“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:
(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.
