2. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-1 Electronics Principles & Applications Eighth Edition Chapter 6 Introduction to Small-Signal Amplifiers Charles A. Schuler

3. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-2 Gain Common-Emitter Amplifier Stabilizing the Amplifier Other Configurations INTRODUCTION

4. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-3 Amplifier Out In Gain = In Out = 3.33 1.5 V 5 V 1.5 V 5 V The units cancel

5. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-4 Gain can be expressed in decibels (dB). The dB is a logarithmic unit. Common logarithms are exponents of the number 10. 10 2 = 100 10 3 = 1000 10 -2 = 0.01 10 0 = 1 10 3.6 = 3981 The log of 100 is 2 The log of 1000 is 3 The log of 0.01 is -2 The log of 1 is 0 The log of 3981 is 3.6

6. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-5 The dB unit is based on a power ratio. dB = 10 x log P OUT P IN 50 W 1 W 501.70 17 The dB unit can be adapted to a voltage ratio. dB = 20 x log V OUT V IN This equation assumes V OUT and V IN are measured across equal impedances.

7. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-6 +10 dB -6 dB +30 dB -8 dB +20 dB dB units are convenient for evaluating systems. +10 dB -6 dB +30 dB -8 dB +20 dB Total system gain = +46 dB

8. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-7 Acoustical sound levels are often measured using dBA units. This instrument also has a dBC scale, which has a different frequency response curve. Exposure to loud sounds is a concern for employers and employees, and citizens in general.

9. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-8 Gain quiz Amplifier output is equal to the input ________ by the gain. multiplied exponents Doubling a log is the same as _________ the number it represents. squaring System performance is found by ________ dB stage gains and losses. adding Logs of numbers smaller than one are ____________. negative Common logarithms are ________ of the number 10.

10. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-9 A small-signal amplifier can also be called a voltage amplifier. Common-emitter amplifiers are one type. C B E Start with an NPN bipolar junction transistor V CC Add a power supply RLRL Next, a load resistor RBRB Then a base bias resistor C A coupling capacitor is often required Connect a signal source The emitter terminal is grounded and common to the input and output signal circuits.

11. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-10 RBRB RLRL V CC C C B E The output is phase inverted.

12. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-11 RBRB V CC C E When the input signal goes positive: B The base current increases. C The collector current increases  times. RLRL So, R L drops more voltage and V CE must decrease. The collector terminal is now less positive.

13. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-12 RBRB V CC C E When the input signal goes negative: B The base current decreases. C The collector current decreases  times. RLRL So, R L drops less voltage and V CE must increase. The collector terminal is now more positive.

14. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-13 350 k  C E B C 1 k  14 V The maximum value of V CE for this circuit is 14 V. The maximum value of I C is 14 mA. I C(MAX) = 14 V 1 k  These are the limits for this circuit.

15. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-14 0 24 6 8 1012 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA 20  A 0  A 100  A 80  A 60  A 40  A The load line connects the limits. SAT. This end is called saturation. CUTOFF This end is called cutoff. LINEAR The linear region is between the limits.

16. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-15 350 k  C E B C 1 k  14 V I B = 14 V 350 k  Use Ohm’s Law to determine the base current: = 40  A

17. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-16 0 24 6 8 1012 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA 20  A 0  A 100  A 80  A 60  A 40  A An amplifier can be operated at any point along the load line. The base current in this case is 40  A. Q Q = the quiescent point

18. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-17 0 24 6 8 1012 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA 20  A 0  A 100  A 80  A 60  A 40  A The input signal varies the base current above and below the Q point.

19. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-18 0 24 6 8 1012 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA 20  A 0  A 100  A 80  A 60  A 40  A Overdriving the amplifier causes clipping. The output is non-linear.

20. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-19 0 24 6 8 1012 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA 20  A 0  A 100  A 80  A 60  A 40  A What’s wrong with this Q point? How about this one?

21. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-20 350 k  C E B C 1 k  14 V I B = 14 V 350 k   = 150 I C =  x I B = 40  A = 150 x 40  A = 6 mA V R L = I C x R L = 6 mA x 1 k  = 6 V This is a good Q point for linear amplification. V CE = V CC - V R L = 14 V - 6 V = 8 V

22. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-21 350 k  C E B C 1 k  14 V I B = 14 V 350 k   = 350 I C =  x I B = 40  A (I B is not affected) = 350 x 40  A = 14 mA (I C is higher) V R L = I C x R L = 14 mA x 1 k  = 14 V (V R L is higher) This is not a good Q point for linear amplification. V CE = V CC - V R L = 14 V - 14 V = 0 V (V CE is lower)  is higher

23. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-22 0 24 6 8 1012 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA 20  A 0  A 100  A 80  A 60  A 40  A The output is non-linear. The higher  causes saturation.

24. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-23 RBRB C E B C RLRL V CC It’s  dependent! This common-emitter amplifier is not practical. It’s also temperature dependent.

25. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-24 Basic C-E amplifier quiz The input and output signals in C-E are phase ______________. inverted The limits of an amplifier’s load line are saturation and _________. cutoff Linear amplifiers are normally operated near the _________ of the load line. center The operating point of an amplifier is also called the ________ point. quiescent Single resistor base bias is not practical since it’s _________ dependent. 

26. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-25 R B1 C E B C RLRL V CC R B2 RERE This common-emitter amplifier is practical. It uses voltage divider bias and emitter feedback to reduce  sensitivity.

27. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-26 +V CC RLRL RERE R B1 R B2 Voltage divider bias { R B1 and R B2 form a voltage divider

28. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-27 +V CC R B1 R B2 +V B Voltage divider bias analysis: V B = R B2 R B1 + R B2 V CC The base current is normally much smaller than the divider current so it can be ignored.

29. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-28 R B1 E B C RLRL V CC R B2 RERE = 220  = 12 V 2.7 k  22 k  = 2.2 k  Solving the practical circuit for its dc conditions: V B = R B2 R B1 + R B2 x V CC V B = 2.7 k  22 k  2.7 k  + x 12 V V B = 1.31 V

30. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-29 R B1 E B C RLRL V CC R B2 RERE = 220  = 12 V 2.7 k  22 k  = 2.2 k  Solving the practical circuit for its dc conditions: V E = V B - V BE V E = 1.31 V - 0.7 V = 0.61 V

31. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-30 R B1 E B C RLRL V CC R B2 RERE = 220  = 12 V 2.7 k  22 k  = 2.2 k  Solving the practical circuit for its dc conditions: I E = RERE VEVE 0.61 V 220  = 2.77 mA I C  I E

32. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-31 R B1 E B C RLRL V CC R B2 RERE = 220  = 12 V 2.7 k  22 k  = 2.2 k  Solving the practical circuit for its dc conditions: V R L = I C x R L V R L = 2.77 mA x 2.2 k  V R L = 6.09 V V CE = V CC - V R L - V E V CE = 12 V - 6.09 V - 0.61 V V CE = 5.3 V A linear Q point!

33. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-32 Review of the analysis thus far: 1. Calculate the base voltage using the voltage divider equation. 2. Subtract 0.7 V to get the emitter voltage. 3. Divide by emitter resistance to get the emitter current. 4. Determine the drop across the collector resistor. 5. Calculate the collector to emitter voltage using KVL. 6. Decide if the Q-point is linear. 7. Go to ac analysis.

34. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-33 R B1 E B C RLRL V CC R B2 RERE = 220  = 12 V 2.7 k  22 k  = 2.2 k  Solving the practical circuit for its ac conditions: The ac emitter resistance is r E : r E = 25 mV IEIE r E = 25 mV 2.77 mA = 9.03 

35. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-34 R B1 E B C RLRL V CC R B2 RERE = 220  = 12 V 2.7 k  22 k  = 2.2 k  Solving the practical circuit for its ac conditions: The voltage gain from base to collector: A V = RLRL R E + r E A V = 2.2 k  220  9.03  = 9.61

36. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-35 R B1 E B C RLRL V CC R B2 RERE = 12 V 2.7 k  22 k  = 2.2 k  Solving the practical circuit for its ac conditions: A V = RLRL rErE 2.2 k  9.03  = 244 An emitter bypass capacitor can be used to increase A V : CECE

37. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-36 Practical C-E amplifier quiz  -dependency is reduced with emitter feedback and voltage _________ bias. divider To find the emitter voltage, V BE is subtracted from ____________. VBVB To find V CE, V RL and V E are subtracted from _________. V CC Voltage gain is equal to the collector resistance _______ by the emitter resistance. divided Voltage gain can be increased by ________ the emitter resistor. bypassing

38. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-37 R B1 E B C RLRL V CC R B2 RERE CECE The common-emitter configuration is used most often. It has the best power gain.

39. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-38 R B1 E B C RCRC V CC R B2 RLRL The common-collector configuration is shown below. Its input impedance and current gain are both high. It’s often called an emitter-follower. In-phase output

40. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-39 R B1 E B C RLRL V CC R B2 RERE The common-base configuration is shown below. Its voltage gain is high. It’s used most at RF. In-phase output

41. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-40 PNP C-E amplifier 47  1 k  1.5 k  22 k  10 k  + 12 V V B = - 3.75 V V E = - 3.05 V I E = 2.913 mA V RL = 4.37 V V CE = - 4.58 V V C = - 7.63 V r E = 8.58  A V = 27

42. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-41 Amplifier configuration quiz In a C-E amplifier, the base is the input and the __________ is the output. collector In an emitter-follower, the base is the input and the ______ is the output. emitter The only configuration that phase-inverts is the ________. C-E The configuration with the best power gain is the ________. C-E In the common-base configuration, the ________ is the input terminal. emitter

43. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-42 REVIEW Gain Common-Emitter Amplifier Stabilizing the Amplifier Other Configurations