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BJTDiffAmpSu10

# BJTDiffAmpSu10 - c Copyright 2010 W Marshall Leach Jr...

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c ° Copyright 2010. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering. The BJT Di ff erential Ampli fi er Basic Circuit Figure 1 shows the circuit diagram of a di ff erential ampli fi er. The tail supply is modeled as a current source I Q . The object is to solve for the small-signal output voltages and output resistances. It will be assumed that the transistors are identical. Figure 1: Circuit diagram of the di ff erential ampli fi er. DC Solution Zero both base inputs. For identical transistors, the current I Q divides equally between the two emitters. (a) The dc currents are given by I E 1 = I E 2 = I Q 2 I B 1 = I B 2 = I Q 2 (1 + β ) I C 1 = I C 2 = αI Q 2 (b) Verify that V CB > 0 for the active mode. V CB = V C V B = ¡ V + αI E R C ¢ μ I E 1 + β R B = V + αI E R C + I E 1 + β R B (e) Calculate the collector-emitter voltage. V CE = V C V E = V C ( V B V BE ) = V CB + V BE 1

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Small-Signal AC Solution using the Emitter Equivalent Circuit This solution uses the r 0 approximations and assumes that the base spreading resistance r x is not zero. (a) Calculate g m , r π , r e , and r ie . g m = I C V T r π = V T I B r e = V T I E r ie = R B + r x + r π 1 + β r 0 = V A + V CE I E (b) Redraw the circuit with V + = V = 0 . Replace the two BJTs with the emitter equivalent circuit. The emitter part of the circuit obtained is shown in Fig. 2(a). Figure 2: (a) Emitter equivalent circuit for i e 1 and i e 2 . (b) Collector equivalent circuits. (c) Using Ohm’s Law, solve for i e 1 and i e 2 . i e 1 = v i 1 v i 2 2 ( r ie + R E ) i e 2 = i e 1 (d) The circuit for v o 1 , v o 2 , and r out is shown in Fig. 2(b). v o 1 = i c 1( sc ) × r ic k R C = α × i e 1 × r ic k R C = α × r ic k R C 2 ( r ie + R E ) ( v i 1 v i 2 ) v o 2 = i c 2( sc ) × r ic k R C = α × i ie 2 × r ic k R C = α × r ic k R C 2 ( r ie + R E ) ( v i 2 v i 1 ) r out 1 = r out 2 = r ic k R C r ic = r 0 · 1 + β (2 R E + r ie ) R B + r π + r x + 2 R E + r ie ¸ + ( R B + r x + r π ) k (2 R E + r ie ) (e) The resistance seen looking into either input with the other input zeroed is r in = R B + r x + r π + (1 + β ) (2 R E + r ie ) The di ff erential input resistance r ind is the resistance between the two inputs for di ff erential input signals. For an ideal current source tail supply, this is the same as the input resistance r in above. Di ff Amp with Non-Perfect Tail Supply Fig. 3 shows the circuit diagram of a di ff erential ampli fi er. The tail supply is modeled as a current source I 0 Q having a parallel resistance R Q . In the case of an ideal current source, R Q is an open circuit. Often a di ff amp is designed with a resistive tail supply. In this case, I 0 Q = 0 . The solutions below are valid for each of these connections. The object is to solve for the small-signal output voltages and output resistances.
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