Problem 4. Figure P-4 shows Problem 5. Figure P Problem 4. Figure P-4 shows
the frequency response of an
amplifier circuit. In the first
decade of frequency, 20Hz,
40Hz, 60Hz, and 80Hz have
been marked with 2, 4, 6, and 8,
respectively. The grid lines of
the succeeding decades follow
the same pattern.
a) What is the gain-bandwidth
product of this amplifier?
b) If this amplifier is operated at
a closed-loop gain of 40 dB,
what will be the bandwidth of
the closed-loop gain?
Figure P- 4
Problem 5. Figure P-5 shows
the frequency response of an
operational amplifier.
a) If this amplifier is operated at a
closed-loop gain of unity, what
phase margin results? Mark the
phase plot on this figure at the
b) If this amplifier is operated at a
closed-loop gain of 40 dB, what
phase margin results?
Figure P-5
Solution 4. (a) Find the gainbandwidth product by multiplying the
coordinates of any point on the straight
line at –20dB/decade. Example points are
marked with a black squares. These are
2kHz  60dB  2MHz (because 60dB
 1000 VV ), 200kHz  20dB  2MHz and
2MHz  0dB  2MHz . Note that all
points lead to the same answer because
they are all on the straight line at
20dB / decade . (b) The open loop gain
is equal to 40 dB (the low-frequency
closed-loop gain) at a frequency of
20KHz. Hence 20kHz will be the closedloop –3dB frequency, which is the same
as the bandwidth. This result could also
be obtained by dividing
Solution 5. (a) For unity closed-loop
gain, the frequency at which A( s)   1
is found where the open loop gain
magnitude crosses 0dB (unity gain). This
frequency is about 1.1MHz, at which the
phase graph indicates about 52 .
The phase axis is actually numbered
as phase margin, since the numbers
go from 90 to 0 . Therefore the
value read is the phase margin of
52 . (b) For 40dB closed-loop
gain, the phase margin is 90 .
Problem 4
90
52
Problem 5