Electronic Devices - Field-Effect Transistors - Discussion
Discussion Forum : Field-Effect Transistors - General Questions (Q.No. 3)
3.
Refer to figure given below. Calculate the value of VDS.
Discussion:
10 comments Page 1 of 1.
Marki said:
3 years ago
@Kvl.
20 - Id (1K+2K) -Vs =0.
Vds = 2V.
Or Vs = Id (1k+2k) = 18.
Vds = Vd-Vs = 20-18 = 2V.
20 - Id (1K+2K) -Vs =0.
Vds = 2V.
Or Vs = Id (1k+2k) = 18.
Vds = Vd-Vs = 20-18 = 2V.
Chinnu said:
6 years ago
Please explain the answer clearly.
Katekyo said:
8 years ago
It's a JFET so basically, if there is no voltage applied to the Gate then the source-to-drain path is complete and will be operating.
Priyanka said:
1 decade ago
It must be 0v because drain to source path is not completed as we are not giving supply to the gate.
Shailesh said:
1 decade ago
Apply KVL to outer circuit we get the required value.
Sunil said:
1 decade ago
Well to find out voltage drop across rs one should apply the voltage division formula. It says. R1(apposite value resistor)/r1+r2 *current flowing through them.
KAVITA RATHORE said:
1 decade ago
vdd-[ID(RD+RS)]=20-[6mA(3K)]=20-18=2V
Hrishikesh said:
1 decade ago
Vdd-[Id(Rd+Rs)]= 20-[6mA(3K)]= 20-18 = 2v
Bnarasimhulu said:
1 decade ago
I want to formula for this question.
Surya said:
1 decade ago
Vdd-[Id(Rd+Rs)]= 20-[6mA(3K)]=18V
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