1. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II DMT 231 / 3 ELECTRONICS II Lecture IV AC Analysis II [BJT Common-Emitter Amplifier]

2. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II RCRC RBRB vsvs vOvO V BB VCCVCC Example I Given :  = 100, V CC = 12V V BE = 0.7V, R C = 6k , R B = 50k , and V BB = 1.2V Calculate the small-signal voltage gain. Small-signal hybrid-  equivalent circuit

3. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II 1. 2. 3. 4. 5. 6. Example I: Solutions

4. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II Basic Common-Emitter Amplifier Circuit vsvs RSRS R1R1 R2R2 RCRC C vOvO V CC Example II Given :  = 100, V CC = 12V V BE(on) = 0.7V, R S = 0.5k , R C = 6k , R 1 = 93.7k , R 2 = 6.3k  and V A = 100V. Calculate the small-signal voltage gain.

5. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II R 1 \\ R 2 VsVs RSRS RCRC rOrO rr gmVgmV VoVo RiRi RoRo Example II: Solution Small-signal equivalent circuit

6. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II The basic common-emitter circuit used in previous analysis causes a serious defect : –If BJT with V BE (on) = 0.7 V is used, I B = 9.5 μA & I C = 0.95 mA –But, if new BJT with V BE (on) = 0.6 V is used, I B = 26 μA & BJT goes into saturation; which is not acceptable  Previous circuit is not practical –So, the emitter resistor is included: Q-point is stabilized against variations in β, as will the voltage gain, A V Assumptions –C C acts as a short circuit –Early voltage = ∞ ==> r o neglected due to open circuit Basic Common-Emitter Amplifier

7. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II Common-Emitter Amplifier with Emitter Resistor CE amplifier with emitter resistor Small-signal equivalent circuit (with current gain parameter, β) inside transistor

8. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II Common-Emitter Amplifier with Emitter Resistor ac output voltage Input voltage loop Input resistance, R ib Input resistance to amplifier, R i Voltage divider equation of V in to V s Remember: Assume V A is infinite,  r o is neglected

9. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II Common-Emitter Amplifier with Emitter Resistor Cont.. So, small-signal voltage gain, A V If R i >> R s and (1 + β)R E >> r π Remember: Assume V A is infinite,  r o is neglected

10. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II Example III Given :  = 100, V BE(on) = 0.7V, V T = 26 mV and V A = ∞.Determine: (i) Base-emitter input resistance, r (ii) transconductance, g m (iii) small-signal transistor output resistance, r o (iv) Input resistance to the base, R ib (v) Input resistance to the amplifier, R i (vi) Small-signal voltage gain, A v. Common-Emitter Amplifier with Emitter Resistor

11. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II RSRS R1R1 R2R2 RERE RCRC vsvs vOvO C V CC CECE VoVo VsVs RCRC RSRS rr roro R 1 || R 2 gmVgmV Emitter bypass capacitor, C E provides a short circuit to ground for the ac signals Common-Emitter Amplifier with Emitter Bypass Capacitor Small-signal hybrid- π equivalent circuit

12. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II Example IV Given :  = 100, V BE(on) = 0.7V, V T = 26 mV and V A = 100.Determine: (i)Quiescent value of base current, I BQ (ii)Quiescent value of collector current, I CQ (iii)Quiescent value of collector-emitter voltage, V CEQ (iv)Base-emitter input resistance, r (v)transconductance, g m (vi)small-signal transistor output resistance, r o (vii)Input resistance seen by the signal source, R in (viii) Output resistance looking back into the output terminal, R o (ix)Small-signal voltage gain, A v. Common-Emitter Amplifier with Emitter Bypass Capacitor

13. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II AC LOAD LINE ANALYSIS DC load line –Visualized the relationship between Q-point & transistor characteristics AC load line –Visualized the relationship between small-signal response & transistor characteristics –Occurs when capacitors added in transistor circuit

14. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II AC LOAD LINE ANALYSIS Common-emitter amplifier with emitter bypass capacitor Note: The next DC & AC load line analysis will be based on this circuit

15. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II KVL on C-E loop AC LOAD LINE ANALYSIS - DC Load Line

16. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II KVL on C-E loop AC LOAD LINE ANALYSIS - AC Load Line AC equivalent circuit

17. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II AC and DC Load Lines

18. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II RSRS R1R1 R2R2 RERE RCRC RLRL vsvs vOvO C C1 C C2 V CC vOvO vsvs R 1  R 2 RSRS RERE RCRC RLRL AC LOAD LINE ANALYSIS Common-emitter amplifier with emitter resistor AC equivalent circuit Note: The DC & AC load line analysis will be based on these circuits

19. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II AC LOAD LINE ANALYSIS - DC Load Line + V CE 0 + I C I CQ V CEQ Q KVL at C-E loop

20. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II + V CE 0 + I C I CQ V CEQ Q AC LOAD LINE ANALYSIS - AC Load Line

21. SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II Self-Reading Chapter 6, page 399 - 402: Basic Common- Emitter Amplifier Circuit Chapter 6, page 402 - 409: Circuit with Emitter Resistor Chapter 6, page 409 - 413: Circuit with Emitter Bypass Capacitor Chapter 6, page 415 - 418: AC Load Line Analysis Donald A. Neamen, MICROELECTRONICS Circuit Analysis and Design, Third Edition