The given circuit

Given, I_DSS = 10 mA, V_p = – 5V, I_D = 6·4 mA, R_S =?
From given circuit

ID = IDSS		1 -	VGS		2
			Vp

Or		ID	=		1 -	VGS			2
		IDSS				Vp

Or		√6.4	= 1 -	VGS
		10		Vp

or 0·8 = 1 –	VGS
	Vp

Or	VGS		= 0.2
	Vp

or V_GS = 0·2 V_p = 0·2 (– 5)
or V_GS = – 1 V
Since V_GS = – I_D R_S
or – 1 = – 6·4 mA × R_S

or R_S ≈ 150 Ω
Hence alternative (A) is the correct choice.

or RS =	1		= 156·25 Ω
	6·4 × 10–3

ID = IDSS		1 -	VGS		2
			Vp

Or		ID	=		1 -	VGS			2
		IDSS				Vp

Or		√6.4	= 1 -	VGS
		10		Vp

or 0·8 = 1 –	VGS
	Vp

Or	VGS		= 0.2
	Vp

or V_GS = 0·2 V_p = 0·2 (– 5)
or V_GS = – 1 V
Since V_GS = – I_D R_S
or – 1 = – 6·4 mA × R_S

or R_S ≈ 150 Ω
Hence alternative (A) is the correct choice.

---

522. An n-channel JFET has a pinch-off voltage of V_P = – 5V, V_DS (max) = 20V and g_m = 2 mA/V. The minimum ‘ON’ resistance is achieved in the JFET for—

1. 1. V_GS = – 7V and V_DS = 0V

   
   
   

   2. V_GS = 0V and V_DS = 0V

   
   
   

   3. V_GS = 0V and V_DS = 20V

   
   
   

   4. V_GS = – 7V and V_DS = 20V

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

---

---

523. For the series connected JFETs, I_DSS = 8 mA and V_PO = – 4V. If V_DD = 15V, R_D = 5 kΩ, R_S = 2 kΩ and R_G = 1 MΩ, find V_DSQ.
     
     
     

1. 1. – 1·22 V

   
   
   

   2. – 2·44 V

   
   
   

   3. 0 V

   
   
   

   4. + 2·44 V

   
   
   

---

1. View Hint View Answer Discuss in Forum

   The given figure
   
   From above circuit arrangement, we observed that V_GSQ2 = V_GSQ1 – V_DSQ1
   But I_DQ1 = I_DQ2
   Since V_GSQ1 = V_{GSQ2
   therefore VDSQ1 = 0V Hence alternative (C) is the correct choice.}

   Correct Option: C

   The given figure
   
   From above circuit arrangement, we observed that V_GSQ2 = V_GSQ1 – V_DSQ1
   But I_DQ1 = I_DQ2
   Since V_GSQ1 = V_{GSQ2
   therefore VDSQ1 = 0V Hence alternative (C) is the correct choice.}

---

or RS =	1		= 156·25 Ω
	6·4 × 10–3