Electronic Devices - Field-Effect Transistors - Discussion

Discussion Forum : Field-Effect Transistors - General Questions (Q.No. 3)

Refer to figure given below. Calculate the value of VDS.

0 V
2 V
4 V
–2 V
Answer: Option
No answer description is available. Let's discuss.
10 comments Page 1 of 1.

Marki said:   1 year ago

20 - Id (1K+2K) -Vs =0.
Vds = 2V.
Or Vs = Id (1k+2k) = 18.
Vds = Vd-Vs = 20-18 = 2V.

Chinnu said:   5 years ago
Please explain the answer clearly.

Katekyo said:   7 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:   9 years 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

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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