Electronics Principles & Applications Fifth Edition Chapter 6 Introduction to Small-Signal Amplifiers ©1999 Glencoe/McGraw-Hill Charles A. Schuler.

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Electronics Principles & Applications Fifth Edition Chapter 6 Introduction to Small-Signal Amplifiers ©1999 Glencoe/McGraw-Hill Charles A. Schuler

Gain Common-Emitter Amplifier Stabilizing the Amplifier Other Configurations INTRODUCTION

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

Gain can be expressed in decibels (dB). The dB is a logarithmic unit. Common logarithms are exponents of the number = = = = = 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

The dB unit is based on a power ratio. dB = 10 x log P OUT P IN 50 W 1 W 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.

+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

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.

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.

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

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.

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.

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.

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.

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

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

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.

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.

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?

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 RL = 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 RL = 14 V - 6 V = 8 V

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 RL = I C x R L = 14 mA x 1 k  = 14 V (V RL is higher) This is not a good Q point for linear amplification. V CE = V CC - V RL = 14 V - 14 V = 0 V (V CE is lower)  is higher

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

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. 

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.

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

+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.

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

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 V = 0.61 V

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

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 RL = I C x R L V RL = 2.77 mA x 2.2 k  V RL = 6.09 V V CE = V CC - V RL - V E V CE = 12 V V V V CE = 5.3 V A linear Q point!

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.

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 

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

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 may be used to increase A V : CECE

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

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.

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

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

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

REVIEW Gain Common-Emitter Amplifier Stabilizing the Amplifier Other Configurations