Physics 160 Lecture 15

Transcription

Physics 160 Lecture 15
Physics 160
Lecture 15
R. Johnson
May 18, 2015
Negative Feedback Formalism
In
+
Out
A

B
Representative of a
non-inverting
voltage amplifier
A= open loop gain (forward gain with no feedback), typically huge (e.g. 100,000)
B= gain of feedback network, typically < 1
AB= loop gain
Feedback
Vout  A  (Vin  B  Vout )

V
A
1
1
 
G  out 
Vin 1  AB B 1  1 AB  
Ideally, you want the loop gain AB to be large,
Ideally
large so that G is determined by
only the feedback network gain B.
May 18, 2015
Physics 160
2
In
3
V+
8
Non-Inverting Amp, More Complete Analysis
+
Out
-
4
2
1
V-
OUT
You may want to add a
9.1k series resistor to this
leg to equalize effects of
non negligible input
non-negligible
current in the case of
bipolar inputs.
R2
90k
R1
10k
0
Using the not-so-simplified analysis:
Suppose the forward gain is A=105 (fairly typical).
H
Here
the
h ffeedback
db k gain
i iis B
B=R
R1/(R1+R
R2)=1/10
) 1/10 ((voltage
l
di
divider).
id )
Then the loop gain is AB=104.
V


1
G  outt  10  
  9.999

4
Vin
 1  10 
May 18, 2015
Physics 160
3
Example
R3
3
V1
10k
15Vdc
445.9uV
+
V+
4
15.00V
U5A
-
LM324
R2
0V
V
11
241.5uV
V-
OUT
2
-206.1mV
1
9990k
V2
1Vac
0Vdc
R1
10k
V3
15Vdc
0


Vout
1
G
 1000  



1

1000
Vin
A


May 18, 2015
Physics 160
4
Non-Inverting Amp Output
90 degree phase change
from capacitive roll-off
(low-pass filter)
filter).
The negative feedback turns
into positive feedback at this
point! The amp will oscillate if
the overall loop gain is greater
than unity at this frequency.
From G=989,
instead of 1000,
we infer that
A=90
A
90,000
000 (not
infinite).
Gain-bandwidth product is evidently about 10001000=106. So
a follower (G=1) would work up to 1 MHz.
May 18, 2015
Physics 160
5
LM324 Data Sheet
http://www.onsemi.com/pub_link/Collateral/LM324-D.PDF
We expect a
frequency limit
of ~1 kHz
G=1000
A follower
(G=1) will work
up to ~1 MHz.
dB  20  log10 G 
20  log
l (90000)  99 dB
May 18, 2015
Physics 160
6
Follower
IN
OUT
Non-inverting amp with 100% feedback. Be
aware that the internal compensation capacitor is
very important here to avoid high-frequency
oscillations!
The LM324, like the LM411, has a large enough
internal capacitor to enhance the Miller effect that
it is stable when operated as a follower.
Some op amps require an external compensation
capacitor, so you can optimize the operation.
May 18, 2015
Physics 160
7
AC Coupled Non-Inverting Amp
(The op amp here is assumed to be operated with dual power
supplies, so that the signal can oscillate about ground.)
This resistor is essential,
to supply
pp y some bias
current to the op amp
input transistors. This is
no different from what we
did with discrete
transistors.
This extra resistor is a good idea,
to make sure that the input bias
current gives the same voltage
shift to both inputs. Not important
for the LM411, but for a BJT op
amp like the LM324 it can be very
important.
May 18, 2015
The input impedance of this
circuit now is only 10 kohms!
(But if you use the LM411 with
JFET inputs
inputs, the resistor can be
much, much bigger.)
Physics 160
8
AC Coupled Follower with Bootstrap
IN
OUT
If necessary, for a follower the bias network can be
bootstrapped as shown here, in exactly the same way as we
did for the emitter-follower. (Note that this example is for an
amplifier with both positive and negative supplies
supplies, so the inputs
and outputs can be centered on ground potential.)
May 18, 2015
Physics 160
9
AC Non-Inverting Amplifier
An alternative to using a coupling capacitor, or
a good idea even if the input is AC coupled.
IN
OUT
This amp will have a gain of 10 at
q
above 1/(2R
(
frequencies
2C1)).
At DC the gain is only unity, which
avoids problems with DC offsets
getting amplified.
May 18, 2015
Physics 160
10
Inverting Amp, More Complete Analysis
R2
Vin  Vout
i
R1  R2
R1
2
V-
In
4
i
-
+
Out
8
3
1
V+
OUT
Without R1 this is a
transresistance
amplifier.
feedback
0
V  Vin  iR1  Vin  (Vin  Vout )  B where
Vout  A  (V  V )   AV
Eliminate V :
May 18, 2015
R1
B
R1  R2
Forward gain
 1 B   1 
Vout
R2
G
 


Vin
R1
1  1 AB 
Physics 160
11
Inverting Amp, Input Current Compensation
R1
2
-
V-
In
4
R2
Add a resistor
here of value
R1R2/(R1+R2)
May 18, 2015
+
Out
8
3
1
V+
OUT
0
A real op amp will have some input bias current
(especially for BJT input)
input). The current flowing
through R1 and R2 will shift the voltage at the
inverting input. We should add a resistor to the
g input
p to compensate:
p
non-inverting
Physics 160
12
Analog Computing
Analog computing:
–
–
–
–
Integrator
Differentiator
Summing amplifier
Log amp
Op-amps provide nearly ideal
performance for these devices, far,
far beyond what can be achieved with
passive components.
(Note:
(N
t historically
hi t i ll analog
l computers
t
were widely used, but by now digital
computers have such high
performance and low cost that analog
computers are largely obsolete.)
May 18, 2015
Physics 160
A 1960 Newmark analog computer, made up
of five units. This computer was used to solve
differential equations and is currently housed
at the Cambridge
g Museum of Technology
gy
(from Wikipedia).
13
Integrator Example
A real circuit will have to
provide a method to zero
the charge on the
capacitor at the start of
the integration!
15.00V
t

1
Vout (t )  Vout (0) 
Vin dt 
R1C1
Q C1
100n
0
U4
3
+
V+
V
Virtual ground
7
-1.321V
40.32nV
V1
0V
OUT
R1
2
V1 = 0
V2 = 0
0.2
2
TD = 0
TR = 0
TF = 0
PW = 1m
PER = 10m
V3
V
1k
B1
5
out
6
1
V
LF411
40.32nV
R2
1k
0V
V2
15Vdc
-15.00V
May 18, 2015
-
V-
15Vdc
B2
4
iin
dV
V
d
dQ
 in 
 C1 out
dt
R1
dt
Physics 160
Note DC offset. This is
hard to control, since there
is no DC feedback. Adding
g
a large resistor in parallel
with C1 can help control
this (but also gives some
error in the integration)
integration).
14
Integration Example
Input Square Pulse
Perfect Integration
(but note the
inversion of sign)
May 18, 2015
Physics 160
15
Differentiator
iin 
dV
dQ
 C 2 in
dt
dt
Vout  iin R1   R1C 2
C1
0.1n
dVin
dt
R1
100k
V2
U1
3
2
V3
.01uF
V
May 18, 2015
B1
6
1
V
LF411
For stability,
stability it
becomes an
integrator at high
frequency.
V1
15Vdc
-
5
4
V1 = 0
V2 = .1
1
TD = 0
TR = .5m
TF = .5m
PW = 0.5m
PER = 2m
1k
B2
OUT
C2
V-
R2
+
V+
+
Virtual ground
7
15Vdc
0
Physics 160
16
Differentiation Example
Slow response on
edges is needed
f stability
for
t bilit
May 18, 2015
Physics 160
17
Stability Issues
•
Integrator:
– voltage gain falls at 6 dB per octave as frequency increases.
– inherently stable against oscillations even with an uncompensated
op-amp.
– the feedback itself looks like Miller-effect frequency compensation.
•
Differentiator:
– voltage gain rises at 6 dB per octave as frequency increases.
– inherently unstable against oscillations.
– the gain must be rolled off at high frequency by introducing an
integration, or oscillation is guaranteed!
1
jRC
Differentiator : G   jRC
Integrator : G  
May 18, 2015
Physics 160
18
Sum (or Subtraction) Amp
What are really being
added here are the 3
currents, which combine
at the virtual ground.
ground
Inverse of the
sum of the 3
voltages V2,
V3 V4
V3,
V4.
Resistor added
in an attempt to
minimize offset
errors due to
input current.
May 18, 2015
Physics 160
19
Diode is to protect against baseemitter breakdown in case input
goes negative.
Log Amp
Emitter voltage is logarithm of
collector current.
D1
D1N914
Q1
Q2N3904
FET inputs
for nearly
zero current
error.
R2
OUT
2
10k
B1
Q2
Q2N3904
6
U2
3
+
7
5
1
B2
OUT
LF411
2
4
VIN
-
V-
R1
0Vdc
B2
V+
V1
-
V-
+
V+
U1
7
15k
3
15Vdc
Source of ~constant
current
current.
B1
5
6
1
V
4
LF411
R3
Do DC scan
of this
source. 15Vdc
Output will
be the log
of this input.
Constant current.
Diode drop here
provides
temperature
compensation.
V2
15k
R4
1k
16 non
non-inverting
inverting amp
With a log amp you can do analog multiplication and division!
Ebers-Moll:
VBE  VT  log( I C  I S )  VT  log I S
I C  Vin / R1
May 18, 2015
But I C  I S
Vout  C  VBE  16  C  0.96 V  log10 I C I S 
Physics 160
0.026  16  ln(10)  0.96
20
Log Amp DC Bias Example
In a real circuit, put a capacitor in
parallel with the diode, for stability.
15.00V
D1
D1N914
Q1
Q2N3904
119 1 A
119.1pA
1.000uA
-1.023uA
23.25nA
987.4uA
R2
0V
15k
R1
10.00mV
1.000uA
5.329mA 0.01Vdc
10k
U2
3
-40.13pA +
B2
OUT
2
-40.13pA
40.13pA LF411
B1
5
6
-189.3uA
1
-2.170mA
In a real circuit, use
pins 1 and 5 of the
op amp to trim the
input offset to zero.
V
3.029V
4
0V
V2
2.171mA
7.398uA
1.000uA
15Vdc
7
189.3mV
980.0uA
-987.4uA
4
VIN
Q2
Q2N3904
V+
15Vdc
V-
V1
V-
U1
3
2.171mA 5
B2
+
-40.32pA
6
OUT
988.4uA
2
1
B1
-40.32pA
LF411
-2.170mA
V+
-1.831uV
7
-515.9mV
R3
189.3mV
189.3uA
189.3uA
15k
R4
1k
0V
4.340mA
-15.00V
15 00V
May 18, 2015
Physics 160
21
Log Amp Simulation
Good logarithmic response
over about
b t 6 or 7 d
decades,
d
but fails for inputs below
about 10 microvolts.
May 18, 2015
Physics 160
22
Behavior, Limitations of Real Op-amps
•
•
•
•
•
•
•
•
•
•
Input offset voltage
Input current, and input-current mismatch
– Design to minimize voltage offsets from this current
– Input offset current
See the data sheet for the
Temperature dependences
LF411 (posted on the course
Slew rate
web page)
page).
– e.g. correcting crossover distortion
Frequency range (gen. purpose op amps are not for high frequencies)
Voltage gain and phase shift
– Stability!
CMRR and Common-mode input range
Input impedance with negative feedback is extremely high
Output impedance with negative feedback is very low
– What is really relevant is the limitation on output current drive
Output range (how close can the output get to the supply rails?)
– Single-supply op amps; can the output slew all the way to zero?
May 18, 2015
Physics 160
23
Offset Voltage and Input Current
•
Offset voltage
– 2 mV max. for the 411
• This is the effective mismatch at the input. At the output it gets
multiplied
lti li d b
by th
the gain!
i !
– Trimming
+
V+
U1
3
7
V1
15
OS2
2
-
V-
OUT
OS1
6
1
R2
4
LM741
5
V2
– Temperature and time drifts
•
10k
15
I
Input
t currents
t (esp.
(
for
f BJTs)
BJT )
– Balancing resistors on inputs
– Input offset current
•
A/C Amp: reduce DC gain to unity to minimize offset effects
May 18, 2015
Physics 160
24
Charge Sensitive Amp
Integrate the pulse of current from a detector to produce an
output voltage proportional to the total charge.
Vout 
R6
10G
C1
0.1p
Virtual ground
U4
3
+
V1
V2
May 18, 2015
VV
2
15Vdc
B2
OUT
15Vdc
-
B1
5
6
1
V
LF411
I1
4
I1 = 0
I2 = 1u
TD = 50n
TR = 1n
TF = 1n
PW = 98n
PER = 100u
Q 0.1pC
0 1pC

1V
C 0.1pF
7
This is not a practical
resistor value in the real
world!
V+
•
C2
10p
R1
R3
100Meg
100Meg
Detector Model,
with 100 fC
pulse
l off charge.
h
Physics 160
Even though the feedback
capacitor is 100 time smaller than
the detector capacitance, it
effectively gets multiplied by A
(forward gain). Therefore, almost
all of the charge ends up on the
feedback capacitor.
Hence the detector ends up with
nearly zero volts on it (virtual
ground).
25
Charge Amp Example
Short pulse of current.
Rise time is non-zero
due to the non-zero
output impedance of the
amplifier.
May 18, 2015
Slow decay from the 10Gohm
resistor in the feedback network.
Physics 160
26
Voltage divider divides the output
by 100, correspondingly reducing
the feedback current. The DC
f db k iis th
feedback
thus almost
l
t
equivalent to a 10 Gohm resistor.
T Network
15.00V
R6
R2
4.024mV
100Meg
100Meg
R4
1Meg
C1
4.028mV
V+
U4
3
7
0 1p
0.1p
+
V1
OUT
2
4.032mV
-
B1
5
406.4mV
6
1
V
LF411
4
I1
V-
15Vdc
I1 = 0
I2 = 1u
TD = 50n
TR = 1n
TF = 1n
PW = 98n
PER = 100u
B2
C2
10p
R1
R3
100Meg
100Meg
0V
V2
15Vdc
-15.00V
May 18, 2015
Physics 160
27
Charge Amp with T Network
Short pulse of current.
Slow decay from the T network, acts
the same as the 10 Gohm resistor.
May 18, 2015
Physics 160
28