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