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Electronics Principles & Applications Eighth Edition

Electronics Principles & Applications Eighth Edition. Charles A. Schuler. Chapter 6 Introduction to Small-Signal Amplifiers. INTRODUCTION. Gain Common-Emitter Amplifier Stabilizing the Amplifier Other Configurations. 5 V. 1.5 V. In. Out. The units cancel. Gain. =. Amplifier.

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Electronics Principles & Applications Eighth Edition

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  1. Electronics Principles & Applications Eighth Edition Charles A. Schuler Chapter 6 Introduction to Small-Signal Amplifiers

  2. INTRODUCTION • Gain • Common-Emitter Amplifier • Stabilizing the Amplifier • Other Configurations

  3. 5 V 1.5 V In Out The units cancel Gain = Amplifier 5 V Out = 3.33 In 1.5 V

  4. Gain can be expressed in decibels (dB). The dB is a logarithmic unit. Common logarithms are exponents of the number 10. Thelog of 100is 2 102 = 100 103 = 1000 10-2 = 0.01 100= 1 103.6 = 3981 Thelog of 1000is 3 Thelog of 0.01is -2 Thelog of 1is 0 Thelog of 3981is 3.6

  5. POUT 50 W PIN 1 W VOUT dB = 20 x log VIN The dB unit is based on a power ratio. 17 1.70 50 dB = 10 x log The dB unit can be adapted to a voltage ratio. This equation assumes VOUT and VIN are measured across equal impedances.

  6. +30 dB -6 dB -8 dB +20 dB +10 dB Total system gain = +46 dB dB units are convenient for evaluating systems. +10 dB -6 dB +30 dB -8 dB +20 dB

  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.

  8. Gain quiz Amplifier output is equal to the input ________ by the gain. multiplied Common logarithms are ________ of the number 10. 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

  9. The emitter terminal is grounded and common to the input and output signal circuits. Next, a load resistor Then a base bias resistor A coupling capacitor is often required Add a power supply Start with an NPN bipolar junction transistor Connect a signal source RB RL VCC C B CC E A small-signal amplifier can also be called a voltage amplifier. Common-emitter amplifiers are one type.

  10. The output is phase inverted. RB RL VCC C B CC E

  11. When the input signal goes positive: The base current increases. The collector current increases b times. So, RL drops more voltage and VCE must decrease. RL C B The collector terminal is now less positive. RB VCC CC E

  12. When the input signal goes negative: The base current decreases. The collector current decreases b times. So, RL drops less voltage and VCE must increase. RL C B The collector terminal is now more positive. RB VCC CC E

  13. The maximum value of VCE for this circuit is 14 V. The maximum value of IC is 14 mA. IC(MAX) = 14 V These are the limits for this circuit. 14 V 1 kW 350 kW 1 kW C B CC E

  14. This end is called saturation. The linear region is between the limits. 100 mA SAT. LINEAR 80 mA 60 mA 40 mA 20 mA 0 mA This end is called cutoff. CUTOFF The load line connects the limits. 14 12 10 IC in mA 8 6 4 2 6 16 2 4 10 12 0 14 18 8 VCE in Volts

  15. Use Ohm’s Law to determine the base current: 14 V = 40 mA IB = 350 kW 350 kW 1 kW 14 V C B CC E

  16. 100 mA 80 mA 60 mA 40 mA 20 mA 0 mA An amplifier can be operated at any point along the load line. The base current in this case is 40 mA. 14 Q 12 10 IC in mA 8 6 4 2 6 16 2 4 10 12 0 14 18 8 VCE in Volts Q = the quiescent point

  17. 100 mA 80 mA 60 mA 40 mA 20 mA 0 mA The input signal varies the base current above and below the Q point. 14 12 10 IC in mA 8 6 4 2 6 16 2 4 10 12 0 14 18 8 VCE in Volts

  18. 100 mA 80 mA 60 mA 40 mA 20 mA 0 mA Overdriving the amplifier causes clipping. 14 12 10 IC in mA 8 6 4 2 6 16 2 4 10 12 0 14 18 8 VCE in Volts The output is non-linear.

  19. 100 mA 80 mA 60 mA 40 mA 20 mA 0 mA What’s wrong with this Q point? 14 12 10 IC in mA 8 6 4 2 6 16 2 4 10 12 0 14 18 8 VCE in Volts How about this one?

  20. 14 V = 40 mA IB = 350 kW IC = b x IB = 150 x 40 mA = 6 mA VRL = IC x RL = 6 mA x 1 kW = 6 V VCE = VCC - VRL = 14 V - 6 V = 8 V This is a good Q point for linear amplification. 350 kW 1 kW 14 V C B CC b = 150 E

  21. 14 V = 40 mA (IB is not affected) IB = 350 kW IC = b x IB = 350 x 40 mA = 14 mA (IC is higher) VRL = IC x RL = 14 mA x 1 kW = 14 V (VRL is higher) VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower) This is not a good Q point for linear amplification. 350 kW 1 kW 14 V C B CC b = 350 E b is higher

  22. 100 mA 80 mA 60 mA 40 mA 20 mA 0 mA The higher b causes saturation. 14 12 10 IC in mA 8 6 4 2 6 16 2 4 10 12 0 14 18 8 VCE in Volts The output is non-linear.

  23. This common-emitter amplifier is not practical. It’s b dependent! It’s also temperature dependent. RB RL VCC C B CC E

  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. 

  25. This common-emitter amplifier is practical. RB1 RL VCC C CC B E RB2 RE It uses voltage divider bias and emitter feedback to reduce b sensitivity.

  26. Voltage divider bias { +VCC RL RB1 RB1 and RB2 form a voltage divider RB2 RE

  27. +VCC Voltage divider bias analysis: RB1 RB2 VB = VCC +VB RB1 + RB2 RB2 The base current is normally much smaller than the divider current so it can be ignored.

  28. Solving the practical circuit for its dc conditions: RB2 VCC = 12 V x VCC VB = RB1 + RB2 RB1 RL 22 kW = 2.2 kW 2.7 kW x 12 V C VB = + 22 kW 2.7 kW B E VB = 1.31 V RB2 2.7 kW RE = 220 W

  29. Solving the practical circuit for its dc conditions: VCC = 12 V RB1 RL 22 kW = 2.2 kW VE = VB - VBE C VE = 1.31 V - 0.7 V = 0.61 V B E RB2 2.7 kW RE = 220 W

  30. Solving the practical circuit for its dc conditions: VCC = 12 V VE IE = RE RB1 RL 22 kW = 2.2 kW C 0.61 V = 2.77 mA IE = 220 W B E RB2 2.7 kW IC@ IE RE = 220 W

  31. Solving the practical circuit for its dc conditions: VRL = IC x RL VCC = 12 V VRL = 2.77 mA x 2.2 kW RB1 RL 22 kW = 2.2 kW VRL = 6.09 V C VCE = VCC - VRL - VE B E VCE = 12 V - 6.09 V - 0.61 V RB2 2.7 kW RE = 220 W VCE = 5.3 V A linear Q point!

  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.

  33. 25 mV rE = IE 25 mV = 9.03 W rE = 2.77 mA Solving the practical circuit for its acconditions: VCC = 12 V The ac emitter resistance is rE: RB1 RL 22 kW = 2.2 kW C B E RB2 2.7 kW RE = 220 W

  34. RL AV = RE + rE 2.2 kW = 9.61 AV = 220 W + 9.03 W Solving the practical circuit for its acconditions: VCC = 12 V The voltage gain from base to collector: RB1 RL 22 kW = 2.2 kW C B E RB2 2.7 kW RE = 220 W

  35. RL AV = rE 2.2 kW = 244 AV = 9.03 W Solving the practical circuit for its acconditions: VCC An emitter bypass capacitor can be used to increase AV: = 12 V RB1 RL 22 kW = 2.2 kW C B E RB2 2.7 kW RE CE

  36. Practical C-E amplifier quiz -dependency is reduced with emitter feedback and voltage _________ bias. divider To find the emitter voltage, VBE is subtracted from ____________. VB To find VCE, VRL and VE are subtracted from _________. VCC Voltage gain is equal to the collector resistance _______ by the emitter resistance. divided Voltage gain can be increased by ________ the emitter resistor. bypassing

  37. The common-emitter configuration is used most often. It has the best power gain. VCC RB1 RL C B E RB2 RE CE

  38. The common-collector configuration is shown below. Its input impedance and current gain are both high. VCC It’s often called an emitter-follower. RB1 RC C B E RB2 RL In-phase output

  39. The common-base configuration is shown below. Its voltage gain is high. It’s used most at RF. VCC RB1 RL C B E In-phase output RB2 RE

  40. 12 V 1.5 kW + 22 kW 10 kW 47 W 1 kW PNP C-E amplifier VB = - 3.75 V VE = - 3.05 V rE = 8.58 W IE = 2.913 mA AV = 27 VRL = 4.37 V VCE = - 4.58 V VC = - 7.63 V

  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

  42. REVIEW • Gain • Common-Emitter Amplifier • Stabilizing the Amplifier • Other Configurations

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