ESE 372 / Spring 2013 / Lecture 14
Last time: Common Emitter Amplifier biased by current source.
Bias
* But make sure that BJT is in FA, i.e. VCEQ > 0.2-0.3V
Equivalent circuit based on Hybrid-π model, caps are
short circuits for signals
1
ESE 372 / Spring 2013 / Lecture 14
Common Emitter Amplifier biased by current source.
Voltage gains
A MAX
 g m  rO
V0
Finite output resistance puts upper limit on amplifier voltage gain
2
ESE 372 / Spring 2013 / Lecture 14
Common Emitter Amplifier biased by current source.
Input/output impedances
Not high enough - problem
Net voltage gain
Depends on β – not too good
3
ESE 372 / Spring 2013 / Lecture 14
Common Emitter Amplifier biased by current source.
Short circuit current gain
Common emitter current gain
4
ESE 372 / Spring 2013 / Lecture 14
Example
npn - BJT with β  100 and VA  100V.
Bias current I E  1mA, VCC  VEE  10V
R B  100k, R C  8k, R S  5k, R L  5k
1. Check if BJT is in FA regime
IE
1mA

 10μ
β  1 101
VB  10μ 100k  1V  VE  1V  0.7V  1.7V  10V.
IB 
2. Calculate small signal parameters
VC  10V  1m  8k  2V  VCE  2V   1.7V   3.7 V  0.3V
3. Find voltage gain
I QC
1mA
mA
gm 

 40
Vth 25mV
V
GV 
rπ 
β
100

 2.5kΩ
g m 40 mA V
rO 
VA 100V

 100kΩ
Q
I C 1mA
R in
RL
 A V0 
R in  R S
R L  R out
A V0  g m  R C || rO   296
R in  R B || r   2.4k, R out
V
V
 R C || rO   7.4k
V
V

G V  0.32    296   0.4  38
V
V

30mV p-p signal corresponds to about 10mV p-p variation of VBE.
This produces about 1.2V p-p across load.
5
ESE 372 / Spring 2013 / Lecture 14
Example – output voltage swing
8V
3.5 V
-1.5
To remain in FA
VCE > 0.2 V
Maximum amplitude of the undistorted sine wave at the output is
limited by maximum negative voltage swing, i.e. by 3.5 V.
For 1mA bias current and 10 V power supply what could we do to
increase the amplitude of undistorted sine wave?
What is the maximum value of this amplitude?
6
ESE 372 / Spring 2013 / Lecture 14
CE amp biased by current source and with RE.
Case 1
RE  0
R in  R B || r   r
A V0  g m  R C || rO 
Case 2 (neglect rO)
RE  0
R in  R B || r  R E    1  r
A V0  
 RC
r  R E    1
Negative feedback resistor RE improves input
impedance at the expense of gain
7
ESE 372 / Spring 2013 / Lecture 14
Common Base amplifier.
No need for CE
!
Equivalent circuit for AC analysis
8
ESE 372 / Spring 2013 / Lecture 14
Common Base amplifier.
Redraw equivalent circuit in more convenient
form and neglect rO for beginning.
9
ESE 372 / Spring 2013 / Lecture 14
Common Base amplifier.
Voltage gain
Noninverting amplifier with open circuit voltage gain value
similar to that of CE amp
Short circuit current gain
i.e. no current gain !
10
ESE 372 / Spring 2013 / Lecture 14
Common Base amplifier.
Input impedance
Very small – PROBLEM
for voltage amplifier
Output impedance
11
ESE 372 / Spring 2013 / Lecture 14
Common Base amplifier.
Net voltage gain
Could we guess
this without
analysis?
Observe current buffer action of CB amp
Equivalent circuit of amp with
current gain equal to one.
12
ESE 372 / Spring 2013 / Lecture 14
CB amplifier with BJT having finite output small signal resistance.
Voltage gain
13
ESE 372 / Spring 2013 / Lecture 14
CB amplifier with BJT having finite output small signal resistance.
Input impedance
When
Impedance
transformation
max
14
ESE 372 / Spring 2013 / Lecture 14
Common Collector amplifier
Again.
No need for CE !
Also
RB and CC1 can be eliminated
Bias current IE will
determine gm, rπ and rO
Redraw equivalent circuit in more
convenient form
15
ESE 372 / Spring 2013 / Lecture 14
Common Collector amplifier
+
When
i.e. no voltage gain !
16