# ECE 3410 – Homework 4 Problem 1. A non ```ECE 3410 – Homework 4
Problem 1. A non-inverting op amp configuration is shown below. The op amp has the following
characteristics:
• DC open-loop gain A0 = 40dB.
• Open-loop 3dB cut-off frequency fc = 1MHz.
• Slew rate SR = 2V/ &micro;s.
• Rail voltages at &plusmn;5V.
• R1 = 1kΩ and R2 = 1kΩ, so the ideal closed-loop gain is 2V/ V.
R2
R1
−
vOUT
+
vIN
−
+
(A) Calculate the unity-gain frequency (ft ) of this amplifier circuit. Give your answer
in Hz.
Solution
ft = fc A0
= (1MHz) (100V/ V)
= 100MHz.
(B) Calculate the 3dB cut-off frequency for the closed-loop circuit.
Solution
ft
G
100MHz
=
2
= 50MHz.
fc =
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ECE 3410 – Homework 4
Solution (cont.)
(C) Sketch the magnitude response (i.e. Bode plot) for this op amp, showing both the
open-loop and closed-loop responses. Identify the unity-gain frequency.
Solution
50
Magnitude [dB]
40
open loop
30
20
10
closed-loop
ft
0
−10
102
103
104
105
106
Frequency [Hz]
107
108
109
(D) Calculate the full-power bandwidth (FPBW) for this op amp.
Solution
SR
2πVR
2 &times; 106 V/ s
=
2π &times; 5V
= 63.662kHz
FPBW ≤
(E) If the circuit amplifies an input signal described by the equation
vIN = VIN sin (2πf t) ,
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ECE 3410 – Homework 4
i. If f = 1MHz, what is the maximum possible amplitude VIN with no slewing?
Solution
SR
2πf
= 0.31831V (zero-to-peak)
VIN ≤
ii. If VIN = 1V, what is the maximum possible f with no slewing?
Solution
SR
2πVIN
= 318.31kHz
f≤
(F) If vIN is a square wave, is it possible to avoid slewing in the circuit’s output?
Solution
No. By definition, square waves change value instantaneously after each halfperiod. The op amp cannot change values instantly.
(G) Carefully study the axes and waveforms shown below. The waveforms are shown
for the ideal behavior (with no slewing). If slewing is present, draw the resulting
waveforms. You may assume that vOUT = 0 at time t = 0.
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ECE 3410 – Homework 4
4
Output [V]
2
0
−2
−4
0
0.5
1
1.5
Time [&micro;s]
2
2.5
3
0
0.5
1
1.5
Time [&micro;s]
2
2.5
3
4
Output [V]
2
0
−2
−4
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ECE 3410 – Homework 4
Problem 2. An inverting Miller integrator circuit is shown below. The component values are
R1 = 10kΩ
R2 = 10kΩ
C = 10&micro;F
C
R2
R1
vIN
−
vOUT
+
(A) When operating at DC, the capacitor C can be viewed as an open circuit (i.e. as
though it isn’t there). What is the DC gain of this circuit?
Solution
GDC = −1V/ V
(B) Using Laplace-transform analysis, show that the steady-state transfer function is
given by
R2
1
H (s) = −
.
R1
1 + sR2 C
You may assume that the op amp is ideal.
Solution
Gi = −
Z2 (s)
Z1 (s)
R2
= − 1+sR2 C2
R1
1
R2
= −
R1
1 + sR2 C2
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ECE 3410 – Homework 4
(C) Since an integrator has transfer function of the form K/s, this circuit can only be
considered an integrator for high-frequency signals. At low frequencies (close to
DC), it is a unity-gain buffer. What is the cut-off frequency that separates the
AC integrating mode from the DC buffering mode?
Solution
1
2πR2 C2
= 1.5915Hz
fc =
(D) The magnitude response of a linear system is
p
|H (f )| = H (j2πf ) H (−j2πf ).
If the circuit has an input sine wave with amplitude 100mV at frequency f = 10Hz,
predict the amplitude of the output waveform.
Solution
Based on the formula, we may solve
(100mV) |H (f )| = (100mV)
R2
R1
s
1
1 + (2πf R2 C)2
!
= 15.718mV
(E) Suppose the op amp has an input bias current Ibias = 1&micro;A and an offset voltage
of Vofs = 10mV. What is the resulting DC offset that appears at the circuit’s
output? (Remember that the offset voltage is modeled as a DC input applied to
the op amp’s non-inverting terminal).
Solution
The DC offset is found by applying the usual formulas. The capacitance C
has no effect, since bias current and offset voltage apply only at DC.
R2
VOUT = Vofs 1 +
+ Ibias R2
R1
= 30mV
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ECE 3410 – Homework 4
Solution (cont.)
Note that you can explore the behavior of this and other op amp configurations
by using simple SPICE simulation models. Here is an example:
∗ Integrator simulation
∗ Input s o u r c e
∗ ( i n p u t i s node 1 ) :
VIN 1 0 SIN ( 0 100m 1 0 )
∗ RC c o n n e c t i o n s :
R1 1 2 10 k
R2 2 3 10 k
C 2 3 10u
∗ Non−i d e a l c h a r a c t e r i s t i c s :
IB 2 0 DC 1u
VOS 4 0 DC 10m
∗ Op amp model ( j u s t a v o l t a g e −c o n t r o l l e d v o l t a g e −s o u r c e )
∗ ( output i s node 4 )
E1 3 0 2 4 1 e6
∗ SIMULATION COMMANDS:
. t r a n 1m 1
. end
After running this simulation in SPICE, you can plot V(3) to look at the
output waveform. It should be very close to what we calculated.
It is also interesting to study the integrator’s behavior with a square-wave
input:
∗ I n t e g r a t o r s i m u l a t i o n with square −wave i n p u t
∗ Input s o u r c e
∗ ( i n p u t i s node 1 ) :
VIN 1 0 PULSE( 1 0 0m −100m 0 1m 1m 0 . 0 5 0 . 1 )
∗ RC c o n n e c t i o n s :
R1 1 2 10 k
R2 2 3 10 k
C 2 3 10u
∗ Non−i d e a l c h a r a c t e r i s t i c s :
IB 2 0 DC 1u
VOS 4 0 DC 10m
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ECE 3410 – Homework 4
Solution (cont.)
∗ Op amp model ( j u s t a v o l t a g e −c o n t r o l l e d v o l t a g e −s o u r c e )
∗ ( output i s node 4 )
E1 3 0 2 4 1 e6
∗ SIMULATION COMMANDS:
. t r a n 1m 1
. end
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