BJTDiffAmpSu10

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

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Unformatted text preview: c Copyright 2010. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering. The BJT Di f erential Ampli f er Basic Circuit Figure 1 shows the circuit diagram of a di f erential ampli f 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 f erential ampli f 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 + ) I C 1 = I C 2 = I Q 2 (b) Verify that V CB > for the active mode. V CB = V C V B = V + I E R C I E 1 + R B = V + I E R C + I E 1 + 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 Small-Signal AC Solution using the Emitter Equivalent Circuit This solution uses the r approximations and assumes that the base spreading resistance r x is not zero. (a) Calculate g m , r , r e , and r ie . g m = I C V T r = V T I B r e = V T I E r ie = R B + r x + r 1 + r = 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 Ohms 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 = i e 1 r ic k R C = 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 = i ie 2 r ic k R C = 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 1 + (2 R E + r ie ) R B + r + r x + 2 R E + r ie + ( R B + r x + r ) 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 + (1 + )(2 R E + r ie ) The di f erential input resistance r ind is the resistance between the two inputs for di f erential input signals. For an ideal current source tail supply, this is the same as the input resistance r in above. Di f Amp with Non-Perfect Tail Supply Fig. 3 shows the circuit diagram of a di f erential ampli f er. The tail supply is modeled as a current source I Q having a parallel resistance R Q . In the case of an ideal current source, R Q is an open circuit. Often a di f amp is designed with a resistive tail supply. In this case, I Q = 0 . The solutions below are valid for each of these connections. The object is to solve for the small-signal outputbelow are valid for each of these connections....
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BJTDiffAmpSu10 - c Copyright 2010. W. Marshall Leach, Jr.,...

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