The BJT Differential Amplifier. Basic Circuit. DC Solution

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1 c Copyright 010. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering. The BJT Differential Amplifier Basic Circuit Figure 1 shows the circuit diagram of a differential amplifier. 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 differential amplifier. 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 1 = = I Q I C1 = I C = I Q (b) Verify that V CB > 0 for the active mode. V CB = V C V B = V + R C I E 1+β R B = V + R C + 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

2 Small-Signal AC Solution using the Emitter Equivalent Circuit This solution uses the r 0 approximations. (a) Calculate g m, r π, r e,andr 0 e. g m = V T r π = (1 + β) V T r e = V T r 0 e = R B + r x 1+β + r e r 0 = V A + V CE (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.. Figure : Emitter equivalent circuit using the r 0 approximations. (c) Using Ohm s Law, solve for i 0 e1 and i0 e. i 0 e1 = v i1 v i ( ) i 0 e = i 0 e1 (d) The circuit for v o1, v o, r out1,andr out isshowninfig.3. Figure 3: Circuits for calculating v o1, v o, r out1,andr out. v o1 = i 0 c1r ic kr C = i 0 e1r ic kr C = r ickr C ( ) (v i1 v i ) v o = i 0 cr ic kr C = i 0 e1r ic kr C = r ickr C ( ) (v i v i1 ) r out1 = r out = r ic kr C r ic = r 0 + r 0 ekr te 1 R te r 0 e + R te 0 R te =R E + re

3 (e) The resistance seen looking into the v i1 (v i ) input with v i =0(v i1 =0)is where r π =(1+β) r e has been used and r in = R B + r x + r π +(1+β) R te = R B + r x +(1+β) R E + R B + r x 1+β + r e = [R B + r x + r π +(1+β) R E ] = R B + rπ 0 r 0 π = r x + r π +(1+β) R E Figure 4: Base equivalent circuit for calculating i b1 and i b. The differential input resistance r id is the resistance seen between the two inputs when v i1 = v id / and v i = v id /, wherev id is the differential input voltage. It can be seen from the figure that it is given by r id =(R B + r 0 π). The Diff AmpwithanActiveLoad Figure5showsaBJTdiff amp with an active load formed by a current mirror with base current compensation. The object is to solve for the open-circuit output voltage v oc, the short-circuit output current i sc, and the output resistance r out. By Thévenin s theorem, these are related by the equation v oc = i sc r out. It will be assumed that the current mirror consisting of transistors Q 3 Q 5 is perfect so that its output current is equal to its input current, i.e. i c4 = i c1. In addition, the r 0 approximations will be used in solving for the currents. That is, the Early effect will be neglected except in solving for r out. For the bias solution, it will be assumed that the tail current I Q splits equally between Q 1 and Q so that 1 = = I Q /. Because the tail supply is assumed to be a current source, the common-mode gain of the circuit is zero when the r 0 approximations are used. In this case, it can be assumed that the two input signals are pure differential signals that can be written v i1 = v id / and v i = v id /. Fordifferential input signals, it follows by symmetry that the signal voltage is zero at the node above the tail current supply I Q. Following the analysis above, the small-signal collector currents in Q 1 and Q are given by i 0 v id c1 = re 0 i 0 c = v id + R E re 0 + R E where re 0 = R B + r x 1+β + r e The short-circuit output current is given by i sc = i 0 c4 i 0 c 3

4 Figure 5: Diff amp with active current-mirror load. With i 0 c4 = i0 c3 = i0 c1 and i0 c = i0 c1, this becomes i sc =i 0 c1 = The output resistance is given by where r ic is given by v id = r out = r 04 kr ic (v i1 v i ) r ic = r 0 + r 0 ekr te 1 R te r 0 e + R te 0 R te =R E + r ie1 =R E + re By Thévenin s theorem, the small-signal open-circuit output voltage is given by v oc = i sc r out = r 04kr ic (v i1 v i ) Example 1 For I Q =ma, R B = 100 Ω, R E =51Ω, V + =15V, V = 15 V, V T =0.05 V, r x =50Ω, β =99, =0.99 V BE1 = V BE =0.65 V, V EB3 = V EB4 = V EB5 =0.65 V, V C = V C4 = 13.7V,andV A =50V,calculatei sc, r out,andv oc. Solution. r e1 = r e = V T I Q =5Ω r 0 e1 = r 0 e = R B + r x 1+β + r e =6.5 Ω R te =R E + r ie1 =R E + r 0 e = 18.5 Ω i sc = 4 (v i1 v i )=0.018 (v i1 v i )

5 r 0 = V A +(V C + V BE ) =65kΩ I Q / This is a db gain of 55.3dB. r ic = r 0 + r 0 ekr te 1 R te r 0 e + R te = 36.7kΩ r 04 = V A +(V + V C4 ) I Q =51.8 kω r out = r 04 kr ic =45.34 kω v oc = i sc r out =579.(v i1 v i ) Diff Amp with Non-Perfect Tail Supply Fig. 6 shows the circuit diagram of a differential amplifier. The tail supply is modeled as a current source I 0 Q having a parallel resistance. In the case of an ideal current source, is an open circuit. Often a diff amp is designed with a resistive tail supply. In this case, I 0 Q =0.Thesolutions below are valid for each of these connections. The object is to solve for the small-signal output voltages and output resistances. Figure 6: BJT Differential amplifier. DC Solutions This solution assumes that I 0 Q is known. If I Q is known, the solutions are the same as above. (a) Zero both inputs. Divide the tail supply into two equal parallel current sources having a current I 0 Q / in parallel with a resistor. The circuit obtained for Q 1 isshownontheleftin Fig. 7. The circuit for Q is identical. Now make a Thévenin equivalent as shown in on the right in Fig. 7. This is the basic bias circuit. 5

6 Figure 7: DC bias circuits for Q 1. (b) Make an educated guess for V BE. Write the loop equation between the ground node to the left of R B and V.Tosolvefor,thisequationis 0 V I 0 Q = 1+β R B + V BE + (R E + ) (c) Solve the loop equation for the currents. = I C =(1+β) I B = V + I 0 Q V BE R B / (1 + β)+r E + (d) Verify that V CB > 0 for the active mode. V CB = V C V B = V + R C I E 1+β R B = V + R C + 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 (f) If =, it follows that 1 = = IQ 0 /. IfI0 Q =0, the currents are given by Small-Signal or AC Solutions Emitter Equivalent Circuit = I C =(1+β) I B = This solution uses the r 0 approximations. V V BE R B / (1 + β)+r E + 6

7 (a) Calculate g m, r π, r e,andr 0 e. g m = V T r π = (1 + β) V T r e = V T r 0 e = R B + r x 1+β + r e r 0 = V A + V CE (b) Redraw the circuit with V + = V =0and IQ 0 =0. Replace the two BJTs with the emitter equivalent circuit. The emitter part of the circuit obtained is shown in 8. Figure 8: Emitter equivalent circuit for the simplified T model.. (c) Using superposition, Ohm s Law, and current division, solve for i 0 e1 and i0 e. i 0 e1 = i 0 e = For =, these become v i1 + k ( ) v i + k ( ) i 0 e1 = v i1 v i ( ) v i re 0 + R E + k (re 0 + R E ) + re 0 + R E v i1 re 0 + R E + k (re 0 + R E ) + re 0 + R E i 0 e = v i v i1 ( ) (d) The circuit for v o1, v o, r out1,andr out isshowninfig.9. Figure 9: Circuits for calculating v o1, v o, r out1,andr out. v o1 = i 0 c1r ic kr C = i 0 e1r ic kr C = r ic kr C re 0 + R E + k (re 0 v i1 v i + R E ) + re 0 + R E v o = i 0 cr ic kr C = i 0 e1r ic kr C = r ic kr C re 0 + R E + k (re 0 v i v i1 + R E ) + re 0 + R E 7

8 r out1 = r out = r ic kr C r ic = r 0 + rekr 0 te 1 R R te = R E + k r 0 e + R E te re 0 + R te (e) The resistance seen looking into the v i1 (v i ) input with v i =0(v i1 =0)is (f) Special case for =. r ib = R B + r x + r π +(1+β) R te v o1 = r ickr C ( ) (v i1 v i ) v o = r ickr C ( ) (v i v i1 ) (g) The equivalent circuit seen looking into the two inputs is similar to that in Fig. 4 with the exception that a resistor representing the effect of must be added. It is shown in Fig. 10. The resistors labeled r 0 π have the same value as the ones in Fig. 4. They are given by r 0 π = r x + r π +(1+β) R E Figure 10: Equivalent circuits for calculating i b1 and i b. The differential input resistance r id is defined the same way that it is defined for Fig. 4. That is, it is the resistance seen between the two inputs when v i1 = v id / and v i = v id /, wherev id is the differential input voltage. In this case, the small-signal voltage at the upper node of the resistor (1 + β) iszerosothatnocurrentflows it. It follows that r id is given by r id =(R B + r 0 π). Hybrid-π Model This solution assumes that r 0 =. Replace the two transistors with the hybrid-π model as shown in Fig. 11. (a) Write the loop equations for the two input loops. Use the relations v π1 = i 0 c1 /g m and v π = i 0 c /g m. v i1 = i0 c1 β (R B + r x )+ i0 c1 + i0 c1 i 0 g m R E + c1 + i0 c These equations are in the form v i = i0 c β (R B + r x )+ i0 c g m + i0 c R E + v i1 =(A + B) i 0 c1 + Bi 0 c 8 i 0 c1 + i0 c

9 Figure 11: Hybrid-π model (r 0 = ). where A = R B + r x β v i = Bi 0 c1 +(A + B) i 0 c R E g m B = (b) Use determinants to solve the two equations simultaneously for i 0 c1 and i0 c. i 0 c1 = (A + B) v i1 Bv i (A + B) = (A + B) v i1 Bv i B A (A +B) i 0 c = (A + B) v i Bv i1 (A + B) = (A + B) v i Bv i1 B A (A +B) Thus the solutions are RB + r x R E i 0 β g m + R Q v i1 v i c1 = RB + r x R E RB + r x R E β g m β g m + R Q RB + r x R E i 0 β g m + R Q v i v i1 c1 = RB + r x R E RB + r x R E β g m β g m + R Q After some algebra, the solutions reduce to those obtained with the emitter equivalent circuit. The output voltages are given by v o1 = i 0 c1r C v o = i 0 cr C For the case of a finite r 0,ther 0 approximation replaces R C with r ic kr C. 9

10 Differential and Common-Mode Gains This solution uses the r 0 approximations. (a) Define the common-mode and differential input voltages as follows: v id = v i1 v i With these definitions, v i1 and v i can be written v icm = v i1 + v i v i1 = v icm + v id v i = v icm v id By linearity, it follows that superposition of v icm and v id canbeusedtosolveforthecurrentsand voltages. (b) Redraw the emitter equivalent circuit as shown in Fig. 1. Figure 1: Emitter equivalent circuit for calculating the common-mode and differential emitter currents. (c) For v i1 = v id / and v i = v id /, it follows by superposition that v a =0and i 0 e1 = v id/ re 0 + R E v o1 = i 0 e1r ic kr C = r ickr C v o = i 0 er ic kr C = +r ickr C The differential voltage gain is given by i 0 e = v id/ v id = r ickr C v id = +r ickr C A d = v o1 = v o = 1 r ic kr C v id v id re 0 + R E (d) For v i1 = v i = v icm, it follows by superposition that i a =0and i 0 e1 = v o1 = i 0 e1r ic kr C = v icm + i 0 e = r ickr C + v icm = v icm + r ickr C + vi1 v i vi1 v i vi1 + v i 10

11 v o = i 0 er ic kr C = The common-mode voltage gain is given by r ickr C + v icm = r ickr C + A cm = v o1 = v o r ic kr C = v icm v icm re 0 + R E + vi1 + v i (e) If the output is taken from the collector of Q 1 or Q, the common-mode rejection ratio is given by CMRR = v o1 /v id v o1 /v icm = v o /v id v o /v icm = 1 re 0 + R E + re 0 = 1 + R E + re 0 + R E ThiscanbeexpressedindB. 1 CMRR db =0log + R Q re 0 + R E Example For I 0 Q =ma, =50kΩ, R B =1kΩ, R E = 100 Ω, R C =10kΩ, V + =0V, V = 0 V, V T =0.05 V, r x =0Ω, β =99, V BE =0.65 V, andv A =50V,calculatev o1, v o, v od, r out,andcmrr. Solution. = ³ 0 V IQ 0 V BE =1.19 ma R B / (1 + β)+r E + V CB = V C V B = V + R C 1+β R B =8.09 V g m = V T =0.047 S r π = (1 + β) V T =.097 kω r e = V T =0.97 Ω r 0 e = R B + r x 1+β + r e =31.17 Ω v o1 = r 0 = V A + V CE I C = kω R te = R E + k = Ω r ic = r 0 + rekr 0 te 1 R = 390.5kΩ te re 0 + R te r ic kr C re 0 + R E + k (re 0 v i1 v i + R E ) + re 0 + R E v o = 36.84v i v i1 r out = r ic kr C =9.75 kω A vd = 1 r ic kr C re 0 = R E A vcm = r ickr C re 0 = R E + CMRR db =0log A vd =51.63 db A vcm = 36.84v i v i 11

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