# need help with Jfets

Discussion in 'Homework Help' started by shikk, Jun 22, 2014.

1. ### shikk Thread Starter New Member

Jun 22, 2014
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Hi, can someone please help. I'm studying for my exam and i have no idea how to solve this Jfet circuit

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Do you manage to identify the JFET type?
Do you know the equation which describe Id vs Vgs ?
Id = K*(Vgs - Vp)^2
And if you look at the circuit Vgs is equal to ??

3. ### shikk Thread Starter New Member

Jun 22, 2014
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Yes i did. its a P-channel Jfet. I do not no what vGS is equal to. I said the current, ID=vs/100 and VGS=Vs(vs is the voltage across the 100 ohm resistor ). plugged both into the equation and got a value for Vs. from that value i got the current ID, and then the voltage Vr. is this correct? is the 20V the value of Vss?

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
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It look like:
VR1+Vsd+VR=20 volts
VR1+Vsd+R2*Id=20 volts
VR1+Vsd+R2*[K*(Vgs-Vp)^2]=20 volts
It looks to me like VR1 is Vgs. If so:
Vgs+Vsd+R2*K*(Vgs-Vp)^2=20 volts
R2 is known
K is known
Vp is known
If we knew Vsd, we could solve for Vgs, then plug Vgs into Id formula, then use Ohm's Law for VR=R2*Id.
I think I am missing some assumption somewhere. Either for Vgs or Vsd.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The idea is to find Id first using the known conditions with the FET in the saturation region - for (say) the case of R2=1K. In this case Id would notionally be independent of R2 for ideal behaviour. Since the circuit behaves very much like a current source one is then required to consider the limiting conditions as the load R2 increases and the FET enters the linear (triode) region.

Last edited: Jun 22, 2014
6. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,516
515
Isn't in saturation Vgs=0?
So Id=0.001(0-4)^2=0.016 A

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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FET's are considered as operating in the active region when in saturation. Vgs is the controlling parameter which in relation to the pinch-off voltage determines the active region saturation current.

At Vgs=0 the device current is the parameter Idss - the saturation current at zero gate-to-source voltage.

Last edited: Jun 22, 2014
8. ### Jony130 AAC Fanatic!

Feb 17, 2009
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1,116
Very good except one small mistake.
Vs = Vss - Id*R1 and Vgs = Vss - Vs so from this we can say that Vgs = Id*RS.

So we have

Id = K*(Vgs - Vp)^2 = Vgs/R1

And after we solve this

http://www.wolframalpha.com/input/?i=10^-4*%28X+-+4%29^2%3DX%2F100

We have Vgs = 0.14835V and Id = 1.4835mA

Last edited: Jun 23, 2014
9. ### Jony130 AAC Fanatic!

Feb 17, 2009
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No; JFET is in saturation when Vds > (Vgs - Vp)

K = 10^-4 = 1/10000 = 0.0001

10. ### shikk Thread Starter New Member

Jun 22, 2014
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What happens if both Vgs values where smaller than vp? how will you know which one to choose?
thanks

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Very unlikely to happen. But maybe you have some example ?