Physical electronics devices and ics miscellaneous

Physical electronics devices and ics miscellaneous

Easy Questions

Physical Electronics Devices and ICs

Physical electronics devices and ics miscellaneous
  1. A JFET is a—








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    NA

    Correct Option: A

    NA

  1. Two perfectly matched silicon transistors are connected as shown in figure below. Assuming the β of the transistor to be very high and forward voltage drop in diodes to be 0·7 V, the value of current I is—










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    The given circuit

    Given, β of a transistor is very high it means IB1 = IB2 ≈ 0,
    VBE = 0·7 V. The given circuit can be redrawn as shown below—

    From figure— – I × 1 kΩ – 0·7 – 0·7 + 5 = 0

    or I =
    5 – 1·4
    1 kΩ

    or I = 3·6 mA
    Hence alternative (B) is the correct choice.

    Correct Option: B

    The given circuit

    Given, β of a transistor is very high it means IB1 = IB2 ≈ 0,
    VBE = 0·7 V. The given circuit can be redrawn as shown below—

    From figure— – I × 1 kΩ – 0·7 – 0·7 + 5 = 0

    or I =
    5 – 1·4
    1 kΩ

    or I = 3·6 mA
    Hence alternative (B) is the correct choice.

  1. The common emitter forward current gain of the transistor shown below is βF = 100. The transistor is operating in—










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    The given circuit

    Since there is no supply connected at the base terminal, therefore the transistor is operating in cut-off region.

    Correct Option: C

    The given circuit

    Since there is no supply connected at the base terminal, therefore the transistor is operating in cut-off region.

  1. A TTL NOT gate circuit is shown below. Assuming VBE = 0·7 V of both the transistors, if Vi = 3 V, then the states of the two transistors will be—










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    NA

    Correct Option: C

    NA

  1. Consider the circuit shown in figure below. If the β of the transistor is 30 and ICBO = 20 nA and the input voltage is + 5 V, the transistor would be operating in—










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    Given, β = 30, ICBO = 20 nA, Vi = 5V

    From above figure
    5 – I1 × 15 kΩ – VBE = 0

    or I1 =
    (5 – 0·7)V
    		= 0·2867 mA
    15 kΩ

    Again from figure
    – 12 + I2 × 100 kΩ – VBE = 0
    or I2 =
    12·7 V
    		= 0·127 mA
    100 kΩ

    And, IB = I1 – I2 = 0·2867 mA – 0·127 mA
    = 0·158 mA
    We know that
    IC = βIB + (1 + β) ICBO
    or IC = 30 × 0·158 + (1 + 30) × 20 × 10–9
    or IC = 4·74 mA + 0·00062 mA
    or IC = 4·74062 mA
    Now, VCE = VCC – ICRC
    or VCE = 12 – 4·7062 × 10–3 × 2·2 × 103
    or VCE = 12 – 10·385 = 1·6146V
    Since VCE > 0·2V. It means transistor is operating in Active region
    Hence alternative (B) is the correct choice.

    Correct Option: B

    Given, β = 30, ICBO = 20 nA, Vi = 5V

    From above figure
    5 – I1 × 15 kΩ – VBE = 0

    or I1 =
    (5 – 0·7)V
    		= 0·2867 mA
    15 kΩ

    Again from figure
    – 12 + I2 × 100 kΩ – VBE = 0
    or I2 =
    12·7 V
    		= 0·127 mA
    100 kΩ

    And, IB = I1 – I2 = 0·2867 mA – 0·127 mA
    = 0·158 mA
    We know that
    IC = βIB + (1 + β) ICBO
    or IC = 30 × 0·158 + (1 + 30) × 20 × 10–9
    or IC = 4·74 mA + 0·00062 mA
    or IC = 4·74062 mA
    Now, VCE = VCC – ICRC
    or VCE = 12 – 4·7062 × 10–3 × 2·2 × 103
    or VCE = 12 – 10·385 = 1·6146V
    Since VCE > 0·2V. It means transistor is operating in Active region
    Hence alternative (B) is the correct choice.