# JFET amplifier problem

Discussion in 'Homework Help' started by PsySc0rpi0n, Jul 6, 2015.

1. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
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I have this new problem to solve.

The attached circuit is an amplifier with a JFET!

We are asked to find Av, Ri and Ro for that circuit with gm = 10mA/V and Rd = ∞ (I think Rd is the same as 1/hoe).

First step is to identify the type of amplifier! I would say that this is a common-source amplifier but I'm not sure!

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2. ### kyka New Member

Jun 7, 2015
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No, it is not common-source, it is common-drain (the input is being applied in the gate and the output is being taken from the source).

With some fast approximate calculations, it turns out that

Av=0.94, because it is a common-drain topology and there is a resistor divider at the input
Ri=180 kOhm, because it is a JFet
Ro=500 Ohm because the source resistance is much smaller than the 10k resistance. In absence, of the 10k resistor the Ro would be 1/gm.

3. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,232
5
kyka, and can you tell me about if my equivalent small AC circuit and hybrid model are correct? And also how to identify common-drain/source/gate amplifiers?

Last edited: Jul 7, 2015
4. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,232
5
With the hybrid model I have that I hope is correct (please anyone confirm it with the screens attached at 1st post), I was trying to calculate the voltage gain but I'm not quite sure how to do it:

Av = Vout/Vin

Vout = gm*Rs//Rload <-- Is this correct?

For Vin, has Ig = 0 I would say that would be:
Vin = (Rg1//Rg2*Vsig)/(Rsig + Rg1//Rg2)

But I don't know Vsig so I can't calculate Av.

Jan 29, 2010
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hi,
E

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6. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,232
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After reading and looking for those schematics in that link, I can't understand what is done for Vout.

For Vout I was multiplying the current gm*Vgs*(Rs//Rload). Wouldn't this be the voltage drop at that parallel that would also be the voltage drop at R_load?

7. ### ericgibbs AAC Fanatic!

Jan 29, 2010
3,223
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Calculate the Output impedance for the Common Drain, then calculate Av [voltage gain]., it will be less than 1

You are given Vin [Vsig], calculate actual Vgate signal level.

You now know the actual Vin' and Av so calculate Vout.

Is this what you are asking.?

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8. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,232
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Kind of...

From the other link I didn't understood what was being done to calculate Vout.

By your reply, do I have (mandatory) to calculate output impedance prior to calculating Av?

9. ### ericgibbs AAC Fanatic!

Jan 29, 2010
3,223
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How could you calculate Av without knowing the total Load resistance ie Rs in || with R_Load, as Zout [Rs] is required in the equation for Av.

10. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,232
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I usually try to calculate Av by means of Vout/Vin. And for Vout I say it would be:

and for Vin I would say maybe:

Vin = Vgs but I think this one is wrong.

11. ### ericgibbs AAC Fanatic!

Jan 29, 2010
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Using the PDF equations and your equations, please post the calculations for Vout , using both methods.

E

12. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,232
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I'm using now the PDF equation for Vout but I will ignore Rds because if this Rds is what I think, 1/hoe, we were learned to ignore it as it is usually too high that will not affect calcs!

Edited;

Attached are my results using both PDF equations and gif equations for the common-drain.

Edited1;
From my class notebook I have that for a common drain setup:

Av = (Rs*gm*Vgs)/(Vgs+gm*Vgs*Rs) = gm*Rs/(1+gn*Rs)

which matches the one from the gif but I don't understand why Vin = Vgs+gm*Vgs*Rs.

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13. ### Jony130 AAC Fanatic!

Feb 17, 2009
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From kvl we have Vi = Vgs + Id*Rs and Id = Vgs*gm
So we have this Vi = Vgs + Vgs*gm*Rs.
Vout = Id*Rs = Vgs*gm*Rs

Vout/Vin = (Vgs*gm*Rs)/(Vgs+Vgs*gm*Rs) = Rs/(1/gm + Rs)

Last edited: Jul 8, 2015
14. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,232
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Ok, thanks Jony1030.

For Ri, I'm thinking about the paths that current can take to ground, so I think Ri = Rg1//Rg2 = 1meg*220k/(1.220meg) = 180.33kΩ.

For Rout I think it's the current flowing through Rs because this Rout is from the R_Load point of view, so Rout = Vout'/Iout.

Vout' = gm*Vgs*Rs
Iout = Id = gm*Vgs

Rout = Vout'/Iout = gm*Vgs*Rs/gm*Vgs = Rs

Gathering all info:

Ri = Rg1//Rg2 = 1meg*220k/(1.220meg) = 180.33kΩ

Rout = Rs = 1kΩ

Feb 17, 2009
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16. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,232
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Ok, sure about Ro. The path for the current generated by imaginary Vs to GND flows through the parallel of gm impedance which is 1/gm and Rs.
So:
Rout = (1/gm)//Rs = 1000*1000/(2000) = 500Ω

But Ri is correct, right?

Feb 17, 2009
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18. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,232
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After noting a small units error at Av, I recalculated it and got:

Av = 0.833

The error was that gm = 10mA/V. So I should use 10*10^(-3) and I used simply 10 which was wrong!

So:

Av = 0.833
Ri = 180.33kΩ
Ro = 90.91Ω

Also Ro was not correct!

19. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,166
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Well for gm = 10mS (milli siemens ) and RL = 1kΩ the JFET gain is
Av =(Rs||RL)/(1/gm + Rs||RL) = 500Ω/(100Ω + 500Ω) = 0.833V/V
But the overall gain is less than 0.833 because Rsig together with Rin form a voltage divider.
Aover = 180.33kΩ/(180.33kΩ + 10kΩ)*0.833 = 0.7892V/V

Rin = 180.33kΩ and Rout = 1k||100Ω = 90.90Ω

20. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,366
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You've chosen Av to be defined as Vsource/Vgate. Are you sure you shouldn't take into account the loading effect of the input impedance on the source impedance, so that Av would be Vsource/Vsig?