain to Source Voltage Figure 93 Drain current iD versus drainitoisource voltage 11135 for constant gatei toisource voltage 1105 214 Copyright 1999 7 2009 by W Marshall Leach Jr 92 Bias Equation Figure 94 shows the MOSFET with the external circuits represented by Th venin dc circuits If the MOSFET is in the pincheoff region the following equations for I D hold ID K V05 7 VTH2 96 VGS VGG V55 IDRSS 97 K K0 1 AVDS 98 VDS VDD IDRDD V55 IDRSS 99 Because this is a set of nonlinear equations a closed form solution for ID cannot be easily written unless it is assumed that K is not a function of VDS and VTH is not a function of V35 The former assumption requires the condition AVDS ltlt 1 With these assumptions the equations can be solved for ID to obtain 1 2 I M1 4KR V 7V 7V 71 910 D Eggs 55 00 SS TH V00 Figure 94 MOSFET dc bias circuit Unless AVDS ltlt 1 and the dependence of VTH on V33 is neglected Eq 910 is only an approximate solution A numerical procedure for obtaining a more accurate solution is to rst calculate ID with K K0 and VTH VTO Then calculate VDS and the new values of K and VTH from which a new value for D can be calculated The procedure can be repeated until the solution for ID converges Alternately computer tools can be used to obtain a numerical solution to the set of nonlinear equations 93 SmallSignal Models There are two smallesignal circuit models which are commonly used to analyze MOSFET circuits These are the hybrid77r model and the T model The two models are equivalent and give identical results They are described below 93 SMALLgSIGNAL MODELS 215 93 1 Hybrid7r Model Let the drain current and each voltage be written as the sum of a dc component and a smallisignal ac component as follows iD ID U05 VGS 1195 912 1135 VBS 1175 913 39UDS VDS Uds 914 If the ac components are suf ciently small7 we can write 81D 81D 81D 915 2d BVGS 195 1 BVBS 39Ubs 1 BVDS 39Uds where the derivatives are evaluated at the dc bias values Let us de ne 81 gm BVD 2K V05 7 VTH 2 KID 916 GS 81D BVTH 39vaID 72K V 7 V 91 gm BVBS GS TH BVBS m Xgm 7 39Y X 918 2v VBS 81D 1 W 2 1 VDS1 r0 7 BVDS 7 3 TA V05 VTH T 919 The smallisignal drain current can thus be written id idg id 920 To where idg gmugs 921 id gmb39ubs 922 The smallisignal circuit which models these equations is given in Fig 95 This is called the hybridi 39 model 932 T Model The T model of the MOSFET is shown in Fig 96 The resistor To is given by Eq 919 The resistors rs and r5 are given by TS 923 m i i i 924 216 Copyright 1999 7 2009 by W Marshall Leach Jr lcl Figure 95 Hybride 39 model of the MOSFET where gm and gm are the transconductances de ned in Eqs 916 and 917 The currents are given by 39U 2d 259 251 E 925 7 0 u 259 g gm39ugs 7 s 39U 251 3 gmbvbs 927 7 51 The currents are the same as for the hybrid77T model Therefore7 the two models are equivalent Note that the gate and body currents are zero because the two controlled sources supply the currents that ow through rs and rsb zoz Figure 96 T model of the MOSFET 94 SmallSignal Equivalent Circuits Several equivalent circuits are derived below which facilitate writing smallesignal lowefrequency equations for the MOSFET We assume that the circuits external to the device can be represented by Th venin equivalent circuits The Norton equivalent circuit seen looking into the drain and the Th venin equivalent circuit 94 SMALLgSIGNAL EQUIVALENT CIRCUITS 217 seen looking into the source are derived Several examples are given which illustrate use of the equivalent circuits 941 Simpli ed T Model Figure 97 shows the MOSFET T model with a Thevenin source in series with the gate and the body connected to signal ground The result derived here can be used for the case where the body connects to the source rather than to signal ground We wish to solve for the equivalent circuit in which the sources 23959 and 23951 are replaced by a single source which connects from the drain node to ground having the value i i We call this the simpli ed T model The rst step is to look up into the branch labeled and form a Thevenin equivalent circuit Superposition of 119 and can be used to write Utg 7 51 Ts Ts u fi r Hrbut r5X 7 lg 5 55 5 awsx 1x 39 my 2 a 7 M 928 7 S 1X 1X9m 1X where 929 Figure 97 T model with Thevenin source connected to the gate and the body connected to signal ground With the circuit seen looking up into the branch labeled replaced with the source utg 1 x in series with the resistance r the simpli ed T model can be drawn It is shown in Fig98 Note that the gate and body leads do not appear in the circuit Compared to the corresponding circuit for the JFET the MOSFET circuit replaces utg with utg 1 x and Q with r19 The circuit is derived with the assumption that the body lead connects to signal ground In the case that the body lead connects to the source lead7 it follows from Fig 97 that 23951 0 In this case7 the MOSFET circuit is identical to the JFET circuit Connecting the body to the source is equivalent to setting X 0 in the MOSFET equations 218 Copyright 1999 7 2009 by W Marshall Leach Jr Figure 98 Simpli ed T model 942 Norton Drain Circuit Figure 99a shows the MOSFET with Th venin sources connected to its gate and source leads and the body lead connected to signal ground The equivalent circuit seen looking into the id branch can be represented by a Norton equivalent circuit consisting of a current source idsc in parallel with a resistor rid as shown in Fig 99b The expressions for idsc and rid are derived below at a b Figure 99 a MOSFET with Th venin sources connected to the gate and source b Norton drain circuit Figure 910 shows the simpli ed T model equivalent circuit The drain current is given by id 2390 930 Using superposition of 11d 119 and 115 we can write 7 lid 7 119 1 Bis 7 1 Tl 20 7 r Uts 931 rorsHRts 1XT5B sllTOTOB s rsHmRtsrsro 11d 7 5 119 1 1 r0 2 7 932 5 r0rgllB 5 B 5rg 1XrgB 5Hr0 UtsrgllroJrBLS rgr0 94 SMALLgSIGNAL EQUIVALENT CIRCUITS 219 Figure 910 Simpli ed T model for the circuit of 99a These equations can be solved for id to obtain an equation of the form Ud 933 2d Md Zdsc where rid and idsc are given by V 7 TO l T ll Rts 7 R755 T ld 7 W 7 7 0 1 Bis r 1 7 0 119 1 2 7 935 56 rgRteromRt51x rgumRts 5 It is convenient to de ne two transconductances Gmg and Gms such that u Zdsc Gmg 7 Gmsuts where 1 1 7 0 G G 936 quot 9 rgRtsHmmRts quot 5 awaits l The Norton equivalent circuit seen looking into the drain is shown in Fig 99b If m is suf ciently large7 we can write Utg X 1 mass Gm i 115 Gm TS Bis 937 In the following7 these equations are referred to as the large m approximation for Z39zisc 943 Th venin Source Circuit Figure 99a shows the MOSFET with a Th venin source connected to its gate and the body lead connected to signal ground The equivalent circuit seen looking into the is branch can be represented by a Th venin equivalent circuit consisting 220 Copyright 1999 7 2009 by W Marshall Leach Jr Rid Rtg id 1 ES 0 W iblt390 quot Mac 7 tg 9 LS Ii v5 3 i 5 Us a b Figure 911 a MOSFET with Th venin source connected to gate b Th venin source circuit of a voltage source 115UC in series with a resistor MS as shown in Fig 99b The expressions for 115UC and n5 are derived below Figure 912 shows the simpli ed T model equivalent circuit The current is 119 1 z 7 1 W 7 Us 75 93g sin su er osi ion 0 u ez39 con ro e curren source an is we can wri e U t f 9 th 9 t ll d t 7 d 7 t given by 11 To Rm EMT 1 X T r0 Bid 7 4 H To T 7 25 r SH TO RM 939 These equations can be solved for US as a function of 119 and is to obtain Us u 9 T0 7ir T0R 5d 940 5 1xmrg row 39 1quot s I 1th vi 1X Figure 912 Simpli ed T model for the circuit of 911a The equation for US is of the form Us Usac 15 941 where tg 7 0 942 was 1 X m T 95 EXAMPLE AMPLIFIER CIRCUITS 221 The equivalent circuit is shown in Fig 911b If m is suf ciently large 1150C and as can be written 39U we 1quotX n5 r 943 In the following these equations are referred to as the large m approximations for 115OC and n5 944 Summary of Models Figure 913 summarizes the equivalent circuits derived above For the case where the body is connected to the source set X 0 in all equations e gls 4 f 3 Figure 913 Summary of the smallisignal equivalent circuits Set X 0 if the body is connected to the source 95 Example Ampli er Circuits This section describes several examples which illustrate the use of the smallisignal equivalent circuits derived above to write by inspection the voltage gain the input resistance and the output resistance of several ampli er circuits 95 1 Common Source Ampli er Figure 914a shows a commonisource ampli er The active device is M1 lts load consists of a currentimirror active load consisting of M2 and M3 The current source IQ sets the drain current in M3 which is mirrored into the drain of M2 Because the sourceitoidrain voltage of M2 is larger than that of M3 the Early effect causes 222 Copyright 1999 7 2009 by W Marshall Leach Jr the dc drain current in M2 to be slightly larger than IQ The input voltage can be written 111 VB 11 where VB is a dc bias voltage which sets the drain current in M1 It must be equal to the drain current in M2 in order for the dc component of the output voltage to be stable In any application of the circuit VB would be set by feedback Looking out of the drain of M1 the resistance to ac signal ground is Rm 702 Figure 914 a CMOS commonesource ampli er b Commonedrain ampli er The shortecircuit output current and output resistance are given by 39U 20sc 211155 Gml l ix 944 Tout TidlllTOZ 945 where Gml is given by Eq 937 ndl is given by Eq 934 and Rtsl R5 The body effect cancels if Rs 0 The openecircuit voltage gain can be written U00c Z390sc 39Ua Gml Gml 7 Tout 11 11139 Mg 1 X 1 X Tidl H702 952 CommonDrain Ampli er Figure 914b shows a commonedrain ampli er The active device is M1 lts load consists of a currentemirror active load consisting of M2 and M3 The current source IQ sets the drain current in M3 which is mirrored into the drain of M2 As with the commonesource ampli er the Early effect makes the drain current in M2 slightly larger than that in M3 The input voltage can be written 111 VB 11 where VB is a dc bias voltage which sets the dc component of the output voltage The shortecircuit output voltage and output resistance are given by 1139 1 Mg 2 51 1 1 71 9mm 947 S Tout T lllrmllroz 948 95 EXAMPLE AMPLIFIER CIRCUITS 223 The openicircuit voltage gain is given by U0OC Z390sc U0OC I gmlraut vi 11139 mm For r191 ltlt 7 01H7 02 this becomes Uaac 1 z 950 U 977117191 1 X1 95 3 Common Gate Ampli er Figure 915a shows a commonigate ampli er The active device is M1 lts load consists of a currentimirror active load consisting of M2 and M3 The current source IQ sets the drain current in M3 Which is mirrored into the drain of M2 As With the commonisource ampli er the Early effect makes the drain current in M2 slightly larger than that in M3 The dc voltage VB is a dc bias voltage Which sets the drain current in M1 Which must be equal to the drain current in M2 in order for the dc component of the output voltage to be stable In any application of the circuit VB would be set by feedback Looking out of the drain of M1 the resistance to ac signal ground is BLdl r02 1quotout J j v 13000 iq o c 19 V V G b Figure 915 a CMOS Commonigate ampli er b CMOS differential ampli er The shorticircuit output current output resistance and input resistance are given by My idusc Gmm 951 Tm Bi n51 952 Tout Mdi H702 953 224 Copyright 1999 7 2009 by W Marshall Leach Jr where Gml is given by Eq 937 ndl is given by Eq 934 and rislis given by Eq 943 The openecircuit voltage gain can be written U00c Z39asc U00c 11 11 i00c Gm1raut Gm1rid1ll702 954 954 Differential Ampli er A MOS differential ampli er with an active currentemirror load is shown in Fig 915b The object is to determine the Norton equivalent circuit seen looking into the output To do this the output is connected to ac signal ground which is indicated by the dashed line It will be assumed that the Early effect can be neglected in all devices in calculating i0sc but not neglected in calculating ram ie we use the To approximations We can write Mg ms id3sc id1sc id4sc ms id2sc 955 Because the tail supply is a current source the currents idl and idg can be calculated by replacing u and 112 with their differential components In this case the ac signal voltage at the sources of M1 and M2 is zero Let 111 MOD2 and 112 711002 where old 111 7 1112 It follows by symmetry that id2sc 72111096 so that i0sc is given by 1139 d Z0sc 22dlsc 1 1 W 2 v 7 v 956 1 X r 2 97711111 U12 where Eq 937 with Bis 0 is used for Gm Note that the body effect cancels This is because the sourceetoebody ac signal voltage is zero for the differential input signals The output resistance is given by Tout 701 H703 957 Note that ndl r01 because Rm R552 0 for the differential input signals The openecircuit output voltage is given by U00c iascraut 9m1 TmHma 1M1 1112 96 SmallSignal HighFrequency Models 961 HybridPi and T Models Figures 916 and 917 show the hybrid77T and T models for the MOSFET with the gateesource capacitance 095 the sourceebody capacitance 051 the drainebody capacitance cbd the drainegate capacitance cgd and the gatebody capacitance cg 96 SMALLgSIGNAL HIGHiFREQUENCY MODELS 225 added These capacitors model charge storage in the device Which affect its high frequency performance The rst three capacitors are given by 2 095 WLCW 958 01750 C S 1 VSB11012 cbd cm 960 1 VDB11012 Where V53 and VDB are dc bias voltages 0150 and 0de are zeroibias values7 and 110 is the builtiin potential Capacitors CW and cg model parasitic capacitances For lC devices7 cgd is typically in the range of 1 to 10fF for small devices and cg is in the range of 004 to 015fF per square micron of interconnect Figure 917 Highifrequency T model 226 Copyright 1999 7 2009 by W Marshall Leach Jr 962 GainBandwidth Product The gainebandwidth product fT is a measure of the available bandwidth of the MOSFET when used as an ampli er It is de ned as the frequency at which the ratio of the shortecircuit phasor drain current Idsc to the phasor gate current 9 satis es lIdSCIgl 1 where Idsc is the value of Id with the drain body and source connected to signal ground At low frequencies the gate current is zero However as frequency is increased a gate current ows in the capacitors in the highefrequency model The solution for fT follows that for the JFET in Sec 842 with 0955 0 CW replaced with cgd 09 and CT 095 cgd 09 It is given by 1 1 1 gm g 961 fT 27T7 SCT 1 7 cgdCT2 27T7 SCT 27139 cgS cgd 091 where the approximation assumes that cgdCT2 ltlt 1 97 Noise Model The principle noise sources in a MOSFET are thermal noise and icker noise in the drain current The icker noise is often much larger than in junction FETs This is because the MOSFET is a surface device The uctuating occupancy of traps in the oxide layer can modulate the conducting surface channel all along the channel causing random trapping and detrapping of the mobile carriers in the traps located at the SLSiOg interface and within the gate oxide Fig 918 shows the device symbols with the noise sources added The polarity of the sources is arbitrary Each has been chosen so as to cause an increase in the drain current The thermal noise and icker noise in the drain bias current ID are modeled by the current source labeled ltd Ifd 1n the band Af id and Ifd have the meanesquare values 23 4kT Af 962 K Iquot A f 2 7 f D W W 93963 where m z 1 and n is usually taken to be unity These values are assumed in the following 971 Equivalent Noise Input Voltage Figure 919a shows the nechannel MOSFET with the drain and body connected to signal ground The external gate and source circuits are modeled by Th venin equivalent circuits With V2 0 the circuit models a commonesource or CS stage With V1 0 it models a commonegate or CG stage The noise sources V and Vtg respectively model the thermal noise in R1 and R2 1n the band Af these have the meanesquare values ufl 4kTR1A f 964 97 NOISE MODEL 227 S D G b Figure 918 MOSFET symbols with noise sources added u 4kTR2A f 965 V D GiB 1 b Figure 919 a MOSFET with Thevenin sources connected to the gate and source b V 7 In noise model To calculate 1 use Eq 937 to obtain V Idsc Gm 13X Vts ltd Ifd 966 where Gm is given by Eq 937 with Bis R2 and V29 and Vtg are given by Vtg V1 th 967 m V2 Vt2 1m Ifd R2 968 The equivalent noise input voltage Vm can be expressed as a voltage in series with either V1 or V2 Because the commonisource ampli er is most often seen7 we will express it as a voltage in series with V1 First7 we factor Gm from Eq 966 to 228 Copyright 1999 7 2009 by W Marshall Leach Jr obtain G 1 V1Vtr1x 1 1 gtlt V2 VtZ ltd Ifd RZl 1 X th fd 969 m The equivalent noise voltage in series With V1 is given by all terms in the brackets except the U1 and V2 terms It is given by VmVt1 1XVt2 I I m 1 GmRZ 970 Gm After some algebra7 this can be reduced to Vm Vt1 1XWzItz rlfd1xr9 th 1XV22Itd1fd 971 1 gm Where r 1 1 x gm has been used To express Vm as a voltage in series With V27 this equation is divided by 1 The meanesquare value of Vm is given by 2392 2392 113139 1121 1122 1 gt02 mg gmfd 972 Which reduces to U51 4kT R1R21X2 Af 3 973 Where X 0 for the case Where the body is connected to the source This expression gives the meanesquare equivalent noise input voltage for the CS ampli er To obtain 11 for the CG ampli er7 this expression is divided by 1 gt02 The noise factor is given by F 113139 4kTR5A f Where R5 R1 for the CS ampli er and R5 R2 for the CG ampli er For minimum noise in the MOSFET7 it can be concluded from Eq 973 that the series resistance in the external gate and source circuits should be minimized and the device should have a high transconductance parameter K and a low icker noise coe icient Kf The component of of due to the channel thermal noise is proportional to 1xID This decreases by 15 dB each time D is doubled 97 NOISE MODEL 229 972 Vn 7 In Noise Model We use Eq 971 to solve for the values of Vn and In in the Vn 7 In noise model Because Vni given by this equation is the voltage in series with the gate R1 must be considered to be the resistance of the signal source In the Vn 7 In model the In noise source connects between the gate and source For this reason R2 must be set to zero in the circuit to solve for In Otherwise the noise contributed by R2 would appear in the model and In would connect from the gate to the lower node of R2 With R2 0 W2 0 and Eq 971 becomes I 1 Vquot V21 1m If 1 w V 19 974 m The above equation is of the form Vni Vts Vn InRS where V25 V and RS R1 It follows that Vn and In are given by I I Vn d T W 975 gm In 0 976 The mean7square values of are 4kTAf K Af 2 f 977 Uquot 3xKID T 4KL20n f if 0 978 Fig 919b shows the model 973 Flicker Noise Corner Frequency The plot of 112 Af as a function of frequency would be identical to that shown for the JFET in Figure 814 The lower frequency at which the plot is twice its high frequency limit is called the icker noise corner frequency which is labeled ffln in the gure At this frequency the thermal noise and the icker noise are equal The icker noise corner frequency can be solved for by equating the thermal and icker noise components in the equation for 112 in Eq 977 It is given by 319 1D 9 9 fflk 16kTLZCnn K 7 For f gt ffln the thermal noise dominates For f lt ffln the icker noise dominates An experimental method for determining the icker noise coe icient is to measure the icker noise corner frequency and use Eq 979 to calculate Kf In some devices the icker noise corner frequency can be as high as 10 MHZ 230 Copyright 1999 7 2009 by W Marshall Leach Jr 974 MOSFET Differential Ampli er The BJT diff amp noise is 3dB greater than the noise for a commoniemitter stage Similarly MOSFET diff amp noise is 3dB greater than the noise of the common source stage This assumes that the signal output from the diff amp is taken diff ferentially Otherwise the commonimode noise generated by the tail current bias supply is not canceled 98 LowFrequency Noise Examples The equivalent noise input voltage at low frequencies is determined below for four example MOSFET circuits It is assumed that the frequency is low enough so that the dominant component of the noise is icker noise In this case the meanisquare noise voltage is approximately given by 2 KfAf KfAf KfAf 4KL200 f 4K0L20m f meowLog f u 980 where K0 is given by Eq 94 Note that 112 depends only on the gate area and is not a function of any dc bias parameters It is straightforward to modify the results for the higher frequency case where the dominant component of the noise is thermal noise or for the more general case where both thermal noise and icker noise are included The circuits are shown in Fig 920 The analysis assumes that each transistor is operated in the saturation region and the noise sources are uncorrelated Because the MOSFET exhibits no current noise the output resistance of the signal source is omitted with no loss in generality 981 CS Ampli er with EnhancementMode Load Figure 920a shows a singlechannel NMOS enhancementimode CS amplifier with an active NMOS enhancementimode load It is assumed that the two MOSFET s have matched model parameters and are biased at the same current With 110 0 the shorticircuit output current can be written 0sc 97771 VH1 7 ngVbZ Note that the body effect in M2 is absent in this equation because the smallisignal voltage from source to body is zero with 110 0 The equivalent noise input voltage is obtained by factoring gml from the expression and retaining all terms except the VS term It has the meanisquare value 2 WI 2 2 2 9m2 2 2 L1 2 Un2 1 982 Um Unl T gml Un2 Unl T 1121171 U51 where gmggm12 LleLng has been used When the lowifrequency approx imation in Eq 980 is used for each 112 term the expression for 112 reduces to K Af L 2 2 f 1 7 1 933 Um QuanmIVlLlf L2 98 LOWiFREQ UENCY NOISE EXAMPLES 231 Figure 920 MOSFET Circuit examples for lowifrequency noise calculations 232 Copyright 1999 7 2009 by W Marshall Leach Jr The value of L1 which minimizes this is L1 L2 The noise can be reduced further by increasing W1 and L2 The noise is independent of W2 982 CS Ampli er with DepletionMode Load Figure 920b shows a singleLchannel NMOS enhancementemode CS ampli er with an active NMOS depletionemode load It is assumed that the two MOSFET s are biased at the same current With V2 07 the expression for 056 is the same as for the circuit of Fig 920a Therefore7 the expression for Vm is the same However7 the two MOSFET s cannot be assumed to have the same icker noise coef cient The lowefrequency expression for of is 2 f1 f2 1 U 7 1 934 m QMHngl 1L1f Kfl L2 The value of L1 which minimizes this is L1 L2 Kflng The noise can be reduced further by increasing W1 and L2 The noise is independent of W2 983 CMOS Ampli er Figure 920c shows a pushepull complementary MOSFET CMOS ampli er It is assumed that the two MOSFET s are biased at the same current With 110 07 the shortecircuit output current is given by 10sc 9m1Vs Vn1 gm2 V Vn2 985 The equivalent noise input voltage is obtained by factoring gml 9mg from the equation and retaining all terms except the V5 term It has the meanesquare value 2 2 2 2 2 7 9m1Un1 ngUTLZ 9 86 Um i 2 39 gm1 gm2 In order for the quiescent output voltage to be midway between the rail voltages7 the circuit is commonly designed with gml 9mg When this is true and the lowe frequency approximation in Eq 980 is used for each 113 term7 of can be written 1 Kf1Af Kf2Af 987 m 4 QuanleLlf QungngLgf The noise can be decreased by increasing the size of both transistors For gml 7 9mg a technique for further reducing of is to increase L for the MOSFET for which Kfp is the largest 984 Differential Ampli er with Active Load Figure 920d shows a diffeamp with a currentemirror active load It is assumed that M1 and M2 have matched model parameters and similarly for M3 and M4 ln 99 CHOPPER SCHEME FOR FLICKER NOISE REDUCTION 233 addition it is assumed that all four transistors are biased at the same current so that gml 9mg and 9mg 974 Because the noise generated by the tail source is a common7mode signal it is cancelled at the output by the current mirror load and is not modeled in the circuit The differential input voltage is given by VM V5 VH1 7 Vng For a differential analysis the voltage Wit2 is applied to the positive input and the voltage 7Wd2 is applied to the negative 7 input By symmetry the small7signal voltage at the sources of M1 and M2 is zero Thus the small7signal voltages from source to body of M1 and M2 are zero and the body effect is absent The component of the short7circuit output voltage due to VM is 0SC gmllQdQ 7 9mg 7Wd2 gmllQd To solve for the component of 0SC due to Who and VH4 the sources V5 VH1 and Vng are set to zero This forces M4 to have zero drain signal current Thus the gate of M4 is a signal ground and 0SC 9mg VH3 7 VH4 The total short7circuit output current is given by 1056 gm1Vs V71 7 VH2 9mg VH3 7 VH4 988 The equivalent noise input voltage is obtained by factoring gml from this equation and retaining all terms except the V term The mean7square value is 2 9 3 vii 1151 1132 m 1153 t 1154 989 97771 When the low7frequency approximation in Eq 980 is used for each 112 term 11 can be written MpCfmlViLif Kfl L3 The noise can be reduced by increasing W1 and L3 and by making L1 L3 Kf1Kf3 The expression is independent of W3 990 99 Chopper Scheme for Flicker Noise Reduction As illustrated in Section 98 the minimization of icker noise can require the fab7 rication of larger devices In applications where the available bias voltages cannot be increased this can force the devices to be operated in the sub7threshold region A method of circumventing these problems is to shift the signal frequency band to a higher frequency band where the icker noise can be neglected compared to the thermal noise This can be accomplished by chopping the signal in a way that causes it to be multiplied by a high7frequency square wave After the signal is ampli ed it is chopped again to shift the signal back to its original frequency band The basic scheme that can be used in CMOS operational ampli ers is illustrated in Fig 921a The op amp is divided into two stages labeled A1 and A2 Chopper switches that are driven by a squarewave chopper signal are shown at the inputs and outputs of stage A1 These are implemented with CMOS switches Let the differential input voltage be denoted by uid UH 7 111 For um a sine wave the mm mm www t Input Stage Stage Q Q Voltage Time a b Figure 921 a Chopper topology b Signal at the output of stage A1 This signal appears inverted at the 7 output of stage A1 waveform at the non7inverting output of the A1 stage is shown in Fig 921b This signal appears inverted at the inverting 7 output of the A1 stage The 3 switches that are shown closed in Fig 921a are closed during the odd half cycles of the chopper signal During the even half cycles all 6 switches change states The effect is to alternately multiply the signals at both the input and the output of stage A1 by 1 and 71 so that the input signal to stage A2 is no longer a chopped signal The switching at the input of A1 converts the signal into a double sideband suppressed carrier signal that is ampli ed by that stage The switching at the output of A1 demodulates the signal back to its baseband frequency Let the signal frequency be f5 and the chopper frequency be fc For symmetri7 cal squarewave switching it can be shown that the spectrum of the chopped signal contains components at the frequencies nfc i f5 where n is an odd positive in7 teger The lowest frequency component is at the frequency fc 7 f5 The chopper frequency must be high enough so that fc 7 f5 lies both above f5 and above the icker noise corner frequency The circuit diagram of an example MOSFET op amp is shown in Fig 922 The circuit contains two differential ampli ers The outputs of the diff amps are switched between the inputs of two common7gate stages The outputs of the common gate stages are switched into the input of a complementary common7source output stage There are two compensation capacitors which are switched between the outputs of the differential ampli ers The purpose of these capacitors is to set the gain7 bandwidth and slew rate of the circuit The tail currents 1 and 2 also affect the gain7bandwidth and slew rate 910 The MOSFET at High Frequencies Figure 923a shows the high7frequency T model of the MOSFET with the source body and drain connected to signal ground and the gate driven by a voltage source having the output impedance ZS R5 jXS The capacitors in the circuit are discussed in Section 96 All noise sources are shown in the gure including icker noise Although icker noise is usually neglected at high frequencies in the BJT and 910 THE MOSFET AT HIGH FREQUENCIES 235 Figure 922 Example op amp circuit that utilizes switching to reduce icker noise The circuits which drive the switches are not shown in the JFET7 it can extend up to frequencies as high as 10 MHZ in the MOSFET Cgd D Zs V GVg 1 01am Vquot l E i i L G 3 V I s cgs39l39cgb 3178 70 chd 1 1 quot tel fol b Figure 923 T model of the CS ampli er at high frequencies Following Eq 7128 for the commonisource JFET at high frequencies7 we can write Idsc Gmg N we Vtsl ltd Ifd ltd Ifd Gmg w Vs st Gmg N 991 236 Copyright 1999 7 2009 by W Marshall Leach Jr where Gmg w is given by 1 7 jwc dgm G g 992 quot 9 w 9 1 jwcTZS l and CT 095 Cg ng It follows that Vm is given by Itd Ifd Gmg W 1 jWCTZs Itd Ifd 9 93 1 7 jwcgdgm 9m I Vm Vts 7 w This has the mean7square value wcTR52 1 7 LUCTXS2 1 mad9m2 4kTAf KfAf u 4kTR5Af n1 gtlt 994 3xKID 4KLZCmf The noise factor is given by 2 U F 995 4kTR5Af Equation 993 is of the form Vm39 Vts V InZS It follows that V and In are given by 1 id Ifd V 996 n 1 7 jwcgdgm gm 39 I I n M jwcTVn 997 1 7 jwcgdgm gm The mean7square values and the correlation coef cient are given by 1 2 K I A u 2 419T Af 1 T Nomigm 39m 977711 Comf 7 1 4kTAf KfAf 9 98 1 wcgdgm2 3xKID 4KLgCaf I ii WT2 u 999 V I v quotquot 70731 9100 unzn The on 7 in noise model is given in Fig 923b The optimum source impedance which minimizes F is given by v j Zapt V 1 7 72 7 m E 9101 TL Because this has no real part7 a conventional noise matching network cannot be designed to optimize the noise factor However7 an inductor in series with the 911 AN INPUT IMPEDANCE MATCHING SCHEME 237 source having the impedance Zap can be used to improve the signalitoinoise ratio at the operating frequency In this case the noise factor is given by wcTR52 1 wcgdgm2 4kTR5A f X 4kTA f K fA f 3KID 4KLZCmf F 9102 911 An Input Impedance Matching Scheme If a transmission line is to be matched at the input to a commonisource amplifier the input impedance must have a positive real part A method for obtaining this at some operating frequency is illustrated in Fig 924a where a series inductor is added to the source lead The object is to calculate the component of the gate input impedance seen looking into the gatetoisource capacitance 095 This impedance ap pears in parallel with the impedance seen looking into the gatetoidrain capacitance cgd The body effect will be neglected ie we assume 1115 0 In this case the gateitoisource capacitance includes the gateitoibody capacitance In addition it will be assumed that To 00 gs Z If V l gt Vs g 9 V9 F1l l l 095 T3 0 98 Ts 9 Ls I gs L 198 L s Figure 924 a Commonisource amplifier with an inductor in series with the source b Equivalent circuit for calculating Z9 Figure 924b shows the MOSFET replaced with the Thevenin equivalent circuit seen looking into the branch of Fig 98 In this circuit the resistor TS is given by TS 91 The equivalent circuit for solving for the component of g which flows through 095 is shown in Fig 924b The current is labeled 195 This circuit can be simplified as shown in Fig 924c Ohm s law and current division can be used to write SL1 T S 9103 L55TSH r5 5L55 0955 95 0Q 0955 r50 55 It follows that the component of the gate input impedance seen looking into 095 is 238 Copyright 1999 7 2009 by W Marshall Leach Jr given by V L 1 Z g S L 9104 95 9 new 55 0955 With 5 jw this is of the form Z95 R95 ngs 9105 where L L 1 5 gm 5 R95 X95 wLS 7 r5095 095 megs 9106 The equivalent circuit for Z95 is shown in Fig 925a A design technique might be to choose LS so that X95 0 and then vary W and L of the MOSFET so as to obtain the desired value of R95 to obtain an impedance match Because the value of 095 is set by the product of W and L7 these parameters must be varied so as to keep their product constant If we let WL S the transconductance can be written Mom W QMCIKL ID 2 I W 9m 2 L D S 7 The corresponding value of R95 is given by L 2 C I 395 5W M 9103 095 S This equation shows how R95 can be varied by varying either W or ID or both while holding the product WL S constant ragS ng gt ZgS gt ng R1173 X98 lt1 03 Figure 925 a Gate series equivalent circuit b Gate parallel equivalent circuit In some applications7 the parallel equivalent input circuit is desired To obtain this7 we solve for the input admittance It is given by 1 R X Y A 9109 95 295 R35 X925 3 R35 X35 It follows that the equivalent parallel circuit is that of a resistor R in parallel with a reactance X given by 2 R2 Rlzs R95 R X s X X95 9110 911 AN INPUT IMPEDANCE MATCHING SCHEME 239 The gate parallel equivalent circuit is shown in Fig 925b If a shunt reactance having a value ngs is added from the gate to ground7 the gate input admittance is real and equal to R 5 Because the gatedrain capacitance cgd has been neglected7 the component of the input impedance due to this capacitance is in parallel with the impedance Z95 To solve for the equivalent noise input voltage7 we use the circuit of Fig 926a to rst solve for the gatetoisource voltage V95 The gate and source voltages can be written Vg V1 V 7 Vgscgss R1 9111 Vs Vgngss ltd Ifd Lss 0955 9 V95 ltd Ifd L55 9112 It follows that Vgs Vg Vs V1 V fun Ifd L55 7 V95 Rlcgs gmLs 5 L5095529113 This can be solved for V95 to obtain V1 V11 ltd Ifd L55 Lngss R1095 gmLs s 1 V95 9114 Idsc nggs 0 ItolIfd Itd1fd 3 b Figure 926 a Circuit for calculating V95 b Circuit for calculating I m Figure 926b gives the circuit for calculating the shorticircuit drain current It is given by Idsc nggs ltd Ifd I I Gm V1 Vt1 tdg fd 1 7 wZLscgs ijlcgs 9115 m where Gm is given by gm G 9116 m Lscgss Rlcgss 1 The equation for Idsc is of the form I m Gm V1 It follows that Vm is given by I I Vm V11 Lscgss Rlcgss 1 9117 m

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.