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SCHOOL OF COMPUTER AND COMMUNICATION ENGINEERINGUNIVERSITI
MALAYSIA PERLIS
Tutorial 2 - SolutionDKT214 Electronic Circuits; Semester 1
2010/2011
1. A certain op-amp has an open-loop gain of 80,000. The maximum
saturated output levels of this particular device are 12 V when the
dc supply voltages are 15 V. if a differential voltage of 0.15 mV
rms is applied between the inputs, what is the peak-to-peak value
of the output?Vout(p) = AolVin = (80,000)(0.15 mV)(1.414) = 17
VSince 12 V is the peak limit, the op-amp saturates.Vout(pp) = 24 V
with distortion due to clipping
2. Determine the output level (maximum positive or maximum
negative) for each comparator in Figure 2.1.
Figure 2.1
(a) Maximum negative(b) Maximum positive(c) Maximum negative
3. Calculate the VUTP and VLTP in Figure 2.2. Vout(max) = 10 V.
Determine also the hysteresis voltage.
Figure 2.2
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V UTP=R2
R1R210 V =18kOhm
65kOhm10V =2.77 V
V LTP=R2
R1R210 V =18kOhm
65kOhm10V =2.77V
V HYS=V UTPV LTP=2.77V2.77 V =5.54 V
4. Determine the output voltage waveform in Figure 2.3
Figure 2.3
5. Refer to Figure 2.4. Determine the following:(a) VR1 and VR2
(b) Current through Rf (c) VOUTThen, find the value of Rf necessary
to produce an output that is five times the sum of the inputs in
that figure.
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Figure 2.4
6. Find the output voltage when the input voltages shown in
Figure 2.5 are applied to the scaling
adder. What is the current through Rf?
Figure 2.5
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7. Determine the rate of change of the output voltage in
response to the step input to the integrator in Figure 2.6.
Figure 2.6
8. A triangular waveform with a peak-to-peak voltage of 2 V and
a period of 1 ms is applied to the differentiator in Figure 2.7
(a). What is the output voltage?
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9. Beginning in position 1 in Figure 2.7 (b), the switch is
thrown into position 2 and held there for 10 ms, then back to
position 1 for 10 ms, and so forth. Sketch the resulting output
waveform if its initial value is 0 V. The saturated output levels
of the op-amp are 12 V.
10. Design an integrator that will produce an output voltage
with a slope of 100 mV/s when the input voltage is a constant 5 V.
Specify the input frequency of a square wave with an amplitude of 5
V that will result in a 5 V peak-to-peak triangular wave
output.
Figure 2.7
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11. A two-input summer is to produce an output voltage V sin2
tvo = when one of the input voltages is V sin5.051 tvi = . Using an
ideal op-amp and a number of resistor, design the summer. All
resistors must be in the range 20 k to 100 k.
The input voltage is comprised of a dc component of 5 V and
sinusoidal component 0.5sint V. If we have a second input for the
dc voltage of 5 V, we can eliminate the dc component in the output
voltage if we make the gain of input 1 equal to the gain of input
2. This is shown in the following circuit.
With the amplitudes of the sinusoidal input and output voltages
of 0.5 V and 2 V respectively, the required gain is;
45.0
2
1
===RRA Fv
Since we are required to use resistors in the range 20 k to 100
k, we select R1 = R2 = 25 k. Then, for the gain of 4, RF must be
100 k.
+
R 1
R 2
R F
v I 1
v I 2
v O
v I 1 = 5 0.5sin t Vv I 2 = 5 V
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+
R 1
R 2
R F
v O
v I 1= 5 0.5sin t V
v I 2 = 5 V
25 k
25 k
100 k
NOTE: We can also use == k 2021 RR and the corresponding = k
80FR which are also in the range of 20 k to 100 k.
12. Design the circuit in Figure 2.8(a) to produce an output
voltage vO in Figure 2.8(c) when the input voltage vI is as shown
in Figure 2.8(b). Assume the op-amp is ideal and use C = 220
pF.
+
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R
C
v I
v OFigure 2.8(a)
t ( s)
t ( s)
v I (V)
5 10 15 2000
0
5
5
8.8
8.8
v O (V)
Figure 2.8(b)
Figure 2.8(c)
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For a differentiator;
( ) ( )dt
tdvCRtv IO =( )
( )[ ]Cdttdvtv
RI
O
/=
50 t( )
s/102
sV 2
6=
= /dt
tdv I
and ( ) V 8.8=tvO
Hence;
1266 102201028.8
1028.8
=
=
CR
= k 20R
13. For the circuit in Figure 2.9(a), C = 0.1 F and R = 10 k.
Both the output and input voltages are zero at t = 0. An input
voltage vI shown in Figure 2.9(b) is applied to the circuit. Sketch
the resulting output waveform vO(t).
Time constant, ms 1101.010 64 === RC
( ) 10for 1 = ttv IThus;
( ) ( ) tdtdttvCR
Vtvtt
IOO3
0
3
010'110''1 === (Since vO = 0 at t = 0)
Hence at V 1ms 1 == Ovt
Also;( ) 21for 1 = ttv I and ms 1at V 1 == tvO
Thus;
t (ms)
v I (V)
-1
1
01 2 3 4
Figure 2.9(b)
R
C
v I
v O
+
A o Figure 2.9(a)
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( ) ( )
[ ][ ]33
103
10
3
10
10101
101'1101
''1
33
3
+=
+=+=
=
t
tdt
dttvCR
Vtv
tt
t
IOO
Hence at ( ) 010-10210 12ms -3-33 =+== Ovt
t (ms)
v O (V)
-1
1
01 2 3 4
