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6-1 Electronics Principles & Applications Eighth Edition Charles A. Schuler Chapter 6 Introduction to Small-Signal Amplifiers McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-2 INTRODUCTION • Gain • Common-Emitter Amplifier • Stabilizing the Amplifier • Other Configurations McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-3 1.5 V In 5V Amplifier Out The units cancel Gain = McGraw-Hill 5V Out InV 1.5 = 3.33 © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Gain can be expressed in decibels (dB). 6-4 The dB is a logarithmic unit. Common logarithms are exponents of the number 10. The log of 100 is 2 2 10 = 100 3 10 = 1000 -2 10 = 0.01 0 10 = 1 3.6 10 = 3981 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-5 The dB unit is based on a power ratio. P 50OUT W 1.7050 dB = 17 10 x log 1PIN W The dB unit can be adapted to a voltage ratio. dB = 20 x log VOUT VIN This equation assumes VOUT and VIN are measured across equal impedances. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-8 Gain quiz Amplifier output is equal to the input ________ by the gain. multiplied Common logarithms are ________ of the exponents number 10. 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-9 A small-signal amplifier can also be called a voltage amplifier. Common-emitter amplifiers are one type. The emitter terminal is grounded Next, load resistor a base bias resistor A coupling capacitor issupply often required Add aapower Start with an NPN junction transistor Connect abipolar signal sourceand andThen common to the input output signal circuits. RB RL C CC McGraw-Hill B VCC E © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-10 The output is phase inverted. RB RL C CC McGraw-Hill B VCC E © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-11 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. The collector terminal is now less positive. RB RL C CC McGraw-Hill B VCC E © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-12 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. The collector terminal is now more positive. RB RL C CC McGraw-Hill B VCC E © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-13 The maximum value of VCE for this circuit is 14 V. The maximum value of IC is 14 mA. These are the limits for this circuit. 14 V IC(MAX) = 1 kW 350 kW 1 kW C CC McGraw-Hill B 14 V E © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-14 The load line connects the limits. This end is called saturation. The linear region is between the limits. 100 mA 14 12 10 IC in mA 8 6 4 2 SAT. LINEAR 80 mA 60 mA 40 mA 20 mA 0 2 4 6 8 10 12 14 16 18 0 mA VCE in Volts This end is called cutoff. McGraw-Hill CUTOFF © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-15 Use Ohm’s Law to determine the base current: 14 V IB = 350 kW 350 kW 1 kW C CC McGraw-Hill = 40 mA B 14 V E © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-16 An amplifier can be operated at any point along the load line. The base current in this case is 40 mA. 100 mA 14 12 10 IC in mA 8 6 4 2 Q 80 mA 60 mA 40 mA 20 mA 0 2 4 6 8 10 12 14 16 18 0 mA VCE in Volts Q = the quiescent point McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-17 The input signal varies the base current above and below the Q point. 100 mA 14 12 10 IC in mA 8 6 4 2 80 mA 60 mA 40 mA 20 mA 0 2 4 6 8 10 12 14 16 18 0 mA VCE in Volts McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Overdriving the amplifier causes clipping. 6-18 100 mA 14 12 10 IC in mA 8 6 4 2 80 mA 60 mA 40 mA 20 mA 0 2 4 6 8 10 12 14 16 18 0 mA VCE in Volts The output is non-linear. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-19 What’s wrong with this Q point? 100 mA 14 12 10 IC in mA 8 6 4 2 80 mA 60 mA 40 mA 20 mA 0 2 4 6 8 10 12 14 16 18 0 mA VCE in Volts How about this one? McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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 6-20 VCE = VCC - VRL = 14 V - 6 V = 8 V This is a good Q point for linear amplification. 350 kW 1 kW C CC McGraw-Hill B E 14 V b = 150 © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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 (VR is higher) 6-21 L 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 C CC B E 14 V b = 350 b is higher McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. The higher b causes saturation. 6-22 100 mA 14 12 10 IC in mA 8 6 4 2 80 mA 60 mA 40 mA 20 mA 0 2 4 6 8 10 12 14 16 18 VCE in Volts 0 mA The output is non-linear. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-23 This common-emitter amplifier is not practical. It’s b dependent! It’s also temperature dependent. RB RL C CC McGraw-Hill B VCC E © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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 b it’s _________ dependent. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-25 This common-emitter amplifier is practical. RB1 RL C CC B RB2 VCC E RE It uses voltage divider bias and emitter feedback to reduce b sensitivity. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-26 Voltage divider bias +VCC RB1 RL RB1 and RB2 form a voltage divider RB2 McGraw-Hill RE © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-27 +VCC Voltage divider bias analysis: RB1 RB2 VB = VCC RB1 + RB2 The base current is normally much smaller than the divider current so it can be ignored. McGraw-Hill +VB RB2 © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-28 Solving the practical circuit for its dc conditions: VCC = 12 V RB1 22 kW B RB2 2.7 kW McGraw-Hill VB = RB2 RB1 + RB2 x VCC RL= 2.2 kW C 2.7 kW x 12 V VB = 2.7 kW + 22 kW E VB = 1.31 V RE = 220 W © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-29 Solving the practical circuit for its dc conditions: VCC = 12 V RB1 22 kW RL= 2.2 kW C B RB2 2.7 kW McGraw-Hill VE = VB - VBE VE = 1.31 V - 0.7 V = 0.61 V E RE = 220 W © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-30 Solving the practical circuit for its dc conditions: VCC = 12 V RB1 22 kW RL= 2.2 kW C B RB2 2.7 kW McGraw-Hill IE = IE = VE 0.61 V 220 W RE = 2.77 mA E RE = 220 W IC @ IE © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-31 Solving the practical circuit for its dc conditions: VRL = IC x RL VCC = 12 V VRL = 2.77 mA x 2.2 kW RB1 22 kW RL= 2.2 kW C B RB2 2.7 kW E VRL = 6.09 V VCE = VCC - VRL - VE VCE = 12 V - 6.09 V - 0.61 V RE = 220 W VCE = 5.3 V A linear Q point! McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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. McGraw-Hill 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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-33 Solving the practical circuit for its ac conditions: VCC = 12 V The ac emitter resistance is rE: RB1 22 kW RL= 2.2 kW C B RB2 2.7 kW McGraw-Hill E RE = 220 W 25 mV rE = IE 25 mV = 9.03 W rE = 2.77 mA © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-34 Solving the practical circuit for its ac conditions: VCC = 12 V The voltage gain from base to collector: RB1 22 kW RL= 2.2 kW C B RB2 2.7 kW McGraw-Hill E AV = AV = RL RE + rE 2.2 kW 220 W + 9.03 W = 9.61 RE = 220 W © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-35 Solving the practical circuit for its ac conditions: VCC = 12 V RB1 22 kW An emitter bypass capacitor can be used to increase AV: RL= 2.2 kW RL AV = rE C B RB2 2.7 kW McGraw-Hill E RE AV = 2.2 kW 9.03 W = 244 CE © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-36 Practical C-E amplifier quiz b-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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-37 The common-emitter configuration is used most often. It has the best power gain. VCC RB1 RL C B RB2 McGraw-Hill E RE CE © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-38 The common-collector configuration is shown below. Its input impedance and current gain are both high. VCC RB1 It’s often called an emitter-follower. RC C B RB2 McGraw-Hill E RL In-phase output © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-39 The common-base configuration is shown below. It’s used most at RF. Its voltage gain is high. VCC RB1 RL C B RB2 McGraw-Hill E RE In-phase output © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-40 PNP C-E amplifier VB = - 3.75 V 1.5 kW 22 kW 10 kW 47 W 12 V + VE = - 3.05 V IE = 2.913 mA rE = 8.58 W VRL = 4.37 V AV = 27 VCE = - 4.58 V 1 kW VC = - 7.63 V McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6-42 REVIEW • Gain • Common-Emitter Amplifier • Stabilizing the Amplifier • Other Configurations McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved.