1 Crore+ students have signed up on EduRev. Have you? |
The following truth table belongs to which of the following four gates? [1997]
The given truth table is of
(OR gate + NOT gate) ≡ NOR gate
A semi-conducting device is connected in aseries circuit with a battery and a resistance. Acurrent is found to pass through the circuit. Ifthe polarity of the battery is reversed, the currentdrops to almost zero. The device may be
In reverse bias, the current through a p-n
junction is almost zero.
The transfer ratio β of a transistor is 50. Theinput resistance of the transistor when used inthe common emitter configuration is 1 kΩ. Thepeak value of the collector A.C. current for anA.C. input voltage of 0.01 V peak is [1998]
Which of the following gates will have an output of 1?
(a)
(b)
(c)
(d)
(a) is a NAND gate so output is
(b) is a NOR gate so output is
(c) is a NAND gate so output is
(d) is a XOR gate so output is
Following is NAND Gate
The truth-table given below is for which gate? [1998]
Relation between A, B and C shows that
The cause of the potential barrier in a p-n diodeis [1998]
During the formation of a junction diode,
holes from p-region diffuse into n-region and
electrons from n-region diffuse into p-region.
In both cases, when an electron meets a hole,
they cancel the effect of each other and as a
result, a thin layer at the junction becomes
free from any of charge carriers. This is called
depletion layer. There is a potential gradient
in the depletion layer, negative on the p-side
and positive on the n-side. The potential
difference thus developed across the
junction is called potential barrier.
In forward bias, the width of potential barrier ina p-n junction diode [1999]
We know that in forward bias of p-n junction
diode, when positive terminal is connected
to p-type diode, the repulsion of holes takes
place which decreases the width of potential
barrier by striking the combination of holes
and electrons.
A depletion layer consists of [1999]
Depletion layer is formed by diffusion of
holes and electrons from p-type
semiconductor to n-type semiconductor and
vice-versa. Hence, donor and acceptor atom
get positive and negative charge leading to
formation of p-n junction. Thus, donor and
acceptor are immobile.
Which of the following when added acts as animpurity into silicon produced n-type semiconductor? [1999]
n-type of silicon semiconductor is formed
when impurity is mixed with pentavalent
atom. Out of given choices only phosphorus
is pentavalent.
In a junction diode, the holes are due to [1999]
Holes are produced due to missing of
electrons.
A gate has the following truth table
The gate is
P, Q and R are related as R = P. Q which is
relation of AND gate.
In the study of transistor as amplifier, if and
, where, IC, IB and IE are the collector, base and emitter currents, then [2000]
As we know that Ie = Ic = Ib
Divide both side by Ie
The forward biased diode is [2000]
As A is at higher potential than B, hence
current will flow from A to B i.e. from p
junction to n junction and diode is forward
biased.
Which gate is represented by the following truth table ?
Which is truth table of NAND gate
The intrinsic semiconductor becomes insulator at
[2001]
At 0K, motion of free electrons stop. Hence
conductivity becomes zero. Therefore, at 0K
intrinsic semiconductor becomes insulator.
For a common emitter circuit if then current gain for common emitter circuit will be [2001]
We have,
Transmission of light in optical fibre is due to [2001]
Light is trasmitted through optical fibre due
to multiple total internal reflections.
A d.c. battery of V volt is connected to a series combination of a resistor R and an ideal diode D as shown in the figure below. The potential difference across R will be [2002]
In forward biasing, the diode conducts. For
ideal junction diode, the forward resistance
is zero; therefore, entire applied voltage
occurs across external resistance R i.e., there
occurs no potential drop, so potential across
R is V in forward biased.
In a p-n junction [2002]
2: The n-side is at a higher electrical potential than the p-side.
In a p-n junction, a p-type semiconductor material is joined to an n-type semiconductor material. This creates a region called the depletion region, which is a region where the concentration of charge carriers (electrons and holes) is significantly lower than in the surrounding materials.
In equilibrium, the electrical potential of the p-side is lower than that of the n-side, due to the difference in the concentration of charge carriers. The electrical potential of the p-side is attracted to the higher concentration of holes on the p-side, while the electrical potential of the n-side is attracted to the higher concentration of electrons on the n-side. This creates an electrical potential difference across the p-n junction, with the n-side at a higher potential than the p-side.
Option 1 is incorrect because the potential of the p and n-sides does not alternate. Option 3 is incorrect because the p-side is at a lower potential than the n-side. Option 4 is incorrect because the p and n-sides are at different potentials. Therefore, the correct answer is 2: The n-side is at a higher electrical potential than the p-side.
In the case of a common emitter transistoramplifier, the ratio of the collector current to theemitter current Ic /Ie is 0.96. The current gain ofthe amplifier is [2002]
If a full wave rectifier circuit is operating from50Hz mains, the fundamental frequency in theripple will be [2003]
In case of full wave rectifier,
Fundamental frequency = 2 × mains
frequency
= 2 × 50 = 100Hz.
An n-p-n transistor conducts when [2003]
When the collector is positive and emitter is
negative w.r.t. base, it causes the forward
biasing for each junction, which causes
conduction of current.
Barrier potential of a p-n junction diode doesnot depend on [2003]
Barrier potential does not depend on diode
design while barrier potential depends upon
temperature, doping density, and forward
biasing.
Following diagram performs the logic function of [2003]
Y = A. B by Demorgan therorem
∴ This diagram performs the function of
AND gate.
The following circuit represents [1999]
Output of upper AND gate =
Output of lower AND gate =
∴ Output of OR gate, Y =
This is Boolean expression for XOR gate.
157 videos|452 docs|213 tests
|
Use Code STAYHOME200 and get INR 200 additional OFF
|
Use Coupon Code |
157 videos|452 docs|213 tests
|
|
|
|
|
|
|
|