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. Increasing the collector supply voltage will increase—








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    NA

    Correct Option: D

    NA

  1. In the active region, the collector current is not changed significantly by—








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    In the active region, the collector current is not significantly by current gain.

    Correct Option: C

    In the active region, the collector current is not significantly by current gain.

  1. Consider the regulated power supply given below:
    Here VZ = 9·1 V and VBE = 0·7 V. Value of V0 is equal to—










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    The given figure

    Given, VZ = 9·1V, VBE = 0·7V, V0 =?
    From above figure, voltage at node A
    VA = VBE + VZ = 0·7 + 9·1 = 9·8V
    Applying polential diviider rule at node A, we get

    VA = V0·
    R4
    R4 + R3

    or V0 = VA·R4 + R3
    R4

    or V0 = 9·8 ×
    (1K + 2K)
    		= 29·4V
    2K

    Hence alternative (B) is the correct choice.

    Correct Option: B

    The given figure

    Given, VZ = 9·1V, VBE = 0·7V, V0 =?
    From above figure, voltage at node A
    VA = VBE + VZ = 0·7 + 9·1 = 9·8V
    Applying polential diviider rule at node A, we get

    VA = V0·
    R4
    R4 + R3

    or V0 = VA·R4 + R3
    R4

    or V0 = 9·8 ×
    (1K + 2K)
    		= 29·4V
    2K

    Hence alternative (B) is the correct choice.

  1. Consider the transistor circuit given below. It is desired to have V0 = Vin. What should be the value of R to achieve the condition. Assume VBE = 0·7 V, VCE(sat) = 0·2V and I = 8·6 mA—










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    The given circuit

    Given, VBE = 0·7V, VCE(sat) = 0·2V and I = 8·6 mA The given circuit can be redrawn as

    From above figure, applying KVL to the loop, we get
    – IR – VBE + 5 = 0
    or – (– 8·6 mA).R + 4·3 = 0

    or R =
    4·3 V
    		= 500 Ω
    8·6 mA

    Hence alternative (B) is the correct choice.

    Correct Option: B

    The given circuit

    Given, VBE = 0·7V, VCE(sat) = 0·2V and I = 8·6 mA The given circuit can be redrawn as

    From above figure, applying KVL to the loop, we get
    – IR – VBE + 5 = 0
    or – (– 8·6 mA).R + 4·3 = 0

    or R =
    4·3 V
    		= 500 Ω
    8·6 mA

    Hence alternative (B) is the correct choice.

  1. The specification for a silicon transistor are: ICQ = 300 mA and fT = 350 MHz, hfe = 150, Cob = 4 pF. The values of rb′e and gm of the transistor will be—








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    Given, Ica = 300 mA, fT = 350 MHZ, hfe = 150, Cob = 4pF

    We know that, gb’e =
    gm
    hfe

    then rb’e =
    1
    		=
    hfe
    		=
    hfe
    		=
    hfe·VT
    gb’egmICQ / VTICQ

    or rb’e
    150 × 25 mV
    		= 12·5 Ω
    300 mA

    or rb’e = 12·5 ohms
    and gm=
    IC
    		=
    300 mA
    		= 12 mhos
    VT25 mV

    Correct Option: C

    Given, Ica = 300 mA, fT = 350 MHZ, hfe = 150, Cob = 4pF

    We know that, gb’e =
    gm
    hfe

    then rb’e =
    1
    		=
    hfe
    		=
    hfe
    		=
    hfe·VT
    gb’egmICQ / VTICQ

    or rb’e
    150 × 25 mV
    		= 12·5 Ω
    300 mA

    or rb’e = 12·5 ohms
    and gm=
    IC
    		=
    300 mA
    		= 12 mhos
    VT25 mV