Electronic Devices - Field-Effect Transistors - Discussion
Discussion Forum : Field-Effect Transistors - General Questions (Q.No. 19)
19.
Refer to figure show below. Calculate the value of VD.
Discussion:
5 comments Page 1 of 1.
Jayant Minz said:
1 decade ago
VDD=+20v
ID=6mA
RD=2Kohm
VD=VDD-IDRD
=>VD=20V-6mA*2kohm
=>VD=20-12
=>VD=8V
ID=6mA
RD=2Kohm
VD=VDD-IDRD
=>VD=20V-6mA*2kohm
=>VD=20-12
=>VD=8V
Katekyo said:
8 years ago
IDRD is not straight up VD? why? please explain.
Kurukawa said:
7 years ago
Vds =Vdd-Id(Rd+Rs)
= 20-6mA(1k+2k) = 2V.
Vds=Vd-Vs,
2V = Vd - (-6mAx1k).
8v = Vd.
= 20-6mA(1k+2k) = 2V.
Vds=Vd-Vs,
2V = Vd - (-6mAx1k).
8v = Vd.
Qwer said:
4 years ago
IdRd is not straight up VD because it is connected in between 2 potentials, VDD and VS, VDD and VS has an effect on the value of VD.
For VS, you can compute directly using IDRS since it is directly connected to the ground.
For VS, you can compute directly using IDRS since it is directly connected to the ground.
(1)
Jefferson said:
4 years ago
By Voltage division theorem.
Vx = Vt(Rx/Rt) 12(2k/2k+1k).
= 8v.
Vx = Vt(Rx/Rt) 12(2k/2k+1k).
= 8v.
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