The BJT Differential Amplifier. Basic Circuit. DC Solution

Transcription

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 B1 = I B = I Q (1+β) I C1 = I C = αi Q (b) Verify that V CB > 0 for the active mode. V CB = V C V B = µ V + α R C I E 1+β R B = V + α 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 and assumes that the base spreading resistance r x is not zero. (a) Calculate g m, r π, r e,andr ie. g m = I C V T r π = V T I B r e = V T r ie = R B + r x + r π 1+β 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. (a). Figure : (a) Emitter equivalent circuit for i e1 and i e. (b) Collector equivalent circuits. (c) Using Ohm s Law, solve for i e1 and i e. v i1 v i ( ) i e = i e1 (d) The circuit for v o1, v o,andr out isshowninfig.(b). v o1 = i c1(sc) r ic kr C = α i e1 r ic kr C = ( ) (v i1 v i ) v o = i c(sc) r ic kr C = α i ie r ic kr C = ( ) (v i v i1 ) r out1 = r out = r ic kr C β (R E + r ie ) r ic = r 0 +(R B + r x + r π ) k (R E + r ie ) R B + r π + r x +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)(R E + r ie ) The differential input resistance r ind is the resistance between the two inputs for differential input signals. For an ideal current source tail supply, this is the same as the input resistance r in above. Diff Amp with Non-Perfect Tail Supply Fig. 3 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.

3 Figure 3: 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. 4. The circuit for Q is identical. Now make a Thévenin equivalent as shown in on the right in Fig. 4. This is the basic bias circuit. (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 IQR 0 Q = 1+β R B + V BE + (R E + ) (c) Solve the loop equation for the currents. = I C α =(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 = µ V + α R C I E 1+β R B = V + α 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 /. If the current source is replaced with a resistor only, the currents are given by = I C α =(1+β) I B = V V BE R B / (1 + β)+r E + 3

4 Small-Signal or AC Solutions This solutions use the r 0 approximations. (a) Calculate g m, r π,andr ie. g m = α V T Figure 4: DC bias circuits for Q 1. r π = (1 + β) V T r ie = R B + r x + r π 1+β r 0 = V A + V CE α (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 5(a). Figure 5: (a) Emitter equivalent circuit. (b) Collector equivalent circuits. (c) Using superposition, Ohm s Law, and current division, solve for i e1 and i e. v i1 + k ( ) v i + k ( ) + 4

5 v i i e = + k ( ) v i1 + k ( ) + For =, these become v i1 v i ( ) i e = v i v i1 ( ) (d) The circuit for v o1, v o, r out1,andr out isshowninfig.6. Figure 6: Circuits for calculating v o1, v o, r out1,andr out. v o1 = i 0 c1 r ic kr C = α i e1 r ic kr C = µ α r ic kr C v i1 v i + k ( ) + v o = i 0 c r ic kr C = α i e1 r ic kr C = µ α r ic kr C v i v i1 + k ( ) + r out1 = r out = r ic kr C βr te r ic = r 0 +(R B + r x + r π ) kr te R te = R E + k ( ) R B + r π + 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 = ( ) (v i1 v i ) v o = ( ) (v i v i1 ) (g) The equivalent circuit seen looking into the two inputs is shown in Fig. 7. The resistors labeled r 0 π are given by r 0 π = r x + r π +(1+β) R E The differential input resistance r id is defined the same way that it is defined for Fig.??. 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 + rπ 0 5

6 Figure 7: Equivalent circuits for calculating i b1 and i b. Differential and Common-Mode Gains (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. 8. Figure 8: Emitter equivalent circuit. (c) For v i1 = v id / and v i = v id /, it follows by superposition that v a =0and v id/ i e = v id/ v o1 = α i e1 r ic kr C = v o = α i e r ic kr C = +α r ickr C 6 v id = v id = +α r ickr C µ vi1 v i µ vi1 v i

7 The differential voltage gain is given by A d = v o1 = v o = 1 v id v id αr ic kr C (d) For v i1 = v i = v icm, it follows by superposition that i a =0and v o1 = α i e1 r ic kr C = v o = α i e r ic kr C = The common-mode voltage gain is given by v icm + i e = + v icm = + v icm = v icm A cm = v o1 = v o = α r ickr C v icm v icm + µ vi1 + v i µ 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 + = 1 + ThiscanbeexpressedindB. µ 1 CMRR db =0log + R Q Example 1 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 + α R C µ 1+β R B =8.09 V g m = α V T =0.047 S r π = (1 + β) V T =.097 kω r e = V T =0.97 Ω r ie = R B + r x 1+β + r e =31.17 Ω r 0 = V A + V CE = kω R te = R E + k ( )=30.83 Ω I C β (R E + r ie ) r ic = r 0 +(R B + r π ) k (R E + r ie ) = 390.5kΩ R B + r π +R E + r ie µ α r ic kr C v o1 = v i1 v i = 36.84v i v i + k ( ) + v o = 36.84v i v i1 7

8 r out = r ic kr C =9.75 kω A vd = 1 α r ic kr C = A vcm = = CMRR db =0log A vd =51.63 db The Diff AmpwithanActiveLoad A vcm Figure9showsaBJTdiff amp with an active load formed by a current mirror with base current compensation. Similar circuits are commonly seen as the input stages of operational amplifiers and audio amplifiers. 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 bias current I Q splits equally between Q 1 and Q so that 1 = = I Q /. Figure 9: Diff amp with active current-mirror load. 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 8

9 supply I Q. Following the analysis above, the small-signal collector currents in Q 1 and Q are given by α v id α v id i c1(sc) = i c(sc) = where r ie = R B + r π 1+β The short-circuit output current is given by With i c4 = i c3 = i c1 and i c = i c1, this becomes i sc =i c1(sc) = The output resistance is given by i sc = i c4 i c α v id = r out = r 04 kr ic α (v i1 v i ) where r ic is given by β (R E + r ie ) r ic = r 0 +(R B + r π ) k (R E + r ie ) R B + r π +R E + r ie By Thévenin s theorem, the small-signal open-circuit output voltage is given by v oc = i sc r out = α (r 04kr ic ) (v i1 v i ) Example 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, α =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. Because r x > 0, weaddittor B in the equations above. Solution. r e1 = r e = V T =5Ω r ie1 = r ie = R B + r x I Q 1+β + r e =6.5 Ω α R te =R E + r ie1 =18.5 Ω i sc = (v i1 v i )=0.018 (v i1 v i ) r 0 = V A +(V C + V BE ) αi Q / This is a db gain of 55.3dB. βr te =65kΩ r ic = r 0 R B + r π + R te 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 ) +(R B + r π ) kr te = 36.7kΩ 9