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Introductory Electronics Laboratory

by: Kris Heathcote

Introductory Electronics Laboratory EL ENG 43

Kris Heathcote

GPA 3.78


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This 14 page Class Notes was uploaded by Kris Heathcote on Thursday October 22, 2015. The Class Notes belongs to EL ENG 43 at University of California - Berkeley taught by Staff in Fall. Since its upload, it has received 33 views. For similar materials see /class/226762/el-eng-43-university-of-california-berkeley in Electrical Engineering at University of California - Berkeley.

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Date Created: 10/22/15
EE 43100 Operational Amplifiers OpAmps Experiment Theory 1 Objective The purpose of these experiments is to introduce the most important of all analog building blocks the operational ampli er opamp for short This handout gives an introduction to these ampli ers and a smattering of the various con gurations that they can be used in Apart from their most common use as ampli ers both inverting and noninverting they also nd applications as buffers load isolators adders subtractors integrators logarithmic ampli ers impedance converters lters lowpass highpass bandpass bandreject or notch and differential ampli ers So let s get set for a fun lled adventure with opamps 2 Introduction Ampli er Circuit Before jumping into opamps let s rst go over some ampli er fundamentals An amplifier has an input port and an output port A port consists of two terminals one of which is usually connected to the ground node In a linear ampli er the output signal A X input signal where A is the ampli cation factor or gain Depending on the nature of the input and output signals we can have four types of ampli er gain voltage gain voltage out voltage in current gain current out current in transresistance voltage out current in and transconductance current out voltage in Since most opamps are voltagevoltage ampli ers we will limit the discussion here to this type of ampli er The circuit model of an ampli er is shown in Figure 1 center dashed box with an input port and an output port The input port plays a passive role producing no voltage of its own and is modelled by a resistive element Ri called the input resistance The output port is modeled by a dependent voltage source AV in series with the output resistance R0 where V is the potential difference between the input port terminals Figure 1 shows a complete ampli er circuit which consists of an input voltage source V5 in series with the source resistance R5 and an output load resistance RL From this gure it can be seen that we have voltagedivider circuits at both the input port and the output port of the ampli er This requires us to recalculate V and V0 whenever a different source andor load is used R V V 1 RSJrRl EE 43100 Operational Amplifiers 2 0 V I I h Q5U fU 5U h a NV 3 INPUT PORT iEIOd indinO AMPLIFIER Figure 1 Circuit model of an ampli er circuit 3 The Operational Ampli er Ideal Op Amp Model The ampli er model shown in Figure 1 is redrawn in Figure 2 showing the standard opamp notation An opamp is a differential to singleended ampli er Le it ampli es the voltage difference VP 7 Vquot V at the input port and produces a voltage V0 at the output port that is referenced to the ground node of the circuit in which the opamp is used Figure 2 Standard opamp Figure 3 Ideal opamp The ideal opamp model was derived to simplify circuit analysis and is commonly used by engineers for rstorder approximation calculations The ideal model makes three simplifying assumptions Gain is in nite A oo 3 Input resistance is in nite R oo 4 EE 43100 Operational Amplifiers Output resistance is zero R0 0 5 Applying these assumptions to the standard opamp model results in the ideal opamp model shown in Figure 3 Because Ri co and the voltage difference Vpi Vquot V at the input port is nite the input currents are zero for an ideal opamp in ip 0 6 Hence there is no loading effect at the input port of an ideal opamp K V 7 In addition because R0 0 there is no loading effect at the output port of an ideal opamp V0 A X V 8 Finally because A ooand V0 must be nite V Vpi Vquot 0 or VP Vquot 9 M Although Equations 35 constitute the ideal opamp assumptions Equations 6 and 9 are used most often in solving opamp circuits Figure 4a Noninverting amplifier Figure 5a Voltage follower Vout VOW Vout Agt1 A1 Vin Vin Alt0 Vin Figure 4b Voltage transfer curve Figure 5b Voltage transfer curve Figure 6b Voltage transfer curve of noninverting amplifier of voltage follower of inverting amplifier EE 43100 Operational Amplifiers A Vout 1 V0111 Vout Vpower Vpower Vpower Agt1 A1 Vn Vin Alt0 Vin Vpower Vpower Vpower Figure 4c Realistic transfer curve Figure 5c Realistic transfer curve Figure 6c Realistic transfer curve of noninverting amplifier of voltage follower of inverting amplifier 4 Non Inverting Ampli er An ideal opamp by itself is not a very useful device since any nite nonzero input signal would result in in nite output For a real opamp the range of the output signal is limited by the positive and negative powersupply voltages However by connecting external components to the ideal op amp we can construct useful ampli er circuits Figure 4a shows a basic opamp circuit the noninverting amplifier The triangular block symbol is used to represent an ideal opamp The input terminal marked with a corresponding to VP is called the non inverting input the input terminal marked with a 7 corresponding to Vquot is called the inverting input To understand how the noninverting amplifier circuit works we need to derive a relationship between the input voltage Vin and the output voltage V0quot For an ideal opamp there is no loading effect at the input so VP V 10 Since the current owing into the inverting input of an ideal opamp is zero the current owing through R1 is equal to the current owing through R2 by Kirchhoff s Current Law which states that the algebraic sum of currents owing into a node is zero to the inverting input node We can therefore apply the voltagedivider formula nd Vquot R V 1 Vm 11 R1R2 EE 43100 Operational Amplifiers From Equation 9 we know that Vin VP Vquot so R V 1 2 V 12 m R J m The voltage transfer curve Vou vs Vin for a noninverting ampli er is shown in Figure 4b Notice that the gain Vom Vin is always greater than or equal to one The special opamp circuit con guration shown in Figure 5a has a gain of unity and is called a voltage follower This can be derived from the noninverting ampli er by letting R1 co and R2 0 in Equation 12 The voltage transfer curve is shown in Figure 5b A frequently asked question is why the voltage follower is useful since it just copies input signal to the output The reason is that it isolates the signal source and the load We know that a signal source usually has an internal series resistance Rv in Figure 1 for example When it is directly connected to a load especially a heavy high conductance load the output voltage across the load will degrade according to the voltage divider formula With a voltagefollower circuit placed between the source and the load the signal source sees a light low conductance load the input resistance of the opamp At the same time the load is driven by a powerful driving source the output of the opamp 5 Invertng Ampli er Figure 6a shows another useful basic opamp circuit the inverting ampli er It is similar to the non inverting circuit shown in Figure 4a except that the input signal is applied to the inverting terminal via R1 and the noninverting terminal is grounded Let s derive a relationship between the input voltage Vin and the output voltage V0quot First since Vquot VP and VP is grounded Vquot 0 Since the current owing into the inverting input of an ideal opamp is zero the current owing through R1 must be equal in magnitude and opposite in direction to the current owing through R2 by Kirchhoff s Current Law V V V V m n 011 n R1 R2 Since Vquot 0 we have R V 2V 14 The gain of inverting ampli er is always negative as shown in Figure 6b EE 43100 Operational Amplifiers 6 Operation Circuit Figure 7 shows an operation circuit which combines the noninverting and inverting ampli er Let s derive the relationship between the input voltages and the output voltage V0quot We can start with the noninverting input node Applying Kirchhoff s Current Law we obtain VBl Vp VBZ Vp4 VB3 Vp Vp 15 R Bl R 82 R 83 R 8 Applying Kirchhoff s Current Law to the inverting input node we obtain VAI Vn VAZ Vn VA3 Vn Vn V0ut 16 R A1 R R R A A2 A3 Since Vquot VP from Equation 9 we can combine Equations 15 and 16 to get ampH Vw L33 jRB ELM m jig 17 RA RBI R82 R83 RAI RAZ RA3 where 1 R I A 1 1 1 1 R RA1 R RA3 and R I 1 Thus this circuit adds V31 V32 and V33 and subtracts VA 1 VA2 and VA3 Different coef cients can be applied to the input signals by adjusting the resistors If all the resistors have the same value then V31 V32 V83VA1 VA2 VA3 18 V 011 Figure 7 Operation circuit EE 43100 Operational Amplifiers 7 Integrator By adding a capacitor in parallel with the feedback resistor R2 in an inverting amplifier as shown in Figure 8 the opamp can be used to perform integration An ideal or lossless integrator R2 oo 1 performs the computat1on V0 Ellm dt Thus a square wave 1nput would cause a tr1angle wave 1 output However in a real circuit R2 lt 00 there is some decay in the system state at a rate proportional to the state itself This leads to exponential decay with a time constant of r R2C Figure 8 Integrator 8 Differentiator By adding a capacitor in series with the input resistor R1 in an inverting ampli er the opamp can be used to perform differentiation An ideal differentiator R1 0 has no memory and performs the W t computation V0 R2C Thus a triangle wave input would cause a square wave output However a real circuit R 1 gt 0 will have some memory of the system state like an lossy integrator with exponential decay of time constant 139 R 1C 9 Differential Ampli er Figure 9 shows the differential ampli er circuit As the name suggests this opamp con guration ampli es the difference of two input signals V V V R 2 20 R1 If the two input signals are the same the output should be zero ideally To quantify the quality of the ampli er the term Common Mode Rejection Ratio CMRR is de ned It is the ratio of the EE 43100 Operational Amplifiers output voltage corresponding to the difference of the two input signals to the output voltage corresponding to common pa of the two signals A good opamp has a high CMRR Figure 9 Differential ampli er 10 Frequency Response of Op Amp The frequency response of any circuit is the magnitude of the gain in decibels dB as a function of the frequency of the input signal The decibel is a common unit of measurement for the relative loudness of a sound or in electronics for the relative magnitude of two power levels A decibel is onetenth of a quotBelquot a seldomused unit named for Alexander Graham Bell inventor of the telephone The gain expressed in dB is 20 loglolGl The frequency response of an opamp is a low pass characteristic passing lowfrequency signals attenuating highfrequency signals Figure 10 n Gain log scale G 3 dB point L B Freq Hz Figure 10 Frequency response of opamp The bandwidth is the frequency at which the power of the output signal is reduced to half that of the maximum output power This occurs when the power gain G drops by 3 dB In Figure 10 the EE 43100 Operational Amplifiers bandwidth is B Hz For all opamps the GainBandwidth product is a constant Hence if the gain of an opamp is decreased its operational bandwidth increases proportionally This is an important tradeoff consideration in opamp circuit design In Sections 3 through 8 above we assumed that the opamp has in nite bandwidth 10 More on Op Amps All of the above opamp configurations have one thing in common There exists a path from the output of the opamp back to its inverting input When the output is not railed to a supply voltage negative feedback ensures that the opamp operates in the linear region as opposed to the saturation region where the output voltage is saturated at one of the supply voltages Ampli cation additionsubtraction and integrationdifferentiation are all linear operations Note that both AC signals and DC offsets are included in these operations unless we add a capacitor in series with the input signals to block the DC component 11 The LMC 6482 Operational Ampli er Guidelines for usage 0 Two powersupply voltages one positive one negative must be connected to the LM741 The input voltage should fall within the range between the two powersupply voltages o The powersupply voltages limit the output voltage swing The voltagetransfer characteristics of the noninverting ampli er the voltage follower and the inverting ampli er are illustrated in Figures 4c 5b and 5c respectively The output swing also depends on the load resistor RL In order for the LMC 6482 to work like an ideal opamp don t connect too heavy a load resistor of low resistance to it Avoid continuous operation under maximum current output otherwise it can burn your nger The input resistance of the LMC 6482 is not tremendously large so resistors like R1 and R2 shown in Figure 4 and 6 should not be too large Of course they should not be too small since neither the external signal source nor the LMC 6482 itself may have suf cient current driving capacity EE 43100 Operational Amplifiers LMC 6482 pins The opamp is packaged in an epoxy miniDIP dual inline package which has 4 pins emerging from each of the two sides of the package The most important part of the identi cation is to note that Pin 1 is marked by a small circular indentation in the lower lefthand corner of the package when viewed from the top side with the manufacturer s marking reading upright The pin numbers on ICs always start with Pin 1 being this labeled pin and increase counterclockwise as read from the top side of the package Thus on this particular package Pin 8 is directly across from Pin 1 and Pin 5 is diagonally across from Pin 1 You will need to use only 5 of the 8 pins These are Inverting Input 7 Pin2 NonInverting Input 7 Pin3 Negative Supply 7 Pin4 Output 7 Pin6 and Positive Supply 7 Pin7 Other pins are available for an optional offset null adjustment Pins 1 amp 5 which is unnecessary for the circuits used in this experiment Do not ground these pins as this will result in failure of the opamp Pin 8 has no internal connection into the silicon chip inside EE 43100 Operational Amplifiers OpAm ps Experiment G u ide In this lab we are going to study operational ampli ers and circuits with opamps The opamp chip that we are going to use is LMC 6482 from National Instrument The con guration of the chip is shown below It has two ampli ers in one chip with 8 pins The pin con guration is also shown in the same gure There is a node on the chip indicating pin 1 The power supply to the chip is 4 V for V39 and 6 V for V in this lab Maximum VV39 is 30V For more information please refer to the device speci cation a uuwm A g 39a 2 v IH39i39ERTIHS IHi L39I gt5 LquotUTFU1 H NIzlnmvmmsi imvmma INPUT a mm B I 5 HOH WERTIHS mm B 6111 7124 Part 1 Noninverting Ampli er a DC measurements 1 Build up the noninverting ampli er as shown in Fig 1 Use 25V channel and 25V channel of the DC power supply for the VDD and Vss the 25V should be set up to 6V and 25V channel should be set up to 4V Use 6V channel of the DC power supply for Vin and measure both input and output using oscilloscope R1 is 5k and R2 is 5k Change Vin from 2V to 3V to verify the proper ampli cation range of DC inputs 2 Fix the DC input 05V measure the ampli er gain VomVin for R2 2 k 5k IOkQ turn R2 and compare with the calculated gain You need to take out the pot from the circuit to measure its value b AC measurement 1 Now set the input signal to a 1 kHz 05 Vpp 0 VDC offset on the function generator display Sine wave from the function generator Use a IOkQ potentiometer as R2 Adjust R2 to see the gain change Can you get a gain less than unity by turning R2 Why EEAKIEIEI Operahunal Amplifiers 2 Tum the potentiometer R2 until the gain is 2 and then adjust the Vpp and DC offset to the input signal Observe the input and output wavefonns as you vary the DC offset for large Vpp say 25V 0 Clipping l Tum the potentiometer until the gain is N2 and then add a DC offset to the input signal 39 10VDC the FG should display OVDC to 05VDC 2 Draw L 39 A a L 39 quotrr39 i the peak tiough and satuiation levels referenced to ground Explain the conditions undei which distortion occurred Fig 1Nonrinver ng Ampli er EE 43100 Operational Amplifiers Part 2 Inverting Ampli er Using the unused opamp of the IC build the inverting ampli er as shown below please use the unused opamp now While you are building a circuit it is safer for the circuits if you turn the DC power supply OUTPUT OFF Let the input signal be a 1 kHz 10Vpp sine wave 0 VDC offset turn R2 to max What s happening to the output signal Adjust the input offset to make the output more complete Now adjust the potentiometer and observe the resulting change in the amplitude and offset of the output Adjust these two parameters until the gain is at its maximum and there s no clipping What range of gain do you have in this circuit Verify the correct amplification of both AC and DC signals What is the phase difference between Vout and Vin and where is it from Is it a delay Ill00ml Part 3 Cascaded connection Now we will study a cascade connection of two amplifiers Connect the output of the inverting amp to act as the input voltage for the noninverting amp Use R2 10k in the inverting circuit and Rz 5k in the noninverting circuit The input signal should be a 1 kHz SOmVpp on the function generator display sine wave and you have to pick the correct offset for the circuit to amplify linearly Adjust the input signal to make sure there is no clipping in the circuit Measure the gain of each stage separately and then the overall gain of this cascaded circuit Part 4 Integrator Put a 01 uF capacitor instead of R2 in a new inverting ampli er Fig 3 and measure the time constant Use a 60 Hz 500mVpp square wave as input After getting the waveforms and triggering correct measure time constant RC how will you measure it Hint your prelab question 4 EE 43100 Operational Amplifiers Compare measured time constant with theory Now change the function generator back to a sine wave input sweep frequency from le to 100kHz and observe the change of the gain with frequency Note on op amp integrator The circuit in gure 4 violates one of the cardinal rules of opamp circuit design there must always be a DC feedback path to the inverting input or the opamp output will go to the railquot The general problem with this integrator circuit is that a small error current input offset current will be integrated by the capacitor to be large output voltages and eventually drive the opamp output into saturation The LMC 6482 opamp you are using has remarkably low input offset currents so that you may not see this effect in a short time If you want to see this effect ask your TA for another pincompatible opamp such as the LM6142 substitute in the integrator circuit and see if you observe any difference in the average DC level of the output Typically a real integrator is made with a zeroreset or a large resistor in parallel with the integrator capacitor Part 5 Differentiator Build the inverting ampli er but put 01 uF capacitor in stead of R1 as shown in Fig 4 Use Rz5k Input a 500 Hz 500 mVpp triangle wave Zoom into the waveform to measure time constant RC Hint prelab question 5 Compare measured time constant with theory Add DC offset to the input signal is there any change on the output signal Why What happens when the input is a triangle wave


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