Transcript Basic BJT Amplifier

ANALOG ELECTRONIC
CIRCUITS 1
EKT 104
Basic BJT Amplifiers (Part 2)
1
Basic Common-Emitter
Amplifier

The basic common-emitter circuit used in
previous analysis causes a serious defect :




If BJT with VBE=0.7 V is used, IB=9.5 μA & IC=0.95 mA
But, if new BJT with VBE=0.6 V is used, IB=26 μA & BJT
goes into saturation; which is not acceptable  Previous
circuit is not practical
So, the emitter resistor is included: Q-point is stabilized
against variations in β, as will the voltage gain, AV
Assumptions


CC acts as a short circuit
Early voltage = ∞ ==> ro neglected due to open circuit
2
Common-Emitter Amplifier
with Emitter Resistor
inside
transistor
CE amplifier with emitter resistor
Small-signal equivalent circuit
(with current gain parameter, β)
3
Common-Emitter Amplifier
with Emitter Resistor

ac output voltage

Input voltage loop

Input resistance, Rib
Vo   I b RC
Vin  I b r  I b   I b RE
Vin
Rib 
 r  1   RE
Ib

Input resistance to amplifier, Ri
Ri  R1 R2 Rib
Remember: Assume VA is infinite, 
ro is neglected

Voltage divider equation of Vin to Vs
 Ri
Vin  
 Ri  RS

Vs

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Common-Emitter Amplifier
with Emitter Resistor
Cont..

So, small-signal voltage gain, AV
Av 
Remember: Assume VA is infinite, 
ro is neglected
Vo   I b RC

Vs
Vs
 Vin  1 
 
   RC 
 Rib  Vs 
 Ri
  RC

Av 
r  1   RE  Ri  RS




If Ri >> Rs and (1 + β)RE >> rπ
  RC
RC
Av 

`1
1   RE
RE
5
6
7
Common-Emitter Amplifier
with Emitter Bypass Capacitor



Sometimes for the purpose of dc biased,
emitter resistor set to be large.
Hence degrades the small-signal voltage gain
too severely.
Apply an emitter bypass capacitor to
effectively short out a portion or all of the
emitter resistance as seen by the ac signals.
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Common-Emitter Amplifier
with Emitter Bypass Capacitor
VCC
Emitter bypass capacitor, CE
provides a short circuit to
ground for the ac signals
RC
R1
vO
RS
Emitter bypass capacitor is used to
short out a portion or all of emitter
resistance by the ac signal. Hence
no RE appear in the hybrid-π
equivalent circuit
CC
vs
R2
RS
Vs
R1|| R2
RE
CE
B
C
r
gmV
ro
Vo
RC
E
Small-signal hybrid-π
equivalent circuit 9
Common-Emitter Amplifier
with Emitter Bypass Capacitor
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Common-Emitter Amplifier
with Emitter Bypass Capacitor
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Common-Emitter Amplifier
with Emitter Bypass Capacitor
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Common-Emitter Amplifier
with Emitter Bypass Capacitor
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Common-Emitter Amplifier
with Emitter Bypass Capacitor
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DC & AC LOAD LINE
ANALYSIS

DC load line


Visualized the relationship between Q-point & transistor
characteristics
AC load line


Visualized the relationship between small-signal response &
transistor characteristics
Occurs when capacitors added in transistor circuit
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Common Emitter Amplifier with
emitter bypass capacitor
Example 1
Common-emitter
amplifier with
emitter bypass
capacitor
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DC Load Line
Solution...

KVL on C-E loop
V   I C RC  VCE  I E ( RE1  RE 2 )  V 
1  
1  

 I C ( RE1  RE 2 )  V , when I E  
 I C
 I C RC  VCE  
  
  
1  
 I C ( RE1  RE 2 )
V   V   VCE  I C RC  
  
1  
  1
For Q - point, when   1, 
  
So, V  V   VCEQ  I CQ ( RC  RE1  RE 2 )
Slope 
-1
RC  RE1  RE 2
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AC Load Line
Solution...

KVL on C-E loop
ic RC  vce  ie RE1  0
Assuming ic  ie
vce  ic RC  ic RE1  ic ( RC  RE1 )
-1
Slope 
RC  RE1
AC equivalent circuit
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DC & AC Load Lines
Full
solution
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AC LOAD LINE ANALYSIS
Determine the dc and ac load line. VBE=0.7V, β=150,
VA=∞
Example 2
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DC Load Line

To determine dc Q-point, KVL around B-E loop
V   I BQ RB  VEB  I E RE  I BQ RB  VEB  (1   ) I BQ RE
I BQ
V   VEB
 5.96A

RB  (1   ) RE
Then I CQ  I BQ  0.894mA & I EQ  (1   ) I BQ  0.9mA
For Q - point, VCEQ  (V   V  )  I CQ RC  I EQ RE  6.53
1
-1

Slope 
RC  RE 15k
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AC Load Line
Small signal
hybrid-π
equivalent
circuit
I CQ  0.894mA ;VECQ  6.53V
r 
VT
gm 
I CQ
I CQ
VT
 4.36k
 34.4mA / V
VA
ro 

I CQ
vo  vec  ( g m v )( RC // RL )  ic ( RC // RL )
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DC & AC Load lines
Full
solution
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Maximum Symmetrical Swing

When symmetrical sinusoidal signal applied to the
input of an amplifier, the output generated is also a
symmetrical sinusoidal signal

AC load line is used to determine maximum output
symmetrical swing

If output is out of limit, portion of the output signal will be
clipped & signal distortion will occur
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Maximum Symmetrical Swing

Steps to design a BJT amplifier for
maximum symmetrical swing:

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Write DC load line equation (relates of ICQ & VCEQ)
Write AC load line equation (relates ic, vce ; vce = - icReq,
Req = effective ac resistance in C-E circuit)
Generally, ic = ICQ – IC(min), where IC(min) = 0 or some
other specified min collector current
Generally, vce = VCEQ – VCE(min), where VCE(min) is
some specified min C-E voltage
Combination of the above equations produce optimum
ICQ & VCEQ values to obtain maximum symmetrical
swing in output signal
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Maximum Symmetrical Swing
Example 3
Determine the maximum symmetrical swing in the output voltage of
the circuit given in Example 2.
Solution:

From the dc & ac load line, the maximum negative swing in the Ic
is from 0.894 mA to zero (ICQ). So, the maximum possible peak-topeak ac collector current:
ic  2( I CQ  I C (min))  2(0.894)  1.79 mA

The max. symmetrical peak-to-peak output voltage:
| vce || ic | Req | ic | ( RC || RL )  (1.79)(5 || 2)  2.56 V

Maximum instantaneous collector current:
iC  I CQ 
1
| ic | 0.894  0.894  1.79 mA
2
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Self-Reading
Textbook: Donald A. Neamen,
‘MICROELECTRONICS Circuit Analysis &
Design’,3rd Edition’, McGraw Hill
International Edition, 2007
 Chapter 6: Basic BJT Amplifiers
 Page: 397-413, 415-424.

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Exercise
Textbook: Donald A. Neamen,
‘MICROELECTRONICS Circuit Analysis &
Design’,3rd Edition’, McGraw Hill
International Edition, 2007
 Exercise 6.5, 6.6, 6.7,6.9
 Exercise 6.10 , 6.11

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