bjt amplifier basic operation
play

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

Nov 12, 2023 •209 likes •1.96k views

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

Recommend

Basic Elec. Engr Basic Elec. Engr. Lab . Lab ECS 204 ECS 204 Asst. Prof. Dr. Prapun Suksompong

Basic Elec. Engr Basic Elec. Engr. Lab . Lab ECS 204 ECS 204 Asst. Prof. Dr. Prapun Suksompong

Basic Elec. Engr Basic Elec. Engr. Lab . Lab ECS 204 ECS 204 Asst. Prof. Dr. Prapun Suksompong prapun@siit.tu.ac.th Operational amplifier Lab 7 Inverting amplifier Summing Amplifier 1 Op-Amp 741 OP erational AMP lifier

256 views • 13 slides
Amplifier Lab Introduction: In this laboratory project you will build an amplifier circuit. This

Amplifier Lab Introduction: In this laboratory project you will build an amplifier circuit. This

K. Disney Engr 003 Mission College Amplifier Lab Introduction: In this laboratory project you will build an amplifier circuit. This amplifier is able to increase the strength of a sound signal. The amplifier can be used to amplify the weak

483 views • 13 slides
RF Power Amplifier Design Markus Mayer & Holger Arthaber Department of Electrical

RF Power Amplifier Design Markus Mayer & Holger Arthaber Department of Electrical

RF Power Amplifier Design Markus Mayer & Holger Arthaber Department of Electrical Measurements and Circuit Design Vienna University of Technology June 11, 2001 Contents Basic Amplifier Concepts Class A, B, C, F, hHCA Linearity

976 views • 53 slides
Class E broadband amplifier w ith C-LC shunt netw ork Basic theory, simulation and prototype A.

Class E broadband amplifier w ith C-LC shunt netw ork Basic theory, simulation and prototype A.

San Diego, CA Jan 09 CLASS E RF/MICROWAVE POWER AMPLIFIERS Class E broadband amplifier w ith C-LC shunt netw ork Basic theory, simulation and prototype A. Mediano 1 , K. Narendra 2 , C. Prakash 2 , 1 I3A, University of Zaragoza, Zaragoza, Spain

198 views • 9 slides
Pr Pre-Columbian Columbian def defor oresta estation as an tion as an amplifier amplifier

Pr Pre-Columbian Columbian def defor oresta estation as an tion as an amplifier amplifier

Pr Pre-Columbian Columbian def defor oresta estation as an tion as an amplifier amplifier of of dr drought in Cent ought in Central al America America Benjamin I Cook Jed O Kaplan, Kevin J Anchukaitis, Michael Puma, Maxwell Kelly Pr

247 views • 10 slides
Basic Single Stage Amplifier Configurations Part I Ref.: Chapter 3 from Razavis book and

Basic Single Stage Amplifier Configurations Part I Ref.: Chapter 3 from Razavis book and

2 Basic Single Stage Amplifier Configurations Part I Ref.: Chapter 3 from Razavis book and chapter 3 from Grays book. IIT-Bombay Lecture 6 M. Shojaei

210 views • 7 slides
Bas Basic ic El Elec. ec. En Engr gr. . Lab Lab ECS EC S 204 04/210 Dr. Prapun

Bas Basic ic El Elec. ec. En Engr gr. . Lab Lab ECS EC S 204 04/210 Dr. Prapun

Bas Basic ic El Elec. ec. En Engr gr. . Lab Lab ECS EC S 204 04/210 Dr. Prapun Suksompong prapun@siit.tu.ac.th Office Hours: BKD 3601-7 Tuesday 9:30-10:30 Friday 14:00-16:00 1 Lab 7 Operational amplifier Inverting

148 views • 13 slides
Nature abhors a void Bertrand Meyer Software Engineering The basic O-O operation x . f ( args

Nature abhors a void Bertrand Meyer Software Engineering The basic O-O operation x . f ( args

Nature abhors a void Bertrand Meyer Software Engineering The basic O-O operation x . f ( args ) Semantics: apply the feature f , with given args if any, to the object to which x is attached and the basic issue studied here: How do we

735 views • 58 slides
Lecture 38 Last time: CMOS cascode transconductance amplifier design example Today

Lecture 38 Last time: CMOS cascode transconductance amplifier design example Today

EECS 105 Spring 2002 Lecture 38 R. T. Howe Lecture 38 Last time: CMOS cascode transconductance amplifier design example Today : BiCMOS voltage amplifier: example of dissection technique for a complicated circuit

263 views • 10 slides
Differential amplifier If A and B has the same voltage there will be no reading on the meter. (

Differential amplifier If A and B has the same voltage there will be no reading on the meter. (

Differential amplifier If A and B has the same voltage there will be no reading on the meter. ( Common Mode ) William Sandqvist william@kth.se Differential amplifier If A and B has different voltages there will be a large reading on the

362 views • 15 slides
Bipolar junction transistor 1 Objectives Describe the basic structure of the bipolar

Bipolar junction transistor 1 Objectives Describe the basic structure of the bipolar

Chapter 4 Bipolar junction transistor 1 Objectives Describe the basic structure of the bipolar junction transistor (BJT) Explain and analyze basic transistor bias and operation operation Discuss the parameters and

392 views • 28 slides
Basic Elec. Engr Basic Elec. Engr. Lab . Lab ECS 204 ECS 204 Asst. Prof. Dr. Prapun Suksompong

Basic Elec. Engr Basic Elec. Engr. Lab . Lab ECS 204 ECS 204 Asst. Prof. Dr. Prapun Suksompong

Basic Elec. Engr Basic Elec. Engr. Lab . Lab ECS 204 ECS 204 Asst. Prof. Dr. Prapun Suksompong prapun@siit.tu.ac.th Operational amplifier Lab 8 V2I and I2V Converters Inverting Integrator 1 V2I and I2V Converters in out

226 views • 11 slides
Problems for Amplifier Section Lecture notes: Sec. 6 F. Najmabadi, ECE65, Winter 2012 Exercise

Problems for Amplifier Section Lecture notes: Sec. 6 F. Najmabadi, ECE65, Winter 2012 Exercise

Problems for Amplifier Section Lecture notes: Sec. 6 F. Najmabadi, ECE65, Winter 2012 Exercise 1: Find the bias point and the amplifier parameters of the circuit below. (Si BJT with = 200, V A = 150 V, ignore Early effect in bias

707 views • 32 slides
325 MHz solid state RF amplifier development at BARC under IIFC Manjiri Pande and Team RFSS,

325 MHz solid state RF amplifier development at BARC under IIFC Manjiri Pande and Team RFSS,

325 MHz solid state RF amplifier development at BARC under IIFC Manjiri Pande and Team RFSS, IADD BARC Outline 1. RF devices 2. Solid State RF Amplifier topology 3. 3. Solid State RF Amplifier (@ 325 MHz) development under IIFC Solid

531 views • 31 slides
Op-amps: introduction * The Operational Amplifier (Op-Amp) is a versatile building block that can

Op-amps: introduction * The Operational Amplifier (Op-Amp) is a versatile building block that can

Op-amps: introduction * The Operational Amplifier (Op-Amp) is a versatile building block that can be used for realizing several electronic circuits. M. B. Patil, IIT Bombay Op-amps: introduction * The Operational Amplifier (Op-Amp) is a versatile

1.87k views • 141 slides
Asynchronous Computing in Sense Amplifier-based Pass Transistor Logic Tsung-Te Liu, Louis P.

Asynchronous Computing in Sense Amplifier-based Pass Transistor Logic Tsung-Te Liu, Louis P.

Asynchronous Computing in Sense Amplifier-based Pass Transistor Logic Tsung-Te Liu, Louis P. Alarcn, Matthew D. Pierson, and Jan M. Rabaey University of California, Berkeley 1 Outline Motivation Sense Amplifier-based Pass Transistor

479 views • 24 slides
Set 10 Search Engines & SEO Outline How do search engines work? Basic operation

Set 10 Search Engines & SEO Outline How do search engines work? Basic operation

IT452 Advanced Web and Internet Set 10 Search Engines & SEO Outline How do search engines work? Basic operation What makes a good one? What makes it difficult? Web Design with search engines in mind Search Engines

628 views • 16 slides
Set 10 Search Engines & SEO Outline How do search engines work? Basic operation

Set 10 Search Engines & SEO Outline How do search engines work? Basic operation

IT452 Advanced Web and Internet Set 10 Search Engines & SEO Outline How do search engines work? Basic operation What makes a good one? What makes it difficult? Web Design with search engines in mind 1 Search Engines

454 views • 8 slides
High Repetition Rate mJ-level Few-Cycle Pulse Laser Amplifier for XUV-FEL seeding . Laser

High Repetition Rate mJ-level Few-Cycle Pulse Laser Amplifier for XUV-FEL seeding . Laser

High Repetition Rate mJ-level Few-Cycle Pulse Laser Amplifier for XUV-FEL seeding . Laser amplifier development: applications at high repetition rate FELs The FLASH-II FEL seeding project Requirements for an XUV seed source and for

603 views • 29 slides
Set11 Search Engines & SEO Outline How do search engines work? Basic operation

Set11 Search Engines & SEO Outline How do search engines work? Basic operation

IT452 Advanced Web and Internet Set11 Search Engines & SEO Outline How do search engines work? Basic operation What makes a good one? What makes it difficult? Web Design with search engines in mind 1 Search Engines

205 views • 6 slides
Chapter 3 Introduce basic device equations Introduce models for manual analysis The Devices

Chapter 3 Introduce basic device equations Introduce models for manual analysis The Devices

Digital IC-Design This chapter will Present intuitive understanding of the device operation device operation Chapter 3 Introduce basic device equations Introduce models for manual analysis The Devices Show secondary and deep-sub-micron

523 views • 28 slides
Basic Mechatronics Workshop Module 2: Sensors LAB-3 Sensor Circuits, Power and Constant voltage,

Basic Mechatronics Workshop Module 2: Sensors LAB-3 Sensor Circuits, Power and Constant voltage,

Basic Mechatronics Workshop Module 2: Sensors LAB-3 Sensor Circuits, Power and Constant voltage, Detector, Amplifier, Display, Output (Conference of Presentation) Dr. Mohamed Abdalbar Lecturer, Mechatronics Department, Egyptian-Korean

639 views • 20 slides
Constructing an Optical Parametric Amplifier (OPA) to allow for wavelength tuning over the

Constructing an Optical Parametric Amplifier (OPA) to allow for wavelength tuning over the

Constructing an Optical Parametric Amplifier (OPA) to allow for wavelength tuning over the visible range for PULSAR Reid Erdwien Travis Severt Advisor: Itzik Ben-Itzhak This work is supported by the Chemical Sciences, Geosciences and

663 views • 18 slides
A 0.5-to-1 V 9-bit 15-to-90 MS/s Digitally Interpolated Pipelined-SAR ADC Using Dynamic Amplifier

A 0.5-to-1 V 9-bit 15-to-90 MS/s Digitally Interpolated Pipelined-SAR ADC Using Dynamic Amplifier

A 0.5-to-1 V 9-bit 15-to-90 MS/s Digitally Interpolated Pipelined-SAR ADC Using Dynamic Amplifier James Lin, Zule Xu, Masaya Miyahara, and Akira Matsuzawa Tokyo Institute of Technology, Japan b. Matsuzawa & Okada Lab y Outline

328 views • 30 slides

More recommend