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For the network shown below, the input swing of 30 volts is equal to the
• During the negative half cycle of input voltage V_{i}, diode is forward biased and acts as shortcircuit. Hence capacitor changes to V_{C} volts.
Using KVL,
5 + V_{C}  15 = 0
or, V_{C}  20 volts
Hence, V_{0} = 5 volts
• During the positive half cycle of input voltage V_{i}, diode is reverse biased and acts as opencircuit.
Applying KVL, we have
15 = 20 + \/_{0}
or V_{0} = 35 volts
The waveforms for V_{i} and V_{0} will be as shown below:
Thus, the input swing of 30 V is equal to the output swing of 30 V.
The output voltage V_{0} for the circuit shown below for the given sinusoidal input signal will be represented as
• When V_{i} is positive:
Diode is ON and acts as short circuit. Output voltage,
V_{0} = 0V
The capacitor charges to 12 volts as shown below and acts like a battery.
• When V_{i} is negative:
Diode is OFF and acts as open circuit. Output voltage,
V_{0} = 1212 = 24 Volt
This gives negatively clamped voltage and the resultant output waveform will be as shown in figure below.
Match ListI (Circuits) with ListII (Applications) and select the correct answer using the codes given below the lists:
ListI
A. Diode comparators
B. Diode dampers
C. Astable multivibrator
D. Diode clippers
ListII
1. Slicer
2. Oscillator
3. DC level to an AC signal
4. Square wave from a sine wave
• WhenV_{i} is positive:
Diode is OFF so that, V_{0} = VR
• When V_{i} is negative:
Diode is OFF so that,.
Diode is ON so that,
Hence, output waveform will be as shown below.
Assertion (A): Clipping circuit controls the shape of the waveform by clipping or removing a portion of the applied wave.
Reason (R): It needs minimum three components namely an ordinary diode, a resistor and a capacitor.
A clamper circuit requires a capacitor not a clipping circuit. Hence, reason is not a correct statement.
Consider the following statements associated with the diode rectifier circuits:
1. The maximum reverse voltage, which can be applied before the breakdown point is reached, is called the “peak inverse voltage”.
2. In a fullwave rectifier, the current in each of the diode flows for the whole input signal cycle.
3. The ratio of direct or average value of the output to the effective value of AC component present in the rectifier output is called the “ripple factor”.
4. Availability of lowcost silicon diodes has made bridge circuit arrangement more economical than centre trapped transformer arrangement despite its requirement of four diodes:
Q. Which of the above statements are correct?
• In a fullwave rectifier each diode conducts during one halfcycle. Hence, statement2 is not correct.
• Ripple factor is the ratio of rms value of AC component to the DC component in the output.
Hence, statement  3 is also not correct.
In the Zener regulator shown below, the output requirements are 5 V, 20 mA. I_{z(min) }and I_{z(max)} are 5 mA and 80 mA, then the value of limiting resistor (R) and the load resistance (R_{L}) will be respectively
Given, minimum Zener current = I_{z(min)} = 5 mA when the input voltage is minimum.
Here, the input voltage varies between 8 V to 12V.
Given, l_{L} = 20 mA and voltage across load,
V_{0} = 5 volts
∴
∴ Limiting resistance,
For the zener voltage regular circuit shown below, the output voltage is required to be maintained at 5 volt with a load current of 10 mA. The zener wattage is 400 mW and input voltage is specified as 10 V ± 2 V.
One of the possible value of series resistance R will be
Given, V_{0} = 5 V, I_{L} = 10 mA
Load resistance,
Maximum Zener current,
(Since P_{z(max)} = 400 mW)
The minimum input voltage required will be when
Under this condition,
I = I_{L} = 10 mA
Minimum input voltage,
V_{i(min)} = V_{0} +IR ...(i)
Given, V_{i} = 10 V ± 2 V
From equations (i) and (ii), we have
8 = 5 + (10 x 10^{3}) R
or,
Now, maximum input voltage,
or 12 = 5 + (90 x 10^{3}) R
or,
Hence, 77.77 Ω < R < 300 Ω
Thus, one of the possible value of out of the given options is R = 266 Ω
The Zener diode V_{Z1} in the figure shown below has the reverse saturation current of 20 mA and reverse breakdown voltage of 100 V whereas the corresponding values for diode V_{Z2} are 40 μA and 40 V.
The current through the circuit is
The Zener diode V_{Z2} is reverse biased and V_{Z1} is forward biased. As both Zener diodes \/_{Z1} and V_{Z2} are connected in series, the reverse saturation current 40 μA of V_{Z2} will flow clockwise in the circuit as 50 V reverse bias appears across the V_{Z2} diode.
Match ListI (Diode Circuits) with Listll (OutputInput Characteristics) and select the correct answer using the codes given below the lists:
ListI
A.
B.
C.
ListII
1.
2.
3.
4.
Codes:
What is the output voltage of clipper circuit shown in figure below for the sinusoidal input voltage with its peak voltage of 100 V, assuming both diodes ideal?
For, V_{i }< 25 volts,
Diode D_{1} will be OFF while diode D_{2} will be ON.
∴
For the circuit shown below, assume that the diodes are ideal. The meter reading would be
The PMMC ammeter will read average value of current. The current through the PMMC ammeter will flow only for positive half cycle of input voltage while current through it will be zero for the negative half cycle.
Hence,
∴
Two identical diodes, D_{1} and D_{2} are connected back to back as shown in figure below. The reverse saturation current I_{0} of each diode is 10^{8} A and the breakdown voltage is 50 V. The voltage drop across diodes D_{1} and D_{2} will be respectively (assume kTlq = 25 mV)
Diode D_{1} is reverse biased and diode D_{2} is forward biased, but as both the diodes form a close loop, the reverse saturation current flows through both the diodes.
or,
or, V_{2} = V_{T}ln 2 = (25 mV) x In 2
= 25 x10^{3 }x 0 .693
= 17.33 mV = 0.01733 V
∴ Voltage drop across D_{2} is
v_{2} ≈ 0.0173 volt
and voltage drop across D_{1} is
V_{1} ≈ 5  0.0173 ≈ 4.983 volts
The 6 V Zener diode shown in figure has zero zener resistance and a knee current of 5 mA.
The minimum value of R so that the voltage across it does not fall below 6 V is
The maximum load current = 80  5 = 75 mA
Hence, minimum load resistance
What are the values of V_{0} and I for the diode circuit shown below?
(Assume the drop across the diodes to be 0.7 V)
If the diode D_{1} conducts, the current I must flow as indicated in figure below.
Here,
Since, I is negative, therefore D_{1} does not conduct.
∴
A DC power supply has noload voltage of 30 V, and a fullload voltage of 25 V at a fullload current of 1 A. Its output resistance and load regulation, respectively, are
Output resistance =
Higher order active filters are used for variable rolloff rate.
In the figure shown below, the diode of Opamps are ideal. For an input V_{in} = sin (ωt) the output voltages V_{0} is
The cutin voltage for each diode in figure is V_{γ} = 0.6 V. Each diode current is 0.5 mA. The value of R_{1} R_{2} and R_{3} will be respectively
(∵ V_{2} = 4.4  5 =  0.6 volt)
In the circuit shown below, diodes D_{1} and D_{2} are ideal. The current i_{1} and i_{2} are respectively
For i_{3} > 0, V_{ab} < 5 volts and V_{D1} < 0
Hence, i = 0
and
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