136. For the circuit shown below each diode has V_r = 0.6 V and r_f = 0. Both diode will be ON if:
     
     
     

1. 1. V_s > 3.9 V

   
   
   

   2. V_s > 4.9 V

   
   
   

   3. V_s > 6.3 V

   
   
   

   4. V_s > 5.3 V

   
   
   

---

1. View Hint View Answer Discuss in Forum

   VA = VS.	500	=	VS
   	500 + 5000		11

   
   and given cut-in voltage i.e., V_γ = 0.6 for each diode, for V_S > 0.6 V, D₁ is ON., while for V_A > 0.6 V, both the diode will be ON.
   KCL at node A.
   

   VA – VS + 0.6	+	VA – VS	+	VA	+	VA – 0.6	= 0
   5 kΩ		5 kΩ		500 kΩ		500 kΩ

   
   

   VA =	2VS + 5.4	> 0.6 V
   	22

   
   ⇒ V_S > 3.9 V
   Hence alternative (A) is the correct choice.

   
   

   Correct Option: A

   VA = VS.	500	=	VS
   	500 + 5000		11

   
   and given cut-in voltage i.e., V_γ = 0.6 for each diode, for V_S > 0.6 V, D₁ is ON., while for V_A > 0.6 V, both the diode will be ON.
   KCL at node A.
   

   VA – VS + 0.6	+	VA – VS	+	VA	+	VA – 0.6	= 0
   5 kΩ		5 kΩ		500 kΩ		500 kΩ

   
   

   VA =	2VS + 5.4	> 0.6 V
   	22

   
   ⇒ V_S > 3.9 V
   Hence alternative (A) is the correct choice.

   
   

---

137. If ν₁ = – 5 V and ν₂ = 10 V, then output voltage ν ₀ is:

1. 1. 9 V

   
   
   

   2. 9.737 V

   
   
   

   3. 9.5 V

   
   
   

   4. 4.5 V

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

---

---

138. If V₁ = V₂ = 10 V, then output voltage ν ₀ is:

1. 1. 0 V

   
   
   

   2. 9.737 V

   
   
   

   3. 9 V

   
   
   

   4. 9.5 V

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Since both the terminal voltage at 10 V i.e., V₁ = V₂ = 10 V, so we assume that both D₁ and D₂ are ON.
   Apply KCL at node A.
   

   V0 – 5	+	V0 – V1	+	V0 – V2	= 0
   9		1		1

   
   

   V0 =	V1	+	V2	+	5	= 9.737 V
   	1		1		9
   	1	+	1	+	1
   	9		1		1

   
   Check condition:
   

   =	V1 – V0	=	V2 – V0	=	10 – 9.737
   	1		1		1

   
   = .627 A > 0
   Therefore both the diodes are ON.
   Hence alternative (B) is the correct choice.

   
   

   Correct Option: B

   Since both the terminal voltage at 10 V i.e., V₁ = V₂ = 10 V, so we assume that both D₁ and D₂ are ON.
   Apply KCL at node A.
   

   V0 – 5	+	V0 – V1	+	V0 – V2	= 0
   9		1		1

   
   

   V0 =	V1	+	V2	+	5	= 9.737 V
   	1		1		9
   	1	+	1	+	1
   	9		1		1

   
   Check condition:
   

   =	V1 – V0	=	V2 – V0	=	10 – 9.737
   	1		1		1

   
   = .627 A > 0
   Therefore both the diodes are ON.
   Hence alternative (B) is the correct choice.

   
   

---

139. If V₁ = – 5 V and V₂ = 5 V then V ₀ is:

1. 1. 9.474 V

   
   
   

   2. 8.943 V

   
   
   

   3. 4.5 V

   
   
   

   4. 9 V

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Since given V₁ = – 5 V and V₂ = 5 V, so we assume that D₁ is OFF and D₂ is ON
   

   V0 =	V2·9	=	5 × 9	= 4.5 V
   	9 + 1		10

   
   Check assumption
   

   ID1 =	V2 – V0	=	5 – 4.5
   	1 kΩ		1 kΩ

   
   = 0.5 mA > 0, therefore D₂ is ON
   V_D1 = V₁ – V₀ = – 5 – 4.5
   = – 9.5 < 0, therefore D₁ is OFF Hence alternative (C) is the correct choice.

   
   

   Correct Option: C

   Since given V₁ = – 5 V and V₂ = 5 V, so we assume that D₁ is OFF and D₂ is ON
   

   V0 =	V2·9	=	5 × 9	= 4.5 V
   	9 + 1		10

   
   Check assumption
   

   ID1 =	V2 – V0	=	5 – 4.5
   	1 kΩ		1 kΩ

   
   = 0.5 mA > 0, therefore D₂ is ON
   V_D1 = V₁ – V₀ = – 5 – 4.5
   = – 9.5 < 0, therefore D₁ is OFF Hence alternative (C) is the correct choice.

   
   

---

---

140. If V₁ = V₂ = 10 V, then output voltage V ₀ is:

1. 1. 9 V

   
   
   

   2. 9.474 V

   
   
   

   3. 4 V

   
   
   

   4. 8.943 V

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Since both the terminal voltage V₁ and V₂ is at 10 V and connected through the same 1 kΩ resistance. So in this case we assume that both the diodes are ON.
   Apply. KCL at node A
   

   V0 – 0	+	V0 – V1	+	V0 – V2	= 0
   9		1		1

   
   

   or V0 =	V1		+	V2
   	1			1
   	1	+	1	+	1
   	9		1		1

   
   

   =	10		+	10		= 9.474 V
   	1			1
   	1	+	1	+	1
   	9		1		1

   
   Check assumption
   

   ID1 = ID2 =	V1 – V0	=	V2 - V0	=	10 – 9·474
   	1 kΩ		1 kΩ		1 kΩ

   
   I_D1 = I_D2 = 0.526 mA > 0
   Therefore Diode D₁ and D₂ both are ON.
   Hence alternative (B) is the correct choice.

   
   

   Correct Option: B

   Since both the terminal voltage V₁ and V₂ is at 10 V and connected through the same 1 kΩ resistance. So in this case we assume that both the diodes are ON.
   Apply. KCL at node A
   

   V0 – 0	+	V0 – V1	+	V0 – V2	= 0
   9		1		1

   
   

   or V0 =	V1		+	V2
   	1			1
   	1	+	1	+	1
   	9		1		1

   
   

   =	10		+	10		= 9.474 V
   	1			1
   	1	+	1	+	1
   	9		1		1

   
   Check assumption
   

   ID1 = ID2 =	V1 – V0	=	V2 - V0	=	10 – 9·474
   	1 kΩ		1 kΩ		1 kΩ

   
   I_D1 = I_D2 = 0.526 mA > 0
   Therefore Diode D₁ and D₂ both are ON.
   Hence alternative (B) is the correct choice.

   
   

---