Chapter 6 BJT Amplifiers - University of Misan · Chapter 6 Basic Electronic Devices and Circuits...

Chapter 6

Basic Electronic Devices and Circuits

EE 111

Electrical Engineering

Majmaah University

2nd Semester 1432/1433 H

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd1

BJT Amplifiers

The Common-Emitter Amplifier

In the common-emitter (CE) amplifier, the input signal is applied to the base and the inverted output is taken from the collector.

The emitter is common to ac input & output signals.

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd

R2

RE

R1

Vin

RC

VCC

Vout

RL

C1

C2

C3

ac short; ZC = 1/( j ω C )[bypass capacitor]

2

=-5

-3

-1

1

3

5

7

9

11

13

AC + DC (in amplifier)

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd

+-5

-3

-1

1

3

5

7

9

11

13

Vout AC (Vp=5V)

-5

-3

-1

1

3

5

7

9

11

13

DC = 8 V (Q-point)

3

The Common-Emitter Amplifier

8.42

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd

2

4

DC Analysis

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd5

The Common-Emitter Amplifier

8.42

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd

2

6

AC Analysis

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd7

Signal (AC) Voltage at the Base

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd8

Input Resistance at the Base

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd9

Output ResistanceThe output resistance is the resistance seen looking back into the output terminal with Vin=0.

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd

(re' is much smaller than rc' )

(rc' is much larger than RC )

10

AmplifierSupply

AmplifierLoad

Why find Output Resistance?Why find Input Resistance?

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd

Voltage division.We prefer high Rin(tot)

Voltage division.We prefer low Rout

11

The Common-Emitter Amplifier

8.42This figure is mentioned in the next slide.

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd

2

12

(actually 3.58 mA on slide 18, but use 3.8 mA)

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd

< 10 mV !

There is significant attenuation (reduction) of the source (supply) voltage due to the voltage division between the source resistance (Rs) and the amplifier’s input resistance (Rin(tot)).

13

The Common-Emitter Amplifier

8.42

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd

2

14

Voltage Gain

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Electronic Devices, 9th edition

Thomas L. Floyd

This is the voltage gain from base to collector.To get the overall gain of the amplifier from the supply voltage to collector, the attenuation of the input circuit must be included.

15

Attenuation is the reduction in signal voltage as it passes through a circuit and corresponds to a gain of less than 1.

gain = 1 / attenuationgain = output / input

attenuation = 1 / gain = input / output

Example: If the signal amplitude is reduced by half,

Attenuation

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd

Example: If the signal amplitude is reduced by half,gain = 0.5or attenuation = 1 / 0.5 = 2

Example:

A source (supply) produces a 10 mV input signal, and the source resistance combined with the load resistance results in a 2 mV output signal.

gain = output / input = 2 / 10 = 0.2attenuation = input / output = 10 / 2 = 5

16

Voltage gain from base to collector = Av = Vc / Vb

Attenuation from source (supply) to base = Vs / Vb

Overall Voltage Gain

(reciprocal of voltage division)

The overall voltage gain of the amplifier, is the

© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.

Electronic Devices, 9th edition

Thomas L. Floyd

of the amplifier, is the voltage gain from base to collector, Vc / Vb, times the reciprocal of the attenuation, Vb / Vs.

≥ 1 < 1 ≥ 1

17