Question

Design a four resistors n-Channel JFET bias circuit for the following specifications: Ip=3mA, Vps = 7V. Assume: Vpp = 14 V, IDss = 6mA, Vp = -5V and I = 0 Find VGs and all resistors values. Draw the designed circuit with all calculated values.

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Answer 1

Solu We have to design a four resistore n-channel JFET voltage divider bias ekt in order to have bias ckl ore ID = 3mA VDS 7 volt Asseme VDD = t4 volt loss - GMA VP ansvolt & IG So the ckt es Given as follows VDD² 14 volt we know that from shockley's equation 2 ID = IDss (1- Vars RE RD VP w 2 D 3 = ex (9- Vio Vors le = 0 A IDss=6MA VP Ja Vp=-5volt Vas ņ NE VP R2 Rs => 1- vas VP z Vus up = ) Vus - 5 x 0.29289 - 1.4644 1014 :. vas - la 4644 volt

The ckt can be redrawn as follows IDD VOD as IG= OA R Ro Po = Is. I did 0 Na & VG VDD X R Peront Vas Is. t RitR2 va Ra RS mm Applying KVL in the loop Hin 3 VG - Vars - Ip Rs = 0 14 R2 3 Rs =-104644 - 00) R + R2 Again Applying Kvi = VDD - PDRO - VDS Rs Rs = 0 14 - 7 Iol Rot Rs) ㅋ 드 ID (Rot Rs) 7 Rot Rs 2.334 K12 3 MA 0 RD + Rs = 2.334KS let The value of Re is the order of IMO 0 8 Rs és IK 2,

0 Rs is as 1 KR RD 1334 KS2. Putting the value of the value of Rs & R is ea ) 1U X R 2 3 Rs - 1.4644 Ret R₂ 3 ) ) 14 R2 Rit R2 3-10 ubuy . 1.5356 π R2 Ret R2 14 - 14 R2 115356 Ret 105357 R2 - 12.4644 R 1a 5356 R Ri 12.4644 2 R la 5 356 =) 8.117 R2 so if R - IMO R 8.117 MO

or in General 14R2 Rit R₂ 3 Rs -1.4644 - 14R₂ = ( 3 R8 -1.4644) (Ret R2, = 14 R₂ - 3 Rs (Ret R2) - 1.4644 (Re AR2) - 14R2 = 3 RSR + 3 R8 R2 -1,4644 R - 1.4644 R2 » R₂ (14-3 Rs + 1.4644). 1.4644 R 3 Rs RI = R = 14+1.4649-3 Rs 3 Rs - 1. ubuy R2 On Ri 15.4644-3RS 3 Rs -1.4644 R R where Rs (2.334 - RD) RQ So the ekt is given by 9 14 volt www. Rp : (2.334 -Re) = 1.334 KN2 8.117M2 R + VDs 7 volt + Vus -1.4644 volt ims - R Rs = Ika