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**Principles & Applications Small-Signal Amplifiers**

Electronics Principles & Applications Seventh Edition Charles A. Schuler Chapter 6 Introduction to Small-Signal Amplifiers (student version) McGraw-Hill © 2008 The McGraw-Hill Companies Inc. All rights reserved.

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**INTRODUCTION Gain Common-Emitter Amplifier Stabilizing the Amplifier**

Other Configurations

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Dear Student: This presentation is arranged in segments. Each segment is preceded by a Concept Preview slide and is followed by a Concept Review slide. When you reach a Concept Review slide, you can return to the beginning of that segment by clicking on the Repeat Segment button. This will allow you to view that segment again, if you want to.

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**Concept Preview Voltage gain is the ratio of Vout to Vin.**

Power gain is the ratio of Pout to Pin. Common logarithms are exponents of 10. Gain or loss in decibels is equal to 10 times the log of the power ratio or 20 times the log of the voltage ratio. dB voltage gain equals dB power gain when the input impedance equals the output impedance. System gain or loss is found by adding dB stage gains or losses.

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

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**Gain can be expressed in decibels (dB).**

The dB is a logarithmic unit. Common logarithms are exponents of the number 10. The log of 100 is 2 102 = 100 103 = 1000 10-2 = 0.01 = 1 103.6 = 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

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**The dB unit is based on a power ratio.**

POUT PIN 50 W 1 W dB = 10 x log 17 1.70 50 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.

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**dB units are convenient for evaluating systems.**

Total system gain = +46 dB

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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

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**Concept Review Voltage gain is the ratio of Vout to Vin.**

Power gain is the ratio of Pout to Pin. Common logarithms are exponents of 10. Gain or loss in decibels is equal to 10 times the log of the power ratio or 20 times the log of the voltage ratio. dB voltage gain equals dB power gain when the input impedance equals the output impedance. System gain or loss is found by adding dB stage gains or losses. Repeat Segment

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Concept Preview In a common emitter amplifier (C-E), the base is the input and the collector is the output. C-E amplifiers produce a phase inversion. The circuit limits are called saturation and cutoff. If a signal drives the amplifier beyond either or both limits the output will be clipped. The operating point (Q-point) should be centered between saturation and cutoff. Beta dependent amplifiers are not practical.

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**A small-signal amplifier can also be called a voltage amplifier.**

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

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

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**RB RL VCC C B CC E When the input signal goes positive:**

The base current increases. C The collector current increases b times. RL So, RL drops more voltage and VCE must decrease. The collector terminal is now less positive. RB VCC CC E

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**RB RL VCC C B CC E When the input signal goes negative:**

The base current decreases. C The collector current decreases b times. RL So, RL drops less voltage and VCE must increase. The collector terminal is now more positive. RB VCC CC E

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**350 kW 1 kW 14 V C B CC E 14 V IC(MAX) = 1 kW**

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

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**The load line connects the limits.**

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

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**350 kW 1 kW 14 V C B CC E Use Ohm’s Law to determine the base current:**

IB = = 40 mA 350 kW 350 kW 1 kW 14 V C B CC E

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**An amplifier can be operated at any point along the load line.**

The base current in this case is 40 mA. 100 mA 14 Q 12 80 mA 10 60 mA IC in mA 8 40 mA 6 4 20 mA 2 0 mA 2 4 6 8 10 12 14 16 18 VCE in Volts Q = the quiescent point

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**The input signal varies the base current above and below the Q point.**

100 mA 14 12 80 mA 10 60 mA IC in mA 8 40 mA 6 4 20 mA 2 0 mA 2 4 6 8 10 12 14 16 18 VCE in Volts

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**Overdriving the amplifier causes clipping.**

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

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**What’s wrong with this Q point?**

100 mA 14 12 80 mA 10 60 mA IC in mA 8 40 mA 6 4 20 mA 2 0 mA 2 4 6 8 10 12 14 16 18 VCE in Volts How about this one?

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**This is a good Q point for linear amplification.**

14 V IB = = 40 mA 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 E b = 150

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**This is not a good Q point for linear amplification.**

14 V IB = = 40 mA (IB is not affected) 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 E b = 350 b is higher

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

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**It’s b dependent! This common-emitter amplifier is not practical.**

It’s also temperature dependent. RB RL VCC C B CC E

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**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.

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Concept Review In a common emitter amplifier (C-E), the base is the input and the collector is the output. C-E amplifiers produce a phase inversion. The circuit limits are called saturation and cutoff. If a signal drives the amplifier beyond either or both limits the output will be clipped. The operating point (Q-point) should be centered between saturation and cutoff. Beta dependent amplifiers are not practical. Repeat Segment

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Concept Preview C-E amplifiers can be stabilized by using voltage divider base bias and emitter feedback. The base current can be ignored when analyzing the divider for the base voltage (VB). Subtract VBE to find VE. Use Ohm’s Law to find IE and VRL. Use KVL to find VCE. IE determines the ac emitter resistance (rE). RL, RE and rE determine the voltage gain. Emitter bypassing increases the voltage gain.

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**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.

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**{ Voltage divider bias +VCC RL RB1 RB1 and RB2 form a voltage divider**

RE

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**Voltage divider bias analysis:**

+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.

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**Solving the practical circuit for its dc conditions:**

VCC RB2 = 12 V VB = x VCC RB1 + RB2 RB1 22 kW RL = 2.2 kW 2.7 kW C VB = x 12 V 2.7 kW + 22 kW B E VB = 1.31 V RB2 2.7 kW RE = 220 W

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**Solving the practical circuit for its dc conditions:**

VCC = 12 V RB1 22 kW RL = 2.2 kW VE = VB - VBE C VE = 1.31 V V = 0.61 V B E RB2 2.7 kW RE = 220 W

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**Solving the practical circuit for its dc conditions:**

VCC = 12 V VE IE = RB1 RL RE 22 kW = 2.2 kW C 0.61 V IE = = 2.77 mA 220 W B E RB2 2.7 kW IE RE = 220 W

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**VCC RB1 RL C B E RB2 RE A linear Q point!**

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

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**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.

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**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 rE = 25 mV IE B E rE = 25 mV 2.77 mA = 9.03 W RB2 2.7 kW RE = 220 W

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**Solving the practical circuit for its ac conditions:**

VCC = 12 V The voltage gain from base to collector: RB1 22 kW RL AV = RL RE + rE = 2.2 kW C AV = 2.2 kW 220 W W = 9.61 B E RB2 2.7 kW RE = 220 W

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**An emitter bypass capacitor can be used to increase AV:**

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

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**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

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Concept Review C-E amplifiers can be stabilized by using voltage divider base bias and emitter feedback. The base current can be ignored when analyzing the divider for the base voltage (VB). Subtract VBE to find VE. Use Ohm’s Law to find IE and VRL. Use KVL to find VCE. IE determines the ac emitter resistance (rE). RL, RE and rE determine the voltage gain. Emitter bypassing increases the voltage gain. Repeat Segment

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**Concept Preview C-E amplifiers are the most widely applied.**

C-E amplifiers are the only ones that provide a phase inversion. C-C amplifiers are also called emitter followers. C-C amplifiers are noted for their low output impedance. C-B amplifiers are noted for their low input impedance. C-B amplifiers are used mostly in RF applications. The analysis procedure for PNP amplifiers is the same as for NPN.

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**The common-emitter configuration is used most often.**

It has the best power gain. VCC Medium output Z RB1 RL C B E Medium input Z RB2 RE CE

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**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 In-phase output B E RB2 RL Low output Z

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**VCC RB1 RL C B E RB2 RE 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 Low input Z

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**PNP C-E amplifier + VB = - 3.75 V VE = - 3.05 V IE = 2.913 mA**

47 W 1 kW 1.5 kW 22 kW 10 kW + 12 V VE = V IE = mA rE = W VRL = V AV = 27 VCE = V VC = V

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**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

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**Concept Review C-E amplifiers are the most widely applied.**

C-E amplifiers are the only ones that provide a phase inversion. C-C amplifiers are also called emitter followers. C-C amplifiers are noted for their low output impedance. C-B amplifiers are noted for their low input impedance. C-B amplifiers are used mostly in RF applications. The analysis procedure for PNP amplifiers is the same as for NPN. Repeat Segment

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**REVIEW Gain Common-Emitter Amplifier Stabilizing the Amplifier**

Other Configurations

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