Physics 160
Lecture 14
Fun with Op Amps
R. Johnson
May 13, 2015
Ideal Op-Amp
Differential gain, of course. Common-mode gain is ideally zero.
Such an ideal op-amp of course does not exist, but a first
analysis of op
op-amp
amp circuits can be done to a good
approximation usually by ignoring the non-ideal behavior.
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A Real Op-Amp: LF411
Differential voltage gain at low frequency is >100,000!
Current sources
15 V
P-channel
JFET-input
differential
amplifier
Class AB pushpull output stage
?
CommonEmitter 2ndstage NPN
amplifier
lifi
15 V
A current source, instead of
resistor is used as a load
resistor,
on the JFET drain, to
produce high voltage gain.
May 13, 2015
Deliberate capacitor enhances the
Miller effect to cause the gain to
decrease at higher frequencies, for
stability.
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(Note: many 741 versions have been made and sold, with varying detailed designs.)
Several current
mirrors provide
bias current to
th various
the
i
stages.
Temperature
stabilized
push-pull
output stage.
A Real Op-Amp (famous 741)
Q12
Q13b
Q13a
Diodes to bias
the output, to
reduce
crossover
distortion
HI
H
2nd stage active load
Q14
Q9
Unusual
input stage:
we hold the
base
constant
and wiggle
the emitter.
Current
source in
diff-amp tail
Q8
Q15
Q19
R6
27k
27
Q18
Vin+
VinQ1
Short-circuit
over-current
protection for
output stage
(~25 mA)
Increases Miller effect
to kill gain at high
frequency (for stability)
Vo
R10
40k
R7
27k
Q2
22
Q21
R5
39k
Q20
Cc
Q3
Q4
Q23
30p
Follower
Q7
Q16
Q17
Q11
Q10
R9
50k
Q6
Q5
R8
100
R3
50k
R4
5k
Unequal
current
mirror;
see Fig.
2.53.
May 13, 2015
R1
1k
R2
1k
Q22
Q24
R11
50k
Current-mirror
C
t i
active load
Physics 160
LO
Sets the
reference
current.
Second
amplification stage.
Common-emitter
amp with an
emitter-follower on
the input
4
Op-Amp Example (Single Supply)
Details of the current references and mirrors are not shown.
High Gain
DC-coupled
diff
differential
ti l
amp.
IIncreases Miller
Mill effect
ff t
to kill gain at high
frequency (for stability)
Short-circuit protection
Pushpull
output
t t
LM324
Darlington
2nd stage amp
(common
(commonemitter)
2 emitter
f ll
followers
May 13, 2015
Physics 160
Helps to pull
the output all
the way to
ground.
5
3
V+
Vin
In
8
Non-Inverting Amplifier
+
2
-
4
Virtual Vin
Out
1
V-
OUT
Note: Zin is very large
(transistor base or gate).
R2
90k
This is good!
R1
10k
Vin  Vout 
R1
R1  R2
0
Simplified analysis of an op-amp with negative feedback:
- Assume infinite gain, so the negative feedback always has to keep the two inputs
equal in order to have a finite output.
- Assume zero current flow into the op-amp inputs.
- Then calculate the relationship between input and output from the feedback
network.
Vout
R2
G
 1
 1  9  10
Vin
R1
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Linear Amp using Negative Feedback
V1
15Vdc
The amp without feedback
is very high gain and nonlinear. With negative
feedback we g
get a g
gain
that depends only on the
feedback network, and we
get excellent linearity. This
is the principle behind opamps.
V1 = 0
V2 = 0.2
TD = 0
TR = 1m
TF = 0.
PW = 0
PER = 1m
V3
R3
R2
1000
1000
Q2N3906
Q2N3906
Q10
Q4
Q9
Q2N3904
+
Q1
Q2
Q2N3904
V

R8
V
90k
Q2N3904
R9
Feedback
network
RLoad
1Meg
10k
0
0
R1
R10
6.8k
V2
15Vdc
18k
Q5
0
Q2N3904
Q6
Q2N3904
Q3
Q2N3904
Example
E
l ffrom L
Lecture
t
10
10.
This is a crude op-amp, since
it is DC coupled, differential,
g g
gain
high
Gain is (10k+90k)/10k=10
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Gain=10 with excellent linearity!
From Lecture 10
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Inverting Amp
Note: Zin
R1=1
1 kohm
i =R
I
100k
R1
2
V-
4
I
In
This is a distinct
disadvantage of this
configuration.
R2
Virtual ground
-
1k
Vin
I in 
R1
Out
+
8
3
1
V+
OUT
Vout   I in  R2  
0
Vin
 R2
R1
S
Simplified
f
analysis off an op-amp with negative feedback:
f
- Assume infinite gain, so the negative feedback always has to keep the
two inputs equal in order to have a finite output.
- Assume zero current flow into the op-amp inputs.
- Then calculate the relationship between input and output from the
feedback network.
Vout
R2
G
May 13, 2015
Vin

R1
 100
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9
Differential Amplifier
Vout  V1  Vout  
R2
R1  R2
V1
OUT
equal
V2
V2 
R2
R1  R2
R
Vout  2 (V2  V1 )
R1
Beware that the CMRR depends almost entirely on the matching
of the resistors, so high precision resistors are essential in order
t make
to
k this
thi work
k well.
ll
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Simple Current Source
V (constant)
This design works
extremely well but is
usually inconvenient,
because the load does not
connect directly to ground
Load
R
No current flows
f
in or out off
the inputs, and the inputs
will be at the same voltage,
so clearly the current
through the load is V/R.
(As long as the op-amp
p doesn’t try
y to exceed
output
the supply voltage!)
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Current Sources
I out 
VCC  Vin
R
Here the control
voltage is
relative to VCC.
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No b
N
base current
error if MOSFET
is used.
I out 
Vin R 2

R1 R3
Here the control
voltage is relative
to ground.
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Example Current Sink
Load would
go here
“Base
Base
current error”
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Current Sink Performance
Emitter current
Z out 
14  9
4.9711  4.969410
3
 2.94 M
Collector current
The output impedance is completely dominated by the base current error and
the Early effect.
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Substitute a MOSFET for the BJT
This is the only place
in your lab course
where you will use a
MOSFET (IRL510).
(IRL510)
No Base
current error
May 13, 2015
Physics 160
Notice how the programmed
current is completely determined
by the resistors and voltage
supplies and completely
supplies,
independent of any transistor
characteristics!
15
Current Sink Performance
The output impedance is so high now that I cannot see any difference
in the simulated current over this 7 volt range! Practically a perfect
current source, as long as the compliance range is not exceeded.
JFET input op amp working
together with a MOSFET
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High-Current Sink
Note, using the IRL510
N-channel power
MOSFET as in the
previous slide is also a
high-current option, as
that device is rated up
to about 4 to 5 amperes
of current.
This also has the advantage
of essentially zero error from
the base current, since the
JFET gate draws negligible
current.
Q2 starts to turn on when
Idrain is about 0.6 mA.
Q1 can put at least 4 mA
i t th
into
the b
base off Q2
Q2, so
Q2 can sink at least
100 mA.
GND
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Correcting Cross-Over Distortion
4
Q1
3
15Vdc
+
OUT
VOFF = 0
VAMPL = .2
FREQ = 100
1k
Q2N3904
V
-
LM324
V
11
2
1
V-
R1
V3
U1A
V+
VCC
Q2N3906
Q2
RL
500
R2
10k
VEE
15Vdc
Note that the
feedback path
is from the
output of the
push pull stage!
push-pull
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f = 100 Hz
The op-amp slews
fast enough to
eliminate most of
the distortion!
Op-Amp output
Push-Pull
Output
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f = 1000 Hz
The op-amp cannot
slew fast enough to
completely correct
the distortion here.
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f = 10,000 Hz
The op-amp cannot
slew fast enough to
make any significant
correction here.
However, a combination of other techniques (diodes)
plus negative feedback will cure crossover distortion.
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