“Everything should be made as simple as possible, but not simpler”
– Albert Einstein
The following multiple choice question on zener diode was included in Kerala Engineering Entrance (KEAM) 2011 question paper. If you identify that R1 is the current limiting resistor and R2 is the load resistor in the simple shunt voltage regulator making use of a zener diode, you won’t have any confusion. Here is the question:
In the circuit given the current through the zener diode is
(a) 10 mA
(b) 6.67 mA
(c) 5 mA
(d) 3.33 mA
(e) 0 mA
Since the zener breakdown voltage (across the zener diode) is 10 V, the voltage across the resistor R2 is 10 V. The current through R2 is 10 V/1500 Ω = 0.00667 A = 6.67 mA.
The voltage drop across R1 is 15 V – 10 V = 5 V.
Therefore, the current through R1 is 5 V/500Ω = 0.01 A = 10 mA
The above current gets divided between the resistor R2 and the zener diode. Therefore, the current through the zener diode is10 mA– 6.67 mA = 3.33 mA.
By clicking on the label ‘zener diode’ below this post you can access similar questions on zener diodes posted on this site.
“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, thevoltage 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.
Questions from electronics, especially those involving the use of diodes are generally simple at the level expected of you. Here are a few questions on diodes:
(1) Currents flowing in each of the following circuits A and B respectively are
(1) 1A, 2 A
(2) 2 A, 1 A
(3) 4 A, 2 A
(4) 2 A, 4 A
This question appeared in EAMCET 2009 (Engineering) question paper.
In circuit A both diodes are forward biased and hence the circuit reduces to two 4 Ω resistors connected across the 8 V battery. Since the parallel combined value of the two resistors is 2 Ω, the current delivered by the battery is 8 V/2 Ω = 4 A.
In circuit B one diode is forward biased and the other diode is reverse biased and hence the circuit reduces to just one 4 Ω resistor connected across the 8 V battery. The current delivered by the battery is therefore 8 V/4 Ω = 2 A. The correct option is (3).
The following questions [No. (2) and (No. (3)] were included in AIEEE 2006 question paper:
(2) In the following, which one of the diodes is reverse biased?
You should note that all potentials are with respect to the ground. Therefore the diode in circuit (1) is reverse biased.
[In circuit (2) the anode of the diode is at a higher positive potential compared to its cathode and hence it is forward biased. In circuit (3) the cathode of the diode is at a higher negative potential and hence it is forward biased. In circuit (4) the cathode of the diode is at a negative potential and hence it is forward biased].
(3)The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
(1) 1.33 A
(2) 1.71 A
(3) 2.00 A
(4) 2.31 A
Since the diode D1 is reverse biased, no current will flow through D1 and the 3 Ω resistor. The current delivered by the battery is limited by the 4 Ω and the 2 Ω resistors only and is equal to 12 V/(4+2)Ω = 2 A.
The following question is meant for checking your grasp of the behaviour of semiconductor diodes:
(4) In the circuit shown the diodes used are silicon rectifier diodes which require a forward bias of 0.7 volt for appreciable conduction. Their leakage current is negligible. The internal resistance of the battery is insignificant. The potential difference between the terminals A and B is very nearly equal to
(a) 6 V
(b) 3 V
(c) 5.3 V
(d) 0.7 V
(e) 0 V
The upper diode is reverse biased and can be ignored. The lower diode which is connected across the terminals A and B is forward biased and hence keeps the voltage across A and B at 0.7 V [Option (d)].
[You can use the voltage drop across a forward biased diode as a small reference voltage in electronic circuits just as you use the relatively larger breakdown voltages of reverse biased zener diodes].
In electronics, occasionally you will get questions meant for high lighting the principle of operation and use of Zener diodes. Questions in this section are usually simple. Consider the following question which appeared in KEAM (Medical) 2008 question paper:
In the given circuit, the current through the resistor 2 kΩ is
(a) 2 mA
(b) 4 mA
(c) 6 mA
(d) 1 mA
(e) 10 mA
This is a very simple question. The voltage across the reverse biased Zener diode is its breakdown voltage, which is indicated in the figure as 12 V. The load resistor 2 kΩ is connected across the Zener diode and hence the current through it is 12 V/2 kΩ = 6 mA.
Now consider the following question pertaining to the points you have to remember in using a Zener diode as a simple shunt voltage regulator (of the type in the above question):
In the simple shunt voltage regulator circuit shown in the adjoining figure, the unregulated supply voltage varies between 8 V and 10 V. If the maximum reverse current that the Zener diode can safely handle is 100 mA and the Zener diode should draw a minimum current of 5mA (for reliable operation of the voltage regulator), what is the expected maximum current drawn by the load resistor RL?
(a) 95 mA
(b) 105 mA
(c) 200 mA
(d) 10 mA
(e) 110 mA
Since the maximum current allowed through the Zener diode is 100 mA, the maximum current drawn by the Zener diode and the load resistor together is 100 mA itself. If the load is disconnected, the entire current will flow through the Zener diode and this is to be limited to 100 mA.
As the Zener diode is expected to draw at least 5 mA (which happens when the load draws the maximum current), the maximum expected load current is (100 mA – 5 mA) = 95 mA.
[Note that in the shunt voltage regulator, the Zener diode and the load together draws the same current irrespective of the input voltage variation and the load current variation (within the design limits)].
In the above question, what is the value of the series resistance R?
(a) 10 Ω
(b) 20 Ω
(c) 40 Ω
(d) 60 Ω
(e) 95 Ω
You have to consider the maximum unregulated input voltage to determine the value of R. Here it is 10 V. Since the regulated voltage (break down voltage of the Zener diode) is 6 V, the voltage drop across R must be 10 V– 6 V = 4 V.
Since the current through R in all situations (within design limits) is 100 mA, the value of R is given by
