Problem 8. Find vo in terms of v1 and v2 . (Assume OpAmp is ideal).
2R
V1
R
The OpAmp to the left is configured
as an inverting amplifier. Therefore,
v10 = −
4R
V’1
−
+
V2
R
−
+
2R
+
Vo
−
2R
v1 = −2v1
R
The OpAmp to the right is configured as an inverting summer. Thus,
vo = −
4R
4R 0
v1 −
v2 = −4v10 − 2v2 = −4(−2v1 ) − 2v2 = 8v1 − 2v2
R
2R
Problem 9. Consider the amplifier circuit below with vi = 0.5 cos(ωt). a) what is the
amplifier gain in dB? b) For what range of frequencies, does this amplifier behave as a linear
amplifier? (Assume a unity-gain bandwith of 106 Hz, ISC = 100 mA, S0 = 1 V/µs and
OpAmps are powered by ±15 V power supplies.)
5kΩ
part a: This is an inverting amplifier:
1kΩ
5k
Vo
=
=5
A=
Vi
1k
AdB = 20 log10 A = 20 log10 (5) = 14 dB
vi
−
+
+
vo
−
Two limits impact the bandwith of this amplifier:
Bandwidth of OpAmp itself:
A0 f0 = fu = (1 + A)f
→
f=
fu
106
=
= 1.66 × 105 = 166 kHz
1+A
6
Slew Rate:
AVi ω ≤ S0
→
5 × 0.5 × ω ≤ 1 V/µs = 106 V/s
106
= 4 × 105 rad/s
2.5
ω
= 64 kHz
f=
2π
ω≤
The slew rate is the most restrictive, so this OpAmp circuit behaves linearly in the frequency
range 0 ≤ f ≤ 64 kHz.
ECE65 Lecture Notes (F. Najmabadi), Spring 2007
72
Problem 10. Consider the inverting amplifier below with RL = 500 Ω. The input signal
is vi = 1 cos(106 t) V. We want the amplifier to behave linearly (i.e., output signal to be
sinusoidal) and Vo /Vi = −K. What is the maximum value of K we can choose? (Assume
a unity-gain bandwith of 106 Hz, ISC = 10 mA, S0 = 4 V/µs and OpAmps are powered by
±15 V power supplies.)
KR
ω = 106 rad/s,
Vi = 1 V,
f = 106 /(2π) Hz
R
|Vo | = KVi = K V
+
Vi
→
RL
-
1) Voltage supply limit (Saturation):
vs− < vo < vs+
+
Vo
K ≤ 15
2) Frequency response limit:
A0 f0 = fu = (1 + A)f = 107
→
(1 + K) ×
106
≤ 107
2π
→
K ≤ 20π − 1 = 62
3) Maximum output current limit:
IL =
Vo
≤ ISC
RL
→
K
≤ 10 × 10−3
500
→
K≤5
4) Slew Rate:
dvo
= AVim ω ≤ S0
dt
→
K × 1 × 106 ≤ 4 × 106
→
K≤4
Slew rate limit is most restrictive with K ≤ 4.
ECE65 Lecture Notes (F. Najmabadi), Spring 2007
73
Problem 11. The input and output signals to an OpAmp circuit are given below. A)
Draw the circuit (assume tr = 0 for this part only). B) If the slew-rate of the OpAmp is
1 V/µs, calculate tr . C) Suppose that we consider a signal as a square wave as long as
tr /(T /2) < 0.1. What is the frequency range that this amplifier produce a square wave?
(Assume a unity-gain bandwith of 106 Hz, ISC = 100 mA, S0 = 1 V/µs and OpAmps are
powered by ±15 V power supplies.)
Vi
Part A: For tr = 0, vo = 5vi . So, the
OpAmp circuit is a non-inverting amplifier
with a gain of 5. Since:
A=1+
R2
=5
R1
→
1V
T
-1 V
R2 = 4R1
5V
Vo
Part B: The departure of the output from
a square wave is due to the slew rate:
S0 = 1 V/µs =
tr =
dv o
dt 10 V
= 10µs
1 V/µs
=
t(s)
−5 − 5 tr
10
=
tr
t(s)
-5 V
tr
Vi
Ii
Part C:
2tr
tr
< 0.1 → T >
T /2
0.1
1
0.05
0.05
f= <
=
= 5 kHz
T
tr
10 × 10−6
ECE65 Lecture Notes (F. Najmabadi), Spring 2007
+
−
Vo
4R
R
74
Problem 12. Consider the amplifier circuit below. The input and output waveforms are
shown. Explain why the output is not a triangular waveform. (Assume a unity-gain bandwith
of 106 Hz, ISC = 100 mA, S0 = 1 V/µs and OpAmps are powered by ±15 V power supplies.)
Vi
The circuit is a non-inverting amplifier with a gain
of 1 + R2 /R1 = 11. Input is a triangular signal with a peak-to-peak amplitude of 1 V. If the
OpAmp was ideal, we would expect that output
to be a triangular signal with a peak-to-peak amplitude of 11 = 11 V.
The output signal, however, is clipped at 5 V.
Output can be clipped by either 1) amplifier saturation and/or 2) maximum output current limit.
We need to examine both limits:
Saturation: As the OpAmp is powered by
±15 V supplies, the saturation voltage for
the OpAmp should be close to ±15 V. As
the output signal is clipped at 5 V, clipping
is NOT due to the OpAmp saturation.
+
−
Vo
50 Ω
100kΩ
10k Ω
vi (V)
1
100
200
300
t ( µ s)
vo(V)
5
t ( µ s)
Maximum output current limit: As the load resistor (50 Ω) is much smaller than the feedback
resistor (100 kΩ), almost all of OpAmp output current will flow in the load. Then:
iL ≤ 100 mA
→
v0 = 50iL ≤ 50 × 0.1 = 5 V
Therefore, maximum output current limit will force the output signal to be clipped at 5 V.
This explains the shape of output signal.
ECE65 Lecture Notes (F. Najmabadi), Spring 2007
75
Problem 13. Design a non-inverting amplifier with a gain of 20 dB to drive a 10 kΩ load.
What is the bandwidth of the circuit you have designed? What is its input impedance?
Vi
Prototype of a non-inverting amplifier is shown with
A = 1 + R2 /R1 . 20 dB gain translates to:
20 dB = 20 log(A)
A= 1+
R2
R1
= 10
→
→
log(A) = 1
R2
=9
R1
→
Ii
+
-
Vo
R
A = 10
R
2
1
As the output impedance of a non-inverting amplifier is “zero,” we can choose R1 and R2
arbitrarily. A reasonable choice is:
R1 = 10 kΩ,
R2 = 91 kΩ
Bandwidth: A × fc = fu = 106 . Thus, fc = 105 Hz= 100 kHz.
Input impedance of the amplifier is infinity as Ii = 0.
Problem 14. Design an active wide-band filter with fl = 20 Hz, fu = 1 kHz, a gain of 4
and an input impedance larger than 5 kΩ.
C2
The prototype of this circuit is shown with:
ωc1
= 20 × 10−3 1
ωc2
1
1
ωu ≈ ωc2 =
ωl ≈ ωc1 =
R2 C2
R1 C1
R2
Zi |min = R1
K≈
R1
R2
vi
R1
C1
−
+
vo
Then, from design condition of Zi ≥ 5 kΩ, R1 > 5 kΩ. To make capacitors small (and also
to make input impedance large) choose R1 and R2 to be large with K = R2 /R1 = 4. A
reasonable set is R1 = 100 kΩ and R2 = 390 kΩ (commercial values). Then:
1
→ C2 = 4 × 10−10 F
R2 C2
1
ωl ≈ 2π20 =
→ C1 = 8 × 10−8 F
R1 C1
ωu ≈ 2π103 =
Commercial values are: C1 = 82 nF and C2 = 390 pF.
We need to consider the impact of the bandwidth of OpAmp chip. A × fc = fu = 106 leads
to fc = 106 /(‘ + 4) = 200 fu = 1 kHz. So the circuit should work properly.
ECE65 Lecture Notes (F. Najmabadi), Spring 2007
76
Problem 15. Design a filter with the transfer function of H(jω) =
−3
and a
1 + jω/5000
minimum input impedance of 50 kΩ.
The transfer function is in the general form for first-order low-pass filters:
H(jω) =
K
1 + jω/ωc
with K = −3 and ωc = 5000 rad/s. As |K| > 1, we need to use an active filter. The
prototype of the circuit is shown below with
C2
H(jω) = −
R2 /R1
1 + jω/ωc
ωc = R 2 C 2
Vi
R1
−
R2
Vo
+
The input impedance of this filter is Zi |min = R1 .
Comparing the transfer function of this prototype
circuit with the desired one, we get:
R2
=3
R1
1
= 5, 000
R2 C2
Zi |min = R1 ≥ 50 kΩ
ωc =
Choosing commercial value of R1 = 51 kΩ, we get:
R2 = 3R1 = 153 kΩ → 150kΩ (Commercial)
1
1
=
= 1.33 × 10−9
C2 =
3
5, 000R2
5 × 10 × 150 × 103
→
1.5 nF (Commercial)
We need to consider the impact of the bandwidth of OpAmp chip. A × fc = fu = 106 leads
to fc = 106 /(1 + 3) = 250 fu = 5/(2π) = 0.8 kHz. So the circuit should work properly.
ECE65 Lecture Notes (F. Najmabadi), Spring 2007
77
Problem 16. We have an oscillator that makes a square wave with a frequency of 1 kHz,
a peak-to-peak amplitude of 10 V, and a DC-offset of 0 V. Design a circuit that coverts this
square wave to a triangular wave with a peak-to-peak amplitude of 10 V, and a DC-offset of
0 V. What would be the frequency of the triangular wave?
Since a triangular wave is the integral of a square wave, we need to use an integrator circuit.
The prototype, the input signal and the desired output signal are shown below. The output
will have the same frequency as input (1 kHz) and same period (1 ms).
R2
vi
R1
v
C2
−
+
v
i
5
5
vo
0.5
1
o
0.5
t (ms)
−5
1
t (ms)
−5
Because the input signal is symmetric (0 DC offset), we can consider only the half period
0 < t < T /2 = 0.5 ms. In this range, vi = 5 V and we want vo = +5 − 2 × 104 t (found from
vo (0) = +5 and vo (t = 0.5 ms) = −5 V). The output of the integrator is given by:
1
vo (t) − vo (0) = −
R1 C2
Z
t
0
vi (t0 )dt0
1
5 − 2 × 10 t − 5 = −
R1 C2
1
−2 × 104 t = −
t →
R1 C2
4
Z
t
5dt0
0
R1 C2 = 2.5 × 10−4
Resistor R2 is needed to discharge the capacitor in long time scale (so that small DC input
does not add up and saturate the OpAmp). This capacitor also ensure that the output has
no DC off-set. Setting:
τ = R2 C2 = 100T = 100 × 10−3 = 0.1
We have two equations in three unknowns R1 , R2 , and C2 and we can choose value of one
arbitrarily. One has to be careful as if we divide the two equations, we get:
R2 C2
R2
0.1
=
=
= 400
R1 C2
R1
2.5 × 10−4
Reasonable choices to keep resistors between 100 and 1 MΩ are R1 = 1 kΩ and R2 = 400 kΩ:
R1 C2 = 2.5 × 10−4
→
C2 =
2.5 × 10−4
= 2.5 × 10−7 = 250 nF
1000
Thus, reasonable commercial choices are R1 = 1 kΩ, R2 = 390 kΩ, and C2 = 0.24 µF.
ECE65 Lecture Notes (F. Najmabadi), Spring 2007
78