B. Tech, Semester-III
Sessional Examination, September -2014
EC301/2EC202 Electronics Devices and Circuits-I
Solution
Q.1
(a)
(b)
Answer the following.
[18]
Apply Poisson’s equation for open circuited PN junction diode and (3)
sketch graph of electric field intensity and built in potential variations
with distance.
ANS:
Match Column A with Column B. Also Justify your answer.
Column A
Column B
I. PN Junction Diode
I. Select a part of transmitted
Signal
II. Peak Detector Circuit
II. As a voltage variable
capacitor
III. Clamper Circuit
III.To change level of
composite
video signal in TV Receiver
IV. Clipper Circuit
IV. For signal amplification
V. Minimum and Maximum
Temperature measurement
ANS:
I=II
II=V
III=III
IV=I
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(8)
Justification:
(i)PN Junction diode can be used as a voltage variable capacitor
(ii) peak detector circuit used as a maximum and minimum temperature
measurement:
As shown in figure, for initial zero voltage across capacitor, as input is
increase, diode is in ON state and capacitor is charge.
When input is decrease, there is no path for capacitor to discharge so
capacitor voltage remain as it is and anode voltage is decrease, so diode
is OFF. Again if input is increase, diode is ON and capacitor charged
again. So maximum voltage across capacitor is equal to peak of input
voltage,that used to measure maximum temperature without
considering intermediate value.
If diode polarity is changed, it is used to detect maximum negative
voltage.
(iii) Clamper Circuit used to to change level of composite video signal
in TV Receiver.
for TV receiver, by changing DC level, the brightness and contrast level is changed.
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(iv) clipper circuit is used to select part of transmitted signal.
(c)
In a communication system, signal is transmitted from transmitting (3)
antenna and received by receiver. The noise components are added
during this process. The received signal as shown in Figure 1. It is
required to suppress the noise in the received signal for original
reproduction of signal. Sketch circuit diagram and expected output
waveform.
Noise
Figure 1
ANS: Series noise clipper(two level)
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If D1 is ON(upper diode), V0= Vin-V1
If D2 is ON(lower diode) V0=Vin+V2
(d)
Analyze circuit as shown in Figure 2, identify its function. Evaluate (4)
voltage V1 and V2 with respect to point A and V3 and V4 with respect
to point B. Take Vin=10sinwt.
Figure 2
ANS:
C1-D1 form clamper, give v1=10V
D2,C4 form peak detector give V4=20V
C2,D3 GIVE V2=20V,
C3D4 GIVE V3=20v
VOLTAGE MULTIPLIER
Explanation:
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For negative half cycle of input, D1 is ON, and voltage across C1 is
10V(same as input).
During positive half cycle, D1 is OFF and voltage across D1 is 20V.
That 20V keep D2 in ON state and capacitor voltage C4 is 20V. similar
way, V2 is 20V and V3 is 20V.
V1: 10V w.r.t to A V3: 40V w.r.t B
V2: 30V w.r.t to A V4: 20V w.r.t B
Q.2
(a)
Answer the following.
[16]
Analyze operation of Figure 3 and identify application of circuit. Sketch (6)
the output waveform Vo if the three signals V1, V2 and V3 are impressed
on the input terminals. Assume that diodes are ideal.
Figure 3
ANS: for common cathode configuration, highest positive voltage decide
which diode conduct first. So if V1>V2>V3, only D1 is ON and output
follow V1. If V2>V1>V3, output follow V2 and same for V3. During
negative half cycle, all diodes are off and output is zero.
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USED FOR DIGITAL OR GATE
(b)
Three phase half wave rectifier.
A diode in series with a resistor RL is forward biased by a voltage Vx. (6)
After a steady state is reached, the input changes to –Vx. Sketch circuit
diagram and analyze operation of circuit. Sketch the voltage across
diode and current through diode as a function of time. Describe
quantitatively the shape of this curves.
ANS: THEORY OF DIODE SWITCHING CHARACTERISTICS
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(c ) Justify Given Statement:
“A ripple factor for half wave rectifier is double than full wave rectifier.”
ANS:
Ripple factor,
𝐼𝑟𝑚𝑠 2−1
)
𝐼𝑑𝑐
For half wave rectifier, Irms=Im/2, Idc=Im/pi
For full wave rectifier, Irms=Im/√2, Idc=2Im/pi
FOR HALF WAVE rectifier R=1.21
FO FULL WAVE rectifier R=0.48
Answer the following.
[16]
In a clamper circuit shown in Figure 5, Rs=Rf = 40Ω, R=10kΩ, and (8)
C=3µF. An input signal has frequency of 4 kHz is applied at t=0. Give
𝑟 = √(
Q.3
(a)
(4)
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mathematical analysis at each point of discontinuity in input waveform
and draw the first three cycles of the output waveform.
Figure 5
ANS: Indicate all calculations
(b)
For the clipping circuit shown in Figure 6, make a plot of Vo versus Vi (8)
for the range of Vi from 0-100V. Indicate all slopes and voltage levels.
Indicate for each region, the diodes which conduct. Sketch output
waveform also.
Figure 6
ANS: Vin < 20 D1 ON D2=OFF…Vo=vin/3
Vin>30,D1=ON, D2=ON
Vo=20(battery voltage)/2= 10V
[Indicate all calculations]
10V
SLOP=0
VIN/3
D1 ON
D1 ON D2 ON
D2 OFF
30V
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