Power electronics and drives miscellaneous Easy Questions and Answers | Page - 4

Power electronics and drives miscellaneous

The fully controlled thyristor converter in the given figure is fed from a single-phase source. When the firing angle is 0°, the dc output voltage of the converter is 300 V. What will be the output voltage for a firing angle of 60°, assuming continuous conduction?

1. 150 V
1. 210 V
1. 300 V
1. 100 π V

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For Fully Controlled Converter (FCC)
Average output voltage, V₀ = 2 √2 V cos α (V = rms)
π

For α = 0, V₀ = 300
∴ 300 = 2 √2 .V
π

⇒ V = 300 π
2 √2

A α = 60 , V₀ = 2 √2 300π .cos 60
π 2 √2

= 300 × 1 = 150 V
2

Correct Option: A

For Fully Controlled Converter (FCC)
Average output voltage, V₀ = 2 √2 V cos α (V = rms)
π

For α = 0, V₀ = 300
∴ 300 = 2 √2 .V
π

⇒ V = 300 π
2 √2

A α = 60 , V₀ = 2 √2 300π .cos 60
π 2 √2

= 300 × 1 = 150 V
2

An SCR is considered to be a semi-controlled device because

1. it can be turned OFF but not ON with a gate pulse
1. it conducts only during one half-cycle of an alternating current wave
1. it can be turned ON but not OFF with a gate pulse
1. it can be turned ON only during one half-cycle of an alternating voltage wave

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NA

Correct Option: C

NA

The following circuit has a source voltage V_s as shown in the graph. The current through the circuit is also shown.

The element connected between a and b could be

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When a forward biased diode is reverse biased, initially it conducts due to the flow of stored charges. After some ti me it goes int o nonconducting state.
Alternately
For 0 < t < 100 ms, V_s = 0,
i_s = 0; diode not conduct
For 100 < t < 200 ms, V_s = 10V,
i = 10 = 1 mA ; diode conducts
10K

For 200 < t < 300 ms, V_s = – 10V, i = 0; diode reverse bias
current is zero after small reverse recovery time.
For 300 < t < 400 ms, V_s = 0;
i = 0; diode not conduct

Correct Option: A

When a forward biased diode is reverse biased, initially it conducts due to the flow of stored charges. After some ti me it goes int o nonconducting state.
Alternately
For 0 < t < 100 ms, V_s = 0,
i_s = 0; diode not conduct
For 100 < t < 200 ms, V_s = 10V,
i = 10 = 1 mA ; diode conducts
10K

For 200 < t < 300 ms, V_s = – 10V, i = 0; diode reverse bias
current is zero after small reverse recovery time.
For 300 < t < 400 ms, V_s = 0;
i = 0; diode not conduct

A three phase fully controlled bridge converter is feeding a load drawing a constant and ripple free load current of 10 A at a firing angle of 30°. The approximate Total Harmonic Distortion (%THD) and the rms value of fundamental component of the input current will respectively be

1. 31% and 6.8 A
1. 31% and 7.8 A
1. 66% and 6.8 A
1. 66% and 7.8 A

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Rms value of load current,
I_s = I_o √2 / 3 = 10 √2 / 3 = 8.165 A
Supply current

Rms value of fundamental current,
i_s1 = 4π₀ sin π = 4 × 10 √3 = 7.8 A
√2π 3 √2π 2

Then, THD (Total harmonic distortion)

= 0.31 = 31%

Correct Option: B

Rms value of load current,
I_s = I_o √2 / 3 = 10 √2 / 3 = 8.165 A
Supply current

Rms value of fundamental current,
i_s1 = 4π₀ sin π = 4 × 10 √3 = 7.8 A
√2π 3 √2π 2

Then, THD (Total harmonic distortion)

= 0.31 = 31%

The power electronic converter shown in the figure given below has a single-pole double-throw switch. The pole P of the switch is connected alternately to throws A and B. The converter shown is a

1. step-down chopper (buck converter)
1. half-wave rectifier
1. step-up chopper (boost converter)
1. full-wave rectifier

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Let Duty cycle be ‘D’.
When at A : t = 0
V_in = L di + V_out , (i_L)_in = (V_in - V₀) DT
dt L

When at B : At t = DT
V_out = -V_L = -Ldi_L
dt

(i_L)_off = -V_out (1 - D)T
L

(V_in – V_out) DT + (– V_out) (1 – D)T = 0
⇒ (V_in – V_out) D = V_out(1 – D)
⇒ V_out = DV_in and D < 1
So, step down chopper.

Correct Option: A

Let Duty cycle be ‘D’.
When at A : t = 0
V_in = L di + V_out , (i_L)_in = (V_in - V₀) DT
dt L

When at B : At t = DT
V_out = -V_L = -Ldi_L
dt

(i_L)_off = -V_out (1 - D)T
L

(V_in – V_out) DT + (– V_out) (1 – D)T = 0
⇒ (V_in – V_out) D = V_out(1 – D)
⇒ V_out = DV_in and D < 1
So, step down chopper.