Question

(a) What are the characteristics of PN junction diode? Point out its specifications. Also point out the significance of dynamic and static resistances. (6)

(b) A 72 KVA transformer supplies a heating and lighting load of 12 KW at unity power factor and a motor load of 70 kVA at 0.766 (lagging) power factor: Calculate the minimum rating of the power-factor improvement capacitors which must be connected in the circuit the ensure that the transformer does not become overloaded. (10)

Answer 1

Part (a)

Characteristics of a PN junction diode, specifications, and static/dynamic resistance:

Characteristics: a PN junction diode conducts current easily in the forward direction (when the anode is positive) once the forward voltage exceeds the threshold (about 0.7 V for silicon, 0.3 V for germanium). In the reverse direction it blocks current until the reverse breakdown voltage is reached, beyond which it conducts heavily (avalanche/Zener breakdown). The forward current rises steeply with voltage; the reverse current is very small (leakage current) until breakdown.
Specifications: maximum forward current (If), maximum reverse voltage / peak inverse voltage (PIV), forward voltage drop (Vf), reverse leakage current, power dissipation rating, reverse breakdown voltage, and switching speed (recovery time).
Static resistance: the ratio of the voltage to the current at a point on the characteristic, R = V/I. It is the resistance of the diode at a particular operating point.
Dynamic (a.c.) resistance: the ratio of a small change in voltage to the corresponding change in current, r = dV/dI. It is the slope of the characteristic at the operating point and is small in the forward conducting region. It is important in small-signal analysis because it determines the a.c. behaviour of the diode.

Part (b)

72 kVA transformer supplies a heating and lighting load of 12 kW at unity p.f. and a motor load of 70 kVA at 0.766 p.f. lagging. Calculate the minimum capacitor rating so the transformer is not overloaded.

Motor: kW = 70 x 0.766 = 53.62 kW. sin phi = sqrt(1 - 0.766^2) = sqrt(1 - 0.5868) = sqrt(0.4132) = 0.6428. Motor kVAr = 70 x 0.6428 = 45.0 kVAr (lagging).
Total kW = 12 + 53.62 = 65.62 kW. Total kVAr = 45.0 kVAr.
Present total kVA = sqrt(65.62^2 + 45^2) = sqrt(4306 + 2025) = sqrt(6331) = 79.6 kVA. This exceeds 72 kVA, so the transformer is overloaded.
To avoid overloading, the total kVA must be reduced to 72 kVA. The kW (65.62 kW) is fixed, so the allowable kVAr is:

kVAr = sqrt(72^2 - 65.62^2) = sqrt(5184 - 4306) = sqrt(878) = 29.6 kVAr.

Capacitor kVAr required = present kVAr - allowable kVAr = 45.0 - 29.6 = 15.4 kVAr.

So the minimum rating of the power-factor improvement capacitors is about 15.4 kVAr (say 16 kVAr).

✓ Verified Model Answer (Text Solution)

Characteristics of a PN junction diode, specifications, and static/dynamic resistance:

72 kVA transformer supplies a heating and lighting load of 12 kW at unity p.f. and a motor load of 70 kVA at 0.766 p.f. lagging. Calculate the minimum capacitor rating so the transformer is not overloaded.