Tuesday, February 08, 2011

Two Questions on Zener Diode Voltage Regulators

“It is unwise to be too sure of one’s own wisdom. It is healthy to be reminded that the strongest might weaken and the wisest might err”

– Mahatma Gandhi

Zener diodes, as you know, are widely used as reference voltage elements in a variety of voltage regulator circuits. Today I give you a couple of multiple choice questions involving the use of zener diodes in simple voltage regulators:

(1) In the circuit shown, D1 is a silicon diode which has a voltage drop of 0.7 V while in full conduction under forward bias. The zener diode D2 has a breakdown voltage of 6.8 V. What is the current through the 450 Ω resistor?

(a) 1 mA

(b) 10 mA

(c) 20 mA

(d) 100 mA

(e) 0 mA

You can use the voltage drop across a forward biased ordinary silicon diode as a reference voltage in voltage regulator circuits. In the circuit shown, a reverse biased zener diode (of breakdown voltage 6.8 V) and a forward biased silicon diode (of voltage drop 0.7 V) in series make a reference voltage of 7.5 V. The output voltage of the circuit is thus 7.5 V.

Since the input voltage to the regulator circuit is 12 V, the voltage drop across the 450 Ω resistor is 12 V – 7.5 V = 4.5 V.

The current through the 450 Ω resistor is (4.5 V)/(450 Ω) = 0.01 A = 10 mA.

(2) The circuit shown in the adjoining figure is the simplest shunt voltage regulator using a zener diode of breakdown voltage 6 V. What is the power dissipated in the zener diode?

(a) 100 mW

(b) 120 mW

(c) 240 mW

(d) 360 mW

(e) 480 mW

Since the input voltage to the regulator circuit is 10 V and the regulated output voltage is 6 V, the voltage drop across the 40 Ω resistor is 10 V – 6 V = 4 V.

Therefore, the current through the 40 Ω resistor (current limiting resistor) is (4 V)/ 40 Ω = 0.1 A = 100 mA. This is the total current flowing into the parallel combination of the zener diode and the100 Ω load resistor.

The current through the load resistor of 100 Ω is (6 V)/(100 Ω) = 0.06 A = 60 mA.

Therefore, the current flowing through the zener diode is 100 mA – 60 mA = 40 mA.

Power dissipated in the zener diode is 6 V×40 mA = 240 mW.

Friday, February 04, 2011

Kerala Medical Entrance 2008 (KEAM 2008) Questions on Communication Systems

The following multiple choice questions on communication systems appeared in the Kerala Medical Entrance 2008 (KEAM-2008) examination question paper:

(1) A 1000 kHz carrier wave is modulated by an audio signal of frequency range 100-5000 Hz. Then the width of the channel in kHz is


(b) 20

(c) 30

(d) 40

(e) 50

Let us assume that the system uses amplitude modulation of the usual double sideband type. (It should have been mentioned in the question)

The channel width is twice the highest modulating signal frequency and is therefore equal to 2×5000 Hz = 10000 Hz = 10 kHz.

[Remember that in the standard AM sound broadcast systems the channel band width allotted to a station is 10 kHz].

(2) If the critical frequency for sky wave propagation is 12 MHz, then the maximum electron density in the ionosphere is

(a) 1.78×1012/m3

(b) 0.178×1010/m3

(c) 1.12×1012/m3

(d) 0.56×1012/m3

(e) 0.148×1012/m3

The critical frequency fc for reflection by the ionosphere is given by

fc = 9 N1/2 where N is the maximum electron number density.

Therefore, N = fc2/81 = (12×106)2 /81 = 1.78×1012/m3