296. The range of possible load current—

1. 1. 5 < i_L < 130 mA

   
   
   

   2. 25 ≤ i_L ≤ 120 mA

   
   
   

   3. 10 ≤ i_L ≤ 110 mA

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   From figure,
   

   I =		6.3 – Vz	=	6.3 – 4.8
   		12Ω		12

   
   

   =	1.5 V	= 125 mA
   	12 kΩ

   
   The range of I_L: when, I_z = 5 mA
   I_L = I – I_z = 125 – 5 = 120 mA
   and range, when, I_z = 100 mA
   I_L = 125 – 100 = 25 mA
   Hence the range of I_L is 25 mA ≤ I_L ≤ 120 mA

   Correct Option: B

   From figure,
   

   I =		6.3 – Vz	=	6.3 – 4.8
   		12Ω		12

   
   

   =	1.5 V	= 125 mA
   	12 kΩ

   
   The range of I_L: when, I_z = 5 mA
   I_L = I – I_z = 125 – 5 = 120 mA
   and range, when, I_z = 100 mA
   I_L = 125 – 100 = 25 mA
   Hence the range of I_L is 25 mA ≤ I_L ≤ 120 mA

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297. In the voltage regulator circuit shown below. The power rating of zener diode is 400 mW. The value of RL that will establish maximum power in Zener diode is—
     
     
     

1. 1. 5 kΩ

   
   
   

   2. 2 kΩ

   
   
   

   3. 10 kΩ

   
   
   

   4. 8 kΩ

   
   
   

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1. View Hint View Answer Discuss in Forum

   Iz (max) =		P	=	400 mW	= 40 mA
   		VZ		10 V

   
   

   Iz + IL =		20 – VZ	=	20 - 10	= 45 mA
   		222		222

   
   I_{L (min)} = 45 – I_z = 45 – 40 = 5 mA
   Now,
   

   RL =		VZ	=	10 V	= 2 kΩ
   		IL (min)		5 mA

   Correct Option: B

   Iz (max) =		P	=	400 mW	= 40 mA
   		VZ		10 V

   
   

   Iz + IL =		20 – VZ	=	20 - 10	= 45 mA
   		222		222

   
   I_{L (min)} = 45 – I_z = 45 – 40 = 5 mA
   Now,
   

   RL =		VZ	=	10 V	= 2 kΩ
   		IL (min)		5 mA

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298. In the voltage regulator shown below the power dissipation in the zener diode is—
     
     
     

1. 1. 1 mV

   
   
   

   2. 1.5 mV

   
   
   

   3. 2 mV

   
   
   

   4. 0.5 mV

   
   
   

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1. View Hint View Answer Discuss in Forum

   Since in the given problem R_Z is zero. so, in order to calculate the power dissipation across the zener diode. We must calculate the current flowing in zener diode
   Where, V_th = Thevenin equivalent voltage
   R_th = Thevenin equivalent resistance
   
   

   Vth =		75	= 50 .	1
   		75 + 150		3

   
   

   =	50	V > VZ
   	3

   
   

   Rth = 150 || 75 =	150 × 75
   	150 + 75

   
   = 50 kΩ
   

   iZ =		Vth	=	50/3
   		Rth + RL		50 kΩ

   
   = 0.33 mA
   Now P_Z = V_Z. i_Z = 15 × 0.33 mA
   = 0.5 mW
   Hence alternative (D) is the correct choice.

   Correct Option: D

   Since in the given problem R_Z is zero. so, in order to calculate the power dissipation across the zener diode. We must calculate the current flowing in zener diode
   Where, V_th = Thevenin equivalent voltage
   R_th = Thevenin equivalent resistance
   
   

   Vth =		75	= 50 .	1
   		75 + 150		3

   
   

   =	50	V > VZ
   	3

   
   

   Rth = 150 || 75 =	150 × 75
   	150 + 75

   
   = 50 kΩ
   

   iZ =		Vth	=	50/3
   		Rth + RL		50 kΩ

   
   = 0.33 mA
   Now P_Z = V_Z. i_Z = 15 × 0.33 mA
   = 0.5 mW
   Hence alternative (D) is the correct choice.

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299. In the voltage regular circuit shown below the maximum load current i_L that can be drawn is—
     
     
     

1. 1. 1.4 mA

   
   
   

   2. 2.3 mA

   
   
   

   3. 1.8 mA

   
   
   

   4. 2.5 mA

   
   
   

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1. View Hint View Answer Discuss in Forum

   At regulated power supply i s =	30 – 9	= 1.4 mA
   	15 × 103

   
   Therefore iL ≤ is = 1.4 mA

   Correct Option: A

   At regulated power supply i s =	30 – 9	= 1.4 mA
   	15 × 103

   
   Therefore iL ≤ is = 1.4 mA

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300. If P is passivation, Q is n-well implant, R is metallization and S is source/drain diffusion, then the order in which they are carried out in a standard n-well CMOS fabrication process is—

1. 1. S - R - Q - P

   
   
   

   2. R - P - S - Q

   
   
   

   3. Q - S - R - P

   
   
   

   4. P - Q - R - S

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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