Voltage Regulators Questions and Answers

11.

Which of the following three pin regulators yield + 9 V output ?

Four three-pin regulator circuits I to IV: I a 7809 with common pin grounded; II a 7805 with a 4 V zener in the common lead, triangle pointing down; III the same with the zener drawn the other way; IV a 7812 with a 5 V zener like II.

A.Options I and III are correct.
B.Options II and IV are correct.
C.Options III and II are correct.
D.Options IV and I are correct.

Exam Year(s)

UGC NET - 2012,

Answer: A

Circuits I and III give +9 V. Circuit I is a 7809, a fixed regulator whose output is 9 V with the common pin at ground. In circuits II and III a 7805 regulator has a 4 V zener diode in its common lead, which lifts the regulator's reference by 4 V, so the output is 5 V plus 4 V, which is 9 V, but only if the zener is reverse biased. In circuit III the zener is drawn with its cathode towards the regulator, so the current from the common pin flows into the cathode and it is reverse biased and holds 4 V. In circuit II the diode is drawn the other way, so it is forward biased, drops only about 0.7 V, and gives about 5.7 V. Circuit IV puts a 5 V zener in a 7812 and would give 12 V plus 5 V, which is 17 V. The answer is that circuits I and III are correct.

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12.

Match List – I with List – II and select the correct answer using codes given below :

List-I	List-II
(a) Gating pulse	(i) BJT
(b) Series pass element	(ii) Monostable multivibrator
(c) Lock range	(iii) Voltage regulator
(d) Base width modulation	(iv) PLL

A.a-i, b-ii, c-iv, d-iii
B.a-ii, b-iii, c-iv, d-i
C.a-iii, b-iv, c-i, d-ii
D.a-iv, b-ii, c-iii, d-i

Exam Year(s)

UGC NET - 2012,

Answer: B

The matching is a-ii, b-iii, c-iv, d-i. A gating pulse is produced by a monostable multivibrator, which gives a pulse of fixed width. The series pass element belongs to a voltage regulator. Lock range is a property of a phase locked loop, PLL. Base width modulation, the Early effect, occurs in a BJT. Option (A) gives the gating pulse as a BJT. Option (C) gives the gating pulse as a voltage regulator. Option (D) gives it as a PLL. Only (B) matches all four.

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Direction for [ Question No: 13 To 14 ] :

Read the passage below and answer the questions, that follows based on your understanding of the passage :

Many instruments require dc power of their operation. This power is available from a portable cell, however, the power output is low besides other problems. AC power is continuously and easily available, so it is convenient to use it. However, ac power must be converted into dc power.

The devices such as half-wave, full wave and bridge rectifiers convert bi-directional voltage to uni-directional voltage. However, the output has large ripple contents. Filters are used to reduce the ripple significantly. Various filters circuits and L, C, LC and CLC. The power supply with rectifier and filter is still unregulated.

The output of such power supplies varies with variations in a.c. mains voltage, load current and temperature.

In order to maintain the output at constant level, voltage or current regulation is necessary. This is obtained by using series or shunt feedback. In addition to regulation, the power supply also requires protection e.g. overload, over current or short circuit protection. Pre- regulation is also carried out to improve the stability.

Regulators can be designed by using discrete components. However, IC regulators are also available. These regulators have facilities like :

(i) voltage / current boosting
(ii) thermal shut-down
(iii) floating point to facilitate higher output voltage.
(iv) switching regulator to reduce the power dissipation.

13.

In low voltage, low current model, IC 723 gives output voltage in the range of

A.2 to 7 V
B.5 to 10 V
C.0 to 10 V
D.0 to 7 V

Exam Year(s)

UGC NET - 2012,

Answer: A

The IC 723 has a reference of about 7.15 V, and in the low-voltage mode, where the output is taken through a divider from that reference, it gives 2 to 7 V (option A). In the high-voltage mode it gives 7 to 37 V.

The ranges 5 to 10 V (B) and 0 to 10 V (C) cross the 7 V reference, which the low-voltage connection cannot do, and 0 to 7 V (D) goes down to zero, which is not possible since the output cannot fall below the reference divider limit of about 2 V.

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14.

In series pass voltage regulator using zener diode, the function of zener diode is a

A.Voltage regulator
B.Voltage comparator
C.Breakdown diode
D.Temperature compensator

Exam Year(s)

UGC NET - 2012,

Answer: B

The key takes the zener diode in a series pass regulator as a voltage comparator (option B). The zener holds a fixed reference voltage at the base of the pass transistor, and the base-emitter junction compares the output with it: \(V_{BE} = V_Z - V_{out}\), so any change in the output changes the drive.

The zener is not the regulator itself (A), since the pass transistor does the regulating, and 'breakdown diode' (C) names the device and not its function. It is not used to compensate temperature (D) in the simplest circuit. The answer rests on the key.

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15.

Consider the following ICs :

1. LM 78 05
2. LM 78 15
3. LM 78 12
The correct sequence of the output noise voltage (at 25 °C and for an operating frequency range of \(10\ Hz \le f \le 100\) KHz) in decreasing order is given by

A.2, 1, 3
B.2, 3, 1
C.1, 2, 3
D.3, 1, 2

Exam Year(s)

UGC NET - 2014,

Answer: B

The decreasing order of output noise is 2, 3, 1. According to the datasheet of the 78xx regulators the output noise voltage grows with the output voltage: about 40 microvolts for the 7805, about 75 for the 7812 and about 90 for the 7815 over 10 Hz to 100 KHz.

The LM 7815 (2) is noisiest, then the 7812 (3), then the 7805 (1). The other sequences put the lower voltage regulators ahead of the 15 V part.

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Voltage Regulators Questions and Answers

Direction for [ Question No: 13 To 14 ] :

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Voltage Regulators Questions and Answers

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