The Common-Emitter Amplifier

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1 c Copyright W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering. The Common-Emitter Amplifier Basic Circuit Fig. shows the circuit diagram of a single stage common-emitter amplifier. The object is to solve for the small-signal voltage gain, input resistance, and output resistance. Figure : Single-stage common-emitter 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 R + R 2 R BB = V EE = V R EE = R E (b) Make an educated guess for V BE. Write the loop equation between the V BB and the V EE nodes. V BB V EE = I B R BB + V BE + I E R EE = I C R BB + V BE + I C R EE (c) Solve the loop equation for the currents. (d) Verify that V CB > 0 for the active mode. I C = I E = I B = V BB V EE V BE R BB / + R EE / 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 /)

2 Figure 2: Bias circuit. Figure 3: Signal circuit. 2

3 Small-Signal or AC Solutions (a) Redraw the circuit with V + = V =0and all capacitors replaced with short circuits as shown in Fig. 3. (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. = R s + R tb = v te =0 R te = R E kr 3 Figure 4: Signal circuit with Thévenin base circuit. Exact Solution This solution is based on the exact equivalent circuits developed in the more advanced notes on the BJT. It treats r 0 as a resistor from collector to emitter without the r 0 approximations. (a) Replace the BJT in Fig. 4 with the Thévenin base circuit and the Norton collector circuit asshowninfig.5. Figure 5: Base and collector equivalent circuits. 3

4 (b) Solve for i c(sc). i c(sc) = G mb = G mb R s + (c) Solve for v o. G mb = r 0 R te / re 0 + R te kr 0 r 0 + R te r 0 e = R tb + r x + + r e v o = i c(sc) r ic kr C kr L = G mb R s + r ic kr C kr L (d) Solve for the voltage gain. r 0 + r 0 ekr te R te / (r 0 e + R te ) = G mb R s + r ic kr C kr L (e) Solve for r in. r in = kr ib r ib = r x + r π + R te ( + ) r 0 + R tc r 0 + R te + R tc (f) Solve for r out. r out = r ic kr C (g) Special Case for R te =0. (h) Special Case for r 0 =. G mb = r 0 e r 0 r ib = r x + r π G mb = r 0 e + R te r ib = r x + r π +(+) R te Example For the CE amplifier of Fig., it is given that R s =5kΩ, R = 20 kω, R 2 = 00 kω, R C =4.3kΩ, R E =5.6kΩ, R 3 = 00 Ω, R L =20kΩ, V + =5V, V = 5 V, V BE =0.65 V, =99, =0.99, r x =20Ω, V A = 00 V and V T =0.025 V. Solve for the gain /,the input resistance r in, and the output resistance r out. The capacitors can be assumed to be ac short circuits at the operating frequency. Solution. For the dc bias solution, replace all capacitors with open circuits. The Thévenin voltage and resistance seen looking out of the base are V BB = V + R 2 + V R R + R 2 =.364 V R BB = =54.55 kω The Thévenin voltage and resistance seen looking out of the emitter are V EE = V and R EE = R E. The bias equation for I E is I E = V BB V EE V BE R BB / ( + )+R EE =2.3 ma 4

5 To test for the active mode, we calculate the collector-base voltage V CB = V C V B = µ V + I E R C V BB I E + R BB =8.52 V Because this is positive, the BJT is biased in its active mode. For the small-signal ac analysis, we need r 0 and r e. To calculate r 0,wefirst calculate the collector-emitter voltage V CE = V CB + V BE =9.7 V It follows that r 0 and r e have the values r 0 = V A + V CE I E =52.8 kω r e = V T I E =.83 Ω For the small-signal analysis, V + and V are zeroed and the three capacitors are replaced with ac short circuits. The Thévenin voltage and resistance seen looking out of the base are given by = R s + =0.96 R tb = R s k =4.58 kω The Thévenin resistances seen looking out of the emitter and the collector are R te = R E kr 3 =98.25 Ω R tc = R C kr L =3.539 kω Next, we calculate r 0 e, G mb, r ic,andr ib. G mb = r 0 e = R tb + r x + + r e =57.83 Ω r 0 R te / re 0 = + R te kr 0 r 0 + R te 57.8 S r 0 + r 0 ekr te R te / (r 0 e + R te ) = 38.6kΩ ( + ) r 0 + R tc r ib = r x +(+) r e + R te r 0 + R te + R tc The output voltage is given by =0.39 kω v o = G mb (r ic kr tc ) = G mb (r ic kr tc ) 0.96 = Thus the voltage gain is A v = The input and output resistances are given by r in = kr ib =8.73 kω r out = r ic kr C =3.539 kω Approximate Solutions These solutions use the r 0 approximations. That is, it is assumed that r 0 = except in calculating r ic.inthiscase,i c(sc) = i 0 c = i 0 e = i b. 5

6 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 e. i 0 e = re 0 = + R te R s + re 0 + R te (b) Solve for i 0 c and r ic. i 0 c = i 0 e = R s + re 0 + R te r 0 + r 0 ekr te R te / (re 0 + R te ) (c) Solve for v o and /. v o = i c(sc) r ic kr C kr L = R s + re 0 + R te = Note that this is of the form (d) Solve for r out. R s + r 0 e + R te = i0 e i0 c i 0 v o e i 0 c r out = r ic kr C (d) Solve for r ib and r in. Because the base node is absorbed, use the formula for r ib. r ib = r x +(+)(r e + R te ) r in = kr ib Example 2 Use the simplified T-model solutions to calculate the values of A v, r in,andr out for Example. A v = = r ib =.03 kω r in =9.73 kω 38.6kΩ r out =4.7 kω 6

7 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. = i b (R tb + r x )+v π + i 0 er te = i0 c (R tb + r x )+ i0 c g m + i0 c R te = i 0 c = R tb + r x + + R te g m (c) Solve for v o. r 0 + r 0 ekr te R te / (r 0 e + R te ) v o = i 0 cr C kr L = (d) Solve for the voltage gain. R tb + r x + + R te g m = R s + R tb + r x + + R te g m = R s + R tb + r x + + R te g m This is of the form (e) Solve for r ib and r in. = i0 c v o i 0 c v b = i b (r x + r π )+i 0 er te = i b (r x + r π )+(+) i b R te = i b [r x + r π +(+) R te ] r ib = v b i b = r x + r π +(+) R te 7

8 r in = kr ib (f) Solve for r out. r out = r ic kr C Example 3 Use the π-model solutions to calculate the values of A v, r in,andr out for Example. g m = r π =.83 kω TModelSolution A v = = = r ib =.03 kω 38.6kΩ r in =9.73 kω r out =4.7 kω (a) After making the Thévenin equivalent circuits looking out of the base and emitter, replace the BJTwiththeTmodelasshowninFig.8. (b) Solve for i 0 c and r ic. Figure 8: T model circuit. = i b (R tb + r x )+i 0 e (r e + R te )= i0 c (R tb + r x )+ i0 c (r e + R te )= i 0 c = R tb + r x + r e + R te (c) Solve for v o. r 0 + r 0 ekr te R te / (r 0 e + R te ) v o = i 0 cr C kr L = R tb + r x + r e + R te = R s + R tb + r x + r e + R te 8

9 (d) Solve for the voltage gain. = R s + R tb + r x + r e + R te Note that this is of the form (e) Solve for r ib and r in. = i0 c v o i 0 c v b = i b r x + i 0 e (r e + R te )=i b r x +(+) i b (r e + R te )=i b [r x +(+)(r e + R te )] r ib = v b i b = r x +(+)(r e + R te ) r in = kr ib (f) Solve for r out. r out = r ic kr C Example 4 Use the T-model solutions to calculate the values of A v, r in,andr out for Example. A v = = = r ib =.03 kω 38.6kΩ r in =9.73 kω r out =4.7 kω 9

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