Download 08-CEAmplifier[1]

Practical Amplifier

To analyse the circuit:



Determine quiescent conditions
Calculate mutual conductance
Calculate small signal performance




Voltage Gain
Input Impedance
Output Impedance
Cut-off frequency
Quiescent Conditions
I B  0  I RB  0  VB  0
VBE  0.5 V  VE  0.5 V
I E  I RE
VE   15

RE
14.5

 0.5 mA  I C
29
VC  15  I C RC
 15  0.5 10  10 V
Small Signal Analysis: Voltage Gain
As before:
iC  g m vBE  g m vIN
vOUT dVOUT

  RC
iC
dI C
vOUT vOUT iC


  RC g m
vIN
iC
vIN
IC
0.5
4
  RC   10 
 200
VT
25
Input and Output Impedance

Unlike the op-amp, transistor amplifiers have
significant output impedances and finite input
impedances


RIN can be comparable with the source resistance of the
input signal
ROUT can be comparable with the load resistance
Input Impedance
iIN
iRB
iB

Input impedance, rIN, is the ratio
of the small signal input voltage
and the small signal input current
vIN
rIN 
iIN  iRB  iB
iIN
iRB
vIN

RB
iB 
iC


vIN g m

Input Impedance (cont)
iIN
iRB
iB
vIN vIN g m
iIN  iRB  iB 

RB

vIN
1

rIN 

 RB ||
iIN 1 / RB  g m / 
gm
iC
vBE
NB. g m 
& rE 
vBE
iE
1
 gm 
rE
Output Impedance

One way to measure rOUT is:


Short the input to 0 V
Output now looks like just rOUT
Output Impedance (cont)
vIN  0  iC  0
Applying Kirchoff’s current law:
iC  iRC  iOUT  0  iOUT  iRC
By Ohm’s law:
vC
vOUT
VC  15  I RC RC 
  RC 
iRC
iRC
rOUT
vOUT
vOUT


  RC   RC
iOUT
iRC
Coupling Capacitors





Capacitor COUT is needed to remove
the d.c. component of the collector
voltage
Capacitor CIN is needed to allow
the base voltage to be offset from
0V
In both cases this is known as
coupling
Both capacitors are chosen to look
like short circuits at operating
frequencies
Their reactance will, however,
become significant at low
frequencies
Equivalent Circuit
rIN
1
v1  vIN
 vIN
rIN  1 / j 2fCIN
1  1 / j 2frIN CIN
v1  vIN
1
1  1 / 2frIN CIN 
2
Cut-Off Frequency
Cut-off frequency, or –3dB point, is when the gain of the
amplifier falls by a factor of 2
v1
1
1


2
vIN
2
1  1 / 2f C rIN C IN 
 1  1 / 2f C rIN CIN   2  2f C rIN CIN  1
2
If the cut-off frequency, fC, is specified and rIN has been
calculated:
C IN
1

2f C rIN
NB. This assumes that COUT still looks like a short circuit
COUT

For the lower cut-off frequency calculation to be
valid, COUT should still look like a short circuit at fC
1
 XC 
 rOUT
2f C COUT

Typically, choose:
COUT 
1
2
fC
rOUT
10
Emitter Capacitor
For the highest voltage gain,
vBE  vIN  vBE  vE
vBE
But, vBE 
vIN
vE
iC
 iE rE
gm
1 VT
where, rE 

gm IC
iE
j 2fCE
vBE
 vE j 2fCE 
rE
Also, vE 
Emitter Capacitor (cont)
For CE to not interfere at fC:
vBE  vE
vBR
vIN
vE
vBE
Where, vE j 2f C C E 
rE
 2f C CE rE  1
To make sure, choose,
1
CE 
fC
2
rE
10
NB. Use rE (=VT/IC) not RE for this calculation!
Summary

In the context of the common-emitter amplifier we
have covered:





Small signal analysis
Mutual conductance
Input/output impedance
Coupling capacitor requirements and cut-off frequencies
Next time:



Applying the same principles to the differential amplifier
It’s actually a much easier circuit to analyse – honest!
Make sure you’re happy with the fundamentals by then!