46. An R–C coupled amplifier is assumed to have a singlepole low frequency transfer function. The Maximum lower-cut -off frequency allowed for the amplifier to pass 50 Hz square wave with no more than 100% till is:

1. 1. 1 kHz

   
   
   

   2. 1 Hz

   
   
   

   3. 1.59 Hz

   
   
   

   4. 1.59 kHz

   
   
   

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1. View Hint View Answer Discuss in Forum

   Fractional tilt =	πfL
   	f

   
   where, f_L = Lower cut-off frequency
   f = applied signal frequency
   

   or fL =	f × fractional tilt
   	π

   
   

   or fL =	50 × 10/100
   	3·14

   
   = 1.59 Hz

   Correct Option: C

   Fractional tilt =	πfL
   	f

   
   where, f_L = Lower cut-off frequency
   f = applied signal frequency
   

   or fL =	f × fractional tilt
   	π

   
   

   or fL =	50 × 10/100
   	3·14

   
   = 1.59 Hz

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47. For the circuit shown below the transistor parameter are V_TN = 0.8 V and kn = 30 µ A/V². If output voltage is V₀ = 0.1 V, when input voltage is V_i = 4.2 V, the required transistor width-to-length ratio is:
    
    
    

1. 1. 1.568

   
   
   

   2. 1.982

   
   
   

   3. 0.731

   
   
   

   4. 2.825

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given that
   V_TN = 0.8 V
   k′_n = 30 µA/V²
   V₀ = 0.1 V
   V_i = 4.2 V
   From fig. V_GS = V_i = 4.2 V
   

   and ID =	5 – V0	=	5 – 0·1
   	10 kΩ		10 × 103

   
   

   =	4·9	= 0.49 mA
   	10 × 103

   
   

   Again ID =	1	. k'n		W			(VGS–VTN)2
   	2			L

   
   = 0.49 × 10^–3 or 0.49 × 10^–3
   

   =	1	× 30 × 10–6.		W			. (4.2 – 0.8)2
   	2			L

   
   

   or		W		=	0·49 × 103	= 2.825
   		L			15 × (3·4)2

   
   Hence alternative (D) is the correct choice.

   
   

   Correct Option: D

   Given that
   V_TN = 0.8 V
   k′_n = 30 µA/V²
   V₀ = 0.1 V
   V_i = 4.2 V
   From fig. V_GS = V_i = 4.2 V
   

   and ID =	5 – V0	=	5 – 0·1
   	10 kΩ		10 × 103

   
   

   =	4·9	= 0.49 mA
   	10 × 103

   
   

   Again ID =	1	. k'n		W			(VGS–VTN)2
   	2			L

   
   = 0.49 × 10^–3 or 0.49 × 10^–3
   

   =	1	× 30 × 10–6.		W			. (4.2 – 0.8)2
   	2			L

   
   

   or		W		=	0·49 × 103	= 2.825
   		L			15 × (3·4)2

   
   Hence alternative (D) is the correct choice.

   
   

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48. The transistors in the circuit of given below have parameter V_TN = 0.8 V, k_n = 40 µA/V² and λ = 0. The width-to-length ratio of M₂ is
    

    	W			= 1.
    	L		2

    
    If V₀ = 0.10 V when V_i = 5 V, then
    

    	W			for M1 is:
    	L		1

    
    
    

1. 1. 47.5

   
   
   

   2. 28.4

   
   
   

   3. 40.5

   
   
   

   4. 20.3

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given that
   V_TN = 0.84
   k′_n = 40 µA/V²
   

   		W			= 1
   		L		2

   
   V₀ = 0.10 V
   V_i = 5 V
   For transistor M₂
   V_DS = 0.1 V
   V_GS1 = V_i = 5 V
   ∴ V_DS < V_GS – V_TN, therefore the transistor M₂ work in the linear region
   I_D2 = I_D1 = I_D
   

   ID2 =	1	k'n		W			. [2(VGS2 – VTn)VDS2 – V2DS2 ]…(A)
   	2			L		2

   
   

   ID1 =	1	kn′		W			(VGS1 – VTn)2 …(B)
   	2			L		1

   
   solving equation (A) and (B)
   

   1. (5 – 0.1 – 0.8)2 =		W			[2 . (5 – 0.8) 0.1 – (0.1)2]
   		L		1

   
   

   (4.1)2 =		W			[0·84 – 0.01]
   		L		1

   
   

   or =		W			=	(4.1)2	= 20.25
   		L		1		0·83

   
   Hence alternative (D) is the correct choice.

   
   

   Correct Option: D

   Given that
   V_TN = 0.84
   k′_n = 40 µA/V²
   

   		W			= 1
   		L		2

   
   V₀ = 0.10 V
   V_i = 5 V
   For transistor M₂
   V_DS = 0.1 V
   V_GS1 = V_i = 5 V
   ∴ V_DS < V_GS – V_TN, therefore the transistor M₂ work in the linear region
   I_D2 = I_D1 = I_D
   

   ID2 =	1	k'n		W			. [2(VGS2 – VTn)VDS2 – V2DS2 ]…(A)
   	2			L		2

   
   

   ID1 =	1	kn′		W			(VGS1 – VTn)2 …(B)
   	2			L		1

   
   solving equation (A) and (B)
   

   1. (5 – 0.1 – 0.8)2 =		W			[2 . (5 – 0.8) 0.1 – (0.1)2]
   		L		1

   
   

   (4.1)2 =		W			[0·84 – 0.01]
   		L		1

   
   

   or =		W			=	(4.1)2	= 20.25
   		L		1		0·83

   
   Hence alternative (D) is the correct choice.

   
   

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49. In the circuit given below the transistor parameters are
    V_TN = 1 V, and k_n = 36 µA/V². If I_D = 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
    	L		1		L		2		L		3

    
    
    

1. 1. 1.75, 6.94, 27.8

   
   
   

   2. 4.93, 10.56, 50.43

   
   
   

   3. 35.4, 22.4, 8.53

   
   
   

   4. 56.4, 38.21, 12.56

   
   
   

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1. View Hint View Answer Discuss in Forum

   For each transistor M₁, M₂ and M₂
   V_GS = V_DS i.e. V_DS > V_GS – V_TN
   Therefore all the transistor are in saturation Given that
   V_TN = 1 V
   k′_n = 36 µA/V²
   I_D = 0.5 mA
   V₁ = 5 V_S V₂ = 2 V
   For transistor M₃
   V₂ = 2V = V_GS3
   I_D = 0.5 × 10^–3
   

   =	1	36 × 10–6		W			. (2 – 1)2
   	2			L		3

   
   

   after simplifying we get		W			= 27.8
   		L		3

   
   For transistor M₂,
   V_GS2 = V₁ – V₂ = 5 – 2 = 3 V
   I_D = 0.5 × 10^–3
   

   = 36 × 10–6		W			(3 – 1)2
   		L		2

   
   

   or		W			= 6.94
   		L		2

   
   For transistor M₁
   VGS₁ = 10 – V₁ = 10 – 5 = 5 V
   I_D = 0.5 × 10^–3
   

   =	1	36 × 10–6		W			. (5 – 1)2
   	2			L		1

   
   

   or		W			= 1.74
   		L		1

   
   Hence alternative (A) is the correct choice.

   Correct Option: A

   For each transistor M₁, M₂ and M₂
   V_GS = V_DS i.e. V_DS > V_GS – V_TN
   Therefore all the transistor are in saturation Given that
   V_TN = 1 V
   k′_n = 36 µA/V²
   I_D = 0.5 mA
   V₁ = 5 V_S V₂ = 2 V
   For transistor M₃
   V₂ = 2V = V_GS3
   I_D = 0.5 × 10^–3
   

   =	1	36 × 10–6		W			. (2 – 1)2
   	2			L		3

   
   

   after simplifying we get		W			= 27.8
   		L		3

   
   For transistor M₂,
   V_GS2 = V₁ – V₂ = 5 – 2 = 3 V
   I_D = 0.5 × 10^–3
   

   = 36 × 10–6		W			(3 – 1)2
   		L		2

   
   

   or		W			= 6.94
   		L		2

   
   For transistor M₁
   VGS₁ = 10 – V₁ = 10 – 5 = 5 V
   I_D = 0.5 × 10^–3
   

   =	1	36 × 10–6		W			. (5 – 1)2
   	2			L		1

   
   

   or		W			= 1.74
   		L		1

   
   Hence alternative (A) is the correct choice.

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50. If the ratio is		W			= 40 and		W			= 15, then V0 is:
    		L		1			L		2

1. 1. 2.91 V

   
   
   

   2. 2.09 V

   
   
   

   3. 3.41 V

   
   
   

   4. 1.59 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   According to question
   

   	W			= 40 and		W			= 15
   	L		1			L		2

   
   Again
   K_n1 (V_GS1 – V_TN1 )2=K_n2 (V_GS2 – V_TN2 )2
   40 (V_GS1 – 0.8)2 = 15 (V_GS2 – 0.8)2 `…(A)
   V_GS1 + V_GS2 = 5 …(B)
   from (A) and (B)
   40 (V_GS1 – 0.8)2 = 15 (5 – V_GS1 – 0.8)2
   8 [V²_GS1 +.64 –1.6 V_GS1] = 3 [17.64 + V²_GS1 –8 V_GS1]
   5 V²_GS1 + 11.2 V_GS1 – 47.8 = 0
   V_GS1 = 2.09
   V₀ = 5 – V_GS1
   = 5 – 2.09 = 2.91 V
   

   Correct Option: A

   According to question
   

   	W			= 40 and		W			= 15
   	L		1			L		2

   
   Again
   K_n1 (V_GS1 – V_TN1 )2=K_n2 (V_GS2 – V_TN2 )2
   40 (V_GS1 – 0.8)2 = 15 (V_GS2 – 0.8)2 `…(A)
   V_GS1 + V_GS2 = 5 …(B)
   from (A) and (B)
   40 (V_GS1 – 0.8)2 = 15 (5 – V_GS1 – 0.8)2
   8 [V²_GS1 +.64 –1.6 V_GS1] = 3 [17.64 + V²_GS1 –8 V_GS1]
   5 V²_GS1 + 11.2 V_GS1 – 47.8 = 0
   V_GS1 = 2.09
   V₀ = 5 – V_GS1
   = 5 – 2.09 = 2.91 V
   

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