Lecture 22. OUTLINE Differential Amplifiers. Reading: Chapter General considerations BJT differential pair

Transcription

1 Lecture 22 OUTLNE Differential Amplifiers General considerations BJT differential pair Qualitatie analysis Large signal analysis Small signal analysis Frequency response Reading: Chapter EE105 Spring 2008 Lecture 22, Slide 1 Prof. Wu, UC Berkeley

2 Humming Noise in Audio Amplifier Consider the amplifier below which amplifies an audio signal from a microphone. f the power supply (V CC ) is time arying, it will result in an additional (undesirable) oltage signal at the output, perceied as a humming noise by the user. Vout = VCC CRC EE105 Spring 2008 Lecture 22, Slide 2 Prof. Wu, UC Berkeley

3 Supply Ripple Rejection Since node X and Y each see the oltage ripple, their oltage difference will be free of ripple. X Y X = = A r Y = in + A r in EE105 Spring 2008 Lecture 22, Slide 3 Prof. Wu, UC Berkeley

4 Ripple Free Differential Output f the input signal is to be a oltage difference between two nodes, an amplifier that senses a differential signal is needed. EE105 Spring 2008 Lecture 22, Slide 4 Prof. Wu, UC Berkeley

5 Common nputs to Differential Amp. The oltage signals applied to the input nodes of a differential amplifier cannot be in phase; otherwise, the differential output signal will be zero. X Y X = A in + = A in + = 0 Y r r EE105 Spring 2008 Lecture 22, Slide 5 Prof. Wu, UC Berkeley

6 Differential nputs to Differential Amp. When the input oltage signals are 180 out of phase, the resultant output node oltages are 180 out of phase, so that their difference is enhanced. X Y X = Y in = A A in + + r = 2A r in EE105 Spring 2008 Lecture 22, Slide 6 Prof. Wu, UC Berkeley

7 Differential Signals Differential signals share the same aerage DC alue and are equal in magnitude but opposite in phase. A pair of differential signals can be generated, among other ways, by a transformer. EE105 Spring 2008 Lecture 22, Slide 7 Prof. Wu, UC Berkeley

8 Single Ended s. Differential Signals EE105 Spring 2008 Lecture 22, Slide 8 Prof. Wu, UC Berkeley

9 Differential Pair With the addition of a tail current, the circuits aboe operate as an elegant, yet robust differential pair. EE105 Spring 2008 Lecture 22, Slide 9 Prof. Wu, UC Berkeley

10 Common Mode Response V = V BE1 BE 2 = = C1 C2 2 EE V = V = V R X Y CC C 2 EE EE105 Spring 2008 Lecture 22, Slide 10 Prof. Wu, UC Berkeley

11 Differential Response V V C1 C2 X Y = = 0 = V = V EE CC CC R C EE EE105 Spring 2008 Lecture 22, Slide 11 Prof. Wu, UC Berkeley

12 Differential Response (cont d) V V C2 C1 Y X = = 0 = V = V EE CC CC R C EE EE105 Spring 2008 Lecture 22, Slide 12 Prof. Wu, UC Berkeley

13 Differential Pair Characteristics A differential input signal results in ariations in the output currents and oltages, whereas a common mode input signal does not result in any output current/oltage ariations. EE105 Spring 2008 Lecture 22, Slide 13 Prof. Wu, UC Berkeley

14 Virtual Ground For small input oltages (+ΔV and ΔV), the g m alues are ~equal, so the increase in C1 and decrease in C2 are ~equal in magnitude. Thus, the oltage at node P is constant and can be considered as AC ground. EE C1 = +Δ 2 EE C 2 = Δ 2 Δ = g ΔV ΔV ( ) ( ) C1 m P Δ = g ΔV ΔV Δ = Δ C2 m P C1 C2 Δ V = 0 EE105 Spring 2008 Lecture 22, Slide 14 Prof. Wu, UC Berkeley P

15 Extension of Virtual Ground t can be shown that if R 1 = R 2, and the oltage at node A goes up by the same amount that the oltage at node B goes down, then the oltage at node X does not change. X = 0 EE105 Spring 2008 Lecture 22, Slide 15 Prof. Wu, UC Berkeley

16 Small Signal Differential Gain Since the output signal changes by 2g m ΔVR C when the input signal changes by 2ΔV, the small signal oltage gain is g m R C. Note that the oltage gain is the same as for a CE stage, but that the power dissipation is doubled. A = 2g mδvr 2ΔV C = g m R C EE105 Spring 2008 Lecture 22, Slide 16 Prof. Wu, UC Berkeley

17 Large Signal Analysis V V = V V in1 in2 BE1 BE 2 C1 C2 = VT ln VT ln S S V ln + = C1 = T C 2 C1 C2 EE V V VT EEe C1 = V V V 1+ e = in1 in 2 in1 in 2 EE C 2 V V V 1+ e in1 in 2 EE105 Spring 2008 Lecture 22, Slide 17 Prof. Wu, UC Berkeley T T

18 nput/output Characteristics V V out1 out 2 = ( V R ) CC C1 C ( V R ) CC C 2 C ( ) = R C2 C1 C R V tanh V in1 in2 = C EE 2V T EE105 Spring 2008 Lecture 22, Slide 18 Prof. Wu, UC Berkeley

19 Linear/Nonlinear Regions of Operation Amplifier operating in linear region Amplifier operating in non-linear region EE105 Spring 2008 Lecture 22, Slide 19 Prof. Wu, UC Berkeley

20 Small Signal Analysis EE105 Spring 2008 Lecture 22, Slide 20 Prof. Wu, UC Berkeley

21 Half Circuits Since node P is AC ground, we can treat the differential pair as two CE half circuits. out1 in1 out 2 in2 = g m R C EE105 Spring 2008 Lecture 22, Slide 21 Prof. Wu, UC Berkeley

22 Half Circuit Example 1 out1 in1 out 2 in 2 = g m r O EE105 Spring 2008 Lecture 22, Slide 22 Prof. Wu, UC Berkeley

23 Half Circuit Example 2 A = g ( ) m1 ro 1 ro 3 R 1 EE105 Spring 2008 Lecture 22, Slide 23 Prof. Wu, UC Berkeley

24 Half Circuit Example 3 A = g ( ) m1 ro 1 ro 3 R 1 EE105 Spring 2008 Lecture 22, Slide 24 Prof. Wu, UC Berkeley

25 Half Circuit Example 4 A = R 1 + g m C R E EE105 Spring 2008 Lecture 22, Slide 25 Prof. Wu, UC Berkeley

26 Differential Pair Frequency Response Since the differential pair can be analyzed using its half circuit, its transfer function, /O impedances, locations of poles/zeros are the same as that of its half circuit. C π 1 C π 2 EE105 Spring 2008 Lecture 22, Slide 26 Prof. Wu, UC Berkeley