Analog electronics circuits miscellaneous

Analog electronics circuits miscellaneous

Easy Questions

Analog Electronic Circuits

Analog electronics circuits miscellaneous
  1. In the circuit given below the transistor parameters are
    VTN = 1 V, and kn = 36 µA/V2. If ID = 0.5 mA, V1 = 5 V and V2 = 2 V then the width-to-length ratio
    i.e.
    W
    		required in each transistor is
    L

    W
    W
    W
    L1L2L3










  1. View Hint View Answer Discuss in Forum

    For each transistor M1, M2 and M2
    VGS = VDS i.e. VDS > VGS – VTN
    Therefore all the transistor are in saturation Given that
    VTN = 1 V
    k′n = 36 µA/V2
    ID = 0.5 mA
    V1 = 5 VS V2 = 2 V
    For transistor M3
    V2 = 2V = VGS3
    ID = 0.5 × 10–3

    =
    1
    		36 × 10–6
    W
    		. (2 – 1)2
    2L3

    after simplifying we get
    W
    		= 27.8
    L3

    For transistor M2,
    VGS2 = V1 – V2 = 5 – 2 = 3 V
    ID = 0.5 × 10–3
    = 36 × 10–6
    W
    		(3 – 1)2
    L2

    or
    W
    		= 6.94
    L2

    For transistor M1
    VGS1 = 10 – V1 = 10 – 5 = 5 V
    ID = 0.5 × 10–3
    =
    1
    		36 × 10–6
    W
    		. (5 – 1)2
    2L1

    or
    W
    		= 1.74
    L1

    Hence alternative (A) is the correct choice.

    Correct Option: A

    For each transistor M1, M2 and M2
    VGS = VDS i.e. VDS > VGS – VTN
    Therefore all the transistor are in saturation Given that
    VTN = 1 V
    k′n = 36 µA/V2
    ID = 0.5 mA
    V1 = 5 VS V2 = 2 V
    For transistor M3
    V2 = 2V = VGS3
    ID = 0.5 × 10–3

    =
    1
    		36 × 10–6
    W
    		. (2 – 1)2
    2L3

    after simplifying we get
    W
    		= 27.8
    L3

    For transistor M2,
    VGS2 = V1 – V2 = 5 – 2 = 3 V
    ID = 0.5 × 10–3
    = 36 × 10–6
    W
    		(3 – 1)2
    L2

    or
    W
    		= 6.94
    L2

    For transistor M1
    VGS1 = 10 – V1 = 10 – 5 = 5 V
    ID = 0.5 × 10–3
    =
    1
    		36 × 10–6
    W
    		. (5 – 1)2
    2L1

    or
    W
    		= 1.74
    L1

    Hence alternative (A) is the correct choice.

Direction: Consider the circuit shown below:
The both transistor have parameter as follows:
VTN = 0.8 V, kn = 30 ΩA/V2

  1. If the ratio is
    W
    		 = 40 and
    W
    		 = 15, then V0 is:
    L1L2








  1. View Hint View Answer Discuss in Forum

    According to question

    W
    		 = 40 and
    W
    		 = 15
    L1L2

    Again
    Kn1 (VGS1 – VTN1 )2=Kn2 (VGS2 – VTN2 )2
    40 (VGS1 – 0.8)2 = 15 (VGS2 – 0.8)2 `…(A)
    VGS1 + VGS2 = 5 …(B)
    from (A) and (B)
    40 (VGS1 – 0.8)2 = 15 (5 – VGS1 – 0.8)2
    8 [V2GS1 +.64 –1.6 VGS1] = 3 [17.64 + V2GS1 –8 VGS1]
    5 V2GS1 + 11.2 VGS1 – 47.8 = 0
    VGS1 = 2.09
    V0 = 5 – VGS1
    = 5 – 2.09 = 2.91 V

    Correct Option: A

    According to question

    W
    		 = 40 and
    W
    		 = 15
    L1L2

    Again
    Kn1 (VGS1 – VTN1 )2=Kn2 (VGS2 – VTN2 )2
    40 (VGS1 – 0.8)2 = 15 (VGS2 – 0.8)2 `…(A)
    VGS1 + VGS2 = 5 …(B)
    from (A) and (B)
    40 (VGS1 – 0.8)2 = 15 (5 – VGS1 – 0.8)2
    8 [V2GS1 +.64 –1.6 VGS1] = 3 [17.64 + V2GS1 –8 VGS1]
    5 V2GS1 + 11.2 VGS1 – 47.8 = 0
    VGS1 = 2.09
    V0 = 5 – VGS1
    = 5 – 2.09 = 2.91 V

  1. If the width-to-length ratios of M1 and M2 are:
    W
    		 =
    W
    		 = 40
    L1L2

    The output V0 is:








  1. View Hint View Answer Discuss in Forum

    For both transistor M1 and M2
    VDS = VGS
    ∴ VDS > VGS – VTN therefore both the transistor are in saturation
    ID1 = ID2
    ID1 = Kn1 (VGS1 – VTN1 )2
    = Kn2 (VGS2 – VTN2)2 …(A)
    Kn1 = Kn2 and VTN1 = VTN2
    on expanding equation
    V2GS1 + V2TN1 – 2VGS1 = V2GS2 + V2TN2 – 2VGS2
    or V2GS1 + V2GS2 = 2 (VGS1 – VGS2) …(B)
    from given figure,
    VGS1 + VGS2 = 5 …(C)
    V2GS1 – (5 – VGS1 )2 = 2 (VGS1 – VGS1 + 5)
    V2GS1 – 25 – V2GS1 + 10 VGS1 = 10
    10 VGS1 = 35
    or VGS1 = 3.5
    VGS2 = 5 – VGS1 = 5 – 3.5 = 1.5 V
    So, V0 = 5 – VGS1 = VGS2 = 1.5 V


    Correct Option: A

    For both transistor M1 and M2
    VDS = VGS
    ∴ VDS > VGS – VTN therefore both the transistor are in saturation
    ID1 = ID2
    ID1 = Kn1 (VGS1 – VTN1 )2
    = Kn2 (VGS2 – VTN2)2 …(A)
    Kn1 = Kn2 and VTN1 = VTN2
    on expanding equation
    V2GS1 + V2TN1 – 2VGS1 = V2GS2 + V2TN2 – 2VGS2
    or V2GS1 + V2GS2 = 2 (VGS1 – VGS2) …(B)
    from given figure,
    VGS1 + VGS2 = 5 …(C)
    V2GS1 – (5 – VGS1 )2 = 2 (VGS1 – VGS1 + 5)
    V2GS1 – 25 – V2GS1 + 10 VGS1 = 10
    10 VGS1 = 35
    or VGS1 = 3.5
    VGS2 = 5 – VGS1 = 5 – 3.5 = 1.5 V
    So, V0 = 5 – VGS1 = VGS2 = 1.5 V


  1. The parameters for the transistor in circuit of fig. below are VTN = 2 V and Kn = 0.2 mA/V2. The power dissipated in the transistor is:










  1. View Hint View Answer Discuss in Forum

    From fig. since gate is connected to the drain. Hence transistor will always in saturation.

    ID =
    10 – VGS
    		= Kn (VGS – VTn)2
    10 kΩ

    10 – VGS = 0.2 × 10–3 × 10 × 103 (VGS – 2)2
    or 10 – VGS = 2 (VGS – 2)2
    After solving quadratic equation, we get
    VGS = – 0.27 V, 3.77 V VGS = – 0.27 is not possible since gate terminal is at 10 V. So, VGS = 3.77 V is taken
    i.e. VGS = VDS = 3.77 V
    ID =
    10 – 3·77
    		= 0.623 mA
    10 kΩ

    Power = ID VDS = 0.623 × 10–3 × 3.77 = 2.35 mW


    Correct Option: B

    From fig. since gate is connected to the drain. Hence transistor will always in saturation.

    ID =
    10 – VGS
    		= Kn (VGS – VTn)2
    10 kΩ

    10 – VGS = 0.2 × 10–3 × 10 × 103 (VGS – 2)2
    or 10 – VGS = 2 (VGS – 2)2
    After solving quadratic equation, we get
    VGS = – 0.27 V, 3.77 V VGS = – 0.27 is not possible since gate terminal is at 10 V. So, VGS = 3.77 V is taken
    i.e. VGS = VDS = 3.77 V
    ID =
    10 – 3·77
    		= 0.623 mA
    10 kΩ

    Power = ID VDS = 0.623 × 10–3 × 3.77 = 2.35 mW


  1. The PMOS transistor shown below has parameters:
    VTP = - 1.2 V,
    W
    		= 20, and KP = 30 µA/V2
    L

    If ID = 0.5 mA and VD = – 3 V, then value of RS and RD are:










  1. View Hint View Answer Discuss in Forum

    NA

    Correct Option: D

    NA