THEORY EXAMINATION (SEM–VI) 2016-17 POWER ELECTRONICS
B.Tech
Electronics Engineering
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POWER ELECTRONICS (EEE602)
Section-wise Solved Answers & Notes
SECTION – A (10 × 2 = 20 Marks)
Very short answers (2 marks each)
(a) Latching current & holding current
• Latching current: Minimum anode current required to keep SCR ON immediately after triggering. • Holding current: Minimum anode current below which SCR turns OFF.
(b) Freewheeling diode
A diode connected across inductive load to provide path for load current when main switch is OFF, reducing voltage spikes.
(c) Positive & negative group of thyristors
• Positive group: Conduct during positive half cycle • Negative group: Conduct during negative half cycle Used in converters and cycloconverters.
(d) Why anti-parallel diodes are connected in inverter circuits
To allow reverse current flow and protect thyristors from reverse voltage during inductive load operation.
(e) Merits & demerits of cycloconverter
Merits: • Direct AC-AC conversion • Variable output frequency
Demerits: • Low output frequency • High harmonic distortion
(f) Firing angle
The angle (α) at which a thyristor is triggered with respect to zero crossing of supply voltage.
(g) Difference between VSI & CSI
VSI
CSI
Voltage source
Current source
Capacitor input
Inductor input
Output voltage controlled
Output current controlled
(h) Applications of power electronics
• Motor drives • SMPS • UPS • HVDC transmission • Electric vehicles
(i) Principle of phase control
Output voltage is controlled by varying firing angle of thyristor.
(j) Step-down chopper principle
Output voltage is controlled by varying duty cycle of switching device.
SECTION – B (Any 5 × 10 = 50 Marks)
(a) Dual converter
Dual converter consists of two converters connected back-to-back. In circulating current mode, both converters operate simultaneously with a reactor limiting circulating current.
(b) Numerical – Single phase full converter
Given: V = 240 V, I = 15 A, α = 45°
(i) RMS supply current
Is(rms)=I=15AI_{s(rms)} = I = 15 AIs(rms)=I=15A
(ii) Fundamental input current
I1=22πI=13.5AI_1 = \frac{2\sqrt{2}}{\pi} I = 13.5 AI1=π22I=13.5A
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