EDC 2 Flashcards
The n-type regions in an npn bipolar
junction transistor are
A. collector, base, and emitter
B. base and emitter
C. collector and base
D. collector and emitter
D. collector and emitter
(look at the figure on your EDC 2 album baby)
For normal operation of an npn
transistor, the base must be
A. positive with respect to the collector
B. negative with respect to the emitter
C. positive with respect to the emitter
D. disconnected
C. positive with respect to the emitter
For NORMAL operation of a transistor,
the B-E junction must be forward-biased. Therefore, for an NPN transistor,since the base is P type and the emitter is N type, for the B-E junction to beforward biased, the base must be more positive with respect to emitter.
Beta (β) is the ratio of…
A. collector current to emitter current
B. collector current to base current
C. emitter current to base current
D. output voltage to input voltage
B. collector current to base current
Current gain for different BJT
configurations are as follows:
1. alpha = collector current/emitter current
2. beta = collector current/base current
3. gamma = emitter current/base current
Where does the transistor act like an
open switch?
A. active region
B. inverted region
C. saturated region
D. cut off region
D. cut off region
In cut off region, both the junctions are
reverse biased. The transistor has
practically zero current because the
emitter does not emit charge carriers to the base. So, the transistor acts as open switch.
Using the load line to answer, under
what condition will be Vce = Vcc?
A. Linear
B. Active
C. Cut-off
D. Saturation
C. Cut-off
At saturation : Vce = Vce(sat) ≈ 0 V
and lc = Ic(sat)
At the active or linear region:
VcE(sat) < Vce < Vcc and 0 < lc <
Ic(sat)
At cut-off: VcE ≈ Vcc and lc ≈0A
(see the fugure at EDC 2 album😚)
If a certain transistor operates at the middle of the load line, an increase in the base resistance will move the Q point
A. Off the load line
B. Down
C. Nowhere
D. Up
B. Down
Increasing Rg → decreases lg→
decreases lc → Q point will be to a
lower position in the load line
(see the figure at EDC 2 album)
An n-type semiconductor material
A. requires no doping
B. has trivalent impurity atoms added
C. is intrinsic
D. has pentavalent impurity atoms added
D. has pentavalent impurity atoms added
N-type semiconductor material, is an
extrinsic semiconductor material that is
the product of doping process that
involves adding donor impurities/
pentavalent atoms such as
Phosphorous, Antimony, Arsenic, or
Bismuth to an intrinsic Silicon or Germanium.
The input impedance of a MOSFET is
A. higher than that of a JFET.
B. No different than that of a JFET
C. lower than that of a JFET.
D. approximately zero ohms.
A. higher than that of a JFET.
Zì(MOSFET) > Zì(JFET)»_space; Zì(BJT)
If a transistor operates at the middle of the load line, an increase in the current gain will move the Q point
A. Down
B. Up
C. Off the load line
D. Nowhere
B. Up
Increasing β → increases lc→Q point will move upward in the load line
(look at EDC 2 album)
The point on the DC load line which is
represented by ‘Q’ is called_____
A. cut off point
B. breakdown point
C. operating point
D. cut in point
C. operating point
The point which represents the values of Ic and Vce that exist in a transistor circuit when no signal is applied is called as operating point. This is also called as working point or quiescent point.
The input of a CB transistor is given
between_____
A. base and collector terminals
B. collector and emitter terminals
C. ground and emitter terminals
D. emitter and base terminals
D. emitter and base terminals
The name of the CB transistor says that
it’s a common based one. The input is
given between the emitter and base
terminals and the output is taken
between collector and base terminals.
The primary difference between the construction of depletion-type and enhancement-type MOSFETs is
A. the size of the transistor
B. the reverse bias junction
C. All of these choices
D. the absence of the channel
D. the absence of the channel
EMOSFET is a like a normally-OFF
switch. A threshold voltage must be
exceeded by the gate to source voltage in order to create a conductive channel between the source and drain.
Therefore, at VGS still at O V there will be no conductive channel that will be created between the source and the drain.
In the saturation region the collector-
base junction is _____ -biased and the base-emitter junction is _____ -biased for a transistor.
A. reverse, forward
B. forward, forward
C. reverse, reverse
D. forward, reverse
B. forward, forward
Where should be the bias point set in order to make transistor work as an amplifier?
A. Cut off
B. Active
C. Saturation
D. Cut off and Saturation
B. Active
To operate transistor as an amplifier, it
requires more current amplification
factor and in cut off and saturation, the current amplification is less, therefore active region is better to fix the Q point.
The base current is the _____ of the emitter and collector currents.
A. difference
B. sum
C. product
D. None of these choices
A. difference
In BJT, the outer layers are the _____
sandwiched layer.
A. the same as
B. much smaller than
C. much larger than
D. None of these choices
C. much larger than
(go to your gallery for more EDC 2)
When does the transistor act like an
open switch?
A. active region
B. saturated region
C. inverted region
D. cut off region
D. cut off region
In cut off region, both the junctions are
reverse biased. The transistor has
practically zero current because the
emitter does not emit charge carriers to the base. So, the transistor acts as open switch.