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# ANAL&DESIGNELECTRONICCIRCT ECE0257

Pitt

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Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Chapter 2 Operational Ampli ers This lecture covers Chapter 2123 1 Function and characteristics of an ideal Opamp 2 The inverting con guration 3 The noninverting con guration Notes 4 This is the rst of 5 lectures to cover chapter 2 5 Study Example 22 Chapter 2 Operational ampli er A few words about opamp Opamp is one of the most popular function blocks to build sophisticate electronic system and instrumentation Today opamps are available at low cost with great varieties of characteristics for instrumentation and circuit development Since the learning of opamp doesn t involve the knowledge of semiconductors it can be treated as a blackbox and easy to learn too Symbol and Characteristics of an ideal 0pamp quotQ U lb The opamp is designed to sense the difference between two input voltage signals An ideal opamp should 0 Has in nite large input impedance doesn t draw current Output impedance is ZERO as an ideal voltage source An ideal opamp should have in nite bandwidth The output is always equal to AvZ 7 1 Terminal one is referred as inverting input terminal terminal 2 is noninverting terminal If input 1 and 2 are the same the output will be zero this is referred as commonmode rejection The gain A is referred as differential gain or openloop gain which is very large gain N A gt 00 Therefore an opamp should never been used in an openloop con guration Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Now let s do the math During this derivation we assume Aaa luwxlmg Inpul Hulpn O 7 F 4 39 In H7 u t rrlui gmuud e PIHACF hllppl common lcrmmnl h Thing to notice virtual ground The inverting con guration The basic con guration for an opamp involves two resistors R1 and R2 shown below 0 Since R2 links terminal 3 and negative input terminal 1 it provides a negative feedback 0 If R2 links terminal 3 and the positive input 2 it will provide positive feedback R V0 R2 VI Rl v v v I I Input res1stance R R1 1source 11 v1 R1 ltlti v 7 I Output res1 stance Row 2 m i 0 z III shonicimuit so Chapter 2 Operational ampli er Course Notes for EE 0257 Introduction to Lasers and Optical Electronics Apparently A is not no we can redo the math by taking a more practical scenario by assuming A is nite but a large numb aquot chapter 2 Operational ampli er Examples The weighted summer The weig ted summer with the same sign Course Notes for EE 0257 Introduction to Lasers and Optical Electronics The Non inverting con guration 0 The Weighted summer With the opposite signs Another commonly used configuration Will yield a positive gain as noninverting con guration 277 e we i T ZN e i Chapter 2 Operational amplifier Course Notes for EE 0257 Introduction to Lasers and Optical Electronics Apparently A is not 00 we can redo the math by taking a more practical scenario by assuming A is nite but a large number Chapter 2 Operational ampli er Application voltage follower Buffers are function block commonly used in microelectronics It has high input impedance and low output impedance In a buffer gain is not important but the ability to drive lowimpedance load is An op am with noninverting con guration can do that this is referred as a voltage follower or unitygain amplifier Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Chapter 2 Operational Ampli ers This lecture covers Chapter 2425 l A single OpAmp difference amplifier 2 Instrumentation amplifier 3 Frequency response of openloop amplifiers 4 Frequency response of closedloop amplifiers Notes Chapter 2 Operational amplifier A few words about difference differential amplifiers Difference amplifiers are designed to respond to the difference between two input signal Difference amplifiers are used to eliminate noise which are commonly identical but sometime much larger than signal level If input of an amplifier has commonmode input VIM applied identically to two input leads of an amplifier And a differential input signal vId also apply to two input leads the output signal of a linear amplifier can be characterized by two gains Ad and Am V0 AdVId Acmvkm An ideal difference amplifier magnifies differential input much larger than the commonmode signal which is characterized by the commonmode rejection ratio CMRR CMRRe 201 WI 7 og A gtgt1 cm 1 Single Op Amp Difference Ampli er The opamp is designed to be a difference amplifier However since the gain of the op amp is too large without any feedback mechanism it is not practical to use it alone In the last lecture we learned two fundamental configuration of an op amp inverting and noninverting configuration shown be ow Inverting Noninverting Course Notes for BB 0257 Introduction to Lasers and Optical Electronics The gains for these two con gurations are negative and positive respectively A A Invertmg R Narmvemng R 1 1 We can therefore combine two con gurations to make an ideal difference ampli er By properly choosing feedback resistance zero commonmode gain can be achieved The resulting circuit is shown below Let s do the math to select resistance In this lecture we will use a different derivation from the textbook Chapter 2 Operational ampli er Using the same procedure we can calculate the commonmode gain Course Notes for BB 0257 Introduction to Lasers and Optical Electronics In additional to rejecting CM signal we wish to have a high input resistance to differential signal We can nd out this by A39 39 A 0 J l Imml lnu mum i M Rid 239Iel The analysis of this circuit is straightforward Two things Di fiGain Ad R 4 1 R3 R1 Input 7 Re sis tan ce 00 The Instrumentation ampli ers However a few problems arise Note from the above analysis if the ampli er needs to have a large 0 The input CM signal will be ampli ed by the ISI stage which differential gain RgR 1 then R has to be small so as the input might saturate the 2quotd stage Even the 2 stage is not saturated by resistance This is a signi cant drawback of the single opamp the CM signal the CMRR will be reduced differential ampli er 0 Two ampli er channels must be perfectly matched otherwise mismatch will appear as a differential signal to spur the real signal This can be resolved by buffering the two input terminals using To overcome this problem we come up with the following voltage followers The additional bene t is to get some additional gain con guration in the rst stage The circuits is shown below Chapter 2 Operational ampli er 3 Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Chapter 2 Operational ampli er Some nonideal characteristics of opamps In this section we consider some nonideal properties of op amps Since this information are important for electronic designers They are normally available in data sheet Frequency response of an openloop opamp For an opamp as it is the typical frequency response openloop is singletimeconstant STC lowpass response shown below lil dB 20 dBI decade UI 76 d3 octave m 101 Ii 10 10 10quot Ii7 MHZ T t Since we only have SINGLE time constant the gain A as a function of frequency 03 can be written as Altcogt A o l C 0 0 or is 3dB corner frequency a or f is the unity frequency Since we only have single time constant the slope is 20 dBdecade For agtgt 01 A Aw 9 60 e 1 wt A060 160 1w Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Frequency response of an closed loop op amp Having familiarizing the closeloop speci cation on frequency response we can calculate the frequency response for closeloop configuration Inverting con uration with nite gain A V0 R2 R1 w A0 K 11R2R1A lsab Substituting1a into the closeloop gain we have V S R2 R1 a Aowb 143 1ilR2Rl A0 a VS A A0 gtgtlt1R2R1 1 4 a 1R2R1 An important conclusion we get out of this derivation is the 3dB corner frequency a w 3 1R2 R1 Non inverting con uration with nite gain A V0 1 R2 R1 A0 m K 11R2R1A lsab If A0 gtgt lR2 R1 we have V0s N 1R2 R1 V16 1 4 a 1R2 R1 Chapter 2 Operational amplifier Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Chapter 2 Operational Ampli ers This lecture covers Section 26 27 l Imperfection of opamp saturation slew rate 2 Fullpower bandwidth 3 Offset voltage offset currents and input bias Notes Chapter 2 Operational amplifier In this lecture we study the limitations on the opamp performance This rst includes output voltage and current saturation Output voltage and current saturation In practical Op amps will operate linearly over a limited range of output voltage and current For a typical opamp 741 the range for voltage and current are around i1 015 V and i1 020 mA Beyond this range the output will be nonlinearly distorted eg cutoff Slew rate and fullpower bandwidth Another nonlinear distortion for large output signals are socalled slewrate limiting This refers to the maximum alow rate of change of signal de ned as dvo dt I WAS An opamp will not respond any signal faster than the maximum slew rate An example is shown below for a stepchange input and it is output for a voltage follower SR One thing to note here is the slewrating limiting is a nonlinear effect which is different from a nite bandwidth distortion The finite bandwidth distortion is a linear effect which does not change the shape of input sinusoid The slewing does change the shape of sinusoid To further clarify this point let s compare the effect of finite bandwidth The transfer function for a voltage follower is V 1 721sa I Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Its step response would be an exponential function U I As long as V is small so that aVltSR the output will follow above as a linear response Another example for nonlinear SR limiting can be illustrated using a sine wave output v0 V0 sin at The changing rate is given by 0 dt So the maximum changing rate is given by 0V0 which depends on both the maximum output voltage and frequency If 0V0 exceeds maximum SR distortion happens as shown below 0V0 cos at I hmlulvml umpm Humm Mien up Imp mm c lmmcd The opamp data sheets provide a frequency fM as the fullpower bandwidth If the maximum output voltage is Vomax then the fM is related to SR as wMV SR amax SR f T 27rV amzx Chapter 2 Operational amplifier DC imperfections offset voltage Since opamp has a very large gain any imbalance between two inputs can instantly saturate output Unfortunately in practical application a number of facts can contribute to the mismatch between two inputs including the opamp designs Given a typical dc gain of gt10000 even a mismatch of a few mV can saturate the opamp This is a must addressing issue To model the dc offset effect an opamp can be modeled as following Generally speaking mmquot quotpmquot 0 V0 range from 15 mV 0 V0 depends on temperature uV C O Ulmm Now let s analyze the impact of offset to performance of a opamp with negative feedback Offsetfree op amp Course Notes for EE 0257 Introduction to Lasers and Optical Electronics One way to overcome the dc offset is by capacitive coupling shown below Since capacitor is an open circuit for DC e opamp won t amplify the V05 however this does not work for an opamp circuits working in dc and low frequency Offset ft ac M 117 Notice the gain of such con guration will become very small at low frequency Here is the analysis Chapter 2 Operational ampli er Input bias and offset currents In a practical opamp both input terminal are supplied with dc currents to mction These two currents can be modeled with two current sources The average of these two currents is referred as input bias current The different between these two currents is referred as input offset current 151152 15 05151 1 2 Given the technology used to build opamp 15 range tom 2A to 100 nA I is one order ofma nitudes smaller than I whatever it is 52 The dc output voltage of a closedloop opamp cue to the input bias currents can be easily found out by considering the an inverting con guration The dc offset voltage becomes V0 15le BRZ The allowable dc offset voltage apparently will be used to determine what is the maximum allowable R2 One way to reduce the dc offset voltage will be connecting the positive input terminal with a resistance R 3 The following analysis justi es the solution and provides a guideline to choose R 3 Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Chapter 2 Operational Amplifiers This lecture covers Section 28 l Inverting con guration with general impedances 2 Inverting integrator 3 Opamp differentiator Notes Chapter 2 Operational ampli er A few words about integrators and differentiators Together with summation subtraction integration and differentiation are two important signal processing algorithms These functions can be readily realized by opamp General Impedances Although you might be still studying RLC circuits the concept of impedance for inductors and capacitors can also be understood without too much difficulty Or you can take the following concept as they are for now For DC a capacitor is an open circuit element with a resistance of 00 An inductor is a short circuit with a resistance of 0 For AC however both capacitors C or inductor L will produce resistance for any AC signal This is referred as impedance For a single frequency sine sig11alAsinat the impedance for A capacitor C 1ij An inductor L j39mL A resistor R R Apparently the total impedance of circuits depends on frequency Sometime fa is replaced by 3 during circuit analysis In the event that we need find out the timedomain response we can convert the frequencyresponse into the transient response by Replace ljoa by J Re lace 39o b i P J y dt Then a frequency response will be converted into a linear differential equation Given the initial condition and input signal the output signal can be readily calculated by the circuit Now let s use an example to illustrate this concept Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Example for the circuit below derive an expression for the transfer function V0sVs show that the circuit is a lowpass circuit Find DC gain f0Hz and 3dB frequency Design the circuit to obtain a dc gain of 40 dB a 3dB frequency of 1 kHz and an input resistance of 1 k9 l397 Ill lf I Chapter 2 Operational amplifier The inverting integrator The basic configuration for n inverting opamp integrators is shown below I 7 mm i u an m mw lhl I We now do the math to perform analysis in both time and frequency domain Time domain the IV crossing the C and the transfer function are 1 t Vct V0 EJOlltdt 1 t Vot 7 0V1tdt7 Va Frequency domain The transfer function is KW 1 Vim ijC V1 kw Vx wRC 1RC is referred as integrator frequency RC is known as integrator time constant Generally speaking an integrator is a lowpass filter with a corner 3 013 frequencv of ZERO 2 Course Notes for BB 0257 Introduction to Lasers and Optical Electronics DC o sel The basic con guration of integrator shown above has a problem at DC Since a capacitor is an opencircuit at DC the opamp will have no negative feedback and can saturated immediately Even at AC any dc offset can be deleterious too The following analysis explains DC o sel voltage DC o sel current Chapter 2 Operational amplifier The solution for the dc offset can be alleviated by connecting a resistor R p Thus the transfer function becomes V s RF R a Vs 71sCRF K To remove the dc offset one would chose low value for RF However the low value for RF will lead to high comer frequency lCRF which will distort the integrator performance Therefore a design tradeoff needs to be carefully entertained R Course Notes for BB 0257 Introduction to Lasers and Optical Electronics The inverting differentiator Interchanging the C and R of the integrator results a differentiator circuit We can perform both timedomain and frequency domain analysis to obtain the transfer functions I R A Chapter 2 Operational amplifier RC is referred as differentiator timeconstant The ideal differentiator can be considered as a highpass filter with a comer frequency at infinity Differentiator output will spike or very sensitive to the sharp change of t e input Differentiators are not stable and should be avoided to use alone in practice Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Chapter 2 Operational ampli er Course Notes for BB 0257 Analysis and Design of Electronic Circuits This lecture covers Section 3133 Chapter 3 Diodes 1 Ideal diode IV characteristics 2 Recti er and logic gates 3 Terminal characteristics 4 Diode models Notes Chapter 3 Diode A few words about diodes 0 Diode is one of the most commonly used electronic elements in microelectronic circuits 0 As the simplest and most fundamental nonlinear circuit element diode can perform many signalprocessing functions that can not be performed by linear circuit such as ideal opamp 0 Knowledge of diode is a must to further our studies on three terminal elements such as transitors MOSFET Ideal diodes The ideal diode can be modeled as the following Atyqu alhmh I gt 0 H 0 7 7 Know um gt liul39wul39ll him 7 7 7 n x 1 a in I I 9 9 C D o O ciao oogo If a voltage applied then the diode is reverse biased no current ows diode behaves like an open circuit If a voltage applied then the diode is forward biased large current ows with ZERO voltage drop diode behaves like a closed circuit To prevent large forward current current limiting resistors are often used to reduce the forward current This simplest nonlinear response is referred as piece wise linear 1 Course Notes for BB 0257 Analysis and Design of Electronic Circuits Terminal characteristics of junction diodes Almost all diodes used today are made of semiconductors ie Si When Si with different dopants were brought together a pmjunction was formed The actual characteristics of diodes were determined by the pnjunction In a real diode the IV curve consists of three distinct regions 0 The forwardbias region vgt0 o The reversebias region vlt0 o The breakdown region vltVZK Forward i Application 1 rectif1er Compressed quot 2x 0 lt Breakdown Reverse l The ForwardBias Region In the forward region the i v follows exponential function closely kT prmdcd scale T iv 4amper 1 V In the above formula I n 0 IS saturation current Tdependant 103915A double every 5 C 39 o VT thermal voltage 25 mV 20 C 0 k138X103923 JK Boltzmann s constant 3 In 0 T 273 Celsius the absolute temperature 0 O 2 AAA vvv q160X103919 coulomb charge carried by single electron Y A B C Y A B C n material dependant constant nNIZ for IC nNI Chapter 3 Diode 2 Course Notes for EE 0257 Analysis and Design of Electronic Circuits From the above curve we see that in forward bias when the forward voltage is less than threshold cutin voltage in this case is 05V the current is negligible However when the forward voltage goes beyond the threshold a small voltage increase can lead to a big jump on current let s do a little math to appreciate it The ReverseBias Region When a reverse voltage is applied to a diode a small but constant current ows through the diode ie saturation current i m S The Breakdown Region When the reverse voltage goes beyond VZK the current increase rapidly with a very little increase of voltage which can be used a voltage regulation Chapter 3 Diode Modeling the diode forward characteristics The most accurate model to describe the diode operation in the forward region is the exponential model described above In the following circuit let s assume VD 13gt 05 V Iv R gt VB 1 Se A xm M IFM T 7 R Now combine above two equations we can precisely determine both 3 and VD by solving VDD 7V0 7 I VV As we see here this is a nonlinear equation that exact solution is not available To nd solution numeric method has to be implemented One common used method is the graphic analysis shown below Dmde churthch Nu Q operating pnim Loml IIHC stom J q Vun Course Notes for BB 0257 Analysis and Design of Electronic Circuits As we see here a precise exponential model is not very convenient to use especially for quick analysis to valid circuit designs Depending on the accuracy requirement we introduce two simple models for rapid analysis ConstantVoltageDrog Model This model is just onestep beyond the ideal diode model instead of treating a diode to be a perfect closecircuit when applying forward bias we take into consideration of the diode threshold 0508V So the diode remains open below the threshold voltage is reached even at forward biasing The model is shown below PiecewiseLinear Mouei m If we need further accuracy we then move one step further trying to emulate the lV relationship beyond the threshold which has been described in the constantvoltagedrop model Now in this model the voltage is not a constant drop Rather we use a linear piecewise approximation to emulate the exponential lV relationship In 4 A we Chapter 3 Diode The model is described by the following equation iD 0 VD S VD0 in VD VDU rD VD gt VDU Example determine VD and I D shown in the following circuit using piecewiselinear model For the diode use VDD5V VDO065V rD2OQ rI kg N 7 0 I l quot klcul lt Im Course Notes for BB 0257 Analysis and Design of Electronic Circuits The SmallSignal Model In many applications the diode is biased at an operational point bias point on the forward i v curve Then a small ac signal is superimposed on that This scenario is shown below Of course we are mostly interested in the characteristic of ac signal when passing the diode this is socalled smallsignal model lauuun u g r A A Ship 7 r39 4L 4t urs mu mu To handle this situation we rst determine the voltage VD and the current I D of the bias point then we linearize the i v curve of the diode around the bias point Therefore the IV curve around bias point will be a straight line and characterized by the small signal resistance 1 rd aiD nVT 8v 11 I The iV relationship of the small ac signal can be described using small signal approximation D D ID nVT Let s derive these relationships in class Chapter 3 Diode 4 Va Course Notes for EE 0257 Analysis and Design ofElectronic Circuits Example Consider the Voltageregulator circuit shown below The Value of R is selected to obtain an output Voltage V0 of 07 a Use the small signal model to show that the change in output Voltage corresponding to a change of 1V in V is AVU nVT mV AV V nVT 707 N b Generalize the expression above for the case ofm diodes connected in series and the Value of R adjusted so that the Voltage across each diode is 07 V and V007 mV V Chapter 3 Diode Course Notes for BB 0257 Analysis and Design of Electronic Circuits Chapter 3 Diodes This lecture covers Section 3436 l Zener diodes 2 Halfwave and fullwave recti ers 3 Peak recti er 4 Super diode 5 Limiting and clamping circuits Notes Chapter 3 Diodes In this lecture we study some of most commonly used circuits involving diodes All these circuits make use of either forward or backward nonlinear characteristics Zener Diodes In the reverse bias when the reverse bias voltage goes beyond the breakdown voltage 7VZK we observe rapidly increase of current as shown below AV 7 A I39 When the diode is used to operate at the breakdown region we refer it as Zener diode Since the slope is extremely large a large current change across the Zener diode will only lead to very small voltage change Therefore Zener is mostly used as a voltage regulator The exact IV curve for Zener is very complicate however we can implement piecewiselinear model and use the incremental resistance rZ and breakdown voltage Vzo as the following VZ VZ0 r212 The model is shown above Course Notes for BB 0257 Analysis and Design of Electronic Circuits Example The 68V zener is shown below with Vz68 V at 5 mA rz20 2 and ZK02 mA The supply voltage W is nominally 10 V but can vary by i1 V a Find V0 with no load and with W10V b Find the change in V0 resulting from ilV change in W known as line regulation c Find the change in V0 resulting from connecting a load resistance RL that draws a current of 1 mA and hence nd the load regulation in mVmA d Find the change in V0 when RL2 kQ e Find the change in V0 when RL05 kQ f What is the minimum value of RL for which the diode still operates in the breakdown regions l llllflV R r 05 H1 Chapter 3 Diodes Course Notes for BB 0257 Analysis and Design of Electronic Circuits Rectifier Circuits Recti er circuit is the most application of diodes A recti er covert AC signal to DC you can nd it everywhere for a wallplug electronic unit In this lecture we study various form of recti er circuits The following scheme shows how the recti er was used in a power supply unit in a wallplug electronic system u ami umier 39 ac lmc The HalfWave Rectifier The halfwave recti er only utilizes halfcycles of the input sinusoid We use batteryplusresistance diode model the equivalent circuit is shown below 0 VS lt VD0 R mi V0 E vs 7 VDO R V0 3 VDO Vs 2 VDO D Rr In the reverse bias when the reverse bias voltage goes beyond the breakdown V D Ideal l39 7 A I W o gt 1 i n R t M A 40 gt u 7 2 II A quotll vvv L Two important characteristics 0 Current handling capability 0 Peak inverse voltage PIV must lt breakdown voltage 0 PIV VS Chapter 3 Diodes The FullWave Rectifier The fullwave recti er can provide a unipolar output The basic con guration is shown below l 4 m hm olmgu 4 2 0 To nd PIV of the diodes in this con guration 0 Considering in the positive cycle the total voltage applied to D2 is reversed at the maXimum VOVS o In the negative cycle the total reverse voltage will be applied on Dl at the maXimum VOVs Therefore PIV2VSVD Course Notes for BB 0257 Analysis and Design of Electronic Circuits The Bridge Recti er Another popular used full wave recti er is shown below and known as the bridge recti er The peak inverse voltage P V V5 7 VD While VS and VD are the maximum or amplitudes of sinusoid signal Here is the analysis 0 Positive cycle the total reverse voltage applied to D3 VD3 reverse V0 VD2 forward 0 Negative cycle the same 0 So the V0 is V0 VS 7 2VDf0rward Therefore the PIV is shown above Chapter 3 Diodes The Recti er with a Filter Capacitor Above recti ers produce DC output with too much voltage variation This variation can be substantially reduced by adding a capacitor Let s rst consider an open circuit scenario which illustrate how a capacitor can level off the voltage variation l w r l Now if we add a load to the circuit shown below the voltage output won t be perfect leveled as shown H 7 u vvv Now let s analyze this circuits by assuming the time constant of RCgtgtT where T is the period of the input sinusoid Course Notes for BB 0257 Analysis and Design of Electronic Circuits The precision halfwave recti er or the super diode Rational When we need super diode I When a weak AC signal need recti cation if the peak voltage is less than 07 V then diode won t work It needs ampli cation I In an instrumentation develop when the lost of 07V due to the turn on voltage of diode is too muc In these situations a precision half wave recti er circuits using an op amp can be develop shown below Supcrdiode Here is explanation on I Why it can rectify less than 07V AC 2 I Why turn on voltage of diode won t offset the output Summa I The peak to peak ripple voltage V VP Conduction interval aJAt m JZV VP Average charging current inV I L 1 7JZVP V Peak charging current z39Dmax IL1 2711 2VP V Chapter 3 Diodes 5 Course Notes for BB 0257 Analysis and Design of Electronic Circuits Limiting and Clamping Circuits Since diodes will not be turned on when the voltage is below VD 07V it can be used to build limiterclipper circuits Figure below show the transformation functions for hard and soft limiters quotH Li 10 We will have a few more design using diode to realize limiter circuits as shown below Chapter 3 Diodes The Clamped capacitor or DC Restorer By using diode we also can design two useful circuits call DC restorer Which provide a pulsesignal with a wellde ned DC level Here is how it works 1 0 1 O V 4 V U 10 V 39 EU 6 V I 0 gt 11 h Cl l I l H1 139 R y 7 30 n l I I Utilizing the similar circuits we also can design a voltage doubler i I M 1 II II L sin 1 D m 393 2 Course Notes for BB 0257 Analysis and Design of Electronic Circuits Chapter 3 Diodes This lecture covers Section 37 Physical operation of diodes pnjunctions Diffusion and dri currents pnjunction under opencircuit conditions pnjunction under reverse and forward biases Limiting and clamping circuits V FP F Jt Notes Chapter 3 Diodes The pnjunction is the foundation for basic microelectronic devices and circuits including diode and bipolar junction transistors BJTs Studying some fundamental semiconductor physics are important for microelectronic designers Basic semiconductor concepts Si is the most commonly used semiconductors to build microelectronic chips Highperformance microchip is built by crystalline silicon shown below polycrystalline and amorphous silicon also gain extensive use lately but it is out of the scope of this course The pure crystalline Si also known as intrinsic Si shown above has the following properties 0 i has 4 bonding electron available to form covalent bonds 0 At absolutely 0K temperature perfect Si crystal has NO free electron won t conduct current However at room temperature 300K thermal excitation can ionize some electron from covalent bonds to form free electron so Si can conduct mall current Si is relatively easy to thermally excite free electron easier than insulators eg quartz that is why it is called SEMI conductors When Si produces a free electron it leaves a hole on bonding site shown above le Course Notes for BB 0257 Analysis and Design of Electronic Circuits 0 Free electron can move around lattice 0 The hole also can move by accepting adjacent electron hopping 0 Once a hole is lled with a FREE ELECTRON not adjacent hopping electron we have a recombination 0 Apparently electron and hole will move at different velocity in Si They are called carriers 0 We can describe the motion of hole as the motion of positive charge so as electron as negative charge 0 Apparently the concentration of electron and holes are identical subscription 139 below refer as Intrinsic n p n1 0 The concentration of intrinsic carriers depends on temperature E n BT 327 B 54gtlt1031Sz39 EG 11zeV k 862x10 SeVK n1 z l5gtlt101 carrierscm3 T 300K n51 5 gtlt 10ZZ atomscm3 Now let s talk about how electron and hole move in Si There are two 39 39 39 39 39 39 and drift nIIuII J u j J I tant in S1 W 0 Diffusion and Drift occur for both electron and holes 0 Diffusion occurs when carriers does NOT distribute uniformly thus diffusion produces net electric current Ifhole and electron concentration are denotes px and nx the diffusion currents D p 126mZ s p 7 dp J qDP J an E D 34cmz s abc Where DP and Dquot are diffusivity of hole and electron for Si Note the and Sign Chapter 3 Diodes Hnlc canccntmtlnn p U 0 Drift occurs under external electric eld the drift velocity is vpidn ypE 1 48001412 Vs vnidn 7ynE y l3500mZVs Where up and p are mobility of hole and electron respectively the sig1 before electron suggests electrons move opposite toward E eld 0 The associated drift currents are indn qpx pE 7 qnx 7 ME qnxuE Jdn qpxp nx E 0 If should note two things 0 Ohm s law s resistivity p 7 l p qlpocm nxyl o Einstein relationship D D n 7 P VT V w 25mV Alp Doped Semiconductors For intrinsic Si electron and hole concentrations are equal However to enhance conductivity both electron and hole concentrations can be Course Notes for BB 0257 Analysis and Design of Electronic Circuits greatly increased by doping Si with donor electron rich and acceptor hole rich I Donors such as phosphorus provides one more electron I Acceptor such as boron provides one more ho e F nt pe Si if donor concentration is ND in thermal equilibrium the electron concentration is close to ND Where the subscript n denotes ntype 0denotes equilibrium Semiconductor physics yields that the product of electron and hole concentration remains constant quotno 39 PHD quotx 2 quotI Pno D Here electron is the majority carrier hole is minority carrier For ptype Si if acceptor concentration is NA in thermal equilibrium the electron concentration is close to NA Ppo NA Where the subscript p denotes ptype 0denotes equilibrium 2 quotp0 39Ppo n 2 in P0 7 quotA Chapter 3 Diodes The pnjunction under opencircuit conditions When ptype and ntype Si are brought together a pnjunction is formed shown below 1 The hole in pregion diffuses into nregion electron in nregion diffuse into pregion So around junction pregion has more net negative charge n region has net positive charge It forms a space charge region or depleted region It exists on both side of pnjunction These space charge region produce Efield the Efield produces a builtin voltage VD VT InN 1 n The space charge width in nregion and pregion are xv 7 NA 7 7 ZiN D Wdepixwarxpi q RNA Typically Wdep around 0 1 1 111 The potential field serves as barrier so majority carrier hole in pside and electron in nside has to overcome to get into the other side Thus diffusion current ID depends on V0 So now we have builtin Efield but how does it stop diffusion current Well don t forget we still have electron minority in p region although it is much less than holes When electron randomly gets into spacecharge region Efield will quickly swift it to nregion by drifting This process occur in nregion too this time is for holes Therefore a dynamic balance is established Diffusion current is egual to drift current I S I D Course Notes for BB 0257 Analysis and Design of Electronic Circuits The pnjunction under reversebias conditions If the pnjunction is reversely biased then 0 Space charge region becomes wide builtin eld get stronger o It will be more difficult to for majority carrier to diffuse through therefore ID drops The drift current 13 will remain relatively as constant since it only depends on Efield strength not voltage Therefore it will lead to a constant reverse current in steady state 115 5 This is where the reverse current of diode comes from gt 1 n 00000 90999 1 U1 The depletion capacitance In additional of reverse biased current as the voltage across the pn junction changes the width of the depletion region changes so as the total charge stored in This behaves like a capacitor To find the capacitance we like to find out the total charge q either positive or negative stored in the junction N N 1 qNDanq AWdEp WeJLLAmn K 1 KNA ND Chapter 3 Diodes As we can see the charge q vs VR is a nonlinear relation therefore the capacitance can NOT be derived by the definition as CQV Rather we define dynamic capacitor around bias point It is a small signal approximation often referred as junction capacitance or depletion capacitance C qu dV R VR V2 We can prove that The more general expression for Cj is c m le3712 Course Notes for BB 0257 Analysis and Design of Electronic Circuits The pnjunction under forwardbias conditions If the pnjunction is forwarded biased shown below then 0 The external voltage will lower then barrier voltage which encourage the diffusion of majority carriers from both sides 0 Now when more holes inject into nregion it increases minority carrier density in nregion ie it increase hole concentration in nregion o For the same token when more electron injects into pregion it increases minority carrier electron density in pregion o The drift current I stay the same since external voltage won t change the strength of Efleld 0 Therefore we have large forward current prcginn yltDcpchinn I 1 region v A cs5 conccmnnion mm Thermal equilibrium aluc o The excess minority carriers have highest concentration around edge of pnjunction shown above The follows an exponential decay using hole p in nregion as an example Pnx pnO Pn xn Pno 19707 mp V pltxgt lameVT 0 LP is the diffusion length for Si since minority carriers will recombine with holes LP is related to excessminority carrier lifetime as Chapter 3 Diodes LP 2 DP TP Diffusion length for crystalline Si l100 pm This NONuniform hole distribution will lead to a diffusion current D V JP q Fpno eV 1 e L F At the edge of pnjunction xx D V J q Fpnoe VT 1 F LP Since the excess minority hole diffuse in nregion it will recombine with electron to reduce the majority carriers electron s concentration However the external current will replenish the majority carriers In eguilibriuma the total current should be the same at any given crosssection charge conservation Therefore at nregion total current is A J DP q pn0 e 17 LP Using the same analysis total current in pregion is n I l Jquot q I 11170 eT J l he total current D D V 1JpJnAqA eVT 4 LP L l n Usingpn0 nlzND np0nlzNA I JP JnA qAni2 DP ie 1E Ise 1 LPND LnNA EG Since n12 BT38 7 therefore Is depends on temperature Course Notes for BB 0257 Analysis and Design of Electronic Circuits The diffusion capacitance In additional of reverse biased current when a forward voltage applies across the pnjunction the excess minority carriers are generated on both side of pnjunction as shown below Wu l i Depletion H region I region I u mgmn 4m 1 mnccmmlion l l l l l l l l l l I i i i i i i lt Thcrnml cqumtwrium These excess carriers are NET charge therefore a capacitance can be defined to describe this charge storage effect This is referred as diffusion capacitance Of course in forwardbias we still have depletion capacitance The total excess minority carrier can be found from the shaded area shown above 0 For nregion Qp Aq gtlt shaded 7 area 7 in 7 n side 0 For pregion Qquot Aq gtlt shaded 7 area7in 7 p side We can prove that L Qpi1p1p1p Qn Dn1nrnln QM TPIP 111quot E 2quot Where TT calls mean transit time Chapter 3 Diodes Now let s prove that Course Notes for BB 0257 Analysis and Design of Electronic Circuits Again the diffusion capacitance is not nonlinear we cannot use the typical de nition of CQV But only can de ne small signal diffusion capacitance or dynamic capacitance around biaspoint 5222 rT V V Now let s prove this too Depletion capacitance or junction capacitance 1 Like a reversebiased situation we also have depletion region capacitance under forwardbias We can nd that by replacing VR with 7V But the accuracy is poor As an alternative we typically use an estimation C m 2C J 10 Chapter 3 Diodes Course Notes for BB 0257 Analysis and Design of Electronic Circuits Chapter 5 Bipolar Junction Transistors BJTs This lecture covers Section 5354 l The transfer characteristics 2 Graphic analysis 3 Operation as a switch 4 DC analysis Notes Chapter 5 Bipolar Junction Transistors BJTs The transfer characteristics The basic con guration of a BJT ampli er is shown below which is referred as commonemitter CE circuit or groundedemitter con guration I v r l 7 Cum Acliw ismmim H mmlc l l I V 1 Sloluc cl 1 l l l R l l u 1 l C l l l o i x 1 Tim th The transfer function shown above can be broken down into three regions and explained here In general the relationship between the output v0 vCE and the input v1 vBE can be rst and universally related by the IV equation for the collector resistance RC v0 VCE VCC RCiC For 0ltv1lt05V the BB is shutoff no current ows into the base so the entire BJT is shutdown and acting as an opencircuit The output will be close to VCC That is from point X to Y When 05Vltv1 the BJT enters the active mode YZ region the iC and v1 become an exponential function Course Notes for BB 0257 Analysis and Design of Electronic Circuits v0 VCC iRclse iC I Se vV This will lead to a sharp exponential drop to the output until v0 drops to 04V below the base v1 The CE is turned on forward biased and the BJT enters saturation In the saturation region as we studied in the previous class the saturation resistance RCEM is very small so the circuit behaves like a closecircuit Since VCEM changes little between 0102V the saturation currents almost remain constant The Ampli er Gain Apparently to operate the BJT as a ampli er it must be biased in the active region YZ It will be ideal to be biased right at the center of the active region so we get the maximum output signal swing The bias point is labeled as Q or quiescent point In this region the transfer function remain relatively linear VBEIV IC 7138 V0 VCC Rclse A V52 A 7 7LIS8V RC ICRC VRC V T dvl VFV VT VT VT The above equation is correct as long as the output amplified signal voltage does not pull the BJT into the cutoff or saturation Otherwise the positive or negative part of the output signal will be clipped off Example Consider a commonemitter circuit using a BJT having 13103915 A39 a collector resistance Rc68k 2 and a power supply Vcc10V a Determine the bias voltage VB needed to operate the BJT at VCE32V what is the value ofIc b Find the voltage gain AV at this bias point If an input sinewave signal of 5mV is superimposed on VBE find the amplitude of the output sinewave signal Chapter 5 Bipolar Junction Transistors BJTs c Find the positive increment of VB above VBE that drivers the BJT to the edge of the saturation where VCE03 V d Find the negative increment in V3 that drives the BJT to within 1 of cutoff Course Notes for BB 0257 Analysis and Design of Electronic Circuits The Graphical Analysis Similar to the MOSFET a graphical analysis is illustrative to see how a BJT works as ampli er and the impact of the bias point to the signal dynamic range A typical B T con guration is shown below v Load Hurt Slope 7E L 0 w 0 First the linear load IV load line of the resistance RB will intersect the eXponential IV curve of the EB diode This will determine the input bias point 0 A superimposing of input AC signal vi will produce a vBE change as shown below 0 Once the bias point VBE is determine we shift the output ic vs vCE again the collector resistance determine the output load line 0 The intersect of the output load line with the IV curve of the BJT at the bias point VBE will yield the output bias point 0 The output current and voltage changes are then determined using the maXimum and minimum values of vBE along the output load line shown below 0 The location of the bias point Q will determine the allowable output signal swing Chapter 5 Bipolar Junction Transistors BJ Ts l lnsluntuncous loud 310 e 7 p R1 lines Almost linear scgmcnl 1 Time l C V m VBE gt 1 mm m l I 1 1 gt I l v V l39ime Time J 9 ii 0pc R39 iulu2 I l N Time kquot V I Vu VCC l r le M Time In Course Notes for EE 0257 Analysis and Design of Electronic Circuits Operation as 21 Switch To operate a BJT as an OnOff switch we need to swing it between the cutoff mode and the saturation mode Given a circuit shown below When the input vlt05V the EB turns off therefore no current ow through the collector the vaCC When 1 goes beyond 05 V the BJT is turned on If the input is a step TTL signal the input will lift the V55 up quickly to reach around 07V At this point a small change of V55 will cause a big increase of to and big voltage drop across Re When the Va is below V55 for about 04V So vaBV the BJT enters the saturation region At the edge of the saturation Edgeof saturation or EOS VC03V we can use this number to calculate both collector and the base currents I VCC 703 C EOS RC 1C 7 50539 15505 Vzaosa 155053R5 V55 At the edge of the saturation we can still use B in active mode to perform this analysis However in a practical situation we need to Chapter 5 Bipolar Junction Transistors BJ Ts drive the El T completely into saturation to ensure safe operations In this case the saturation voltage VCEWWJV rather than 03 V And we have to use the in the saturation region famed I 7 Vcc VCEmt Cm c I Cmt farced beyond Similar to the MOSFET a graphical analysis is illustrative to see how a JT Example The transistor shown below has B from 50 to 150 Find the value ofRB resulting in saturation with an overdrive factor of 10 T10 V IN Course Notes for BB 0257 Analysis and Design of Electronic Circuits Example 2 perform analysis of the following BJT circuits the Example 3 perform analysis of the following BJT circuits minimum 530 for this BJT r r A t T is v N W gnu we quot 1 wg r g39 H g i H 39 I 7 7 i I 10m gr m g H gm i lt4m 4 Chapter 5 Bipolar Junction Transistors BJTs Course Notes for EE 0257 Analysis and Design of Electronic Circuits Example 4 perform analysis of the following BJT circuit 3100 lJV In Ln 5 V o i707 tku 7 IUA UH gt HHOI1IJ Chapter 5 Bipolar Junction Transistors BJTs lav Biasing in BJT ampli er circuits Like MOS ampli er the biasing of BJT ampli er must lead to predictable and stable The circuits must be in sensitive of temperature drifting It also need to tolerate large variation of 3s in circuits consist of different type of BJTs Let s rst exam two con gurations that is sensitive to temperature and 3 changes 4 Rm We see some drawback on these con gurations 0 The sharp exponential relationship Itst will lead to a large drifting ofIc and VCE So this bias is not temperature stable 0 For the second con guration11S VCC 7 07R1S IC IB however if a circuit unit consists of a lot of BJT the variation of 3 make this con guration unpredictable One power supply bias scheme Vt L V t I i 39ie A R R R i I I a i Ixl39 Aquot l w KL 3 RL 4 m 1b Course Notes for BB 0257 Analysis and Design of Electronic Circuits First let s see how the introduction of RE increase the l stability R R R Z Vac R5 1 2 R1 R2 R1 R2 1 V55 7 V55 E R 7R5 H1 To make the circuit insensitive to temperature and 3 we just need to V55 R V55 gtgt VBE 07V RE gtgt 1 Which is quite achievable some more design consideration 0 If VBB is too high then the voltage drop across RC and VCR is limited we will not have enough voltage spans for signal The general rule of thumb is V33 N V513 VCC VCBI3 VCC and chI3 Vcc To satisfy the 2nd requirement we need R3 to be small which also leads to large drain of current across the voltage divider and small input resistance The tradeoff leads to the current ow through the voltage divider R1 and R2 to be 01 1 5 RE will also provide feedback to stabilize the bias current If two power supplies are available the biasing circuit is shown below 1 1 LL Chapter 5 Bipolar Junction Transistors BJTs Collectortobase feedback bias scheme L 11 V00 ERG T 5R5 T VBE A A I in INTy IERC 1R5VBE R R l I v 7 Mn INN IVE V55 7 V55 E R iR Jrl l I39ll The resistance R 3 determines the signal span becomes R V05 IBRB E fl The feedback resistance R 3 also provides negative feedback which also stabilize the circuit We will eXplain this in class Constantcurrent bias scheme The best way to bias a circuit is of course provide constant and never change current I 5 this can be done by a constantcurrent source LL 9 R1 tgk Ed 7 1 REFi R Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Chapter 2 Operational Ampli ers This lecture covers Chapter 2123 1 Function and characteristics of an ideal Opamp 2 The inverting con guration 3 The noninverting con guration Notes 4 This is the rst of 5 lectures to cover chapter 2 5 Study Example 22 Chapter 2 Operational ampli er A few words about opamp Opamp is one of the most popular function blocks to build sophisticate electronic system and instrumentation Today opamps are available at low cost with great varieties of characteristics for instrumentation and circuit development Since the learning of opamp doesn t involve the knowledge of semiconductors it can be treated as a blackbox and easy to learn too Symbol and Characteristics of an ideal 0pamp quotQ U lb The opamp is designed to sense the difference between two input voltage signals An ideal opamp should 0 Has in nite large input impedance doesn t draw current Output impedance is ZERO as an ideal voltage source An ideal opamp should have in nite bandwidth The output is always equal to AvZ 7 1 Terminal one is referred as inverting input terminal terminal 2 is noninverting terminal If input 1 and 2 are the same the output will be zero this is referred as commonmode rejection The gain A is referred as differential gain or openloop gain which is very large gain N A gt 00 Therefore an opamp should never been used in an openloop con guration Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Now let s do the math During this derivation we assume Aaa luwxlmg Inpul Hulpn O 7 F 4 39 In H7 u t rrlui gmuud e PIHACF hllppl common lcrmmnl h Thing to notice virtual ground The inverting con guration The basic con guration for an opamp involves two resistors R1 and R2 shown below 0 Since R2 links terminal 3 and negative input terminal 1 it provides a negative feedback 0 If R2 links terminal 3 and the positive input 2 it will provide positive feedback R V0 R2 VI Rl v v v I I Input res1stance R R1 1source 11 v1 R1 ltlti v 7 I Output res1 stance Row 2 m i 0 z III shonicimuit so Chapter 2 Operational ampli er Course Notes for EE 0257 Introduction to Lasers and Optical Electronics Apparently A is not no we can redo the math by taking a more practical scenario by assuming A is nite but a large numb aquot chapter 2 Operational ampli er Examples The weighted summer The weig ted summer with the same sign Course Notes for EE 0257 Introduction to Lasers and Optical Electronics The Non inverting con guration 0 The Weighted summer With the opposite signs Another commonly used configuration Will yield a positive gain as noninverting con guration 277 e we i T ZN e i Chapter 2 Operational amplifier Course Notes for EE 0257 Introduction to Lasers and Optical Electronics Apparently A is not 00 we can redo the math by taking a more practical scenario by assuming A is nite but a large number Chapter 2 Operational ampli er Application voltage follower Buffers are function block commonly used in microelectronics It has high input impedance and low output impedance In a buffer gain is not important but the ability to drive lowimpedance load is An op am with noninverting con guration can do that this is referred as a voltage follower or unitygain amplifier Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Chapter 2 Operational Ampli ers This lecture covers Chapter 2425 l A single OpAmp difference amplifier 2 Instrumentation amplifier 3 Frequency response of openloop amplifiers 4 Frequency response of closedloop amplifiers Notes Chapter 2 Operational amplifier A few words about difference differential amplifiers Difference amplifiers are designed to respond to the difference between two input signal Difference amplifiers are used to eliminate noise which are commonly identical but sometime much larger than signal level If input of an amplifier has commonmode input VIM applied identically to two input leads of an amplifier And a differential input signal vId also apply to two input leads the output signal of a linear amplifier can be characterized by two gains Ad and Am V0 AdVId Acmvkm An ideal difference amplifier magnifies differential input much larger than the commonmode signal which is characterized by the commonmode rejection ratio CMRR CMRRe 201 WI 7 og A gtgt1 cm 1 Single Op Amp Difference Ampli er The opamp is designed to be a difference amplifier However since the gain of the op amp is too large without any feedback mechanism it is not practical to use it alone In the last lecture we learned two fundamental configuration of an op amp inverting and noninverting configuration shown be ow Inverting Noninverting Course Notes for BB 0257 Introduction to Lasers and Optical Electronics The gains for these two con gurations are negative and positive respectively A A Invertmg R Narmvemng R 1 1 We can therefore combine two con gurations to make an ideal difference ampli er By properly choosing feedback resistance zero commonmode gain can be achieved The resulting circuit is shown below Let s do the math to select resistance In this lecture we will use a different derivation from the textbook Chapter 2 Operational ampli er Using the same procedure we can calculate the commonmode gain Course Notes for BB 0257 Introduction to Lasers and Optical Electronics In additional to rejecting CM signal we wish to have a high input resistance to differential signal We can nd out this by A39 39 A 0 J l Imml lnu mum i M Rid 239Iel The analysis of this circuit is straightforward Two things Di fiGain Ad R 4 1 R3 R1 Input 7 Re sis tan ce 00 The Instrumentation ampli ers However a few problems arise Note from the above analysis if the ampli er needs to have a large 0 The input CM signal will be ampli ed by the ISI stage which differential gain RgR 1 then R has to be small so as the input might saturate the 2quotd stage Even the 2 stage is not saturated by resistance This is a signi cant drawback of the single opamp the CM signal the CMRR will be reduced differential ampli er 0 Two ampli er channels must be perfectly matched otherwise mismatch will appear as a differential signal to spur the real signal This can be resolved by buffering the two input terminals using To overcome this problem we come up with the following voltage followers The additional bene t is to get some additional gain con guration in the rst stage The circuits is shown below Chapter 2 Operational ampli er 3 Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Chapter 2 Operational ampli er Some nonideal characteristics of opamps In this section we consider some nonideal properties of op amps Since this information are important for electronic designers They are normally available in data sheet Frequency response of an openloop opamp For an opamp as it is the typical frequency response openloop is singletimeconstant STC lowpass response shown below lil dB 20 dBI decade UI 76 d3 octave m 101 Ii 10 10 10quot Ii7 MHZ T t Since we only have SINGLE time constant the gain A as a function of frequency 03 can be written as Altcogt A o l C 0 0 or is 3dB corner frequency a or f is the unity frequency Since we only have single time constant the slope is 20 dBdecade For agtgt 01 A Aw 9 60 e 1 wt A060 160 1w Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Frequency response of an closed loop op amp Having familiarizing the closeloop speci cation on frequency response we can calculate the frequency response for closeloop configuration Inverting con uration with nite gain A V0 R2 R1 w A0 K 11R2R1A lsab Substituting1a into the closeloop gain we have V S R2 R1 a Aowb 143 1ilR2Rl A0 a VS A A0 gtgtlt1R2R1 1 4 a 1R2R1 An important conclusion we get out of this derivation is the 3dB corner frequency a w 3 1R2 R1 Non inverting con uration with nite gain A V0 1 R2 R1 A0 m K 11R2R1A lsab If A0 gtgt lR2 R1 we have V0s N 1R2 R1 V16 1 4 a 1R2 R1 Chapter 2 Operational amplifier Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Chapter 2 Operational Ampli ers This lecture covers Section 26 27 l Imperfection of opamp saturation slew rate 2 Fullpower bandwidth 3 Offset voltage offset currents and input bias Notes Chapter 2 Operational amplifier In this lecture we study the limitations on the opamp performance This rst includes output voltage and current saturation Output voltage and current saturation In practical Op amps will operate linearly over a limited range of output voltage and current For a typical opamp 741 the range for voltage and current are around i1 015 V and i1 020 mA Beyond this range the output will be nonlinearly distorted eg cutoff Slew rate and fullpower bandwidth Another nonlinear distortion for large output signals are socalled slewrate limiting This refers to the maximum alow rate of change of signal de ned as dvo dt I WAS An opamp will not respond any signal faster than the maximum slew rate An example is shown below for a stepchange input and it is output for a voltage follower SR One thing to note here is the slewrating limiting is a nonlinear effect which is different from a nite bandwidth distortion The finite bandwidth distortion is a linear effect which does not change the shape of input sinusoid The slewing does change the shape of sinusoid To further clarify this point let s compare the effect of finite bandwidth The transfer function for a voltage follower is V 1 721sa I Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Its step response would be an exponential function U I As long as V is small so that aVltSR the output will follow above as a linear response Another example for nonlinear SR limiting can be illustrated using a sine wave output v0 V0 sin at The changing rate is given by 0 dt So the maximum changing rate is given by 0V0 which depends on both the maximum output voltage and frequency If 0V0 exceeds maximum SR distortion happens as shown below 0V0 cos at I hmlulvml umpm Humm Mien up Imp mm c lmmcd The opamp data sheets provide a frequency fM as the fullpower bandwidth If the maximum output voltage is Vomax then the fM is related to SR as wMV SR amax SR f T 27rV amzx Chapter 2 Operational amplifier DC imperfections offset voltage Since opamp has a very large gain any imbalance between two inputs can instantly saturate output Unfortunately in practical application a number of facts can contribute to the mismatch between two inputs including the opamp designs Given a typical dc gain of gt10000 even a mismatch of a few mV can saturate the opamp This is a must addressing issue To model the dc offset effect an opamp can be modeled as following Generally speaking mmquot quotpmquot 0 V0 range from 15 mV 0 V0 depends on temperature uV C O Ulmm Now let s analyze the impact of offset to performance of a opamp with negative feedback Offsetfree op amp Course Notes for EE 0257 Introduction to Lasers and Optical Electronics One way to overcome the dc offset is by capacitive coupling shown below Since capacitor is an open circuit for DC e opamp won t amplify the V05 however this does not work for an opamp circuits working in dc and low frequency Offset ft ac M 117 Notice the gain of such con guration will become very small at low frequency Here is the analysis Chapter 2 Operational ampli er Input bias and offset currents In a practical opamp both input terminal are supplied with dc currents to mction These two currents can be modeled with two current sources The average of these two currents is referred as input bias current The different between these two currents is referred as input offset current 151152 15 05151 1 2 Given the technology used to build opamp 15 range tom 2A to 100 nA I is one order ofma nitudes smaller than I whatever it is 52 The dc output voltage of a closedloop opamp cue to the input bias currents can be easily found out by considering the an inverting con guration The dc offset voltage becomes V0 15le BRZ The allowable dc offset voltage apparently will be used to determine what is the maximum allowable R2 One way to reduce the dc offset voltage will be connecting the positive input terminal with a resistance R 3 The following analysis justi es the solution and provides a guideline to choose R 3 Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Chapter 2 Operational Amplifiers This lecture covers Section 28 l Inverting con guration with general impedances 2 Inverting integrator 3 Opamp differentiator Notes Chapter 2 Operational ampli er A few words about integrators and differentiators Together with summation subtraction integration and differentiation are two important signal processing algorithms These functions can be readily realized by opamp General Impedances Although you might be still studying RLC circuits the concept of impedance for inductors and capacitors can also be understood without too much difficulty Or you can take the following concept as they are for now For DC a capacitor is an open circuit element with a resistance of 00 An inductor is a short circuit with a resistance of 0 For AC however both capacitors C or inductor L will produce resistance for any AC signal This is referred as impedance For a single frequency sine sig11alAsinat the impedance for A capacitor C 1ij An inductor L j39mL A resistor R R Apparently the total impedance of circuits depends on frequency Sometime fa is replaced by 3 during circuit analysis In the event that we need find out the timedomain response we can convert the frequencyresponse into the transient response by Replace ljoa by J Re lace 39o b i P J y dt Then a frequency response will be converted into a linear differential equation Given the initial condition and input signal the output signal can be readily calculated by the circuit Now let s use an example to illustrate this concept Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Example for the circuit below derive an expression for the transfer function V0sVs show that the circuit is a lowpass circuit Find DC gain f0Hz and 3dB frequency Design the circuit to obtain a dc gain of 40 dB a 3dB frequency of 1 kHz and an input resistance of 1 k9 l397 Ill lf I Chapter 2 Operational amplifier The inverting integrator The basic configuration for n inverting opamp integrators is shown below I 7 mm i u an m mw lhl I We now do the math to perform analysis in both time and frequency domain Time domain the IV crossing the C and the transfer function are 1 t Vct V0 EJOlltdt 1 t Vot 7 0V1tdt7 Va Frequency domain The transfer function is KW 1 Vim ijC V1 kw Vx wRC 1RC is referred as integrator frequency RC is known as integrator time constant Generally speaking an integrator is a lowpass filter with a corner 3 013 frequencv of ZERO 2 Course Notes for BB 0257 Introduction to Lasers and Optical Electronics DC o sel The basic con guration of integrator shown above has a problem at DC Since a capacitor is an opencircuit at DC the opamp will have no negative feedback and can saturated immediately Even at AC any dc offset can be deleterious too The following analysis explains DC o sel voltage DC o sel current Chapter 2 Operational amplifier The solution for the dc offset can be alleviated by connecting a resistor R p Thus the transfer function becomes V s RF R a Vs 71sCRF K To remove the dc offset one would chose low value for RF However the low value for RF will lead to high comer frequency lCRF which will distort the integrator performance Therefore a design tradeoff needs to be carefully entertained R Course Notes for BB 0257 Introduction to Lasers and Optical Electronics The inverting differentiator Interchanging the C and R of the integrator results a differentiator circuit We can perform both timedomain and frequency domain analysis to obtain the transfer functions I R A Chapter 2 Operational amplifier RC is referred as differentiator timeconstant The ideal differentiator can be considered as a highpass filter with a comer frequency at infinity Differentiator output will spike or very sensitive to the sharp change of t e input Differentiators are not stable and should be avoided to use alone in practice Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Chapter 2 Operational ampli er Course Notes for BB 0257 Introduction to Lasers and Optical Electronics A few words about ampli er circuits Chapter 1 Introduction of Amplifiers 0 Amplifier circuits are the core of any analog circuits 0 Input signal from transducers convert signal in electric form are week 0 Signal amplification is a must before AD conversion filtering noise cancellation encryption FFT modulationdemodulation broadcasting etc etc 0 Now let s discuss in some details on what are those index to characterize performance of a given amplifier Amp circuits This lecture covers Chapter 1416 1 Signal amplification 2 Characteristics used to analyze and gauge amplifier circuits gain linearity saturation nonlinear transfer biasing frequency response 3 Circuit model for amplifiers Characteristics of an Ampcircuit Notes 0 Amplifier circuits symbol 4 This is the only lecture to cover chapter 1 5 You need read section 1113 input Oulpm 6 Section 17 Will not be cover nor required a Circuit symbol for amplifier b An ampli er with a common terminal ground between the input and output ports 1 Voltage gain power gain and current gain o Voltag gain AV E quot I i 0 Current ga1n Ai E quot i P v 139 0 Power ga1n AP E L 0 1 VIII Chapter 1 Introduction to amplifier 1 Course Notes for BB 0257 Introduction to Lasers and Optical Electronics Expressing gain in Decibels 0 Voltage gain in dB 201ogAV dB 0 Current gain in dB 2010gAT dB 0 Voltage gain in dBlOlogAP dB 2 Power supplies An amp needs to draw DCpower to amplify signal signal get more power A more realistic ampcircuit symbol is shown below o In the above figure an amp needs a DC to signal signal so as the need of DC supply This is not convenient in micro electronic circuits sometime that is why many amps need to be biased The Fig b is commonly used as a simplified symbol Pdc VIII V212 Pub 131 13L Pdissipafed 0 Since the power drawn from the signal is usually small the amplifier power efficiency can be defined 77 P L gtlt 100 Pdc 3 Saturation Ideally we hope our amp can remain linear to amplify signal with any strength with the same gain Practically an amp operated from two DC supply can pour out a output voltage more than the voltage of DC supplies So there is a saturation issue illustrated in the follow figure Chapter 1 Introduction to amplifier 39 ll 7quot l H hv a mm L quotv a G3 gt lt g gt 39 5 To avoid distortion of the output signal the input signal must be kept with the linear range L L Sv1 S Nonlinearity and Biasing Practically the transfer function Voutput vs Vinput is not linear at all only part of the transfer function is linear Therefore we need operate the amp at the middle of the linear region shown above this point is referred as DC bias point Q point or operating point or quiescent point This can be done by adding your signal on the top of DC bias voltage Of course both DC bias signal cannot exceed linear range 2 Course Notes for EE 0257 Introduction to Lasers and Optical Electronics vI V Vot Vot Avvit dvo AVd VI Q Example A transistor amplifier has the transfer characteristic v0 10 lO e40VI Find L L dc bias voltage V that results in V0 5 V and gain muquot ID I 0 new mar 11 m 5 Some notations 0 Total instantaneous quantities DC AC signal 139At vct DC quantity IA Vc Power suplly VDD DD Signal quantity pure AC 1395 t vct Amplitude of sine signal In V0 Chapter 1 Introduction to amplifier Frequency response If an amp operates in linear region then an ampcircuit is a linear circuit In we feed a sine signal to a linear circuit the output will be the sine signal with the same frequency however the amplitude and phase Will be different 0 lug I up i lam m r gtl I I I 2 I To characterize the frequency response we need the ratio of output amplitude to the input amplitude and the phase change V a T 0 ltwgt K m 4M w o Bandwidth the band of frequencies over which the gain of the amplifier is ahnost constant to within a certain number of dB usually 3 dB or 50 Anal sis of fre uenc res onse low ass LP and hi h ass HP 7 Couise Notes for BB 0257 Introduction to Lasers and Optical Electronics 0 Lowpass filter Bode plots o Highpass lter Bode plots awlliml In Chapter 1 Introduction to ampli er Circuit models As always any amp circuits must be integrated into a system to perform For any functional electronic circuits it consists tens to thousands unit To analyze the performance and characteristic of the system we must simulate or model individual ampcircuit 1 Voltage amplifier models Course Notes for BB 0257 Introduction to Lasers and Optical Electronics For an ideal amp one needs a few things 0 Large input resistance 0 Respectable voltage gain 0 Low output resistance buffering o Desired power gain These cannot be accomplished by singlestage amp circuit rather we need cascade or multistage ampli er Example evaluate the vltage current and power gain of the following cascade amp J 1 mix 1 r mi til 1kg n ma 1 rota4 Chapter 1 Introduction to ampli er 2 Other ampli er types Although voltageamp is the most popular one in various applications we might wish our amp can amplify different quantity such as robust current ampli cation high output resistance Therefore in different situations we wish to apply different circuit models using either voltage source or current sources TABLE 11 The Four Ampll er Types Circuit Model Gain Parameter Ideal Characteristics 9 quot pevalrcuxl Voltage oim R vi R1 A RF Type Vollagc Amplifier tinmm Ampli er Transconducmnce mpli cl Slinl i ircml Transmnducwnc R m Gm E 1 AN R39 39r n Trunsresismme fl R v Am i 1 gt quotr OpenCil uullTmnsreslsiimte R lt 0 i i Rm 5 L VIA R n i

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