Electronic Devices - DC Biasing-FETs - Discussion
Discussion Forum : DC Biasing-FETs - General Questions (Q.No. 23)
23.
What are the voltages across RD and RS?
Discussion:
6 comments Page 1 of 1.
Millize said:
4 years ago
k = Id(on)/(Vgs(on)-Vgs(th))^2 = 0.12 mA/V^2.
Vg = (40V)(18M)/(18M+220M) = 3.03 V,
KVL loop.
Vg-Vgsq-IdRs=0; Id = k(Vgsq-Vgs(th))^2.
Substituting;
3.03-Vgsq-((0.12mA/V^2)(Vgsq=5)^2)(0.82k) = 0,
Solving for Vgsq we get 2.32 V,
For E-MOSFET to turn on Vgsq should be greater than Vgsth,
But here 2.32<5 therefore E-MOSFET would not turn on.
Vg = (40V)(18M)/(18M+220M) = 3.03 V,
KVL loop.
Vg-Vgsq-IdRs=0; Id = k(Vgsq-Vgs(th))^2.
Substituting;
3.03-Vgsq-((0.12mA/V^2)(Vgsq=5)^2)(0.82k) = 0,
Solving for Vgsq we get 2.32 V,
For E-MOSFET to turn on Vgsq should be greater than Vgsth,
But here 2.32<5 therefore E-MOSFET would not turn on.
Kimi said:
4 years ago
@Rdatthin.
How can you get 2.32? Explain please.
How can you get 2.32? Explain please.
RDatthin said:
4 years ago
Vgs = 2.32 when calculated which is less than the Vth , therefore Id=0.
Abdul jakul said:
5 years ago
According to me, 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.
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.
Exynos said:
6 years ago
k = ID(on)/(Vgs(on)-Vth)^2 ,
Vth = Vgs(th),
Therefore Id = 0,
V=0.
Vth = Vgs(th),
Therefore Id = 0,
V=0.
Shashanka said:
8 years ago
How it is 0?
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers