NEET Questions on Electronics

"I object to violence because when it appears to do good, the good is only temporary; the evil it does is permanent"

    – Mahatma Gandhi

Here are three questions on electronics. These were included in the NEET 2016 (July) question paper:

(1) For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 4V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kΩ, then the input signal voltage is

(1) 30 mV

(2) 15 mV

(3) 10 mV

(4) 20 mV

The output signal current i_c through the collector resistance R_c is the ratio of output signal voltage v_c to the collector resistance R_c:

            i_c = v_c/R_c = 4V/2 kΩ = 2 mA

The ratio of the collector current to the current amplification factor β_ac gives the signal current i_b in the base circuit:

             i_b = i_c/β_ac = 2 mA/100 = 0.02 mA

[Note that β_ac is the amplification factor for small signal currents]

The input signal voltage v_i is the voltage drop across the base resistance R_b due to the base current:

             v_i = i_bR_b = 0.02 mA×1 kΩ = 0.02 volt = 20 mV

[You can obtain the answer in no time if you remember the expression for the voltage gain A_v of a common emitter amplifier given by

            A_v = v_o/v_i = β_acR_o/r_i  = 100×(2 kΩ/1kΩ) = 200

Note that R_o is the output resistance and r_i is the input resistance.

This gives v_i = v_o/200 = 4V/200 = 0.02 V = 20 mV]

 (2) The given circuit has two ideal diodes connected as shown in the figure below.

 

The current flowing through the resistance R₁ will be

(1) 1.43 A

(2) 3.13 A

(3) 2.5 A

(4) 10.0 A

Since D₁ is reverse biased, no current will flow through it. We can therefore forget about the branch containing D₁ and R₂. The battery sends a current through the series combination of R₁ and R₃ since the diode D₂ is forward biased.

            Current through R₁ = 10 V/(2Ω + 2Ω) = 2.5 A

(3) What is the output Y in the following circuit, when all the three inputs A,B,C are first 0 and then 1?

(1) 1,0

(2) 1,1

(3) 0,1

(4) 0.0

When all the three inputs are 0 the NAND gate following the AND gate has both inputs at 0 level. The output Y of the NAND is therefore 1. When all the three inputs are 1, the NAND gate has both inputs at 1 and hence the output is 0. The correct option is (1).

Kerala Engineering Entrance (KEAM -Engineering) 2011 Questions (MCQ) on Transistor Amplifier and Oscillator

Today we will discuss two multiple choice questions involving transistors which appeared in Kerala Engineering Entrance (KEAM -Engineering) 2011 question paper. These questions are simple as is the case with questions in this section at the level expected in the case of 12^th grade (Plus Two) students.

(1) In a common emitter transistor amplifier, the output resistance is 500 KΩ and the current gain β = 49. If the power gain of the amplifier is 5×10⁶, the input resistance is

(A) 325 Ω

(B) 165 Ω

(C) 198 Ω

(D) 225 Ω

(E) 240 Ω

Power gain A_p is given by

A_p = β_ac A_v where β_ac is the current gain and A_v is the voltage gain (for small a.c. signals).

Since A_v = β_acR_o/R_i where R_o is the output resistance and R_i is the input resistance, we have

A_p = β_ac²R_o/R_i

Therefore R_i = β_ac²R_o/A_p = 49²×(500×10³) /(5×10⁶) = 240 Ω, very nearly.

(2) A transistor oscillator is (i) an amplifier with positive feedback (ii) an amplifier with reduced gain (iii) the one in which dc supply energy is converted into ac output energy. Then

(A) all (i), (ii) and (iii) are correct

(B) only (i) and (ii) are correct

(C) only (i) and (iii) are correct

(D) only (ii) and (iii) are correct

(E) only (ii) is correct

Statements (i) and (iii) are correct where as statement (ii) is incorect. So the correct option is (c).

Multiple Choice Questions (including KEAM 2009) on Transistor Amplifiers

Questions from electronics are generally simple and interesting at the level expected from you. Here are a few questions:

(1) A transistor amplifier circuit is operated with an emitter current of 2 mA. The collector current is 1.98 mA. The common emitter current gain (β_dc) of the transistor used in the circuit is

(a) 45

(b) 50

(c) 100

(d) 125

(e) 200

Current gain (β_dc) in the common emitter configuration is given by

β_dc = I_C /I_B = 2 mA/(2 – 1.98) mA = 100

(2) A common emitter low frequency amplifier has an effective input resistance of 1 kΩ. The collector load resistance is 5 kΩ. On applying a signal, the base current changes by 10 μA and the collector current changes by 1 mA. What is the power gain of this amplifier?

(a) 100

(b) 500

(c) 1000

(d) 10000

(e) 50000

The power gain or power amplification (A_P) is the product of the current gain β_ac and the voltage gain A_v:

A_P = β_ac× A_v

Here β_ac = ∆I_C /∆I_B = (1×10^–3)/(10×10^–6) = 100 and

A_v = – β_ac×R_L/R_i where R_L is the load resistance and R_i is the input resistance. The negative sign just indicates that in the common emitter configuration the output signal is phase shifted by 180º.

[We use the load resistance R_L instead of the output resistance R_o (of the amplifier stage) in the above expression since the load resistance is small compared to the transistor output resistance. (The output resistance of the amplifier stage is the parallel combined value of R_L and the transistor output resistance)].

Ignoring the negative sign, A_v = 100×5/1 = 500

Therefore, power gain A_P = β_ac× A_v = 100×500 = 50000.

(3) For a common emitter amplifier, the audio signal voltage across the collector resistance 2 kΩ is 2V. If the current amplification factor of the transistor is 200 and the base resistance is 1.5 kΩ, the input signal voltage and base current are

(a) 0.1 V and 1 μA

(b) 0.15 V and 10 μA

(c) 0.015 V and 1 A

(d) 0.0015 V and 1 mA

(e) 0.0075 V and 5 μA

This MCQ appeared in KEAM (Engineering) 2009 question paper.

The collector signal current i_c is given by

i_c = v_c/R_c = 2 V/ 2 kΩ = 1 mA.

Since the current amplification (β_ac) of the transistor is 200, the base signal current i_b is given by

i_b = i_c /β_ac = 1 mA /200 = 0.005 mA = 5 μA

The input signal is the voltage drop (i_bR_b) produced across the base resistance by the flow of the base current.

Therefore, the input signal voltage = i_bR_b = 5 μA×1.5 kΩ = 7.5×10^–3 V = 0.0075 V.

(4) In an NPN transistor 108 electrons enter the emitter in 10^–8 s. If 1% electrons are lost in the base, the fraction of current that enters the collector and current amplification are respectively

(a) 0.8 and 49

(b) 0.9 and 90

(c) 0.7 and 50

(d) 0.99 and 99

(e) 0.88 and 88

This MCQ appeared in KEAM (Medical) 2009 question paper.

The number of electrons (108) entering the base and the time (10^–8 s) given in this question are not required to solve the problem. They can serve as distraction while attempting this simple question!

Since 1% electrons are lost in the base, 99% reach the collector so that the fraction of current that enters the collector is 99/100 = 0.99.

The current amplification is the ratio of the collector current to the base current and is equal to 99/1 = 99.

To access all questions from electronics on this blog you may click on the label ‘electronics’ below this post.

AIPMT 2008 Question on Feed back Amplifier

The following question on negative feed back amplifier is simple even though Plus Two students will normally be unprepared to answer it:

The voltage gain of an amplifier with 9% negative feed back is 10. The voltage gain without feed back will be

(1) 10

(2) 1.25

(3) 100

(4) 90

If the voltage gain without feed back is A_v and the feed back factor (fraction of output voltage fed back to the input) is β, the voltage gain (A_fb) with feed back is given by

A_fb = A_v/(1– βA_v)

In the case of negative feed back the sign of the feed back factor β is negative so that the voltage gain with feed back is given by

A_fb = A_v/(1+ βA_v)

Since A_fb = 10 and the magnitude of β is 9% = 0.09, we have on substituting,

10 = A_v/(1+ 0.09A_v)

This gives A_v = 100.

Here is another question on feed back amplifiers:

Pick out the wrong statement:

When negative feed back is applied in a transistor amplifier

(1) its voltage gain is decreased

(2) its band width is decreased

(3) distortion produced by the amplifier is decreased

(4) the transistor current gain is unchanged

The second option alone is incorrect. The band width of a negative feed back amplifier will be greater than that of the same amplifier without the feed back. Since there is a reduction in the voltage gain consequent on the negative feed back, the 3 dB down frequency on the lower side will be shifted towards lower frequency and the 3 dB down frequency on the upper side will be shifted towards higher frequency.