Question

(a) Explain the applications of PN junction diode (6)

(b) A full-wave, 1-phase rectifier employs a double diode valve, the internal resistance of each element of which may be assumed constant at 500W. The transformer r.m.s secondry voltage from teh centre-tap to each anode is 300V and the load has a resistance of 2000W. Evaluate (10)

(i) Mean load current.

(ii) r.m.s value of load current

(iii) The d.c. output power

(iv) The input power to the anode circuit

(v) The rectification efficiency

Answer 1

Part (a)

Applications of a PN junction diode:

Rectification: converting a.c. to d.c. in power supplies (half-wave, full-wave, bridge rectifiers).
Demodulation/detection: extracting the signal from a modulated carrier in radio receivers.
Clipping and clamping: limiting voltage levels and shifting d.c. levels in waveforms.
Voltage regulation: Zener diodes maintain a constant voltage.
Protection: freewheeling diodes across inductive loads, and reverse-polarity protection.
Logic and switching: diodes in logic gates and as switches in electronic circuits.
Light emission: LEDs (light-emitting diodes) for indication and lighting.
Photodiodes: light detection in sensors.

Part (b)

Full-wave, 1-phase rectifier with a double diode valve, internal resistance of each element 500 ohm, transformer r.m.s. secondary voltage from centre-tap to each anode 300 V, load resistance 2000 ohm.

Peak voltage per half Vm = 300 x root 2 = 424.3 V.
Total resistance per half cycle = internal resistance + load = 500 + 2000 = 2500 ohm.
Peak current I_peak = Vm/2500 = 424.3/2500 = 0.1697 A.

(i) Mean load current: for a full-wave rectifier, I_dc = (2 Vm)/(pi x 2500) = (2 x 424.3)/(3.1416 x 2500) = 848.6/7854 = 0.108 A.

(ii) r.m.s. value of load current: I_rms = I_peak/root 2 = 0.1697/1.414 = 0.120 A.

(iii) D.C. output power: P_dc = I_dc^2 x R_load = 0.108^2 x 2000 = 0.01166 x 2000 = 23.3 W. (Also = I_dc x V_dc, where V_dc = I_dc x 2000 = 216 V.)

(iv) Input power to the anode circuit: P_in = I_rms^2 x (total resistance) = 0.120^2 x 2500 = 0.0144 x 2500 = 36.0 W.

(v) Rectification efficiency: eta = P_dc/P_in = 23.3/36.0 = 0.647 = 64.7%.

So mean load current = 0.108 A, r.m.s. current = 0.120 A, d.c. output power = 23.3 W, input power = 36.0 W, rectification efficiency = 64.7%.

✓ Verified Model Answer (Text Solution)

Applications of a PN junction diode:

Full-wave, 1-phase rectifier with a double diode valve, internal resistance of each element 500 ohm, transformer r.m.s. secondary voltage from centre-tap to each anode 300 V, load resistance 2000 ohm.