Figure 1: Common-base amplifier.

Transcription

1 The Common-Base Amplifier Basic Circuit Fig. 1 shows the circuit diagram of a single stage common-base amplifier. The object is to solve for the small-signal voltage gain, input resistance, and output resistance. Figure 1: Common-base amplifier. DC Solution (a) Replace the capacitors with open circuits. Look out of the 3 BJT terminals and make Thévenin equivalent circuits as shown in Fig. 2. V BB = V + R 2 + V R 1 R 1 + R 2 R BB = R 1 kr 2 V EE = V E = V CC = V + R CC = R C (b) Make an educated guess for V BE. Write the loop equation between the V BB and the V EE nodes. To solve for I C,thisequationis V BB V EE = I B R BB + V BE + I E E = I C β R BB + V BE + I C E (c) Solve the loop equation for the currents. (d) Verify that V CB > 0 for the active mode. I C = I E = βi B = V BB V EE V BE R BB /β + E / V CB = V C V B =(V CC I C R CC ) (V BB I B R BB )=V CC V BB I C (R CC R BB /β) 1

2 Small-Signal or AC Solutions Figure 2: DC bias circuit. (a) Redraw the circuit with V + = V =0and all capacitors replaced with short circuits as shown in Fig. 3. Figure 3: Signal circuit. (b) Calculate g m, r π, r e,andr 0 from the DC solution.. g m = I C V T r π = V T I B r e = V T I E r 0 = V A + V CE I C (c) Replace the circuits looking out of the base and emitter with Thévenin equivalent circuits asshowninfig.4. v tb =0 R tb =0 v te = v s R te = R s k 2

3 Figure 4: Signal circuit with Thévenin emitter circuit. Exact Solution (a) Replace the BJT in Fig. 4 with the Thévenin emitter circuit and the Norton collector circuit asshowninfig.5. Figure 5: Emitter and collector equivalent circuits. (b) Solve for i c(sc). (c) Solve for v o. i c(sc) = G me v te = G me v s 1 r 0 + re 0 G me = R te + rekr 0 0 r 0 + re 0 re 0 = 1+β + r e v o = i c(sc) r ic kr C kr L = G me v s r ic kr C kr L r 0 + r 0 ekr te 1 R te / (r 0 e + R te ) 3

4 (d) Solve for the voltage gain. (e) Solve for r in. A v = v o v s = G me r ic kr C kr L r in = R 1 kr 2 kr ie r ie = re 0 r 0 + R tc re 0 + r 0 + R tc / (1 + β) (f) Solve for r out. r out = r ic kr C Example 1 For the CB amplifier in Fig. 1, it is given that R s = 100 Ω, R 1 = 120 kω, R 2 = 100 kω, R C =4.3kΩ, =5.6kΩ, R 3 = 100 Ω, R L =20kΩ, V + =15V, V = 15 V, V BE =0.65 V, β =99, =0.99, =20Ω, V A = 100 V and V T =0.025 V. SolveforA v, r in,andr out. Solution. Because the dc bias circuit is the same as for the common-emitter amplifier example, thedcbiasvalues,r e, g m, r π,andr 0 are the same. In the signal circuit, the Thévenin voltage and resistance seen looking out of the emitter are given by v te = v s =0.9825v s R te = R s k =98.25 Ω The Thévenin resistances seen looking out of the base and the collector are R tb =0 R tc = R C kr L =3.539 kω Next, we calculate r 0 e, G me, r ic,andr ie. r 0 e = R tb + 1+β + r e =12.03 Ω G me = 1 R te + r 0 ekr 0 r 0 + r 0 e r 0 + r 0 e = S r 0 + r 0 ekr te 1 R te / (re 0 + R te ) = 442.3kΩ r ie = re 0 r 0 + R tc re 0 + r 0 + R tc / (1 + β) =12.83 Ω The output voltage is given by v o = G me (r ic kr tc ) v te = G me (r ic kr tc ) v s =30.97v s Thus the voltage gain is The input and output resistances are A v =30.97 r in = R 1 kr 2 kr ib =12.81 Ω r out = r ic kr C =4.259 kω Approximate Solutions These solutions assume that r 0 = except in calculating r ic.inthiscase,i c(sc) = i 0 c = i 0 e = βi b. 4

5 Figure 6: Simplified T model circuit. Simplified T Model Solution (a) After making the Thévenin equivalent circuits looking out of the base and emitter, replace the BJT with the simplified T model as shown in Fig. 6. (b) Solve for i 0 c and r ic. 0 v te = i 0 e r 0 i 0 e + R te = c r 0 e + R te = i 0 c = v te re 0 + R te (c) Solve for v o. r 0 + r 0 ekr te 1 R te / (r 0 e + R te ) v o = i 0 cr ic kr C kr L = v te re 0 r ic kr C kr L = v s + R te re 0 r ic kr C kr L + R te (d) Solve for the voltage gain. A v = v o v s = R s r 0 e + R te r ic kr C kr L (e) Solve for r ie and r in. (f) Solve for r out. 0 v e = i 0 er 0 e = i 0 e = v e r 0 e r ie = v e i 0 e = r 0 e r in = r 0 ek r out = r ic kr C Example 2 For Example 1, use the simplified T-model solutions to calculate the values of A v, r in, and r out. 5

6 A v = =30.97 r in =12Ω r out =4.259 kω π Model Solution (a) After making the Thévenin equivalent circuits looking out of the base and emitter, replace the BJT with the π model as shown in Fig. 7. (b) Solve for i 0 c and r ic. Figure 7: Hybrid-π model circuit. 0 v te = i b + v π + i 0 er te = i0 c β + i0 c g m + i0 c R te = i 0 c = v te β R te g m (c) Solve for v o. v o = i 0 cr ic kr C kr L = v te β R te g m r 0 + r 0 ekr te 1 R te / (r 0 e + R te ) 1 r ic kr C kr L = v s β R r ic kr C kr L te g m (d) Solve for the voltage gain. A v = v o v s = 1 β R r ic kr C kr L te g m (e) Solve for r out. r out = r ic kr C 6

7 (f) Solve for r ie and r in. 0 v e = i b ( + r π )= i0 e 1+β ( + r π )= i 0 e = v e 1+β + r π r ie = v e i 0 e = + r π 1+β r in = r ie k Example 3 For Example 1, use the π-model solutions to calculate the values of A v, r in,andr out. TModelSolution A v = =30.97 r in =12Ω r out =4.259 kω (a) After making the Thévenin equivalent circuits looking out of the base and emitter, replace the BJTwiththeTmodelasshowninFig.??. (b) Solve for i 0 c and r ic. Figure 8: T model circuit. 0 v te = i b + i 0 e (r e + R te )= i0 c β + i0 c (r e + R te )= i 0 c = v te β + r e + R te (c) Solve for v o. v o = i 0 cr ic kr C kr L = v te β + r e + R te r 0 + r 0 ekr te 1 R te / (r 0 e + R te ) 1 r ic kr C kr L = v s β + r r e + R ic kr C kr L te 7

8 (d) Solve for the voltage gain. (e) Solve for r ie and r in. A v = v o v s = 0 v e = i b + i 0 er e = i0 e 1+β + i 0 er e = i 0 e (f) Solve for r out. 1 β + r r e + R ic kr C kr L te r ie = v e i 0 e µ rx 1+β + r e = i 0 e = = 1+β + r e r in = kr ie r out = r ic kr C v e 1+β + r e Example 4 For Example 1, use the T-model solutions to calculate the values of A v, r in,andr out. A v = =30.97 r in =12Ω r out =4.259 kω 8