BEE 332 Lab3 Multi-Transistor Configurations Group’s Name 1 Abraham Asnashe Lab1-Activity Collecting data & analysis, Graphing, Circuit construction, Answers to questions and Commenting Collecting data& analysis, Graphing, Circuit construction Answers to questions and Commenting Collecting data& analysis, Graphing, Circuit construction Answers to questions and Commenting 2 Monyrith Chhen 3 Retta Dechassa Due Nov 2, 2012 Instructor: Lawrence Lam, PhD BEE 332 Lab # 3 Multi-Transistor Configurations Page | 1 OBJECTIVE In Multi-Transistor lab we experiment with several common multiple transistor circuits and observe their characteristics. Among these circuits are a current mirror, different current sources, and a differential amplifier. During this lab we also experiment with the design of a three-transistor circuit, attempting to fit given criteria. In addition we show how two transistors are used to achieve amplifiers with improved performance, we analysis also multiple transistor amplifiers using resistive loads and we continue to build the amplifier concepts necessary to consider integrated circuit amplifiers. INTRODUCTION This experiment are to examine the operating characteristics of several of the most common multi-transistor configurations, including current mirrors, current sources and sinks, and differential amplifiers. EQUIPMENT AND MATERIALS USED: Equipment name DC power supply Equipment/Materials image DMM Volt-Meter Amp-Meter Power Generator Oscilloscope Breadboard CA3046 transistor Resistors 100 kΩ 5% 1/4 W 100 kΩ potentiometer BEE 332 Lab # 3 Multi-Transistor Configurations Page | 2 Procedure 1 Current mirror Figure E3.1 Measurement-1 Set up We connect PPS1 power supply to implement the VCC = +6.0 V. and we use PPS2 power supply to implement the VEE = -6.0 V DC power supply rails. And then we Adjust the R2 potentiometer to vary the collector voltage of Q2 from +6.0 V down to about –5.0 V. so we take pairs of readings of the current measured on one DMM versus the collector voltage measured on the other DMM or oscilloscope. The Suggested voltage values are +6 V, +3 V, 0.0 V, -3 V, and – 5 V. Record these pairs of readings. Finally we record the DC voltages at the base and collector of Q1.Please sees the measurement-1 Data below Measurement-1 Data Suggested value is 100Kohms but actual measured value for R1 = 98.5kohms Table for the measured value for VC2 and IC2 at different voltages: Voltage Value 6V 3V 0V -3 V -5 V Measured Vc2 (V) 5.96V 3.02V 0.094.4 V - 3.02 V - 5.00 V Ic2 (µA) 133.37 uA 129.91 uA 125.84 uA 121.48 uA 117.81 uA Page | 3 BEE 332 Lab # 3 Multi-Transistor Configurations Below are some screenshots for different voltage value(Vc2): At 6V At 3 V BEE 332 Lab # 3 Multi-Transistor Configurations Page | 4 . At 0 V At -3 V BEE 332 Lab # 3 Multi-Transistor Configurations Page | 5 . 8497 MΩ (b) Explain what purpose the connection between the base and collector of Q1 (pins 6 and 8) serves. The reason why Our Vc cannot go all the way to -6 because of the voltage drop on the Q2 itself i. Rout= (Vc21-Vc22)/ (Ic21-Ic22) = (5.0 V.e.87V-0V=0. (c) Explain why the collector voltage of Q2 cannot be brought all of the way down to the lower power supply rail of VEE = -6.13V. The Value of Vbe1 Vbe1 = VB1-VE1= -5.37 -129.on voltage for the transistors 0. From our table results we can calculate the output resistance of the current sources or the current mirror. taken from your recorded data.87V > Vce1 Sat this insures Q1 is FAR. VCE2. Answer for Question-1 (a) Calculate the output resistance of this current source as Rout = ΔV/ΔI. page 3) (d) Explain if and how the current source is sensitive to the values of the power supply voltages.There is an Vbe.02)/ (133.87 V Vce1=Vbe1-Vcb1=0.1 above. BEE 332 Lab # 3 Multi-Transistor Configurations Page | 6 . (please see figure E3. The connection between the collector and the base of the Q1 serves to keep VBC1 at the value of zero this will keep Q1 at the Forward Active region since BE-junction would stay in a forward biased mode.At -5 V We also measured Voltage at base and collector of Q1.91) µA = 0. Vbg1=Vcg1=-5.87V in our case so this prevent the transistors’ Vc to be at the same potential with –Vee=-6V.13-(-6) = .96 -3. An Ideal current source would not be affected by the changes in the power supply. BEE 332 Lab # 3 Multi-Transistor Configurations Page | 7 . This change in the value of Ic is due to Early effect.the change in collector current is minor and can be consider as constant current source. However . but in this case changing the value of the voltage caused a change in the output current even though the voltage on the base of Q2 remains constant. The change of VCE value changes the value of Ic. 2a Measurement Set up 2a: First we verify that the two power supply rails are at ±6. And Replace R3 on the emitter of Q2 with a wire.0 V.0 V down to about -5.0 Volts. And record the DC voltages at the bas BEE 332 Lab # 3 Multi-Transistor Configurations Page | 8 . probably around a collector voltage of –4. Add a new resistor R3 in series with the emitter of Q1.Then we turn the dual DC power supply ON.0 Volts.0 to –5. Verify once more that the two power supply rails are at ±6. and then we take pairs of readings of the current measured on the one DMM versus the collector voltage measured on the other DMM and record. then we take readings only up to the point where Q2 saturates.0 V down to about +5.0 V. And Adjust the R2 potentiometer to vary the collector voltage of Q2 from +6. Finally we record the DC voltages at the base and collector of Q1. Measurement Set up 2b: First we Turn the dual DC power supply OFF. finally we take pairs of readings of the current measured on the one DMM versus the collector voltage measured on the other DMM or oscilloscope and record.Procedure 2 Widlar current sources Figure E3.0 V. Also we Adjust the R2 potentiometer to vary the collector voltage of Q2 from +6. 6 5.068 K-ohms: Actual measured Vc2 (v) Ic2 (µA) 6 5.39 -5.40 V 10.8 5. Voltage Value Vc2 Vb1 (v) Vc1 (v) 6 5.2 5 -5.4 -5.00V)/(10.025 * ln (114µA /10µA) ) / 10 µA = 6084 Ω = 6.8 5.6 5.83 uA 5.99 V 10.2 5.81 uA 5 5.39 -5.43/100k = 114µA Ic2 = 10 µA R3 = (VT * ln (Ic1/Ic2)) / Ic2 = = (0.988 k Ω = 6.79 uA Rout=change V/change I =(5.88uA-10.88 uA 5.60 V 10.99V-5.08k Ω Pick R3 value = 5.79uA)=11Mohms.4 -5.4 -5.39 -5.068 k Ω Table for measured value for VC2 and IC2 at different voltages with R3 = 6.87 uA 5.00V 10.08 k Ω + 0.4 5.39 -5.4 -5.39 Voltage Value Vc2 BEE 332 Lab # 3 Multi-Transistor Configurations Page | 9 .4 -5.4 5.4 -5.85 uA 5. Table for base and collector voltages with respect to ground at different voltages: The base and collector voltage with respect to ground at Q1 is remain almost constant for different values Vce2.Measurement-2a Ic1 = Vc1 / R1 = 11.81 V 10.39 -5.20 V 10. 8 V BEE 332 Lab # 3 Multi-Transistor Configurations Page | 10 .Below are some screenshots for different voltage values: At 6 V Below TEK machine graph show that we took At 5. 4 V BEE 332 Lab # 3 Multi-Transistor Configurations Page | 11 .Below TEK machine graph show that we took At 5.6 V Below TEK machine graph show that we took At 5. 2 V Below TEK machine graph show that we took At At 5 V BEE 332 Lab # 3 Multi-Transistor Configurations Page | 12 .Below TEK machine graph show that we took At 5. 1 uA 2V 1.35 -5.140 670.35 -5.35 -5.2b Measurement-2b Ic1 = (Vc1 / R1 )= 11.7 uA -3 V -3. Ic2 = 1 mA R3 = [(VT * ln (Ic2/Ic1) ) / Ic1 ]= [(0.09 673.15 uA -4 V Table for base and collector voltages with respect to ground at different voltages: Voltage value Vb1 (v) Vc1 (v) 6 5 4 3 2 1 0 -1 -2 -3 -4 -5.9 uA 4V 3.00 674.08 672.35 -5.35 -5.35 -5.35 -5.35 -5.35 -5.1 uA 0V -1.77 678.99 658.01 674.35 -5.02 662.04 666.35 -5.35 -5.35 -5.6 uA -2 V -3.35 -5.36 uA -1 V -1.025 * ln (1mA/114uA) ) / 114µA ]= 476.35 -5.98 663.3 uA 3V 2.35 -5.35 -5.7 uA 5V 4.22 µ Pick resistor value = 470 Ω.98 676.43/100k = 114µA. Measured R3 = 458 Ω Table for measured value for VC2 and IC2 at different voltages with R3 = 458 Ω: Voltage value Vc2 (v) Ic2 (µA) 5.35 -5.Figure E3.35 -5.4 uA 6V 4.35 -5.35 Page | 13 BEE 332 Lab # 3 Multi-Transistor Configurations .19 uA 1V 0.35 -5. Below are some screenshots for different voltage value: TEK machine graph show that we took At 6 V Below TEK machine graph show that we took At 5 V BEE 332 Lab # 3 Multi-Transistor Configurations Page | 14 . Below TEK machine graph show that we took At 4 V Below TEK machine graph show that we took At 3 V BEE 332 Lab # 3 Multi-Transistor Configurations Page | 15 . Below TEK machine graph show that we took At 2 V Below TEK machine graph show that we took At 1 V BEE 332 Lab # 3 Multi-Transistor Configurations Page | 16 . Below TEK machine graph show that we took At 0 V Below TEK machine graph show that we took At -1 V BEE 332 Lab # 3 Multi-Transistor Configurations Page | 17 . Below TEK machine graph show that we took At -2 V Below TEK machine graph show that we took At -3 V BEE 332 Lab # 3 Multi-Transistor Configurations Page | 18 . Rout= (Vc1-Vc2)/(Ic1-Ic2) = (5.2a.99-5.00)/(10. E3. The presence of a R3 in reducing Widlar circuit (please see figure E3.98)/( 678. But we do not have this negative feedback in boosting widlar case at Q2.88-10.4-676. Therefore the output resistance is higher in reducing widlar compare to boosting Widlar. page 11) and an increase in current at Q2 increases creates an increase in voltage drop across R3 which in turn reduces a voltage drop across baseemittor of Q2 which in turn counteract the increase in current( contributes to the stability of the Q2) and hence.2b.77-4.Below TEK machine graph show that we took At -4 V Answer for Question-2 (a) Calculate the output resistance of the reducing Widlar current source of Fig.2a above. provides a negative feedback to the input loop and increases the output resistance. E3.465 MΩ In our circuit above the output resistance for the boost Widlar current source (c) Explain why the output resistance of the reducing Widlar current source is higher than for the boosting case. BEE 332 Lab # 3 Multi-Transistor Configurations Page | 19 .79) µA = 11 MΩ (b) Calculate the output resistance of the boosting Widlar current source of Fig.7) µA = = 0. Rout= (Vc21-Vc22)/(Ic21-Ic22) = (5. 4 V -4.4 V -5.72 V -4.72 V -4. and then we take pairs of readings of the current measured on the one DMM versus the collector voltage measured on the other DMM or oscilloscope and record.05 V 0.02 V -3.72 V -4.00 V -4.72 V -4.4 V 106. Measurement-3 Table for measured value for VC2 and IC2 at different voltages: Suggested Actual measured CollectorCollector Current (IC2 µA) Voltage value Emitter Voltage (VC2) 6V 4V 2V 0V -1 V -3 V -5 V 5.8 uA 105. Finally we record the DC voltages at the base and collector of Q1.1 uA 106.72 V -4.0 V.4 V -5.3 uA 105.140 V -1.0 V and PPS2 to -6.72 V Page | 20 BEE 332 Lab # 3 Multi-Transistor Configurations .05 V 2.0 V.Procedure 3 Wilson current source Measurement Set up 3: We Set PPS1 to +6. And we verify that the two power rails are at ±6.6 uA 106.4 V -5.4 V -5.4 V -5.72 V -4.1 uA 104.3 uA 105.0 V. Adjust the R2 potentiometer to vary the collector voltage of Q2 from +6.6 uA Table for base and collector voltages with respect to ground at different voltages: Suggested Voltage value Base Voltage (VB1) Collector Voltage (Vc1) 6V 4V 2V 0V -1 V -3 V -5 V -5.4 V -5.99 V 4.0 V down to about -5. Below are some screenshots for different voltage value:At 6 V Below TEK machine graph show that we took At 4 V BEE 332 Lab # 3 Multi-Transistor Configurations Page | 21 . Below TEK machine graph show that we took At 2 V Below TEK machine graph show that we took At 0 V BEE 332 Lab # 3 Multi-Transistor Configurations Page | 22 . Below TEK machine graph show that we took At -1 V Below TEK machine graph show that we took At -3 V BEE 332 Lab # 3 Multi-Transistor Configurations Page | 23 . The output node in this circuit is the BEE 332 Lab # 3 Multi-Transistor Configurations Page | 24 . The reference current is modeled by the multimeter XMM1. We modeled this circuit in Multisim. Output resistance: ∆V= V2-V1= ∆I= I2-I1= Rout=∆V/∆I = (V2-V1)/( I2-I1 )= (b) Design a current source for which the output current is exactly twice that of the reference current by using three of the BJTs on the CA3046 npn array.Below TEK machine graph show that we took At -5 V Answer for Question-3 (a) Using your measured data. Through experiment we found that this circuit produced the desired output current of twice the reference current. Your data may be such that you can only say that the output resistance is greater than a certain value. and the output current is XMM2. Explain in your lab notebook how this circuit achieves the factor of two scaling between the reference and output currents. calculate the output resistance for this Wilson current source. BEE 332 Lab # 3 Multi-Transistor Configurations Page | 25 .08%.collector of Q2. therefore I would deem this a sufficient model. which is less than that of the 5% tolerance resistors that we are using to construct out circuit.The chosen resistor value for R2 produces an output with an error of about 0. which is correct.716 V. VE1 = VE2 = -0.386 V. Measured output voltage 3.68 V Screenshot of the sin wave that shows 180 degree shift: BEE 332 Lab # 3 Multi-Transistor Configurations Page | 26 . VC1 sin wave is 180 degree shift.38 V Vc Should be around 3V.716 V. therefore it is in forward active mode.Procedure 4 Differential amplifier Verification: Vc1 = 3. Vc2 = 3. which means VBE = 0. 5 Screenshot of the sin-wave input and output before it start clipping: BEE 332 Lab # 3 Multi-Transistor Configurations Page | 27 . Gain = Vo / Vin = 28.52 V Screenshot of the sin wave that shows input and output are in phase: Measurement-4 First part: Before the sin-wave starts to clip we measured Vout = 4. No shift: measure output voltage 3.VC2 sin wave is in phase. Therefore.56 V and Vin = 160 mVpp. 56 * 0.19 V After increasing the frequency until the output voltage reaches 3.86 M-Hz: Second part of measurement: (common-mode input signal) VC2 is out of phase 180 degree to the input voltage.86 M-Hz. Screenshot of 70% of Vmax at 1.Calculating 70 % of the maximum voltage.7 = 3. we obtained 1. 4.19 V. Screenshot of the sin wave that shows 180 degree shift: BEE 332 Lab # 3 Multi-Transistor Configurations Page | 28 . Screenshot of the sin wave that shows 180 degree shift: We calculated the gain dividing output voltage by the input voltage and we expect to obtain a gain less than 1 in which is exactly what we got. BEE 332 Lab # 3 Multi-Transistor Configurations Page | 29 . Gain = Vout / Vin = 2.Vc1 is out of phase by 180 degree to the input voltage. we increased the amplitude until we reached the largest possible outpu Before the sin-wave starts to clip we measured Vout = 2.72 < 1 Then.23 Vpp.52 V and Vin = 5.16 V / 3 V = 0. 7 = 1.76 V After increasing the frequency until the output voltage reaches 1.99M-Hz: Final part: Both input and output are in phase. Gain = Vo / Vin = 2.Therefore.52 V / 5. we obtained 0.99 M-Hz.485 Screenshot of the sin-wave input and output before it start clipping: Calculating 70 % of the maximum voltage.52 * 0. 2. Common-mode gain = Vo/Vin = 186mV/186mV = 1 Screenshot of the sin wave that shows input and output are in phase: BEE 332 Lab # 3 Multi-Transistor Configurations Page | 30 .76 V.20 V = 0. Screenshot of 70% of Vmax at 0. BEE 332 Lab # 3 Multi-Transistor Configurations Page | 31 .DC Offset by 4V. expressing it in both as a pure ratio and in decibels (dB). but also increasing the potential resistance using two additional transistors..Vb=0 since there is no dc offset.Question-4 (a) Calculate the common-mode rejection ratio (CMRR). The common-mode gain would jump to 1 since now there will be an offset at Vb so then Our interpretation would be that the transistors would share a gain of 1 now that there is an offset voltage and the transistors are now in the “ON” mode and ready to amplify. we would see that the circuit is symmetrical due to the bases being shorted together. maintaining the current. The common-mode gain is ½ and inverting because examining the circuit in the small signal model. That is why the output would be less than the input (c) Explain why the common-mode gain jumps to approximately 1. CONCLUSION BEE 332 Lab # 3 Multi-Transistor Configurations Page | 32 . We believe by using the principle of the Wilson Current source we would be able to increase RE using the current source. And Vbe=Vb.0 and becomes noninverting when a positive DC offset is applied from the signal generator.5 and inverting when no DC offset is applied from the signal generator. Try to figure this one out---it will add greatly to your ability to troubleshoot analog circuits. Hint: think about how ideal the current source RE is and how it might be improved. = (b) Explain why the common-mode gain is approximately 0.Ve=0-Ve …………………. We would like to increase all the resistors in order to create a proportional increase within the CMRR since they are all factored on the resistors. This is a tricky question! But it represents a situation which often occurs in the lab and which confuses people. (d) Suggest a way to increase the CMRR of this differential amplifier. BEE 332 Lab # 3 Multi-Transistor Configurations Page | 33 .After we completing this experiment which is Multi-Transistor Configurations . We have reinforced our understanding of the qualities of an ideal current source and seen how to properly design a circuit to achieve a desired current output. we now have a much greater understanding of the uses and properties of multiple transistor circuits.
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