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BJT amplifier: basic operation V CC I C I C R C V CC I C R C V o I E - PowerPoint PPT Presentation

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BJT amplifier: basic operation V CC I C I C R C V CC I C R C V o I E V B V B I B t I E 0 0.2 0.4 0.6 0.8 V BE t M. B. Patil, IIT Bombay BJT amplifier: basic operation V CC I C I C R C V CC I C R C V o I E V B V B I B t I E 0 0.2

  1. BJT amplifier saturation 5 linear V CC / R C I B5 4 5 V V CC I B4 V o (Volts) 3 R C I B3 I C V o C 2 B I B2 V i R B 1 E I B1 0 0 V CC 0 0 1 2 3 4 5 V CE ( V o ) V i (Volts) * The key challenges in realizing this amplifier in practice are M. B. Patil, IIT Bombay

  2. BJT amplifier saturation 5 linear V CC / R C I B5 4 5 V V CC I B4 V o (Volts) 3 R C I B3 I C V o C 2 B I B2 V i R B 1 E I B1 0 0 V CC 0 0 1 2 3 4 5 V CE ( V o ) V i (Volts) * The key challenges in realizing this amplifier in practice are - adjusting the input DC bias to ensure that the BJT remains in the linear (active) region with a certain bias value of V BE (or I C ). M. B. Patil, IIT Bombay

  3. BJT amplifier saturation 5 linear V CC / R C I B5 4 5 V V CC I B4 V o (Volts) 3 R C I B3 I C V o C 2 B I B2 V i R B 1 E I B1 0 0 V CC 0 0 1 2 3 4 5 V CE ( V o ) V i (Volts) * The key challenges in realizing this amplifier in practice are - adjusting the input DC bias to ensure that the BJT remains in the linear (active) region with a certain bias value of V BE (or I C ). - mixing the input DC bias with the signal voltage. M. B. Patil, IIT Bombay

  4. BJT amplifier saturation 5 linear V CC / R C I B5 4 5 V V CC I B4 V o (Volts) 3 R C I B3 I C V o C 2 B I B2 V i R B 1 E I B1 0 0 V CC 0 0 1 2 3 4 5 V CE ( V o ) V i (Volts) * The key challenges in realizing this amplifier in practice are - adjusting the input DC bias to ensure that the BJT remains in the linear (active) region with a certain bias value of V BE (or I C ). - mixing the input DC bias with the signal voltage. * The first issue is addressed by using a suitable biasing scheme, and the second by using “coupling” capacitors. M. B. Patil, IIT Bombay

  5. BJT amplifier: a simple biasing scheme 15 V V CC R C R B 1 k C B E “Biasing” an amplifier ⇒ selection of component values for a certain DC value of I C (or V BE ) (i.e., when no signal is applied). M. B. Patil, IIT Bombay

  6. BJT amplifier: a simple biasing scheme 15 V V CC R C R B 1 k C B E “Biasing” an amplifier ⇒ selection of component values for a certain DC value of I C (or V BE ) (i.e., when no signal is applied). Equivalently, we may bias an amplifier for a certain DC value of V CE , since I C and V CE are related: V CE + I C R C = V CC . M. B. Patil, IIT Bombay

  7. BJT amplifier: a simple biasing scheme 15 V V CC R C R B 1 k C B E “Biasing” an amplifier ⇒ selection of component values for a certain DC value of I C (or V BE ) (i.e., when no signal is applied). Equivalently, we may bias an amplifier for a certain DC value of V CE , since I C and V CE are related: V CE + I C R C = V CC . As an example, for R C = 1 k, β = 100, let us calculate R B for I C = 3 . 3 m A , assuming the BJT to be operating in the active mode. M. B. Patil, IIT Bombay

  8. BJT amplifier: a simple biasing scheme 15 V V CC R C R B 1 k C B E “Biasing” an amplifier ⇒ selection of component values for a certain DC value of I C (or V BE ) (i.e., when no signal is applied). Equivalently, we may bias an amplifier for a certain DC value of V CE , since I C and V CE are related: V CE + I C R C = V CC . As an example, for R C = 1 k, β = 100, let us calculate R B for I C = 3 . 3 m A , assuming the BJT to be operating in the active mode. I B = I C β = 3 . 3 m A = 33 µ A = V CC − V BE = 15 − 0 . 7 100 R B R B M. B. Patil, IIT Bombay

  9. BJT amplifier: a simple biasing scheme 15 V V CC R C R B 1 k C B E “Biasing” an amplifier ⇒ selection of component values for a certain DC value of I C (or V BE ) (i.e., when no signal is applied). Equivalently, we may bias an amplifier for a certain DC value of V CE , since I C and V CE are related: V CE + I C R C = V CC . As an example, for R C = 1 k, β = 100, let us calculate R B for I C = 3 . 3 m A , assuming the BJT to be operating in the active mode. I B = I C β = 3 . 3 m A = 33 µ A = V CC − V BE = 15 − 0 . 7 100 R B R B → R B = 14 . 3 V 33 µ A = 430 kΩ . M. B. Patil, IIT Bombay

  10. BJT amplifier: a simple biasing scheme (continued) 15 V V CC R C R B 1 k C B E With R B = 430 k, we expect I C = 3 . 3 m A , assuming β = 100. M. B. Patil, IIT Bombay

  11. BJT amplifier: a simple biasing scheme (continued) 15 V V CC R C R B 1 k C B E With R B = 430 k, we expect I C = 3 . 3 m A , assuming β = 100. However, in practice, there is a substantial variation in the β value (even for the same transistor type). The manufacturer may specify the nominal value of β as 100, but the actual value may be 150, for example. M. B. Patil, IIT Bombay

  12. BJT amplifier: a simple biasing scheme (continued) 15 V V CC R C R B 1 k C B E With R B = 430 k, we expect I C = 3 . 3 m A , assuming β = 100. However, in practice, there is a substantial variation in the β value (even for the same transistor type). The manufacturer may specify the nominal value of β as 100, but the actual value may be 150, for example. With β = 150, the actual I C is, I C = β × V CC − V BE = 150 × (15 − 0 . 7) V = 5 m A , R B 430 k which is significantly different than the intended value, viz., 3.3 m A . M. B. Patil, IIT Bombay

  13. BJT amplifier: a simple biasing scheme (continued) 15 V V CC R C R B 1 k C B E With R B = 430 k, we expect I C = 3 . 3 m A , assuming β = 100. However, in practice, there is a substantial variation in the β value (even for the same transistor type). The manufacturer may specify the nominal value of β as 100, but the actual value may be 150, for example. With β = 150, the actual I C is, I C = β × V CC − V BE = 150 × (15 − 0 . 7) V = 5 m A , R B 430 k which is significantly different than the intended value, viz., 3.3 m A . → need a biasing scheme which is not so sensitive to β . M. B. Patil, IIT Bombay

  14. BJT amplifier: improved biasing scheme 10 V V CC R C 3.6 k R 1 10 k I C I B I E R 2 R E 2.2 k 1 k

  15. BJT amplifier: improved biasing scheme 10 V V CC R C R C 3.6 k R 1 10 k I C I C R 1 V CC I B I B I E I E R 2 V CC R 2 R E R E 2.2 k 1 k

  16. BJT amplifier: improved biasing scheme 10 V V CC R C R C R C 3.6 k R 1 10 k I C I C I C R 1 R Th V CC V CC I B I B I B I E I E I E R 2 V CC R 2 V Th R E R E R E 2.2 k 1 k M. B. Patil, IIT Bombay

  17. BJT amplifier: improved biasing scheme 10 V V CC R C R C R C 3.6 k R 1 10 k I C I C I C R 1 R Th V CC V CC I B I B I B I E I E I E R 2 V CC R 2 V Th R E R E R E 2.2 k 1 k R 2 2 . 2 k V Th = V CC = 10 k + 2 . 2 k × 10 V = 1 . 8 V , R Th = R 1 � R 2 = 1 . 8 k R 1 + R 2 M. B. Patil, IIT Bombay

  18. BJT amplifier: improved biasing scheme 10 V V CC R C R C R C 3.6 k R 1 10 k I C I C I C R 1 R Th V CC V CC I B I B I B I E I E I E R 2 V CC R 2 V Th R E R E R E 2.2 k 1 k R 2 2 . 2 k V Th = V CC = 10 k + 2 . 2 k × 10 V = 1 . 8 V , R Th = R 1 � R 2 = 1 . 8 k R 1 + R 2 Assuming the BJT to be in the active mode, KVL: V Th = R Th I B + V BE + R E I E = R Th I B + V BE + ( β + 1) I B R E M. B. Patil, IIT Bombay

  19. BJT amplifier: improved biasing scheme 10 V V CC R C R C R C 3.6 k R 1 10 k I C I C I C R 1 R Th V CC V CC I B I B I B I E I E I E R 2 V CC R 2 V Th R E R E R E 2.2 k 1 k R 2 2 . 2 k V Th = V CC = 10 k + 2 . 2 k × 10 V = 1 . 8 V , R Th = R 1 � R 2 = 1 . 8 k R 1 + R 2 Assuming the BJT to be in the active mode, KVL: V Th = R Th I B + V BE + R E I E = R Th I B + V BE + ( β + 1) I B R E V Th − V BE β ( V Th − V BE ) → I B = , I C = β I B = . R Th + ( β + 1) R E R Th + ( β + 1) R E M. B. Patil, IIT Bombay

  20. BJT amplifier: improved biasing scheme 10 V V CC R C R C R C 3.6 k R 1 10 k I C I C I C R 1 R Th V CC V CC I B I B I B I E I E I E R 2 V CC R 2 V Th R E R E R E 2.2 k 1 k R 2 2 . 2 k V Th = V CC = 10 k + 2 . 2 k × 10 V = 1 . 8 V , R Th = R 1 � R 2 = 1 . 8 k R 1 + R 2 Assuming the BJT to be in the active mode, KVL: V Th = R Th I B + V BE + R E I E = R Th I B + V BE + ( β + 1) I B R E V Th − V BE β ( V Th − V BE ) → I B = , I C = β I B = . R Th + ( β + 1) R E R Th + ( β + 1) R E For β = 100, I C =1.07 m A . M. B. Patil, IIT Bombay

  21. BJT amplifier: improved biasing scheme 10 V V CC R C R C R C 3.6 k R 1 10 k I C I C I C R 1 R Th V CC V CC I B I B I B I E I E I E R 2 V CC R 2 V Th R E R E R E 2.2 k 1 k R 2 2 . 2 k V Th = V CC = 10 k + 2 . 2 k × 10 V = 1 . 8 V , R Th = R 1 � R 2 = 1 . 8 k R 1 + R 2 Assuming the BJT to be in the active mode, KVL: V Th = R Th I B + V BE + R E I E = R Th I B + V BE + ( β + 1) I B R E V Th − V BE β ( V Th − V BE ) → I B = , I C = β I B = . R Th + ( β + 1) R E R Th + ( β + 1) R E For β = 100, I C =1.07 m A . For β = 200, I C =1.085 m A . M. B. Patil, IIT Bombay

  22. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C 3.6 k R 1 10 k I C I B I E R 2 R E 2.2 k 1 k With I C = 1 . 1 m A , the various DC (“bias”) voltages are

  23. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C 3.6 k R 1 10 k I C I B I E R 2 R E 2.2 k 1 k With I C = 1 . 1 m A , the various DC (“bias”) voltages are V E = I E R E ≈ 1 . 1 m A × 1 k = 1 . 1 V ,

  24. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C 3.6 k R 1 10 k I C I B 1.1 V I E R 2 R E 2.2 k 1 k With I C = 1 . 1 m A , the various DC (“bias”) voltages are V E = I E R E ≈ 1 . 1 m A × 1 k = 1 . 1 V ,

  25. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C 3.6 k R 1 10 k I C I B 1.1 V I E R 2 R E 2.2 k 1 k With I C = 1 . 1 m A , the various DC (“bias”) voltages are V E = I E R E ≈ 1 . 1 m A × 1 k = 1 . 1 V , V B = V E + V BE ≈ 1 . 1 V + 0 . 7 V = 1 . 8 V ,

  26. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C 3.6 k R 1 10 k I C 1.8 V I B 1.1 V I E R 2 R E 2.2 k 1 k With I C = 1 . 1 m A , the various DC (“bias”) voltages are V E = I E R E ≈ 1 . 1 m A × 1 k = 1 . 1 V , V B = V E + V BE ≈ 1 . 1 V + 0 . 7 V = 1 . 8 V ,

  27. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C 3.6 k R 1 10 k I C 1.8 V I B 1.1 V I E R 2 R E 2.2 k 1 k With I C = 1 . 1 m A , the various DC (“bias”) voltages are V E = I E R E ≈ 1 . 1 m A × 1 k = 1 . 1 V , V B = V E + V BE ≈ 1 . 1 V + 0 . 7 V = 1 . 8 V , V C = V CC − I C R C = 10 V − 1 . 1 m A × 3 . 6 k ≈ 6 V ,

  28. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C 3.6 k R 1 10 k I C 6 V 1.8 V I B 1.1 V I E R 2 R E 2.2 k 1 k With I C = 1 . 1 m A , the various DC (“bias”) voltages are V E = I E R E ≈ 1 . 1 m A × 1 k = 1 . 1 V , V B = V E + V BE ≈ 1 . 1 V + 0 . 7 V = 1 . 8 V , V C = V CC − I C R C = 10 V − 1 . 1 m A × 3 . 6 k ≈ 6 V ,

  29. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C 3.6 k R 1 10 k I C 6 V 1.8 V I B 1.1 V I E R 2 R E 2.2 k 1 k With I C = 1 . 1 m A , the various DC (“bias”) voltages are V E = I E R E ≈ 1 . 1 m A × 1 k = 1 . 1 V , V B = V E + V BE ≈ 1 . 1 V + 0 . 7 V = 1 . 8 V , V C = V CC − I C R C = 10 V − 1 . 1 m A × 3 . 6 k ≈ 6 V , V CE = V C − V E = 6 − 1 . 1 = 4 . 9 V . M. B. Patil, IIT Bombay

  30. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C R 1 3.6 k 10 k I C I B I E R 2 R E 2.2 k 1 k A quick estimate of the bias values can be obtained by ignoring I B (which is fair if β is large). In that case, M. B. Patil, IIT Bombay

  31. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C R 1 3.6 k 10 k I C I B I E R 2 R E 2.2 k 1 k A quick estimate of the bias values can be obtained by ignoring I B (which is fair if β is large). In that case, R 2 2 . 2 k V B = V CC = 10 k + 2 . 2 k × 10 V = 1 . 8 V . R 1 + R 2 M. B. Patil, IIT Bombay

  32. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C R 1 3.6 k 10 k I C I B I E R 2 R E 2.2 k 1 k A quick estimate of the bias values can be obtained by ignoring I B (which is fair if β is large). In that case, R 2 2 . 2 k V B = V CC = 10 k + 2 . 2 k × 10 V = 1 . 8 V . R 1 + R 2 V E = V B − V BE ≈ 1 . 8 V − 0 . 7 V = 1 . 1 V . M. B. Patil, IIT Bombay

  33. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C R 1 3.6 k 10 k I C I B I E R 2 R E 2.2 k 1 k A quick estimate of the bias values can be obtained by ignoring I B (which is fair if β is large). In that case, R 2 2 . 2 k V B = V CC = 10 k + 2 . 2 k × 10 V = 1 . 8 V . R 1 + R 2 V E = V B − V BE ≈ 1 . 8 V − 0 . 7 V = 1 . 1 V . I E = V E = 1 . 1 V = 1 . 1 m A . R E 1 k M. B. Patil, IIT Bombay

  34. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C R 1 3.6 k 10 k I C I B I E R 2 R E 2.2 k 1 k A quick estimate of the bias values can be obtained by ignoring I B (which is fair if β is large). In that case, R 2 2 . 2 k V B = V CC = 10 k + 2 . 2 k × 10 V = 1 . 8 V . R 1 + R 2 V E = V B − V BE ≈ 1 . 8 V − 0 . 7 V = 1 . 1 V . I E = V E = 1 . 1 V = 1 . 1 m A . R E 1 k I C = α I E ≈ I E = 1 . 1 m A . M. B. Patil, IIT Bombay

  35. BJT amplifier: improved biasing scheme (continued) 10 V V CC R C R 1 3.6 k 10 k I C I B I E R 2 R E 2.2 k 1 k A quick estimate of the bias values can be obtained by ignoring I B (which is fair if β is large). In that case, R 2 2 . 2 k V B = V CC = 10 k + 2 . 2 k × 10 V = 1 . 8 V . R 1 + R 2 V E = V B − V BE ≈ 1 . 8 V − 0 . 7 V = 1 . 1 V . I E = V E = 1 . 1 V = 1 . 1 m A . R E 1 k I C = α I E ≈ I E = 1 . 1 m A . V CE = V CC − I C R C − I E R E = 10 V − (3 . 6 k × 1 . 1 m A ) − (1 k × 1 . 1 m A ) ≈ 5 V . M. B. Patil, IIT Bombay

  36. Adding signal to bias V CC R C R 1 v B R 2 R E

  37. Adding signal to bias V CC R C R 1 v B R 2 R E * As we have seen earlier, the input signal v s ( t ) = � V sin ω t (for example) needs to be mixed with the desired bias value V B so that the net voltage at the base is v B ( t ) = V B + � V sin ω t .

  38. Adding signal to bias V CC R C R 1 v B C B v s R 2 R E * As we have seen earlier, the input signal v s ( t ) = � V sin ω t (for example) needs to be mixed with the desired bias value V B so that the net voltage at the base is v B ( t ) = V B + � V sin ω t . * This can be achieved by using a coupling capacitor C B .

  39. Adding signal to bias V CC R C R 1 v B C B v s R 2 R E * As we have seen earlier, the input signal v s ( t ) = � V sin ω t (for example) needs to be mixed with the desired bias value V B so that the net voltage at the base is v B ( t ) = V B + � V sin ω t . * This can be achieved by using a coupling capacitor C B . * Let us consider a simple circuit to illustrate how a coupling capacitor works. M. B. Patil, IIT Bombay

  40. RC circuit with DC + AC sources v C R 2 A v A v s (t) R 1 V 0 (DC) V m sin ω t We are interested in the solution (currents and voltages) in the “sinusoidal steady state” when the exponential transients have vanished and each quantity x ( t ) is of the form X 0 (constant) + X m sin( ω t + α ). M. B. Patil, IIT Bombay

  41. RC circuit with DC + AC sources v C R 2 A v A v s (t) R 1 V 0 (DC) V m sin ω t We are interested in the solution (currents and voltages) in the “sinusoidal steady state” when the exponential transients have vanished and each quantity x ( t ) is of the form X 0 (constant) + X m sin( ω t + α ). There are two ways to obtain the solution: M. B. Patil, IIT Bombay

  42. RC circuit with DC + AC sources v C R 2 A v A v s (t) R 1 V 0 (DC) V m sin ω t We are interested in the solution (currents and voltages) in the “sinusoidal steady state” when the exponential transients have vanished and each quantity x ( t ) is of the form X 0 (constant) + X m sin( ω t + α ). There are two ways to obtain the solution: (1) Solve the circuit equations directly: v A ( t ) + v A ( t ) − V 0 = C d dt ( v s ( t ) − v A ( t )) . R 1 R 2 M. B. Patil, IIT Bombay

  43. RC circuit with DC + AC sources v C R 2 A v A v s (t) R 1 V 0 (DC) V m sin ω t We are interested in the solution (currents and voltages) in the “sinusoidal steady state” when the exponential transients have vanished and each quantity x ( t ) is of the form X 0 (constant) + X m sin( ω t + α ). There are two ways to obtain the solution: (1) Solve the circuit equations directly: v A ( t ) + v A ( t ) − V 0 = C d dt ( v s ( t ) − v A ( t )) . R 1 R 2 (2) Use the DC circuit + AC circuit approach. M. B. Patil, IIT Bombay

  44. Resistor in sinusoidal steady state v R (t) R i R (t) M. B. Patil, IIT Bombay

  45. Resistor in sinusoidal steady state v R (t) R i R (t) Let v R ( t ) = V R + v r ( t ) where V R = constant, v r ( t ) = � V R sin ( ω t + α ), = constant, i r ( t ) = � i R ( t ) = I R + i r ( t ) where I R I R sin ( ω t + α ). M. B. Patil, IIT Bombay

  46. Resistor in sinusoidal steady state v R (t) R i R (t) Let v R ( t ) = V R + v r ( t ) where V R = constant, v r ( t ) = � V R sin ( ω t + α ), = constant, i r ( t ) = � i R ( t ) = I R + i r ( t ) where I R I R sin ( ω t + α ). Since v R ( t ) = R × i R ( t ), we get [ V R + v r ( t )] = R × [ I R + i r ( t )]. M. B. Patil, IIT Bombay

  47. Resistor in sinusoidal steady state v R (t) R i R (t) Let v R ( t ) = V R + v r ( t ) where V R = constant, v r ( t ) = � V R sin ( ω t + α ), = constant, i r ( t ) = � i R ( t ) = I R + i r ( t ) where I R I R sin ( ω t + α ). Since v R ( t ) = R × i R ( t ), we get [ V R + v r ( t )] = R × [ I R + i r ( t )]. This relationship can be split into two: V R = R × I R , and v r ( t ) = R × i r ( t ). M. B. Patil, IIT Bombay

  48. Resistor in sinusoidal steady state v R (t) R i R (t) Let v R ( t ) = V R + v r ( t ) where V R = constant, v r ( t ) = � V R sin ( ω t + α ), = constant, i r ( t ) = � i R ( t ) = I R + i r ( t ) where I R I R sin ( ω t + α ). Since v R ( t ) = R × i R ( t ), we get [ V R + v r ( t )] = R × [ I R + i r ( t )]. This relationship can be split into two: V R = R × I R , and v r ( t ) = R × i r ( t ). In other words, a resistor can be described by V R v r (t) R R I R i r (t) DC AC M. B. Patil, IIT Bombay

  49. Capacitor in sinusoidal steady state v C (t) i C (t) C M. B. Patil, IIT Bombay

  50. Capacitor in sinusoidal steady state v C (t) i C (t) C Let v C ( t ) = V C + v c ( t ) where V C = constant, v c ( t ) = � V C sin ( ω t + α ), = constant, i c ( t ) = � i C ( t ) = I C + i c ( t ) where I C I C sin ( ω t + β ). M. B. Patil, IIT Bombay

  51. Capacitor in sinusoidal steady state v C (t) i C (t) C Let v C ( t ) = V C + v c ( t ) where V C = constant, v c ( t ) = � V C sin ( ω t + α ), = constant, i c ( t ) = � i C ( t ) = I C + i c ( t ) where I C I C sin ( ω t + β ). Since i C ( t ) = C dv C dt , we get [ I C + i c ( t )] = C d dt ( V C + v c ( t )). M. B. Patil, IIT Bombay

  52. Capacitor in sinusoidal steady state v C (t) i C (t) C Let v C ( t ) = V C + v c ( t ) where V C = constant, v c ( t ) = � V C sin ( ω t + α ), = constant, i c ( t ) = � i C ( t ) = I C + i c ( t ) where I C I C sin ( ω t + β ). Since i C ( t ) = C dv C dt , we get [ I C + i c ( t )] = C d dt ( V C + v c ( t )). This relationship can be split into two: I C = C dV C = 0, and i c ( t ) = C dv c dt . dt M. B. Patil, IIT Bombay

  53. Capacitor in sinusoidal steady state v C (t) i C (t) C Let v C ( t ) = V C + v c ( t ) where V C = constant, v c ( t ) = � V C sin ( ω t + α ), = constant, i c ( t ) = � i C ( t ) = I C + i c ( t ) where I C I C sin ( ω t + β ). Since i C ( t ) = C dv C dt , we get [ I C + i c ( t )] = C d dt ( V C + v c ( t )). This relationship can be split into two: I C = C dV C = 0, and i c ( t ) = C dv c dt . dt In other words, a capacitor can be described by V C v c (t) I C i c (t) C DC AC M. B. Patil, IIT Bombay

  54. Voltage sources in sinusoidal steady state DC voltage source: v S (t) V S v s (t) i S (t) I S i s (t) v S (t) = V S + 0 DC AC M. B. Patil, IIT Bombay

  55. Voltage sources in sinusoidal steady state DC voltage source: v S (t) V S v s (t) i S (t) I S i s (t) v S (t) = V S + 0 DC AC AC voltage source: v S (t) v s (t) V S i S (t) I S i s (t) v S (t) = 0 + v s (t) DC AC M. B. Patil, IIT Bombay

  56. RC circuit with DC + AC sources DC circuit AC circuit v C V C v c R 2 R 2 R 2 A A A v A v a V A v s (t) v s (t) R 1 R 1 R 1 V 0 (DC) V 0 V m sin ω t V m sin ω t M. B. Patil, IIT Bombay

  57. RC circuit with DC + AC sources DC circuit AC circuit v C V C v c R 2 R 2 R 2 A A A v A v a V A v s (t) v s (t) R 1 R 1 R 1 V 0 (DC) V 0 V m sin ω t V m sin ω t V A + V A − V 0 DC circuit: = 0. (1) R 1 R 2 M. B. Patil, IIT Bombay

  58. RC circuit with DC + AC sources DC circuit AC circuit v C V C v c R 2 R 2 R 2 A A A v A v a V A v s (t) v s (t) R 1 R 1 R 1 V 0 (DC) V 0 V m sin ω t V m sin ω t V A + V A − V 0 DC circuit: = 0. (1) R 1 R 2 v a + v a = C d AC circuit: dt ( v s − v a ). (2) R 1 R 2 M. B. Patil, IIT Bombay

  59. RC circuit with DC + AC sources DC circuit AC circuit v C V C v c R 2 R 2 R 2 A A A v A v a V A v s (t) v s (t) R 1 R 1 R 1 V 0 (DC) V 0 V m sin ω t V m sin ω t V A + V A − V 0 DC circuit: = 0. (1) R 1 R 2 v a + v a = C d AC circuit: dt ( v s − v a ). (2) R 1 R 2 V A + v a + V A + v a − V 0 = C d Adding (1) and (2), we get dt ( v s − v a ). (3) R 1 R 2 M. B. Patil, IIT Bombay

  60. RC circuit with DC + AC sources DC circuit AC circuit v C V C v c R 2 R 2 R 2 A A A v A v a V A v s (t) v s (t) R 1 R 1 R 1 V 0 (DC) V 0 V m sin ω t V m sin ω t V A + V A − V 0 DC circuit: = 0. (1) R 1 R 2 v a + v a = C d AC circuit: dt ( v s − v a ). (2) R 1 R 2 V A + v a + V A + v a − V 0 = C d Adding (1) and (2), we get dt ( v s − v a ). (3) R 1 R 2 Compare with the equation obtained directly from the original circuit: v A + v A − V 0 = C d dt ( v s − v A ). (4) R 1 R 2 M. B. Patil, IIT Bombay

  61. RC circuit with DC + AC sources DC circuit AC circuit v C V C v c R 2 R 2 R 2 A A A v A v a V A v s (t) v s (t) R 1 R 1 R 1 V 0 (DC) V 0 V m sin ω t V m sin ω t V A + V A − V 0 DC circuit: = 0. (1) R 1 R 2 v a + v a = C d AC circuit: dt ( v s − v a ). (2) R 1 R 2 V A + v a + V A + v a − V 0 = C d Adding (1) and (2), we get dt ( v s − v a ). (3) R 1 R 2 Compare with the equation obtained directly from the original circuit: v A + v A − V 0 = C d dt ( v s − v A ). (4) R 1 R 2 Eqs. (3) and (4) are identical since v A = V A + v a . M. B. Patil, IIT Bombay

  62. RC circuit with DC + AC sources DC circuit AC circuit v C V C v c R 2 R 2 R 2 A A A v A v a V A v s (t) v s (t) R 1 R 1 R 1 V 0 (DC) V 0 V m sin ω t V m sin ω t V A + V A − V 0 DC circuit: = 0. (1) R 1 R 2 v a + v a = C d AC circuit: dt ( v s − v a ). (2) R 1 R 2 V A + v a + V A + v a − V 0 = C d Adding (1) and (2), we get dt ( v s − v a ). (3) R 1 R 2 Compare with the equation obtained directly from the original circuit: v A + v A − V 0 = C d dt ( v s − v A ). (4) R 1 R 2 Eqs. (3) and (4) are identical since v A = V A + v a . → Instead of computing v A ( t ) directly, we can compute V A and v a ( t ) separately, and then use v A ( t ) = V A + v a ( t ). M. B. Patil, IIT Bombay

  63. Common-emitter amplifier coupling R C capacitor R 1 coupling capacitor V CC C C C B R L v s R 2 load R E resistor C E bypass capacitor

  64. Common-emitter amplifier coupling R C R C capacitor R 1 R 1 coupling capacitor V CC C C V CC C B R L v s R 2 R 2 load R E R E resistor C E bypass DC circuit capacitor

  65. Common-emitter amplifier coupling R C R C R C capacitor R 1 R 1 R 1 coupling capacitor V CC C C C C V CC AND C B C B R L R L v s R 2 R 2 v s R 2 load R E R E R E resistor C E C E bypass DC circuit AC circuit capacitor

  66. Common-emitter amplifier coupling R C R C R C capacitor R 1 R 1 R 1 coupling capacitor V CC C C C C V CC AND C B C B R L R L v s R 2 R 2 v s R 2 load R E R E R E resistor C E C E bypass DC circuit AC circuit capacitor * The coupling capacitors ensure that the signal source and the load resistor do not affect the DC bias of the amplifier. (We will see the purpose of C E a little later.)

  67. Common-emitter amplifier coupling R C R C R C capacitor R 1 R 1 R 1 coupling capacitor V CC C C C C V CC AND C B C B R L R L v s R 2 R 2 v s R 2 load R E R E R E resistor C E C E bypass DC circuit AC circuit capacitor * The coupling capacitors ensure that the signal source and the load resistor do not affect the DC bias of the amplifier. (We will see the purpose of C E a little later.) * This enables us to bias the amplifier without worrying about what load it is going to drive. M. B. Patil, IIT Bombay

  68. Common-emitter amplifier: AC circuit R C R 1 C C C B R L v s R 2 R E C E

  69. Common-emitter amplifier: AC circuit R C R 1 C C C B R L v s R 2 R E C E * The coupling and bypass capacitors are “large” (typically, a few µ F ), and at frequencies of interest, their impedance is small. For example, for C = 10 µ F , f = 1 kHz, 1 Z C = 2 π × 10 3 × 10 × 10 − 6 = 16 Ω, which is much smaller than typical values of R 1 , R 2 , R C , R E (a few kΩ). ⇒ C B , C C , C E can be replaced by short circuits at the frequencies of interest.

  70. Common-emitter amplifier: AC circuit R C R C R 1 R 1 C C C B R L R L v s R 2 v s R 2 R E C E * The coupling and bypass capacitors are “large” (typically, a few µ F ), and at frequencies of interest, their impedance is small. For example, for C = 10 µ F , f = 1 kHz, 1 Z C = 2 π × 10 3 × 10 × 10 − 6 = 16 Ω, which is much smaller than typical values of R 1 , R 2 , R C , R E (a few kΩ). ⇒ C B , C C , C E can be replaced by short circuits at the frequencies of interest.

  71. Common-emitter amplifier: AC circuit R C R C R 1 R 1 C C C B R L R L v s R 2 v s R 2 R E C E * The coupling and bypass capacitors are “large” (typically, a few µ F ), and at frequencies of interest, their impedance is small. For example, for C = 10 µ F , f = 1 kHz, 1 Z C = 2 π × 10 3 × 10 × 10 − 6 = 16 Ω, which is much smaller than typical values of R 1 , R 2 , R C , R E (a few kΩ). ⇒ C B , C C , C E can be replaced by short circuits at the frequencies of interest. * The circuit can be re-drawn in a more friendly format.

  72. Common-emitter amplifier: AC circuit R C R C R 1 R 1 C C C B R L R L R C R L v s R 2 v s R 2 v s R 1 R 2 R E C E * The coupling and bypass capacitors are “large” (typically, a few µ F ), and at frequencies of interest, their impedance is small. For example, for C = 10 µ F , f = 1 kHz, 1 Z C = 2 π × 10 3 × 10 × 10 − 6 = 16 Ω, which is much smaller than typical values of R 1 , R 2 , R C , R E (a few kΩ). ⇒ C B , C C , C E can be replaced by short circuits at the frequencies of interest. * The circuit can be re-drawn in a more friendly format.

  73. Common-emitter amplifier: AC circuit R C R C R 1 R 1 C C C B R L R L R C R L v s R 2 v s R 2 v s R 1 R 2 R E C E * The coupling and bypass capacitors are “large” (typically, a few µ F ), and at frequencies of interest, their impedance is small. For example, for C = 10 µ F , f = 1 kHz, 1 Z C = 2 π × 10 3 × 10 × 10 − 6 = 16 Ω, which is much smaller than typical values of R 1 , R 2 , R C , R E (a few kΩ). ⇒ C B , C C , C E can be replaced by short circuits at the frequencies of interest. * The circuit can be re-drawn in a more friendly format. * We now need to figure out the AC description of a BJT. M. B. Patil, IIT Bombay

  74. BJT: AC model C V m = 10 mV i C 1.1 V m = 5 mV V m = 2 mV B i C (mA) 0.9 i B i E v BE 0.7 E v BE (t) = V 0 + V m sin ω t 0.5 0 0.2 0.4 0.6 0.8 1 V 0 = 0.65 V, f = 1 kHz t (msec) M. B. Patil, IIT Bombay

  75. BJT: AC model C V m = 10 mV i C 1.1 V m = 5 mV V m = 2 mV B i C (mA) 0.9 i B i E v BE 0.7 E v BE (t) = V 0 + V m sin ω t 0.5 0 0.2 0.4 0.6 0.8 1 V 0 = 0.65 V, f = 1 kHz t (msec) * As the v BE amplitude increases, the shape of i C ( t ) deviates from a sinusoid → distortion. M. B. Patil, IIT Bombay

  76. BJT: AC model C V m = 10 mV i C 1.1 V m = 5 mV V m = 2 mV B i C (mA) 0.9 i B i E v BE 0.7 E v BE (t) = V 0 + V m sin ω t 0.5 0 0.2 0.4 0.6 0.8 1 V 0 = 0.65 V, f = 1 kHz t (msec) * As the v BE amplitude increases, the shape of i C ( t ) deviates from a sinusoid → distortion. * If v be ( t ), i.e., the time-varying part of v BE , is kept small, i C varies linearly with v BE . How small? Let us look at this in more detail. M. B. Patil, IIT Bombay

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