Electronic Devices - DC Biasing-FETs - Discussion
Discussion Forum : DC Biasing-FETs - General Questions (Q.No. 31)
31.
What is the new value of RD when there is 7 V across VDS?
3 k
3.3 k
4 k
5 k
Discussion:
4 comments Page 1 of 1.
Severina said:
3 years ago
in FET, Ig = 0 and Id = Is.
1. Vgs = Vg - Vs.
so, Vgs = 0 - RsId.
2. using Shockley's Equation, substitute Vgs
Id = Idss(1-(Vgs/Vp))^2.
Id = 10m (1- (-IdRs)/-8))^2.
Shift solve
Id = 2.59 mA.
3. by KVL, considering Vds = 7V,
-20 + Rd (2.59 mA) + 7 + (1k)(2.59 mA) = 0.
Rd (new) = 4.023 kΩ => answer.
1. Vgs = Vg - Vs.
so, Vgs = 0 - RsId.
2. using Shockley's Equation, substitute Vgs
Id = Idss(1-(Vgs/Vp))^2.
Id = 10m (1- (-IdRs)/-8))^2.
Shift solve
Id = 2.59 mA.
3. by KVL, considering Vds = 7V,
-20 + Rd (2.59 mA) + 7 + (1k)(2.59 mA) = 0.
Rd (new) = 4.023 kΩ => answer.
Jay said:
3 years ago
Why solve for the value of RD?
Aren't these resistors already fixed in value? Then why the answer is 3.3k.
Aren't these resistors already fixed in value? Then why the answer is 3.3k.
(1)
Anonymous said:
5 years ago
@Anomy.
Yes, you're right. The value of Id is rounded off to 3mA. So, instead of approximate analysis, we should've used the exact analysis. The correct answer is 4.023K.
Yes, you're right. The value of Id is rounded off to 3mA. So, instead of approximate analysis, we should've used the exact analysis. The correct answer is 4.023K.
Anomy said:
5 years ago
I think, Id = 2.5876mA and drop near Rd as 10.41. Therefor 10.41V/2.5876mA resulting in Rd=4.023K.
Please, anyone, help me what have I done wrong in this. I used Shockley's equation and substituted Vgs=Id*Rs.
Please, anyone, help me what have I done wrong in this. I used Shockley's equation and substituted Vgs=Id*Rs.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers