Electronic Devices - DC Biasing-FETs - Discussion
Discussion Forum : DC Biasing-FETs - General Questions (Q.No. 4)
4.
For what value of RD is the voltage across VDS zero?
2.400 k
5.167 k
6.167 k
6.670 k
Discussion:
7 comments Page 1 of 1.
Rayashinex said:
7 years ago
Vg= R2Vdd/R1+R2.
Vg=1.8V,
Vgs= Vg-IdRs.
Vgs= 1.8- Idss(1-Vgs/Vp)^2 (Rs),
Vgs = - 1.8V.
Then get Id using eqn Vgs=Vg-IdRs,
which I s equal to 2.4mA.
Vds= Vdd-Id (Rd+Rs),
0=16-2.4mA (Rd +1.5k).
Rd= 5.167 k ohms.
Vg=1.8V,
Vgs= Vg-IdRs.
Vgs= 1.8- Idss(1-Vgs/Vp)^2 (Rs),
Vgs = - 1.8V.
Then get Id using eqn Vgs=Vg-IdRs,
which I s equal to 2.4mA.
Vds= Vdd-Id (Rd+Rs),
0=16-2.4mA (Rd +1.5k).
Rd= 5.167 k ohms.
(1)
MR CRISPY said:
7 years ago
@Katekyo.
VGS is not equal to VG in voltage divider circuit id ID is not equal to zero.
VGS is not equal to VG in voltage divider circuit id ID is not equal to zero.
Samuel said:
8 years ago
Vgs = 1.82V (Voltage Divider)
Vgs = Vg - Vs = Vg - Id*Rs.
Substituting it in Shockley's Equation I get the quadratic equation as follows:-
1.41*10^5 Id^2 - 1220 Id + 2.13 = 0.
Id = 2.43mA, 6.23mA (which to choose as both are below Idss).
Also using 2.43mA I get Rd = 5.08k.
whereas using 6.23mA I get Rd = 1.07k.
None of the answers match.
Vgs = Vg - Vs = Vg - Id*Rs.
Substituting it in Shockley's Equation I get the quadratic equation as follows:-
1.41*10^5 Id^2 - 1220 Id + 2.13 = 0.
Id = 2.43mA, 6.23mA (which to choose as both are below Idss).
Also using 2.43mA I get Rd = 5.08k.
whereas using 6.23mA I get Rd = 1.07k.
None of the answers match.
Samuel said:
8 years ago
Vgs = 1.82V (Voltage Divider)
Vgs = Vg - Vs = Vg - Id*Rs.
Substituting it in Shockley's Equation I get the quadratic equation as follows:-
1.41*10^5 Id^2 - 1220 Id + 2.13 = 0.
Id = 2.43mA, 6.23mA (which to choose as both are below Idss).
Also using 2.43mA I get Rd = 5.08k.
whereas using 6.23mA I get Rd = 1.07k.
None of the answers match.
Vgs = Vg - Vs = Vg - Id*Rs.
Substituting it in Shockley's Equation I get the quadratic equation as follows:-
1.41*10^5 Id^2 - 1220 Id + 2.13 = 0.
Id = 2.43mA, 6.23mA (which to choose as both are below Idss).
Also using 2.43mA I get Rd = 5.08k.
whereas using 6.23mA I get Rd = 1.07k.
None of the answers match.
Katekyo said:
8 years ago
I don't get it. My answer is 5.251k.
vgs=16*270k/(2.1M+270k) = 1.823V.
Then solving for Id = 8mA * (1-(-1.823/-4))^2 = 2.37mA.
Then KVL with Rd unknown,
16V-2.37mA * Rd-1.5k*2.37mA = 0.
Rd = 5.251k.
Where did I messed up?
vgs=16*270k/(2.1M+270k) = 1.823V.
Then solving for Id = 8mA * (1-(-1.823/-4))^2 = 2.37mA.
Then KVL with Rd unknown,
16V-2.37mA * Rd-1.5k*2.37mA = 0.
Rd = 5.251k.
Where did I messed up?
Hazem said:
9 years ago
How? Please prove it.
Jee said:
1 decade ago
What is the looping?
(1)
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