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How does this circuit (with JFET) work?

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mtkee2003

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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.
 

hr_rezaee

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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.
 

mtkee2003

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simple JFET Question

hi hr_rezaee

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

see its simulated file for more info.

regards
 

flatulent

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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.
 

zhi_yi

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Re: simple JFET Question

hi, please help me to describe how do the circuit above works..

thank you.. :)
 

mtkee2003

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simple JFET Question

hi

download its simulation file and see how it works

regards
 

hr_rezaee

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Re: simple JFET Question

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

Added after 5 minutes:

zhi_yi said:
hi, please help me to describe how do the circuit above works..

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

    mtkee2003

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zhi_yi

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Re: simple JFET Question

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

hr_rezaee

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simple JFET Question

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

mtkee2003

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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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