OP-Amp OPEN-LOOP RESPONSE
The frequency response indicates how the voltage
gain changes with frequency,and the phase response
indicates how the phase shift between the input and
output signal changes with frequency.
Frequency Response
Op-amps that have a constant -20dB/decade roll-off
from fc to unity gain are called compensated op-amp.A
compensated op-amp has only one RC network that
determines its frequency charateristic.
An op-amp that has more than one critical frequency
is called an uncompensated op-amp
Aol
-
R1
AV1
+
C1
Op-amp
Av2
R2
C2
R3
Av3
+
C3
Op-amp
Representation of an op-amp with three internal stages
A’v (dB)
Av1
Av2
Av3
0
fc1
fc2
Individual Response
fc3
f
Av(dB)
Av1+Av2+Av3
-20dB/decade
-40dB/decade
75
50
25
0
-60dB/decade
1
fc1
fc2
fc3
Composite response for an uncompensated op-amp
f(Hz)
Op-Amp have a frequency range with starts
at 0Hz.At the upper end,the frequency
range is limited by the bandwidth and the
slew rate.
The gain decreases with increasing
frequency.
• The gain decreases after first critical
frequency with a slope of 20dB/decade,after the second critical
frequency with a slope of -40dB/decade
and etc.
PHASE RESPONSE
Phase response indicates how the phase shift between
i/p and o/p signal changes with frequency.
With increasing frequency,a growing phase shift will occur
between i/p & o/p
In a multistage amplifier,each stage contributes to the
total phase lag.Each RC lag network can produce up
to a -90o phase shift.
 f 
−1 f 
−1 f 
φtotal = −tan   − tan   − tan  
 fc1 
 fc2 
 fc3 
−1
Example
A certain op-amp has three internal amplifier
stages with midrange gains of 30dB,40dB and
20dB.Each stage also has a critical frequency
associated with it as follows:
fc1=600Hz,fc2=50kHz and fc3=200kHz.
(a) What is the midrange open-loop gain of the opamp,expressed in dB
(b) What is the total phase shift through the
amplifier,including inversion,when the signal
frequency is 10 kHz.
Answer
(a ) Aol ( mid ) = 30dB + 40dB + 20dB = 90dB
 f 
 10kHz 
0
(b) φ1 = − tan −1   = − tan −1 
 = −86.6
 600 Hz 
 fc 
φ 2 = − tan
−1
 f 
10 kHz 
0

 = − tan −1 
 = − 11 .3
 50 kHz 
 fc 
 f 
−1  10kHz 
0
φ3 = − tan   = − tan 
2
.
9
=
−

 200kHz 
 fc 
−1
φtot = −86.6 0 − 11.30 − 2.9 0 − 180 0 = −2810
Op-AMP CLOSED-LOOP RESPONSE
RECALL
FOR NON INVERTING AMPLIFIER
ACL ( NI ) =
Rf
Ri
+ 1
For the Voltage Follower ,
Acl (VF ) ≅ 1
For the inverting amplifier,
Rf
Acl ( I ) ≅ −
Ri
Effect of Negative Feedback on
Bandwidth.
The closed-loop critical frequency of an opamp is
f c(cl) = f c(ol) (1 + BAol(mid) )
fc(cl) is higher than the open loop critical frequency fc(ol)
by the factor 1+BAol(mid).Recall that B is the feedback
attenuation, Ri/(Ri +Rf).Since fc(cl)equals to bandwidth
for the closed-loop amplifier,the bandwidth is also
increased by the same factor.
BWc ( cl ) = BWol (1 + BAol ( mid ) )
EXAMPLE
A certain amplifier has an open-loop gain in
midrange of 180,000 and an open-loop
critical frequency of 1500Hz. If the
attenuation of the feedback is 0.015,what
is the closed-loop bandwidth?
Solution
BWcl = BWol (1 + BAol ( mid ) )
= 1500 Hz[1 + (0.015)(180,000)]
= 4.05 MHz
Closed-Loop vs Open-loop Gain
Av
Open-loop gain
Aol(mid)
Closed-loop gain
Acl(mid)
fc(ol)
fc(cl)
f
When the open-loop gain of an op-amp is
reduced by negative feedback,the
bandwidth is increased. The closed-loop
gain is independent of the open-loop gain.
The point of intersection is the critical
frequency,fc(cl),for the closed-loop
response.The closed-loop gain has the
same roll-off rate as the open-loop gain.
Gain –Bandwidth product
An increase in closed-loop gain causes a
decrease in the bandwidth and viceversa,such that the product of gain and
bandwidth is a constant.This true as long
as the roll-off rate is fixed -20dB/decade
If Acl represents the gain of any of the
closed-loop configurations and fc(cl)
represents the closed-loop critical
frequency.
Aclfc(cl)=Aolfc(ol)=Unity gain bandwidth
The gain-bandswidth product is always
equal to the frequency at which the opamp’s open-loop gain is unity
Aclfc(cl)=unity-gain bandwidth
Op-Amp Bandwidth
• Open-loop bandwidth: BWol = fc(ol)
• Closed-loop critical frequency:
fc(cl) = fc(ol)(1 + BAol(mid))
• Since fc(cl) = BWcl , the closed-loop
bandwidth is: BWcl = BWol(1 + BAol(mid))
• Gain Bandwidth Product is a constant as
long as the roll-off rate is fixed:
Aclfc(cl) = Aolfc(ol) = unity-gain
bandwidth
Example
For each amplifier determine the closed-loop gain and bandwidth.
The op-amps in each circuit exhibit an open-loop gain of 125dB and
a unity-gain bandwidth of 2.8MHz.
Rf
Vin
Vin
−
Vout
+
Rf=12k
Ri=1.0k
(a)
Ri
5.6k
1.0M
−
Vout
+
(b)
−
Vout
Rf
+
Ri
Vin
2.2K
Vin
100K
−
Vout
+
2.2K
(c )
(d)
SOLUTION
BWAcl=UNITY GAIN BANDWIDTH
12k
= 13
(a ) Acl ( NI ) = 1 +
1k
2.8MHz
BW =
13
(b ) Acl ( I )
1M
=−
= − 179
5 .6 k
2 .8 MHZ
BW =
= 15 .7 KHZ
178 .6
(C) ACL(VF) = 1
BW = fC(CL)
Unity− gain BW 2.8MHz
=
=
= 2.8MHz
1
Acl
100K
= −45.5
(d ) Acl ( I ) = −
2.2K
2.8MHz
BW =
= 61.6kHz
45.5
POSITIVE FEEDBACK AND STABILITY
Vin
−
Vout
Vf
+
Internal inversion
Makes Vf 1800 out of
Phase with Vin.
B
With negative feedback,the
Signal fedback to the input
Of an amplifier is out of
phase with
The input signal,the amplifier
is stable.
When the signal fedback from output to
input is in phase with the output signal,a
positive feedback condition exits and the
amplifier can oscillate.That is,positive
feedback occurs when the total phase shift
through the op-amp and feedback network
is 3600,which is equivalent to no phase
shift(00).
Positive Feedback & Stability
• Positive feedback, where the output signal being
fed back is in-phase to the input, will cause the
amplifier to oscillate when the loop gain, AolB >
1.
• Phase margin, θpm , is the amount of additional
phase shift required to make the total phase shift
around the feedback loop 360o.
• To ensure stability for all midrange frequencies,
an op-amp must be operated with an Acl such
that the roll-off rate beginning at fc is -20
dB/decade.
LOOP Gain
For instability to occur:(a) There must be positive feedback
(b) The loop gain of the closed-loop amplifier
must be greater than 1.
The loop gain of a closed-loop amplifier is
defined to be the op-amp’s open-loop
gain times the attenuation of the
feedback network.
Loop Gain =AolB
PHASE MARGIN
For each amplifier configuration as shown below,the
feedback loop is connected to the inverting input.The is
an inherent phase shift of 1800 because of the inversion
between input and output.
Vin
−
+
180 0 + φtotal
R1
R2
Additional phase shift produced by the RC lag network
within the amplifier.
Rf
Ri
Vin
180
0
180
−
+ φ total
−
+
+
Vout
Vin
0
+ φ total
Vout
The phase margin, φ pm is the amount of
additional phase shift required to make the total
phase shift around the loop 3600 (3600 is
equivalent to 00)
180 + φtotal + φ pm
0
φ pm = 180 − φtotal
0
If the phase margin is positive,the total phase shift is less than
3600,then amplifier
Is stable.If the phase margin is zero or negative,the amplifier is
potentially unstable
Because the signal fed back can be phase with the input.
Stability Analysis
Av(dB)
Aol(mid)
-20dB/decade
120
-40dB/decade
100
60
25
-60dB/decade
f1
0
1
10
100
1k
f2
f3
10k
100k
f(Hz)
1M
We use an uncompensated three stage opamp with an open-loop response.For this case
,three are three different critical
frequencies,which indicate three internal RC lag
networks.At first critical frequency,fc1,the gain
begins rolling off at -20dB/decade.When the
second critical frequency,fc2,is reached,the gain
decreases at -40dB/decade;and when the third
critical frequency,fc3,is reached,the gain drops
at -60dB/decade.
To analyze an uncompensated closed-loop
amplifier for stability,the phase margin must be
determined.
CASE 1.
Av(dB)
Aol(mid)
120
Open loop gain
Acl(mid)
106
75
50
25
0
1
10
100
1K 5k 10k
100k
1M
f(Hz)
Case1
The closed-loop gain intersects the openloop response on the -20dB/decade
slope,the midrange closed-loop gain is
106dB,and the closed loop critical frequency
is 5kHz.
 f 
−1  f 
−1  f 
φtotal = − tan   − tan   − tan  
 f c1 
 f c2 
 f3 
−1
Where f=5kHz, fc1=1kHz,fc2=10kHz and
fc3=100kHz.
 5 kHz 
− 1  5 kHz
tan
−



 10 kHz
 1 kHz 
= − 78 . 7 0 − 26 . 6 0 − 2 . 9 0 = − 108
φ total = − tan
−1

 − tan

.1 0
−1
φTOTAL = 180 − 108.1 = +72
 5 kHz 


 100 kHz 
CASE 2
The closed-loop gain is lowered to where it
intersects the open-loop response on the 40dB/decade slope,the midrange closed-loop
gain in this case is 80dB,and the closed-loop
critical frequency is approximately 30kHz.The
total phase shift at f=30kHz due to three lag
network is calculated as follows;
 30 kHz
φ total = − tan 
 1kHz
= − 88 .1 0 − 71 .6 0
Phase M arg in is
−1

−1  30 kHz 
−1  30 kHz 

 − tan 
 − tan 
 100 kHz 
 10 kHz 

− 16 .7 0 = − 176 .4 0
φ pm = 180 0 − 176 .4 0 = + 3 .6 0
Av(dB)
Aol(mid)
Open loop gain
120
106
Closed-loop gain
Acl(mid)
80
-40dB/decade
50
25
0
1
10
100
1K
10k 30k
100k
1M
Case closed-loop gain intersects open-loop gain on 40db/decade
f(Hz)
The phase margin is positive,so the
amplifier is still stable for frequencies
in its midrange,but a very slight
increase in frequency above fc would
cause it to oscillate.
Av(dB)
Aol(mid)
Open loop gain
120
80
50
Closed-loop gain
25
-60dB/decade
Acl(mid)
18
0
1
10
100
1K
10k
100k 500k 1M
f(Hz)
CASE 3
The closed-loop gain is further decreased until it intersects the
open-loop response on the -60dB/decade slope,the midrange
closed-loop gain is 18dB,and the closed-loop critical frequency is
500kHz.The total phase shift at f = 500kHz due to three lag
networks is;
 500 kHz 
−1  500 kHz 
−1  500 kHz 
φ total = − tan 
 − tan 
 − tan 

 1kHz 
 10 kHz 
 100 kHz 
= − 89 .9 0 − 88 .9 0 − 78 .7 0 = −257 .5 0
Phase M arg in is
−1
φ pm = 180 0 − 257 .5 0 = −77 .5 0
Phase Margin is negative and the amplifier is unstable
at the upper end of its midrange.
EXAMPLE 1
(a)It has been determined that the op-amp circuit in Figure below has
three internal critical frequencies as follows: 1.2kHz,50kHz,250kHz.If
the midrange open-loop gain is 100dB,is the amplifier configuration
stable,marginally stable or unstable?
(b) Determine the phase margin for each value of phase lag.
(i) 300 (ii)600 (iii) 1200 (iv) 1800
(v) 2100
Vin
Ri
2.7k
10M
−
+
Vout
SOLUTION
Rf
Vin
Ri
2.7k
10M
−
Vout
+
10 MΩ
= 3704
Acl =
2.7 kΩ
Acl (dB) = 20 log 3704 = 71.37 dB
At 50kHz,the midrange gain has dropped from
100dB at a -20dB/decade rate to
A ol =
A ol ( mid
)
 f
1+ 
 f
 c ( ol )




2
=
100
 50 KhZ 
1+ 

 1 . 2 KhZ 
2
= 2399
At Aol(dB)=20log(2399)=67.6dB
Since Acl>Aol(71.37dB>67.7dB),the closed-loop
gain intersects the open-loop gain on the 20dB/decade slope.Therefore the amplifier is
stable
(b) i) θ pm = 180 − φtotal = 180 − 30 = 150
0
0
0
0
ii) θ pm = 180 − φtotal = 180 − 60 = 120
0
(iii ) 60
(iv) 0
0
0
(v) − 30
0
0
0
0
Examples 2
A certain op-amp has the following internal
critical frequencies in its open-loop
response:125Hz,25kHz and 180kHz.What
is the total phase shift through the
amplifier when the signal frequency is
50kHz?
SOLUTION

φ 1 = − tan 

−1 
φ 2 = − tan 

−1 
φ 3 = − tan 

−1

−1  50 kHz 
0
 = − tan 
 = − 89 .9
 125 kHz 

f 
−1  50 kHZ 
0
 = − tan 
=
−
63
.
4

fc 
 180 kHz 
f 
−1  50 kHz 
0
 = − tan 
=
−
15
.
5

fc 
 180 kHz 
f
fc
φ total = − 89 . 9 0 − 63 .4 0 − 15 .5 0 = − 169 0
EXAMPLES 3
Each graph show both the open-loop and the closed-loop
gain response of a particular op-amp configuration.Analyze
the case for stability
Av(dB)
Aol(mid)
-20dB/decade
Acl(mid))
-40dB/decade
f
Av(dB)
Aol(mid)
-20dB/decade
-40dB/decade
Acl(mid))
-60dB/decade
f
Av(dB)
Aol(mid)
-20dB/decade
Acl(mid))
-40dB/decade
f
SOLUTION
(a)The closed-loop gain intersects the openloop curve in the region of -20dB/decadestable.
(b) The closed-loop gain intersects the
open-loop curve in the region of 60dB/decade – unstable
© The closed-loop gain intersects the openloop curve in the region of -40dB/decadeMarginally stable.
Aol
Phase Compensation
Uncompensated Aol
-20
dB
/de
c
-20
dB
/de
c
0
fc1
With some
compensation
With more
compensation
fc2
fc3
f
Compensating Circuit
• Compensation is used to either eliminate openloop roll-off rates greater than -20 dB/dec or
extend the -20 dB/dec rate to a lower gain.
• Two basic methods of compensation for IC opamps: internal and external.
• In either case an RC series circuit is added so
that its critical frequency is less than the
dominant (i.e. lowest) fc of the internal lag
circuits of the op-amp.
Phase Lag Compensation
When phase shift equal or exceed 1800,the
amplifier can oscillate.
For uncompensated op-amp the danger of
instability.
COMPENSATION is used to either
eliminate open-loop roll-off rates greater
than -20dB/decade or extend the 20dB/decade rate to a lower gain.To allow
op-amps to be operated at low closed-loop
gain.
Compensating Network
There are two basic methods of
compensation for integrated circuit opamp;
i) Internal
ii) External
Compensating Network
Vout
R1
Vo
Vout
Vin
Rc
Cc
Vout
-20dB/decade
 R2

 R1 + R2
0

Vin

XCc ≅ ∞
XCc ≅ 0
fc
Basic Compensating Network Action
f
At low frequencies where the reactance of the
compensating capacitor,Xc is extremely
large,the output voltage is equal to input
voltage.
When the frequency reaches its critical
value,fc=1/ [2 π (R1+Rc)C],the output voltage
decreases at -20dB/decade.This roll-off rate
continues until Xc ≅ 0
Stage 1
Vin
Stage 2
R1
AV1
+
C1
Op-amp
A
Av2
R2
Vout
C2
Rc
Compensating network
Cc
Representation of op-amp with compensation
The critical frequency of the compensating
network is set to a value less than the
dominant(lowest) critical frequency of the
internal lag networks.This causes the 20dB/decade roll-off to begin at the
compensating network’s critical frequency.
The roll-off of the compensating network
continues up to the critical frequency of the
dominant lag network.
The roll-off the compensating network must end
at 1kHz.
Avol(dB)
Aol(mid)
Original open-loop gain
120
Compensated open-loop gain
100
60
25
Uncompensated BW
Compensate
BW
0
1
10
100
f(Hz)
1k
10k
100k
1M
What is the uncompensated
bandwidth? and what is the
compensated bandwidth?
Critical frequency of compensating network
1K
0
f (Hz)
-10
-20
Acom(dB)
Attenuation of compensating network
Example of compensated op-amp frequency response
Example
A certain op-amp has the open-loop as
shown in figure below,As you can see,the
lowest closed-loop gain for stability is
assured is approximately 40dB(where the
closed-loop gain line still intersects the 20dB/decade slope).In a particular
application, a 20dB closed-loop gain is
required.
(a)Determine the critical
frequency for the compensating
network.
(b) Sketch the ideal response
curve for the compensating
network
(C)Sketch the total total ideal
compensated open-loop
response.
Avol(dB)
100
90
-20dB/decade
80
70
60
50
40
30
-40dB/decade
20
10
f(Hz)
fc
0
1
10
100
1k
10k
100k
1M
SOLUTION
(a) The gain must be dropped so that the 20dB/decade roll-off extends down to
20dB rather than to 40dB.Therefore,the
critical frequency of the compensating
network is 10Hz.
(b) The roll-off of the compensating network
must end at 100Hz,as shown in figure 1
© The total open-loop response resulting
from compensating as shown in figure2
0
10
100
1K
10K
-10
-20
Acom(dB)
Compensating network response
1M
f (Hz)
Avol(dB)
100
90
-20dB/decade
80
70
60
50
40
30
20
10
f(Hz)
0
1
10
100
1k
10k
100k
1M