Electronic Devices - DC Biasing-FETs - Discussion
Discussion Forum : DC Biasing-FETs - General Questions (Q.No. 25)
25.
Calculate the value of VDSQ.
Discussion:
2 comments Page 1 of 1.
Millize said:
4 years ago
Since, Vgs is 2.32 which is less than Vgsth that is 5 V,
the E-MOSFET would not turn on and therefore Id = 0 so by KVL Vdd-Id(Rd+Rs)-Vgsq=0 ; Vdd=Vdsq.
the E-MOSFET would not turn on and therefore Id = 0 so by KVL Vdd-Id(Rd+Rs)-Vgsq=0 ; Vdd=Vdsq.
Kolii said:
7 years ago
From what I understand, Vdsq = 40V - Id*Rd.
Id= k(Vgs-Vt)^2 & k=(Idon/(Vgson - Vt)^2.
Vgs=(40*18*10^6)/(18*10^6+126*10^6) = 5V,
Id = 1.2 x 10^-4(5V - 5V)^2= 0A.
Vdsq = 40V - 0A*3x10^3 = 40V.
Hope, I did it correct.
Id= k(Vgs-Vt)^2 & k=(Idon/(Vgson - Vt)^2.
Vgs=(40*18*10^6)/(18*10^6+126*10^6) = 5V,
Id = 1.2 x 10^-4(5V - 5V)^2= 0A.
Vdsq = 40V - 0A*3x10^3 = 40V.
Hope, I did it correct.
(2)
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