Test: Diode Circuits- 2 - Electrical Engineering (EE) MCQ

• During the negative half cycle of input voltage V_i, diode is forward biased and acts as short-circuit. Hence capacitor changes to V_C volts.

Using KVL,
-5 + V_C - 15 = 0
or,  V_C - 20 volts
Hence, V₀ = 5 volts
• During the positive half cycle of input voltage V_i, diode is reverse biased and acts as open-circuit.
Applying KVL, we have
15 = -20 + \/₀
or V₀ = 35 volts
The waveforms for V_i and V₀ will be as shown below:

Thus, the input swing of 30 V is equal to the output swing of 30 V.

• When V_i is positive:
Diode is ON and acts as short circuit. Output voltage,
V₀ = 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,
V₀ = -12-12 = -24 Volt
This gives negatively clamped voltage and the resultant output waveform will be as shown in figure below.

• WhenV_i is positive:
Diode is OFF so that, V₀ = -VR
• When V_i is negative:
 Diode is OFF so that,.
 Diode is ON so that,
Hence, output waveform will be as shown below.

A clamper circuit requires a capacitor not a clipping circuit. Hence, reason is not a correct statement.

• In a full-wave rectifier each diode conducts during one half-cycle. Hence, statement-2 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.

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,
V₀ = 5 volts
∴ 
∴ Limiting resistance,

Given, V₀ = 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) = V₀ +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_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.

For, V_i < 25 volts,
Diode D₁ will be OFF while diode D₂ will be ON.
∴ 

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,


∴ 

Diode D₁ is reverse biased and diode D₂ is forward biased, but as both the diodes form a close loop, the reverse saturation current flows through both the diodes.

or, 
or, V₂ = V_Tln 2 = (25 mV) x In 2
= 25 x10^-3 x 0 .693
= 17.33 mV = 0.01733 V
∴ Voltage drop across D₂ is
v₂ ≈ 0.0173 volt
and voltage drop across D₁ is
V₁ ≈ 5 - 0.0173 ≈ 4.983 volts

The maximum load current = 80 - 5 = 75 mA
Hence, minimum load resistance

If the diode D₁ conducts, the current I must flow as indicated in figure below.
Here,


Since, I is negative, therefore D₁ does not conduct.
∴ 

Output resistance = 

Higher order active filters are used for variable roll-off rate.

(∵ V₂ = 4.4 - 5 = - 0.6 volt)

For i₃ > 0, V_ab < 5 volts and V_D1 < 0
Hence, i = 0
and