Physics 160
Lecture 7
Current Sources
C
Common-Emitter
E itt A
Amplifier
lifi
R. Johnson
April 20, 2015
Offsetting the Input from Ground
This way we
need only a
single supply
supply.
15V
2.5 mA
~7 5V
~7.5V
f>100 Hz
•
•
•
•
•
Choose Vout to be biased to about one half of VCC.
Choose
C
oose RE to
og
give
e the
e des
desired
ed b
bias
as cu
current.
e
Current in the divider should be >> IB.
But if R1 and R2 are too small, the input impedance will be low.
C Rin must be large enough not to attenuate the lowest
C·R
frequencies of interest.
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Physics 160
2
Emmitter-Follower Exercise
Design an emmitter-follower with a single 15 V supply and A/C
input coupling, to operate in the frequency range above 100 Hz.
The output should be biased at roughly half of the supply
voltage. Assume a 1 kohm source impedance. Design to a
2.5 mA quiescent current. Check that the gain with the stated
source impedance is at least 95%.
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Standard 5% Resistor Values
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Standard 10% Capacitor Values
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Emitter-Follower Example (Single Supply)
f 3dB 
1
1

 98 Hz
2 68n  24k
What are the input and output impedances of this amp?
1
1
1
1
1




Z in 51k 56k   2.7k 24k
April 20, 2015
Z out  1k   10
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AC Small-Signal Voltage Gain
The gain is less than
unity because of the
voltage division
between the source
impedance and the
parallel combination
of the bias resistors.
24k
24k
 0.96
1k  24k
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Physics 160
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Current Sources
•
The collector of a bipolar transistor is a natural current source:
the collector current is nearly independent of its voltage!
VC
Z out 
I C
•
Since the change in collector current is very small, even for a
large change in voltage
voltage, the output impedance is extremely high
(hundreds of k or M).
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Ramping supply down from 15V to <2V
Z out 
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Physics 160
12V
 2.5 M
4.7 A
9
Current Sources
VCC
•
R1
Load
Very high (but not infinite) output
impedance. Close to ideal current
source with 1 transistor!
– We will see ways to improve this
performance even more.
IC
•
Biasing
– Base current error ( dependent)
– See text for other options.
•
R2
RE
•
RE provides “negative feedback”
Compliance: VCE cannot fall below
about 0.2 volts!
NPN (sink) vs. PNP (source)
– This example can be called a current
“sink”, but it is a semantic detail based
on the arbitrary definition of the
direction of conventional current.
What current has been programmed in this schematic?
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Simple Current Sink Simulation
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Sweep of the Load Resistance
Compliance limit
Z out 
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V R  I 8k 1mA


 2 M
I
I
4 A
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Current Source Biasing Schemes
Sink
NPN
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Sink
Source
NPN
PNP
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Current Sources Temperature Compensation
•
VBE and  depend on temperature
One way to compensate for changing VBE
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Voltage Controlled Current
A current source for which we let the base voltage vary.
“T
“Transconductance”
d t
”
I out  g m  Vin
IC  I E
I C
VE

RE
Iout
Vin
RE
1

VB
RE
 gm 
April 20, 2015
1
RE
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Common-Emitter Inverting Amplifier
VCC
RC
vout   RC  iC   g m RC  vin
vout  
RC
 vin
RE
15k
Like a voltage controlled
current sink, but we take the
collector voltage to be the
output.
Vout
Q1
Vin
R
Gain   C
RE
Q2N3904
RE
1k
VEE
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Physics 160
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Common-Emitter Amp Example
Design a common-emitter amplifier with a voltage gain of about 15
that operates from a single 15V supply.
Roughly maximize the output dynamic range (i.e. center the output
bias halfway between the power rails).
AC couple the input with a 3dB point of around 100 Hz.
Design for a bias current of about 0.5 mA.
Chapter
p
2, Problem 3
Design a CE amp with:
Gain=15, VCC=15 V, VEE=0, IC=0.5 mA
Output bias=7.5V,
bias=7 5V f3dB=100 Hz
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Standard 5% Resistor Values
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Standard 10% Capacitor Values
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Common-Emitter Amplifier
10k
15  1.15
10k  120k
in
R1
RC
120k
15k
C1
out
0.2u
0.22
C2
Q1
V2
10u
V
15Vdc
RS
Q2N3904
10
RL
V1
R2
1Vac
0Vdc
10k
RE
10Meg
1k
0
•
•
Transconductance
T
d t
~ 1/RE
Voltage gain = gmRCRC/RE
– In practice the gain with no load will be a bit lower due to the non-ideal
performance of the transistor (finite ).
)
– Of course RS and RL will affect the apparent gain (Zin is not infinite, and Zout
is not zero, but is actually quite large). More on this later.
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PSpice bias solution:
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Common-Emitter Amplifier
• The gain is not quite 15
15, but only about 14
14.
• Note that the upper frequency cutoff is worse than 10 times lower than for our
emitter-follower design. We will understand the reason for this soon (Miller
Effect).
Effect)
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Input and Output Impedance
Input
R1
RC
120k
15k
Output
p
C1
C2
0.2u
0.22
10u
Q1
V2
V
15Vdc
RS
Q2N3904
10
RL
V1
R2
1Vac
0Vdc
10k
RE
10Meg
1k
0
•
Input impedance (well above the 3dB point, so that C1 is not relevant).
1
1
1
1



Z in R1 R2  RE
•
Output impedance
Since Zoutt tends to be rather
high, usually you follow this
amplifier by an EF stage!
April 20, 2015
Z out  RC
Physics 160
Z in  10k
Remember, the impedance looking
into the collector is so large
compared with RC that it can be
neglected as a parallel contribution.
23
The input impedance is rather poor, because of the bias circuit. So if the
source impedance is 1k, for example, we lose about 10% of our gain.
With 1k source
impedance
We will soon learn how to “bootstrap” the bias circuit to greatly
increase the input impedance
impedance.
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Measuring Zin and Zout
•
•
Think in terms of simple voltage dividers.
This also highlights the importance of understanding these two
quantities!
Measuring Zin
Measuring Zout
Source
Impedance
RS
Z out
scope
scope
RL
Z in
Load
Impedance
V
RL
VS
RL  Z out
April 20, 2015
V
Physics 160
Z in
 VS
RS  Z in
25
Grounded-Emitter Amp
15 00V
15.00V
119.6uA
R1
496.6uA
RC
120k
15k
C1
0V
C2
0.2u
0.22
Q1
V2
7.551V
496.6uA
15Vdc
RS
0A
0V
0A
V1
3.802uA
0V
Q2N3904
-500.4uA
10
616.2uA
0V
645.1mV
10u
V
0V
RL
115.8uA
R2
1Vac
0Vdc
10Meg
0A
5.57k
0
•
•
V gain of ~260, but this circuit is a disaster waiting to happen! You can make
it work in PSpice by tuning the biasing, but in real life
– The bias point will not be stable with temperature or changing parts!
– Also, the linearity will be poor, except with very small input signals (<<10mV
input).
– Input impedance of the transistor base is low (rE) and unstable.
B adding
By
ddi an emitter
i
resistor
i
we iintroduce
d
“
“negative
i feedback.”
f db k At
A the
h cost off
reduced overall voltage gain, we greatly improve linearity and stability.
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Linearity with and without Negative Feedback
Voutt
Vout
With 1k emitter resistor.
resistor
400 mV sine wave input
April 20, 2015
Without emitter resistor.
resistor
19 mV sine wave input
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Common-Emitter Amp with High Gain
15 00V
15.00V
115.7uA
R1
472.6uA
RC
120k
15k
C1
1.000V
C2
0.2u
Q1
V2
7.910V
10u
472.6uA
V
15Vdc
RS
1.120V
3.648uA
588.3uA
0A
1.000V
0A
V3
VOFF = 1
VAMPL = 0.019
0V FREQ = 10k
0V
Q2N3904
-476.3uA
10
476.3mV
112.0uA
R2
10k
476.3uA
R7
1k
RL
10Meg
C3
0A
10u
0V
0
•
•
•
Use off th
U
the bipass
bi
capacitor
it around
d th
the emitter
itt resistor
i t ensures th
thatt the
th
biasing is stable, even with large gain.
High gain is achieved above 100 Hz. Below 100 Hz we don’t care.
But still,
still the linearity is poor,
poor except with very small input signals
signals, and the
input impedance of the transistor base is still low.
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