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# differential amplifier for electronic

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### differential amplifier for electronic

1. 1. Chapter 5Differential amplifier By -ISMET- Edited by Nazirah Mohamat Kasim & Shahilah Nordin
2. 2. • An arrangement of transistors which allows the difference between two signals source to be amplified.• The output is proportional to the difference between these two inputs.• The best direct coupled stages available to the IC designer.
3. 3. • Differential amplifiers +VCC are used in low and high frequency RC RC amplifiers, analog modulators and digital Vi1 V01 V02 Vi2 logic states. Q1 Q2• The basic differential amplifier is shown in RT Figure 5.1:- -VEE Figure 5.1
4. 4. +VCCFor the dc analysis ofdifferential amplifier RC RCcircuit, all inputs are V01 V02set to zero as shown in Q1 Q2Figure 5.2. IT/2 IT/2 RT IT -VEE Figure 5.2
5. 5. From Figure 5.2, IT IC  2and V B  0therefore, V E  V B - V BE  0  0.7V   0.7V V EE - 0.7VEmitter current, I T  RT ITCollector voltage, V C  V CC - RC 2
6. 6. Three possible input signal combinations for differentialamplifier: Differential modeSingle-Ended common Mode mode
7. 7. •Single ended mode – an active signal is applied to onlyone input while the other is grounded.•Differential mode – two opposite polarity active signalsare applied to the amplifier.•Common mode -Two signals of the same amplitude, frequency and phase are applied to the differential amplifier. - The output of the amplifier is ideally zero when measured the difference between the output terminals.
8. 8. Definition The input is applied and the output is measured at one of the output terminal.The input is in differential mode and the output is measured between thedifferences of two output terminals.
9. 9. +VCCThe input signal isapplied to one RC RCinput with the V0other input is Q1 Q2 Vi2= +connected to the Vi1 0ground, as shown - RTin Figure 5.3: -VEE Figure 5.3
10. 10. The ac equivalent circuit of Figure 5.3 isshown in Figure 5.4. Vo 1 Vo 2 bIb RC RC bIb r? r? + Vi 1 - RT Figure 5.4
11. 11. b VTAssume: Ib1  Ib2  Ib , r  I CQ and β 1  β 2  β Assume RT is very large and using KVL we obtain V i1 - 2I b r  0 V i1  2I b r The output at the collector will be V o  - b IbR C Therefore, Voltage gain, VO RC Av   Vi  VT  2 I    CQ 
12. 12. The mode of this +VCCoperation isassumed by letting RC RCa single sine wave V0signal is connectedbetween the two Q1 Q2 + +inputs. Vi1 Vi2Refer Figure 5.5 - - RT(Double ended -VEEdifferentialamplifier circuit) Figure 5.5
13. 13. The ac equivalent circuit of Figure 5.5is shown in Figure 5.6. Vo 1 Vo 2 bIb RC RC bIb r? r? + + Vi 1 Vi 2 - - RT Figure 5.6
14. 14. Figure 5.6 shows an ac connection of a differential amplifier. Weseparate the input signals as Vi1 and Vi2, with separate outputsresulting as VO1 and VO2 with the transistors replaced with its acequivalent circuit. Similarly as for the single ended operation, Assume R T is very large and it becomes open circuit V i1 - V i 2  Ib r  Ib r V i1 - Vi2  V d Therefore, V d  2I b r The output at the collector terminal, V o  - b Ib R C Therefore, Differenti al voltage gain, VO RC Av   Vd V  2 T  I   CQ 
15. 15. Refer Figure 5.6, Vd Input impedance Z i(diff )  Ii V d  2I b r and Ii  Ib Therefore, Z i(diff)  2r 
16. 16. The same input signal is applied to the +VCCtwo input terminals of differentialamplifier with the same magnitude andphase. RC RCAn ac connection showing commoninput to both transistors is shown in V01Figure 5.7 VCM Q1 Q2 Ideal differential amplifier -The output voltage Vo is expected to be zero. RT -because the difference -VEE between the two outputs at each collector is opposite to each other and they are Figure 5.7 cancels out each other.
17. 17. However …practically there is +VCCan output at thecollectors but the RCvalue is small. To V01we analyze, Q1 treat the amplifier + VCM 2RTas symmetry, that is -RT is made to beparallel and be -VEEreplaced with 2RT asshown in Figure 5.8 Figure 5.8
18. 18. To determine the common mode gain, the ac equivalent circuit of figure 5.8 is drawn as given in Figure 5.9VCM Vo r βIb RC 2RT From the circuit of Figure 5.9, V CM  V i  Ib r   IE 2R T V   b Ib  T  2R T  I  Figure 5.9  CQ  and V o  - b IbR C Therefore, the common mode gain, A CM of the differenti al amplifier, VO RC RC A CM      Vi VT 2R  2R T T I CQ
19. 19. The ratio of the magnitude of its differential gain, Ad tothe magnitude of its common mode gain, ACM. AdCMRR  A CMThe value of the CMRR is often given in dB, Ad CMRR (dB)  20 log 10 A CM
20. 20. Example 1.The circuit given in Figure 5.10 has the following parameters:hfe1 = hfe2 = 120, VT = 26mV, VBE1 = VBE2 = 0.7VCalculate: +10Va) ICQ1, ICQ2 and VCE1b) Differential gain, Ad 3.9kΩ 3.9kΩc) Common mode gain, ACM V01d) CMRR in dB Vi1 Vi2 Q1 Q2e) Differential input impedance, Zi(diff)f) Output impedance, Zo 5.6kΩ -10V Figure 5.10
21. 21. SolutionDC analysis:Since VB = 0, VE = -0.7VUsing KVL around loop A: VE – IERE + 10V = 0 10  V E 10  (-0.7)Then IE    1.66mA RT 5.6k
22. 22. Solution IEa) I CQ1 , I CQ2   0.83mA 2 V CE1  V CC - I C1 R C1 - V E  0  10  (0.83m)(3. 9k)  (  0.7)  7.46Vb) From the ac equivalent circuit of differential mode: VO RC VT 26mV A d   ; but   31.32 Ω Vd V  IC 0.83mA 2 T  I    C  3.9k Then A d     62.26 2(31.32)
23. 23. Solutionc) From the ac equivalent circuit of common mode: VO RC A CM    Vi VT  2R T IC 3.9k     0.35 31.32  2(5.6k)d) Ad 62.26 CMRR(dB)  20log 10  20log 10  45dB A CM 0.35e)  VT  Z i(diff)  2   I   2(120)(31. 32)  7.5k Ω  C f) ZO  R C  3.9k Ω
24. 24. Good differential amplifier should have high CMRR Ad  CMRR  A CM High CMRR means the differential amplifier circuit has theability to reject common mode signal (noise).Ideally, ACM = 0 not amplify the noise signal.
25. 25. In practical way there is still small signal at the output ofcommon mode signal. to  CMRR, we have A CM   RC  A CM  VT  2R T  ICTo reduce common mode gain, A CM  we can increase R T Popular method to increase  R T is by using :Constant Current SourcePractical current source is a current supply with a resistance.
26. 26. An ideal current source, R=∞Whereas practical current source resistance, R is very large.An ideal current source provides a constant currentregardless of the load connected to it. Practical Current Source IT RT Figure 5.11
27. 27. Constant Current Sourcecircuit can be built using:FET devices BJT devices Combination of FET & BJT devices Three popular Constant- Current Sources for differential amplifier Bipolar Transistor Constant Current Source Transistor / Zener Constant Current Source Current Mirror circuit.
28. 28. BJT Constant Current Source Figure 5.12a shows that an npn transistor together withresistors operate as a constant current source. IT C3 B3 E3 R1 R2 RE -VEE RT Figure 5.12a
29. 29. BJT Constant Current Source Figure 5.12b and Figure 5.12c shows simplified circuit from Figure5.12a IT C3 IT C3 B3 RTH Q3 B3 E3 E3 VTH R1 R2 RE LOOP1 RE VEE VEE VEE RT RT Figure 5.12b Figure 5.12c R1 V TH    VEE  R TH  R 1 // R 2 R1  R 2 By using KVL at loop1 in Figure 5.12c IT  IC3  IE3 - V TH  I B R TH  V BE 3  I E 3 R E  0 V TH  V EE - V BE3 V TH  V EE - V BE3 IE3  IT  R TH R TH  RE  RE b1 b1
30. 30. BJT Constant Current Source Figure 5.12d and Figure 5.12e shows ac equivalent circuit for circuitin Figure 5.12a b3 rbb c3 rce3 ib bi b b3 rbb r e3 c3 r rce3R1 R2 e3 ie ib bi b ie R1//R2 RE RE RT RT Figure 5.12d Figure 5.12c R T  rce 3 1  b    R b  rce 3 1  b  where Re R b   R 1 // R 2   rbb  r   Re  R b Re  RE
31. 31. Example 2If R1 = 4.7kΩ, R2: 4.7kΩ, RE = 2.2kΩ and VEE = 20V in figure5.12: IT Figure 5.12 R1 R2 RE -VEECalculate current, I.Answer: 4.23mA
32. 32. Zener Constant Current Source•From BJT constant current source, R2 is replaced withzener diode as in Figure 5.13. IT C3 B3 + Q3 VBE3 Figure 5.13a + _ E3 VZ loop1 R1 - RE IE3•Using KVL at loop 1: -VEE - V Z  V BE3  IE3 R E  0 V Z  V BE3  IE3 R E I T  IC3  IE3 V Z - V BE3 V Z - V BE3 IE3  IT  RE RE
33. 33. Zener Constant Current Source Figure 5.13b and Figure 5.13c shows ac equivalent circuit for circuit in Figure 5.13a b3 rbb c3 rce3 rbb r e3 c3 ib bi b b3 r rceR1 ib ie e3 bi b ie Re RE RT RT Figure 5.13b Figure 5.13c R T  rce 3 1  b    R b  rce 3 1  b  where Re R b  rbb  r   Re  R b Re  RE
34. 34. Example 3If R1 = 3.3kΩ, RE: 2.2kΩ, and VEE = 15V in figure 5.13: IT + Figure 5.13 VZ R1 - RE -VEECalculate current, I.Answer: 2.64mA
35. 35. Current Mirror Circuit Used primarily in IC. Circuit in Figure 5.14 is an example of constant currentsource circuit that can be used for differential amplifier. For current mirror circuit, +VCC I C  I C3  I C4 IX IB  IB3  IB4 IE  IE3  IE4 RX IT=IC IC I X  I C  2I B But IB  b so, IC 2IB IC I X  IC  2 Q4 Q3 b IB + IB  b 2  IC     IE VBE IE  b  - RT  b  IC  IT    IX  IX   b  2 Therefore I T  IC  I X Figure 5.14a
36. 36. Current Mirror CircuitFigure 5.14b shows ac equivalent circuit for circuit in Figure 5.14a c4 b4 b3 c3 Ib4 rbb4 rbb3 ib3 rce3RX rce3 bi b 4 bi b3 r 4 r 4 e3 e3 RT Figure 5.13b R T  rce 3
37. 37. Example 4Given Rx = 1.5kΩ, VCC = 18V in figure 5.14, calculate themirrored current, I in the circuit.Answer: 11.53mA +18V IXSolutionUsing KVL around the loop: RX I 18 – IXRX – VBE = 0 Q1 Q2 + IX = 11.53mA VBE - Figure 5.14
38. 38. Example 5For the differential amplifier of Figure 5.15, determine thefollowing: +15Va) ICQ, VCEQ 4.7kΩ 4.7kΩb) ACM V0c) Ad Vi1 Q1 Q2 Vi2d) CMRR in dBe) Differential input impedance, Zi(diff) 1.5kΩ Q3f) Output impedance, ZoGiven:rce = 1/1μS 1.2kΩ β1 = β2 = β3 = 150 -15V VT = 26mV Figure 5.15
39. 39. Solutiona) DC analysis: V Z  V BE IE3   3.67mA RE IE3 ICQ   1.84mA 2 V CEQ  V C - VE  V C - ICQ R C - V E  7.05V VT and  14.1 Ω ICQ
40. 40. Solutionb) VO -RC RC A CM   - Vi VT 2R T  2R T IC Where R T  rce 3 1  b  Then R T  80.635 MΩ A CM  - 29.14  10 -6 Therefore
41. 41. Solution VO RCc) A d    -166.67 Vd  VT  2  I    C d) Ad CMRR(dB)  20log 10  135.147dB A CMe) Z i(diff)  2r   4.23k Ωf) Z O  R C  4.7k Ω
42. 42. TUTORIAL 5
43. 43. Question 1Figure 5.16 shows an emitter coupled pair differential amplifierwith Vi1 and Vi2 as the input and VO as the output. Q1 and Q2have the following parameters:β1 = β2 = 150, VT = 26mV, VBE1 = VBE2 = 0.7V +15VCalculate: 4.7kΩ 4.7kΩa) ICQ1, ICQ2b) Differential gain, Ad Vi1 V01 Vi2c) Common mode gain, ACM Q1 Q2d) Differential input impedance, Zi(diff)e) Output impedance, Zo 4.9kΩ -15V Figure 5.16
44. 44. Question 2Figure 5.17 is a differential amplifier with β1=β2= β3=100,VBE=0.7V, VT=26mV and 1/ro =40μS. Both diodes are made ofsilicon. Determine: +15Va) ICQ1, ICQ2 15kΩ 15kΩb) Ad, ACM and CMRR in dB V0 Vi1 Vi2c) Zi(diff) and Zo Q1 Q2 2kΩ Q3 470kΩ -15V Figure 5.17
45. 45. Question 3Figure 5.18 shows an emitter coupled pair differential amplifierwith Vi1 and Vi2 as the input and VO as the output. Determine: +12Va) The quiescent point of the 1.5k Ω 1.5k Ω differential amplifier for transistor Q1b) CMRR in dB Vi1 V0 Vi2 Q1 Q2c) Output resistance (Ro)d) Differential input resistance, Ri(diff) 1.5k Ω Q3 Q4 -12V Figure 5.18
46. 46. The EndSee you in the next lesson!Don’t forget to do revision. -ISMET-edited by Nazirah Mohamat Kasim & Shahilah Nordin