IEEE Solid-States Circuits Magazine - Spring 2021 - 128

IIT Madras, Chennai, India, in 2016. He
is currently pursuing his Ph.D. degree at
the University of Twente, Enschede, The
Netherlands. He was a design engineer
with Texas Instruments India Ltd., Bengaluru,
from 2015 to 2017.
Student Circuit Contest Solution
Consider a unity gain feedback system
illustrated in Figure 2(a) consisting
of an amplifier with the transfer
function G(s). Figure 2(b) depicts the
Bode magnitude plot of the loop gain
of the system. Let us apply a unit
step input to this feedback system
and observe its immediate output
response from 0 to
t 1s.UGB~ /=
t 1sUGB~ /=
After
a change at the input, the feedback
system's response until
is weak. This is because, in this time
interval, the input can be thought of
as made of frequencies higher than
~ UGB
for which the loop gain is 11 ,
as seen in the Bode plot. If the output
reaches close to the input value
(1 V for unit step input) by the time
t /,
=1sUGB~
21sUGB~
the feedback system
functions properly. Otherwise, after
t /,
the input can be thought
of as containing low- frequency content
as well, for which the loop gain is
high, as displayed in Figure 2(b). This
leads to an overshoot in the output
above the given input voltage.
To calculate the immediate output
response of the feedback system
for a unit step input, observe that,
for /,t 1sUGB
1 ~
Vo
is the same as
the open-loop response since the
feedback is very weak in this time
interval. Hence, we calculate ()
from the open-loop
Vto
for t 1sUGB
1 ~
/
response and extrapolate this function
until
t 1sUGB~ /=
and call it fast
loop response [FLR(t)].
FLR(t) can be calculated by takVi
+
-
After
t > 1/ωUGBS
of Input Change,
Amplification Is High
ω(rad/s)
(a)
ωUGB
For t < 1/ωUGBS
of Input Change,
Amplification Is Low
FIGURE 2: (a) A block diagram of the unity gain feedback system and (b) the Bode
magnitude plot of its loop gain.
G(s)
Vo
ing the inverse Laplace transform of
G(s). Since we are looking at the fast
response, i.e., frequencies close to
~UG ,B
we simplify the poles of G(s) as
;,
1+=
p
s
11 pfor
+= 2~UGB
p
s
p
s
for 1~UGB
p
;.
The zeros are also simplified similarly.
Now, we apply this technique to
() ()
Gs As=
late FLR(t):
As L
()
of the contest and calcuL
110
s
Vi(t) at 2 mHz
1
0.5
-0.5
-1
500 1,000 1,500
Time (s)
×10-4 Vi(t) - Vo(t) at 2 mHz
1
-1
500 1,000 1,500
Time (s)
(a)
2,000
0.02
-0.02
10
20
Time (s)
(b)
30
(b)
FIGURE 3: Vi(t) and Vi(t) - Vo(t) are (a) in phase at 2 mHz and (b) out of phase at ~180
(100 mHz).
128 SPRING 2021
IEEE SOLID-STATE CIRCUITS MAGAZINE
40
2,000
1
0.5
-0.5
-1
10
20
Time (s)
Vi(t) - Vo(t) at 100 mHz
30
40
Vi(t) at 100 mHz
-ccmm
s
sss
ss
11 10
#
## #
==#
~UGB
#t.
~UGB of A(s) can be calculated to
be 10 rad/s. Hence, FL () /11R
=
in this case. Thus, the feedback system
does not overshoot for a unit
step input.
Any input can be thought of as
continuous changes at the input
of the feedback system. As long
as Vo
=1sUGB~
any changes in the input
reaches the desired value in
t /,
do not cause overshoot, and hence the
system is stable. Any other time constants
of the system that take more
time to respond than
t 1sUGB~ /=
have less or no impact on the stability.
Since, for the given A(s),
~18 ,0
the angular frequency at which the
Bode phase response reaches 180c is
slower than
~UG ,B predicting the stability
based on the Bode plot at 180~ is
incorrect as it has less or no effect on
stability. The effect of
~ 180 appears
only in the settled behavior. This
is depicted in Figure 3, where ()
Vti
|G( jω)| (dB)
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