Electronic Circuits I Set 3

Electrical Engineering

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This set of Electronic Circuits 1 Multiple Choice Questions & Answers (MCQs) focuses on Electronic Circuits I Set 3

Q1 | 7 BIASING BJT SWITCHING CIRCUITSJFET - DC LOAD LINE AND BIAS POINT, VARIOUS BIASING METHODS OF JFET - JFET BIAS CIRCUIT DESIGN
  • rd < 6kΩ
  • rd > 6kΩ
  • rd > 4kΩ
  • rd < 4kΩ
Answer: rd < 6kΩ
Explaination:
in given circuit, vgs = -5v vds = vdd – idsrd
Q2 | Consider the circuit shown. VDS=3 V. If IDS=2mA, find VDD to bias circuit.
  • -30v
  • 30v
  • 33v
  • any value of voltage less than 12 v
Answer: 33v
Explaination:
vds = vdd – ids(10k + 5k) 3 = vdd – 2(15)
Q3 | To bias a e-MOSFET                        
  • we can use either gate bias or a voltage divider bias circuit
  • we can use either gate bias or a self bias circuit
  • we can use either self bias or a voltage divider bias circuit
  • we can use any type of bias circuit
Answer: we can use either gate bias or a voltage divider bias circuit
Explaination:
to bias an e-mosfet, we cannot use a self bias circuit because the gate to source voltage for such a circuit is zero.
Q4 | Consider the following circuit. Process transconductance parameter = 0.50 mA/V2, W/L=1, Threshold voltage = 3V, VDD = 20V. Find the operating point of circuit.
  • 20v, 25ma
  • 13v, 22ma
  • 12.72v, 23.61ma d) 20v, 23.61ma
Answer: 12.72v, 23.61ma d) 20v, 23.61ma
Explaination:
ids = [k’w/l(vgs – vt)2]/2
Q5 | Consider the following circuit. IDSS = 2mA, VDD = 30V. Find R, given that VP = – 2V.
  • 10kΩ
  • 4kΩ
  • 2kΩ
  • 5kΩ
Answer: 4kΩ
Explaination:
idss = 2ma
Q6 | Process transconductance parameter is 40μA/V2. Find drain to source current in saturation.
  • 0.10 ma
  • 0.05ma c) – 0.05ma
  • d) – 50a
Answer: d) – 50a
Explaination:
isd = k’w(vsg –
Q7 | Consider the following circuit. Given that VDD = 15V, VP = 2V, and IDS = 3mA, to bias the circuit properly, select the proper statement.
  • rd < 6kΩ
  • rd > 6kΩ
  • rd > 4kΩ
  • rd < 4kΩ
Answer: rd < 6kΩ
Explaination:
in given circuit, vgs = -5v vds = vdd – idsrd
Q8 | Consider the following circuit. Process transconductance parameter = 0.50 mA/V2, W/L=1, Threshold voltage = 3V, VDD = 20V. Find the operating point of circuit.
  • 20v, 25ma
  • 13v, 22ma
  • 12.72v, 23.61ma d) 20v, 23.61ma
Answer: 12.72v, 23.61ma d) 20v, 23.61ma
Explaination:
ids = [k’w/l(vgs – vt)2]/2 vgs = 20 x 35 / 55 = 12.72 v
Q9 | Which of the following relation is true about gate current?
  • ig=id+is
  • id=ig
  • is= ig
  • ig=0
Answer: ig=0
Explaination:
the fet physical structure which contains silicon dioxide provides infinite resistance. hence no current will flow through the gate terminal.
Q10 | For a fixed bias circuit the drain current was 1mA, what is the value of source current?
  • 0ma
  • 1ma
  • 2ma
  • 3ma
Answer: 2ma
Explaination:
we know that for an fet same current flows through the gate and source terminal, hence source current=1ma.
Q11 | For a fixed bias circuit the drain current was 1mA, VDD=12V, determine drain resistance required if VDS=10V?
  • 1kΩ
  • 1.5kΩ
  • 2kΩ
  • 4kΩ
Answer: 2kΩ
Explaination:
vds=vdd-id rd
Q12 | Which of the following equation brings the relation between gate to source voltage and drain current in Self Bias?
  • vgs=vdd
  • vgs=-id rs
  • vgs=0
  • vgs=1+id rs
Answer: vgs=-id rs
Explaination:
vrs=id rs
Q13 | For a self-bias circuit, find drain to source voltage if VDD=12V, ID=1mA, Rs=RD=1KΩ?
  • 1v
  • 2v
  • 10v
  • 5v
Answer: 10v
Explaination:
vds=vdd-id (rd+rs)
Q14 | Find the gate voltage for voltage divider having R1=R2=1KΩ and VDD=5V?
  • 1v
  • 5v
  • 3v
  • 2.5v
Answer: 2.5v
Explaination:
vg = r2×vdd/r1+r2
Q15 | Find the gate to source voltage for voltage divider having R1=R2=2KΩ and VDD=12V, ID=1mA and RS=4KΩ?
  • 3v
  • 2v
  • 0v
  • 1v
Answer: 2v
Explaination:
vg = r2×vdd/r1+r2
Q16 | What will happen if values of Rs increase?
  • vgs increases
  • vgs decreases
  • vgs remains the same
  • vgs=0
Answer: vgs decreases
Explaination:
increasing values of rs result in lower quiescent values of id and more negative values of vgs.
Q17 | What is the current flowing through the R1 resistor for voltage divider (R1=R2=1KΩ, VDD=10V)?
  • 5ma
  • 3ma
  • 1ma
  • 2ma
Answer: 5ma
Explaination:
ir1=ir2 =vdd/r1+r2
Q18 | The h-parameters analysis gives correct results for                      
  • large signals only
  • small signals only
  • both large and small
  • not large nor small signals
Answer: small signals only
Explaination:
every linear circuit is associated with h –parameters. when this linear circuit is terminated with load rl, we can find input impedance, current gain, voltage gain, etc in terms of h-parameters.
Q19 | For what type of signals does a transistor behaves as linear device?
  • small signals only
  • large signals only
  • both large and small signal
  • no signal
Answer: small signals only
Explaination:
the small variation in the total voltage and current due to an application of signal moves the point up and down just by a bit and that whole up and down dynamics of the operating point from its dc value point can be approximated to be along a straight line. whole analysis can be done with same assumption of linearity with the limit of signal being in the same vicinity of the dc operating point.
Q20 | How many h-parameters are there for a transistor?
  • two
  • three
  • four
  • five
Answer: four
Explaination:
a transistor has four h- parameters –
Q21 | The dimensions of hie parameters are
  • mho
  • ohm
  • farad
  • ampere
Answer: ohm
Explaination:
hie = vbe/ib; common emitter input impedance
Q22 | The hfe parameter is called                in CE arrangement with output short circuited.
  • voltage gain
  • current gain
  • input impedance
  • output impedance
Answer: current gain
Explaination:
hfe in ce arrangement is given as
Q23 | What happens to the h parameters of a transistor when the operating point of the transistor changes?
  • it also changes
  • does not change
  • may or may not change
  • nothing happens
Answer: it also changes
Explaination:
it is very difficult to get exact values of h parameters for a particular transistor. it is because these parameters are subject to considerable variation unit to unit variation, variation due to change in temperature and variation due to the operating point.
Q24 | If temperature changes, h parameters of a transistor            
  • also change
  • does not change
  • remains same
  • may or may not change
Answer: also change
Explaination:
it is very difficult to get exact values of h parameters for a particular transistor. it is because these parameters are subject to considerable variation unit to unit variation, variation due to change in temperature and variation due to the operating point.
Q25 | In CE arrangement, the value of input impedance is approximately equal to            
  • hie
  • hib
  • hoe
  • hre
Answer: hie
Explaination:
hie = vbe/ib; common emitter input impedance
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