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. The gm of the MOS transistor is 4 mA/V and the common base current gain of the bipolar transistor is 0.99. The transconductance of the composite transistor as shown in the figure below will be—










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    Emitter current (IE) = gm Vin (in the case of BJT)

    α (bipolar) =
    IC
    IE

    IC = α IE
    IC = α. gm Vin …(i)
    gm (overall) =
    IC
    		…(ii)
    Vin

    from (i) and (ii)
    gm (overall) =
    α·gm Vin
    Vin

    = α. gm = .99 × 4 m A/V
    = 3.96 m A/V
    Hence alternative (B) is the correct answer.

    Correct Option: B

    Emitter current (IE) = gm Vin (in the case of BJT)

    α (bipolar) =
    IC
    IE

    IC = α IE
    IC = α. gm Vin …(i)
    gm (overall) =
    IC
    		…(ii)
    Vin

    from (i) and (ii)
    gm (overall) =
    α·gm Vin
    Vin

    = α. gm = .99 × 4 m A/V
    = 3.96 m A/V
    Hence alternative (B) is the correct answer.

  1. In figure both transistors are identical and have a high value of beta. Take the do base emitter voltage drop as 0.7 volt and kT/q = 25 mV. The small signal low frequency voltage gain (V0 / Vi) is equal to—










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    Given RC = 1.2 kΩ, RE = 1 kΩ from the given

    The gain AV =
    V0
    		= -
    RC
    		(by using direct formula)
    Viβ

    or AV =
    1.2 kΩ
    		= – 1.2
    1 kΩ


    Note: Usually AV =
    –hfe · RC
    (1 + hfe) RE + hie

    assuming very high gain factor i.e. hfe >> 1 and hie is very small as compare to RE.
    AV ≈ -
    RC
    RE

    Hence alternative (D) is the correct answer.

    Correct Option: D

    Given RC = 1.2 kΩ, RE = 1 kΩ from the given

    The gain AV =
    V0
    		= -
    RC
    		(by using direct formula)
    Viβ

    or AV =
    1.2 kΩ
    		= – 1.2
    1 kΩ


    Note: Usually AV =
    –hfe · RC
    (1 + hfe) RE + hie

    assuming very high gain factor i.e. hfe >> 1 and hie is very small as compare to RE.
    AV ≈ -
    RC
    RE

    Hence alternative (D) is the correct answer.

  1. The equivalent circuit of a CE transistor is represented by—








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    The equivalent circuit of a CE transistor is represented by figure (B).

    Correct Option: B

    The equivalent circuit of a CE transistor is represented by figure (B).

  1. The drain-source output V-I characteristics of an n-channel depletion FET has—








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    We know that drain source output V-I characteristics of an n-channel depletion FET is given by relation ID = IDSS 1 – VGS VP 2 from this relation it is clear that IDS will be positive and maximum when VGS = 0, and on increasing VGS the drain current decreases. Hence alternative (B) is the correct answer.

    Correct Option: B

    We know that drain source output V-I characteristics of an n-channel depletion FET is given by relation ID = IDSS 1 – VGS VP 2 from this relation it is clear that IDS will be positive and maximum when VGS = 0, and on increasing VGS the drain current decreases. Hence alternative (B) is the correct answer.

  1. The small-signal model of common-base transistor configuration is shown in the given figure.

    The current source (I) and resistor (R) in the output side are (α is the CB current gain of the transistor)—








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    IC = α Ie (since the output current is multiplied by gain factor R0 = re of CB configuration) Hence alternative (D) is the correct answer.

    Correct Option: D

    IC = α Ie (since the output current is multiplied by gain factor R0 = re of CB configuration) Hence alternative (D) is the correct answer.