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# How does this circuit (with JFET) work?

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#### mtkee2003

##### Full Member level 2
hi

*** >> i say: because of the V_GS of J1 is equal to ZERO thus we must say the Drain current is equal to I_D=I_DSS=8mA.
but it is not true, because of I_D is equal to 1mA. ( because of J2 is a current source).
but why *** is not true?

assume that: BETA in JFETs is equal to I_DSS/(VP*VP)
regards.

simple JFET Question

j1 is in triode , because VGD<Vp. it acts such as a resistor. you can compute the impedance of it. it is load of Q2.

simple JFET Question

hi hr_rezaee

it is not true. the V_DG= -0.13 that is J1 works like a current source.

regards

Re: simple JFET Question

J1 is labeled with the wrong polarity of threshold voltage. It should be a positive number since it is a p channel type.

Re: simple JFET Question

thank you..

simple JFET Question

hi

regards

Re: simple JFET Question

hi
V_DG>-|Vp| and V_GS=0 => triode
it is in triode and acts as resistor.

zhi_yi said:

thank you..

it is a diffrential amp. j2 is current source and j1 is resistor , therefor we have a simple diffrential amp.

### mtkee2003

Points: 2
Re: simple JFET Question

thank you,
when do the transistor Q1 and Q2 would turned ON?

simple JFET Question

hi
Q1 and Q2 always turne on because VB~0v and VEE=-10v.

simple JFET Question

hi
thank you my friend

you are right. you can help me and i donate 3 points.
because of its symmetrical circuit, IC1 and IC2 must be equal and then J1 is in the Triod.

regards.
Mostafa, Iran

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