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ECE306 Lab 2, Feedback amplifiers 04.12.2018 LAB 2 FEEDBACK AMPLIFIERS PRELAB ASSIGNMENT 1. Read Sedra and Smith 5th ed.[1] ,sections 8.1,8.3,8.4 and 8.6.1 before coming to the lab. Read the laboratory instructions and compare to S&S. 6th ed. sections 10.1,10.3, 10.4 and 10.6. 2. Calculate the theoretical value of the voltage gain V2 /V1 for the circuit shown in Figure 2 of the notes. Note that there is no load resistor. 3. Calculate the theoretical value of the open loop voltage gain Vo'/Vs for the circuit shown in Figure 5 of the notes. 4. Calculate the theoretical value of the open loop feed-forward gain A for the circuit shown in Figure 5 of the notes. 5. Calculate the theoretical open loop output resistance Ro without feedback for the circuit shown in Figure 5 of the notes. 6. Calculate the theoretical loop gain Aβ for the circuit shown in Figure 5 of the notes. 7. Calculate the theoretical value of the closed loop voltage gain Vo/Vs for the shunt-shunt amplifier in Figure 4 of the notes. 8. Calculate the theoretical closed loop output resistance with feedback Rof for the shunt-shunt amplifier in Figure 4 of the notes. 9. Calculate the theoretical open loop feed-forward gain A for the series-shunt amplifier in Fig. 9 of the notes. 10. Calculate the theoretical open loop output resistance without feedback Ro for the series-shunt amplifier in Fig. 9 of the notes. 11. Calculate the theoretical loop gain Aβ for the series-shunt amplifier in Fig. 8 of the notes. 12. Calculate the theoretical value of the closed loop voltage gain Vo/Vs using the circuit in Fig. 8 of the notes. 13. Calculate the theoretical closed loop output resistance with feedback Rof using the circuit in Fig. 8 of the notes. 1 ECE306 Lab 2, Feedback amplifiers 04.12.2018 LABORATORY INSTRUCTIONS In this lab you will investigate the effects that negative feedback has on amplifiers. You will start with a basic building block amplifier with know parameters. You will apply feedback to improve the parameters. You will study two different types of amplifiers in this lab using similar evaluation techniques. Part 1, part 2 and part 3 have a theory section and lab procedures. Introduction: Feedback Amplifiers Basic Feedback Amplifiers, 1. A block and β block Amplifier Topologies 1. voltage-mixing voltage-sampling 2. current-mixing current-sampling 3. voltage-mixing current-sampling 4. current-mixing voltage-sampling Part 1: Basic amplifier module Procedure 1: Build the basic amplifier. Measure the voltage gain and output impedance. Part 2: Shunt-shunt amplifier Procedure 2: Build the A block of a shunt-shunt amplifier and measure the gain. Procedure 3: Measure the open loop output impedance of the shunt-shunt amplifier. Procedure 4: Measure the closed loop voltage gain of the shunt-shunt amplifier. Procedure 5: Measure the closed loop output impedance of the shunt-shunt amplifier. Part 3: Series-shunt amplifier Procedure 6: Build the A block of a series-shunt amplifier and measure the gain. Procedure 7: Measure the output impedance of the series-shunt amplifier. Procedure 8: Measure the closed loop voltage gain of the series-shunt amplifier. Procedure 9: Measure the closed loop output impedance of the series-shunt amplifier. List of Figures Figure 1: The basic block diagram of a negative feedback amplifier. Figure 2: The circuit schematic for the ‘Basic’ amplifier. Figure 3: ‘Basic’ amp model. Figure 4: The shunt–shunt topology model using the ‘basic’ amp. Figure 5: The A circuit model, note the feedback resistor, β, is disconnected. Figure 6: The circuit schematic for the shunt-shunt A block. Figure 7: The circuit schematic for the shunt-shunt closed loop amp. Figure 8: The series–shunt topology model using the ‘basic’ amp. Figure 9: The A circuit model for the series–shunt amplifier. Figure 10: The circuit schematic for the series-shunt A block. Figure 11: The circuit schematic for the series-shunt closed loop amp. 2 ECE306 Lab 2, Feedback amplifiers 04.12.2018 Introduction Negative feedback is used to improve the parameters of a basic amplifier building block. The simple op-amp would be almost useless without it. Negative feedback has a number of desirable effects. Among these are: controlled, well defined gain, extended bandwidth, improved input impedance, and improved output impedance. Feedback amplifiers are generalized into the form shown in Fig. 1. The A block usually has a significant amount of gain. The β block provides the negative feedback required to achieve those desirable effects listed earlier. The amplifier topology presented in Fig. 1 allows similar techniques to be used to evaluate the parameters and the performance of different types of amplifiers. The different types are listed in table 1. Negative feedback Figure 1 shows the basic block diagram for negative feedback. The x signals can be voltage or current. The signal types define the amplifier type as listed in Table 1. The gain of the feed-forward network is A, and the gain of the feedback network is β. xs xi Source A xo Load xf β Figure 1: The basic block diagram of a negative feedback amplifier. The text makes use of the idea of separating the amplifier function A from the feedback function β so that the general amplifier feedback equation can be used for any combination of x in voltage or current. This closed loop gain, Af, equation is: Af xo A x s 1 Aβ (1) Feedback topologies The feedback network samples from the output and mixes it back into the input. If the output is a voltage, then it must be sampled in shunt. If it is a current, then it must be sampled in series. If the input is a voltage, then it must be mixed in series. If it is a current, then it must be mixed in shunt. A feedback amplifier is denoted either by its input–output type (e.g. voltage-to-voltage) or its mixing– sampling type (e.g. series–shunt). The two descriptions are equivalent. Figures 8.XX taken from [1]. helps understand the series / shunt configurations. Typical amplifier type Voltage Amplifier Current Amplifier Transconductance Amp Gain VO/VIN IO/IIN IO/VIN Reference topology name voltage-mixing voltage-sampling current-mixing current-sampling voltage-mixing current-sampling 3 Alternate topology name series–shunt shunt–series series–series ECE306 Lab 2, Feedback amplifiers Transimpedance Amp VO/IIN current-mixing voltage-sampling 04.12.2018 shunt–shunt Table 1: The four basic feedback amplifier topologies. series- shunt shunt-series series-series shunt-shunt Figures 8.XX: The four basic feedback topologies In this lab, we will investigate two different feedback amplifier topologies: a current-mixing voltagesampling (shunt–shunt) and a voltage-mixing voltage-sampling (series–shunt). See Table 1. In voltage-sampling current-mixing, "voltage-sampling" refers to sampling the feedback voltage at the output and "current-mixing" refers to summing the source and feedback currents at the input. In (series–shunt), "series" refers to the mixing configuration at the input and "shunt" refers to the sampling configuration at the output. Basic amplifier Figure 2 shows the ‘basic’ amplifier block we will use throughout this lab. This basic building block amplifier is used instead of simply an op-amp. The reason for using the ‘basic’ amp is because a simple op-amp has near ideal parameters. That means it is almost impossible to measure them or the 4 ECE306 Lab 2, Feedback amplifiers 04.12.2018 parameters of a feedback amplifier constructed with it. In other words we will use a poor op-amp for the following experiments. The parameters we are concerned about measuring are the input and output impedance and the gain. This building block will be used in conjunction with other circuitry to form a feedback amplifier. The other circuitry connected to this ‘basic’ amplifier will determine both the A and the β of the feedback amplifier. As you will see the ‘basic’ amp becomes the A block when you include the loading affects of the source, load and β impedances. First let’s analyze then build the ‘basic’ amp. We need to determine the parameters of interest, namely, the input and output impedance and the gain. We can determine them from analyzing the circuit within the rectangle which is the ‘basic’ amp. Then we will build it and measure them. Rf Vplus 100K Vminus 4 5 Vminus 2 V1 GND U1 Vo V2 R1 1K C1 0.1 C2 0.1 1K 7 1 3 Ro 6 LF356 + 1K - Rin GND Vplus Basic Amplifier Ceramic Bypass Caps Caps in uF 1.0 uF Bypass Caps are better Mount close to power pins Figure 2: The circuit schematic for the ‘Basic’ amplifier. Inside the basic amp is a standard inverting op-amp circuit with additional resistors, R1 and RO. We will get to RO later. R1 does not influence the parameters that we are concerned with. It is there to keep the DC offset of the basic amplifier low. Most op-amp circuits result in a lower DC offset when each of the two input terminals is loaded by the same impedance. This is from ‘op-amps 301’. On to the parameters of interest. You probably know that the closed loop gain of this circuit is usually given as in Eq. 2. VO Rf (2) V1 Rin This gain is given as a ratio of VO to V1 like the voltage amp topology. This equation is correct. Note VO is inside the basic amp block. This amp can also be represented as a trans-impedance amp. We can also call it a shunt–shunt or a current-mixing voltage-sampling circuit. Trans-impedance has a VO/IIN gain equation. Well it does have this equivalent gain but we like to think about the inverting amp gain as VO/V1. The trans-impedance gain equation for the basic amp is given in Eq. 3 and the input and output impedances are given in equations 4 & 5. VO Rf Iin (3), V1 Rin Iin (4), V2 RO IX V10 See the appendix if you want to know how we get these equations. It’s easy. 5 (5) ECE306 Lab 2, Feedback amplifiers 04.12.2018 So let’s calculate the gain as given in Eq. 1 using A and β for the inverting op-amp block to see if we get Eq. 3 as a final result. A = -100,000,000 and β = -1/100,000. See the appendix to understand where these values come from. It’s hard but worth the effort to understand it. Equation 1 gives: Af A 100M 100M 100M 99,900 VA 100K VA 100M 1 Aβ 1 100K 1 1000 1001 (6) Equation 6 demonstrates that the gain equation and analysis technique (appendix) do agree. We can draw a new model for our ‘basic’ amp using the parameters we just determined in our analysis. Figure 3 shows the new ‘basic’ amp model. This amp model makes it easier to analyze our feedback amplifier configurations. Ro 1K V1 Rin + V1 GND 1K V2 -100V1 GND Basic Amplifier Figure 3: ‘Basic’ amp model. From now on we will present circuits with the basic amp model shown in Fig. 3 followed by the actual circuit built for you on the proto boards from Fig. 2. We will not show the bypass capacitors in the following circuits but they are used. Note picture (1) to the right is the proto-board with the basic amplifier circuit. V+ (1) and V- (7) are the power supply pins. +Vin (6) is grounded in Fig. 2. –Vin (3) is the input V1 in Fig. 2. Picture 1. Basic Amp Prototype GND (2) is where the bypass capacitors are grounded. NC (4,5) are not connected to anything on the board. Finally VO‘ (8) is V2 in Fig. 2 and is usually referred to as VO’ in the text. There are 2 small wires sticking up from the board acting as test points for the power supply voltages that appear at the VCC (7) and VDD (4) pins of the op-amp. 1. The “basic amp” circuit shown in Fig. 2 is already built for you. See Picture 1. Measure the voltage gain V2/V1 at 1 kHz and compare with the theoretical value. Measure the -3dB frequency, with respect to 1KHz, and the unity gain frequency for the basic amplifier. Measure the output resistance of the basic amplifier, as described below. It is easy to measure the output impedance. Measure V2 (VO’) for a 1KHz sine wave input, pick an input level that doesn’t clip the output signal, with no load. Make sure you are not clipping with the scope. Now measure V2 (VO’) with a resistor loading the output of the ‘basic’ amp. Keep the load resistor somewhere between 100Ω and 10KΩ. Assume the voltage source inside the ‘basic’ amp model is ideal. i.e. 0Ω with a real output impedance. With the above measurements you have everything you need to calculate Ro using Kirchoff’s laws. You are not going to be told how to calculate this. Figure it out. You should also be able to guess RO from Figure 3. Measure it even 6 ECE306 Lab 2, Feedback amplifiers 04.12.2018 though you can guess. You will use this output impedance measurement technique later in the lab and it will be good to know you are doing it right. Shunt–shunt amplifier For the first part of the experiment, we use the ‘basic’ amp to form a current-to-voltage (shunt–shunt) feedback amplifier as shown in Figure 4. The closed loop gain of this current-to-voltage feedback amplifier, From table 1, is voltage out divided by current in. A transconductance equation as given in Eq 7. This equation comes from the analysis technique used in Fig. 8.20 in [1]. See the appendix of this lab. A Vo Is (7) where IS is the current delivered by the source voltage VS. Rf b 100K Rs 1K Ro 1K Vs + Rin Vo V1 1K GND -100V1 GND Basic Amplifier Figure 4: The shunt–shunt topology model using the ‘basic’ amp. A circuit To measure the open loop gain of the feed-forward, A, network, we must form the A circuit as shown in Fig. 5. This consists of the basic amplifier loaded by the feedback and source impedances but with the feedback function disconnected. (Fig. 8.20[1]) Rs 1K Ro 1K Vs Rin GND Rf b 100K + V1 1K Vo' -100V1 - Rf b 100K Basic Amplifier GND Figure 5: The A circuit model, note the feedback resistor, β, is disconnected. 7 ECE306 Lab 2, Feedback amplifiers Rf 04.12.2018 100K 2 Vs Rf b 100K GND R1 Ro 1K 6 LF356 U1 Vo' Vo Rf b 7 1 3 + 1K - Rin 4 5 Vminus Rs 1K 100K 1K Vplus GND Basic Amplifier Figure 6: The circuit schematic for the shunt-shunt A block. We will use Fig. 5 with to calculate the theoretical open loop gain of the A circuit. We start with VS and end up with VO’. This amplifier has the gain of VO’/IS. Where IS is the current delivered by the source to the amplifier. If we define the voltage source as VS with a source impedance of RS then we can generate an IS of VS/RS with an impedance of RS. With this definition the current into the amplifier is IS, the current feeds RS, RFB and RIN in parallel to generate V1. A VO ' IS VO ' VO and Rf b R f b RO IS VS RS (9) V1 IS REQ Rin V1 IS RSR f b RSRin RSR f b RinR f b so A VO ' RS VS and VO 100V1 REQ is RS, Rfb and Rin in parallel REQ (8) (10) RSRinR f b RSRin RSR f b RinR f b so (11) Express 8 using 9, 10 and 11 A VO ' V RSR f b V RSR f b I RSR f b RSR f b RS O 100 1 100 S VS VS R f b RO IS R f b RO IS RSRin RSR f b RinR f b R f b RO A 100 RSR f b RSR f b RSRin RSR f b RinR f b R f b RO (13) for Rfb >> RO and Rin A 100 RS 50Rin 50,000 VA 2 Note From 8 IS (14) VS I 1 so S VS R S RS (15) lets see what VO’/VS is 8 (12) ECE306 Lab 2, Feedback amplifiers 04.12.2018 VO ' VO ' IS I RSR f b RSR f b 1 A S 100 VS IS VS VS RSRin RSR f b RinR f b R f b RO RS For Rfb >> RO and Rin (16) VO ' RSR f b RSR f b 1 RS 100 50 VV VS (R S Rin )R f b R f b RS RS Rin (17) Figure 8.20[1] shows the Norton equivalent of Figure 5 in which IS divides between RS, Rfb, and Rin and thus reduces the gain to about Rin/Rfb of what it would be if Rin were infinite. Thus for small Rin a reasonable, close enough, gain of the feed-forward network, A, is given in Eq.14 2. Connect the circuit shown in Figure 6. Set the frequency of Vs to 1 kHz. Measure the open loop voltage gain Vo'/Vs. Open loop output resistance Ro (using the circuit in Fig. 6) 3. Use the method described above (p5). Calculating Aβ The feedback network samples voltage in shunt and injects current into RS||Rin in shunt, Fig. 8.XX. The gain of the feedback network is given by: β 1 1 Rf b 100,000 A V (17) Note: The theoretical loop gain Aβ, is much lower than the usual gain for most op-amp based feedback amplifiers. We have chosen Aβ low so that we will be able to measure changes caused by using feedback and compare measured with theoretical values. For the basic amplifier if we substituted an operational amplifier with a typical gain of 100,000, Aβ would be very large and it would be difficult to measure output resistance with feedback Ro etc. With an A of - 50,000 VA and β 1 A we get an Aβ = 0.5. 100kΩ V An alternate way to calculate the loop gain, Aβ, from Fig. 4 is to short circuit the input and break the feedback loop between VO and Rfb. Then the gain around the loop is: Aβ RS || Rin 1KΩ || 1KΩ 500 1 100 100 100 100 0.498 0.5 RS || Rin R f b 1KΩ || 1KΩ 100KΩ 100,500 201 Closed loop voltage gain 9 ECE306 4. Lab 2, Feedback amplifiers 04.12.2018 Measure the closed loop voltage gain Vo/Vs using Fig. 7. Rf b 100K Rf 100K Rin 4 5 Vminus Rs 1K 1K 2 6 LF356 3 + Vo U1 7 1 R1 GND Ro 1K - Vs 1K GND Vplus Basic Amplifier Figure 7: The circuit schematic for the shunt-shunt closed loop amp. Closed loop output resistance Rof The theoretical closed loop output resistance with feedback Rof from Eq. 8.37 [1] Rof Ro 1 Aβ (18) 5. Using the circuit in Fig. 7, measure the closed loop output resistance Rof by measuring the open circuit and loaded output voltages as you did before. Series–shunt amplifier For the second part of the experiment, we use the basic amplifier to form a voltage-to-voltage (series–shunt) feedback amplifier as shown in Figure 8. T1 Vg 1 1:1 3 Ro 1K + Vs GND 2 -4 Rin Vo + V1 1K -100V1 - R2 Basic Amplifier 10K R1 1K GND Figure 8: The series–shunt topology model using the ‘basic’ amp. Here we must use a transformer to isolate the signal generator reference from the ground of the amplifier and the oscilloscope. The closed loop gain of this voltage-to-voltage feedback amplifier is 10 ECE306 Lab 2, Feedback amplifiers Vo Vs Af 04.12.2018 (19) Note VS is not the output voltage of the signal generator. The methodology used to find A and β is presented in Fig. 8.11 [1] shown in the appendix of this lab. The gain of the feedback network is: β R1 R1 R 2 (20) The A circuit To find the gain of the feed-forward network, A, we must form the A circuit as shown in Fig. 9. This consists of the ‘basic’ amplifier loaded with the feedback impedances but with the feedback function disconnected. The feed-forward gain is: A Vo ' Vs (21) 6. Connect the circuit shown in Fig. 10. Set the frequency of VS to 1 kHz and measure the open loop voltage gain A. Remember you can’t ground either side of the VS signal or you will change the circuit significantly when measuring it. Use the DVM or V1-V2 on the scope. Pin 1 on the transformer is the red wire, 2 is black, 3 is blue and 4 is yellow. T1 Vg 1 1:1 3 Ro 1K + Vs GND 2 -4 Rin Vo' + V1 1K -100V1 - R2 Basic Amplifier 909 R1||R2 10K R1 1K GND Figure 9: The A circuit model for the series–shunt amplifier. Note that V1<Vs because of the attenuation caused by Rin and R1||R2. To calculate the theoretical gain in Fig. 8, Vs is divided by the attenuator formed by R1||R2 and the input resistance of the basic amplifier Rin. We also need to know the attenuation factor for VO’. Input attenuation is: V1 Rin 1KΩ 0.524 VS Rin R1 || R2 1KΩ 0.909kΩ And the output attenuation is: 11 (22 ECE306 Lab 2, Feedback amplifiers 04.12.2018 VO ' R1 R2 11KΩ 0.916 VO RO R1 R2 12KΩ (23) Now we can evaluate Eq. 21. A Vo ' 0.917VO 0.917( 100)V1 91.7 48.05 VV Vs V1 0.524 V1 0.524 1 0.524 (24) Collecting terms from 21, 22 and 23 we get Eq. 25: A VO' Rin R1 R2 (-100) 0.524 0.917 ( 100) 48.05 VV VS Rin R1 || R2 Ro R1 R2 (25) Same answer that Eq. 24 gives how nice. Rf 100K 3 2 3 Vs GND 2 4 R3 Ro 1K 6 LF356 U1 Vo' Vo 7 1 1:1 + 1 1K - Vg Rin 4 5 Vminus T1 1K R2 Vplus 10K Basic Amplifier R1 R1||R2 909 1K GND Figure 10: The circuit schematic for the series-shunt A block. Open loop output resistance Ro (using the circuit in Fig. 10) 7. Measure the open loop output resistance RO as you did before by measuring VO’ and VOL’. Once again measure VO’ with the 10KΩ and 1KΩ resistors connected to the output then add an additional RL to get the new VO’. Calculating Aβ Again note the theoretical loop gain Aβ, is much lower than that for most feedback amplifiers. We have chosen it low so that we will be able to measure changes caused by using feedback and compare measured with theoretical values. The feedback network samples voltage in shunt and injects voltage into Vs in series. The gain of the feedback network is given by 12 ECE306 Lab 2, Feedback amplifiers 04.12.2018 R1 1kΩ 0.09 R1 R 2 1kΩ 10kΩ Aβ = -48.05(-0.0909) = 4.37 β (26) An alternative way to find the loop gain Aβ is to short circuit the input Vs and break the feedback loop between Vo and R2. Then the gain around the loop is βA R1 || Rid 1kΩ || 1kΩ (-100) (-100) 0.0437( 100) 4.37 R1 || Rid R2 1kΩ || 1kΩ 10kΩ (27) Closed loop voltage gain (using the circuit in Fig. 11) 8. At 1 kHz, measure the closed loop voltage gain Vo/Vs. Rf 100K 3 2 3 Vs GND 2 4 R3 Ro 1K 6 LF356 1K Vo U1 7 1 1:1 + 1 1K - Vg Rin 4 5 Vminus T1 R2 Vplus 10K Basic Amplifier R1 1K GND Figure 11: The circuit schematic for the series-shunt closed loop amp. Closed loop output resistance Rof (using the circuit in Fig. 11) The theoretical closed loop output resistance with feedback Rof from Sedra and Smith equation (8.19) Ro Rof (28) 1 Aβ 9. Measure the closed loop output resistance ROf as you did before by measuring VO’ and VOL’. Once again measure VO’ with the 10KΩ and 1KΩ resistors connected to the output then add an additional RL to get the new VOL’. 13 ECE306 Lab 2, Feedback amplifiers 04.12.2018 POST LAB QUESTIONS 1. For the circuit shown in Fig. 2 compare the measured voltage gain V2/V1 at 1 kHz with the theoretical value. Also compare the measured output resistance of the basic amplifier with the theoretical value. Plot the “open-loop” gain of the basic amp building block. 2. Compare the measured value of the open loop voltage gain Vo'/Vs to the theoretical value of the open loop voltage gain of the circuit shown in Fig. 5. 3. Compare the measured value of the open loop output resistance Ro without feedback to the theoretical value of same for the circuit shown in Fig. 5. 4. Compare the measured value of the closed loop voltage gain Vo/Vs for the shunt-shunt amplifier to the theoretical value of same for the circuit shown in Fig. 4. 5. Compare the measured value of the closed loop output resistance with feedback Rof for the shunt-shunt amplifier to the theoretical value of same for the circuit shown in Fig. 4. 6. Compare the measured value of the open loop feed-forward gain A for the series-shunt amplifier to the theoretical value of same for the circuit shown in Fig. 10. 7. Compare the measured value of the open loop output resistance without feedback Ro for the series-shunt amplifier to the theoretical value of same for the circuit shown in Fig. 10. 8. Compare the measured value of the closed loop voltage gain Vo/Vs for the series-shunt amplifier to the theoretical value of same for the circuit shown in Fig. 9. 9. Compare the measured value of the closed loop output resistance with feedback Rof for the series-shunt amplifier to the theoretical value of same for the circuit shown in Fig. 9. [1] A. S. Sedra and K. C. Smith, Microelectronic circuits, Fifth edition. Saunders, Philadelphia, 2004 14 ECE306 Lab 2, Feedback amplifiers 04.12.2018 Appendix Inverting op-amp configuration gain, Rin and RO discussion. The inverting op-amp is really a trans-impedance amp even though the gain is often specified with the voltage amp gain equation. Let’s look closely at what is going on. Rf 100K 4 5 Vminus 2 V1 U1 Vo V2 1K 7 1 3 Ro 6 LF356 + 1K - Rin GND R1 1K GND Vplus Figure A1 The positive input terminal of the op-amp is at ground potential because there is no current going into or out of it and thus no voltage drop across R1. This tells us that the negative input terminal will also be at ground potential, the same voltage as the positive terminal, because the negative feedback tries to keep the negative terminal at the same voltage as the positive terminal. If you don’t believe this statement then glance at pages 64 – 76 [1]. This is true when the op-amp is operating linearly, when the output is not clipping, and we are assuming this. So the input impedance is simply Rin since the input voltage sees is a resistor to ground. So the input voltage V1 divided by the input resistance Rin is the input current Iin, Eq. A1. Iin V1 Rin (A1) Solving for V1 gives: V1Iin Rin Substituting A2 into Eq. 2 yields: VO R f IinRin Rin (A3) multiply both sides by Rin and you get Eq. 3. VO Rf Iin (3) (A2) The output impedance of the basic amp is RO. We have a voltage source, VO, with near 0Ω output impedance driving the output node, V2, through RO. This one we can get by inspection. The above discussion is for the mid band of the op-amp circuit. Note that you will always be in the mid band of the inverting op-amp circuit in these experiments. A and β determination for the inverting op-amp configuration. 15 ECE306 Lab 2, Feedback amplifiers 04.12.2018 We will use the technique described in Fig 8.20 of [1], shown in appendix. The basic amp must be loaded by R11, R22 and RS to calculate the gain. RS is Rin. We will assume the output impedance of the op-amp is 0Ω and the open loop gain of the op-amp is 100,000V/V. The A gain is expressed in terms of VO/Iin. β we know is -1/Rf = -1/100,000 from b in Fig. 8.20. This makes R11 and R22 = 100KΩ (Fig. 8.20). In the shunt-shunt topology of Fig. 8.XX the A circuit shows the input parameters Vi, Ii and Ri. The output parameters are VO and RO. VO is generated by a voltage source with a gain of A*Ii. We know from the gain of the op-amp that 1V in yields -100,000V out when operated open loop. 1V in yields 1/Ri = 1/1000 = 1mA Ii for our circuit. So A*Ii must be -100,000 for Ii = 0.001. A = -100,000/0.001 = 100M. Figure 8.20 [1] Finding A and β for shunt-shunt. 16 ECE306 Lab 2, Feedback amplifiers Figure 8.11 [1] Finding A and β for series-shunt. 17 04.12.2018