Showing posts with label common emitter. Show all posts
Showing posts with label common emitter. Show all posts

Friday, November 23, 2012

Multiple Choice Questions on Transistors including EAMCET Engineering 2004 and 2005 Questions




“The pursuit of truth and beauty is a sphere of activity in which we are permitted to remain children all our lives.”
– Albert Einstein


A very important semiconductor device you need to study is the transistor, the invention of which brought about a revolution in electronics. Questions involving transistors have been discussed earlier on this site. You can access them by trying a search for ‘transistor’ using the search box provided on this page or by clicking on the label ‘transistor’ below this post. Today we shall discuss a few more questions on transistors
(1) In an amplifier circuit using a transistor the collector current changes by1.224 mA when the emitter current changes by 1.228 mA. What is the common emitter small signal current gain of the transistor used in the circuit?
(a) 0.997 (very nearly)
(b) 1.003 (very nearly)
(c) 122 (very nearly)
(d) 307
(e) 306
The small signal common emitter current gain is the ratio of the change in collector current to the change in base current. Since the emitter current is the sum of the collector current and the base current, the change in base current in the present case is 1.228 mA – 1.224 mA which is equal to 0.004 mA.
Therefore common emitter small signal current gain (βac) = 1.224/0.004 = 306.
(2) In an n-p-n transistor in CE configuration
(i) the emitter is more heavily doped than the collector
(ii) emitter and collector can be interchanged
(iii) the base region is very thin but is heavily doped
(iv) the conventional current flows from base to emitter
(a) (i) and (ii) are correct
(b) (i) and (iii) are correct
(c) (i) and (iv) are correct
(d) (ii) and (iii) are correct
The above question appeared in EAMCET Engineering 2004 question paper.
Option (c) is the answer. Note that the configuration deoes not matter.
(3) An n-p-n transistor power amplifier in CE configuration gives
(a) voltage amplification only
(b) current amplification only
(c) both current and voltage amplifications
(d) only power gain of unity
This question appeared in EAMCET engineering 2005 question paper. In common emitter configuration a transistor (n-p-n as well as p-n-p) gives both current and voltage amplifications so that there will be appreciable power gain. Therefore the correct option is (c).

(4) The adjoining figure shows an n-p-n transistor operated in the common emitter configuration in which the potential divider resistors R1 and R2 provide the base voltage required for forward biasing the base emitter junction. A student wrongly selected too small a value for R1 so that the transistor got saturated. (When a transistor is saturated, its emitter to collector voltage is nearly zero). The collector load resistor RC is 5 KΩ and the emitter resistor RE (used for bias stabilization) is 1 KΩ. If βdc of the transistor is 200 what is the minimum base current required for saturating the transistor?
Even though this question may appear to be difficult at the first glance, it is really simple. You have to first calculate the current I which will produce a voltage drop of 12 volts across 6 KΩ.
[Since the emitter to collector voltage is nearly zero when the transistor is saturated, the entire supply voltage of 12 volts must be dropped across the resistors RC and RE].
Therefore we have
            I × 6 KΩ = 12 V so that I = 2mA
The emitter current and collector current are nearly the same (because of large current gain βdc) so that we can take the collector current to be 2 mA. The base current required for producing a collector current (IC) of 2 mA is the saturating base current in this circuit. Therefore the minimum base current IB required for saturating the transistor is given by
            IB = IC/βdc = 2 mA/200 = 0.01 mA = 10 μA.

Saturday, April 14, 2012

KEAM (Engineering) 2010 Questions (MCQ) on Electronics

"You may never know what results come of your actions, but if you do nothing, there will be no results."
–Mahatma Gandhi

Questions on electronics will be generally interesting to most of you. Today we will discuss questions in this section which appeared in Kerala engineering entrance (KEAM - Engineering) 2010 question paper. Here are the questions with their solution:
 
(1) A full wave rectifier with an a.c. input is shown:

The output  voltage across RL is represented as

The rectified output voltage will be a direct voltage but there will be very large amount of ripples. The capacitor C acts as a filter to remove the ripples; but there will still be a small amount of  ripples in the output. Therefore the correct option is (e).
(2) In the given circuit the current through the battery is
(a) 0.5 A
(b) 1 A
(c) 1.5 A
(d) 2 A
(e) 2.5 A
Since the diode D1 is reverse biased, no current will flow through the D1 branch. Diodes D2 and D3 are forward biased and hence the battery drives currents through the 20 Ω resistor and the series combination of the two 5 Ω resistors.
The current driven through the 20 Ω resistor is 10 V/20 Ω = 0.5 A.
The current driven through the 10 Ω resistor is 10 V/10 Ω = 1 A.
Therefore, total current through the battery is 0.5 A + 1 A = 1.5 A
(3) The collector supply voltage is 6 V and the voltage drop across a resistor of 600 Ω in the collector circuit is 0.6 V, in a transistor connected in common emitter mode. If the current gain is 20, the base current is
(a) 0.25 mA
(b) 0.05 mA
(c) 0.12 mA
(d) 0.02 mA
(e) 0.07 mA
We have ICRC = 0.6 V where IC is the collector current and RC is the resistance in the collector circuit.
Therefore,  IC×600 Ω = 0.6 V from which IC = 0.6/600 A = 10–3 A = 1 mA.
Since the current gain β is given by
            β = IC/IB where IB is the base current, we have
            IB = IC/β = 1 mA/20 = 0.05 mA.
(4) A pure semiconductor has equal electron and hole concentration of 1016 m–3. Doping by indium increases nh to 5×1022 m–3. Then the value of ne in the doped semiconductor is
(a) 106 m–3
(b) 1022 m–3
(c) 2×106 m–3
(d) 1019 m–3
(e) 2×109 m–3
According to the law of mass action we have
            ni2 = nenh where ni is the electron concentration as well as the hole concentration in the intrinsic (pure) semiconductor, ne is the electron concentration in the doped semiconductor and nh is the hole concentration in the doped semiconductor.
Therefore ne = ni2/nh = (1016)2/(5×1022) = 2×109 m–3