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Electronics Principles & Applications Sixth Edition Chapter 6 Introduction to Small-Signal Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler

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Gain Common-Emitter Amplifier Stabilizing the Amplifier Other Configurations INTRODUCTION

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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 V out to V in. Power gain is the ratio of P out to P in. 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 Gain = In Out = 3.33 1.5 V 5 V 1.5 V 5 V The units cancel

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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. 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

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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.

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+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

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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.

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Concept Review Voltage gain is the ratio of V out to V in. Power gain is the ratio of P out to P in. 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. 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.

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RBRB RLRL V CC C C B E The output is phase inverted.

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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.

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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.

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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.

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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.

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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

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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

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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.

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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.

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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?

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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

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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

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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.

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RBRB C E B C RLRL V CC It’s dependent! This common-emitter amplifier is not practical. It’s also temperature dependent.

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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 (V B ). Subtract V BE to find V E. Use Ohm’s Law to find I E and V R L. Use KVL to find V CE. I E determines the ac emitter resistance (r E ). R L, R E and r E determine the voltage gain. Emitter bypassing increases the voltage gain.

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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.

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+V CC RLRL RERE R B1 R B2 { R B1 and R B2 form a voltage divider Voltage divider bias

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+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.

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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

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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

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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

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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!

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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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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

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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

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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

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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

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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 (V B ). Subtract V BE to find V E. Use Ohm’s Law to find I E and V R L. Use KVL to find V CE. I E determines the ac emitter resistance (r E ). R L, R E and r E 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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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. Medium output Z Medium input Z

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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 Low output Z

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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 Low input Z

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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

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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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