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JFET Amp need to finish design

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bits1bytes2

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I am having issues with finishing the design for a JFET amp. I am only given the following:
Capacitors - 22uF
RG - 100K ohm
RL = 100K ohm
VDD - 15V
AV - 8 @ 1KHz
JFET - 2N5485

From the above I need to design the Amp and perform all of the calculations to find RD, RS1 and RS2 for the circuit diagram at the following link: **broken link removed**

The issue that I am having is that I can't seem to figure out the RD without knowing the RS values and vice versa.

Please show me the method I should use to find these values. So far my class has been stumped and we weren't able to get any method that works from our professor. The method he used turned out to not work (by his own admission) and he never offered up another method or the actual values that we should end up with. We were only told that the actual values should be different for each person because JFETS don't all share the same Q-Points.

Thanks in advance for any help that anybody can offer. Please let me know if you need more clarification, but I'm not sure I have much more to give as that is about all that is on my sheet.
 
It's not to complicated - however, you need some JFET parameters from the data book, which have rather large tolerances.
According to my data book nominal values are Vgs,off=-4 volts and Iss=10mA.. Based on these values there are several alternatives:

1.) One possible choice is Vgs=-2 volts. Then you can derive the corresponding Id from the data sheet (should be some milliamps).
2.) Compute Rs=Rs1+Rs2=|Vgs|/Id. That is the required dc-effective Rs.
3.) The selected Id corresponds to a certain transconductance g (see datasheet or calculate based on the known quadratic relationship Id=f(Vgs)
4.) Select Rd in such a way that the dc drop across Rd equals approx. Vds; Use Vcc=V(Rd)+Vds+V(Rs)
5.) Calculate the gain using the formula with feedback: AV=-g*R,eff/(1+g*Rs) with R,eff=Rd||RL (approx. R,eff=Rd because RL very large).
6.) Set AV=-8 and find the new ac-effective value for Rs1, which should be smaller than the dc-effective value Rs=Rs1+Rs2.
7.) Use a large capacitor across Rs2 - thus, only Rs1 determines the gain as desired.
 
Ok.. sorry this took me so long, life and a C program got in the way last night.

I already completed 1, 2 and 3, those weren't an issue. The values that I came up with using a Geometric mean for the IDSS and VGS(off) are:

IDQ 2.122 mA
VGS -0.6V

#4 kinda confused me, what do you mean by VCC? I thought that was limited to BJT transistors? And are you trying to say that I need to calculate the Vdrops across Rd and Rs and add them with Vds?
Sorry, I am still learning this, so I am just not sure what you are referring to yet.

I'm trying to work through the DC analysis and then the AC stuff, but I had forgotten that Rs2 would be removed in the AC analysis, so thanks for that reminder.

#7 is not an option as all my capacitors are stuck at 22uF per my assignment, I asked if I could change it and my professor said "No."

Thanks again for your help.
 
#4 kinda confused me, what do you mean by VCC? I thought that was limited to BJT transistors? And are you trying to say that I need to calculate the Vdrops across Rd and Rs and add them with Vds?
Sorry, I am still learning this, so I am just not sure what you are referring to yet.

Sorry for confusing you: Vcc should be read as VDD.
All dc voltages are added: VDD=Id*(Rd+Rs) + Vds.
 
I have all of my calculations completed now. I ended up with the following:

Rd => 1.25K Ohm
Rs1 => 100 Ohm
Rs2 => 33 Ohm
Av => 7.83

My Calculation for VDD (not VCC) ened up to be about 15.42V which is still close, and I'm sure I'm not taking all the Resistance or absolute correct Current into consideration.

My Rs calculation varied between 131 and 136 Ohm, I figured 133 was about as close as real world numbers as I could really create. And my gain ended up a little lower than 8, but it should be close enough. I am not sure if it will work when I build it in the lab but I hope so.

So, my only other question is: Do my above assumptions and calculations seem to be accurate enough and correct?

I will update again when I've completed the actual build of the Amp.

Thanks again for your help.
 
You should be aware that the real circuit (hardware) definitely will deviate from the above results because the JFET used will have other properties than assumed for calculation (tolerances!!)
 
Just wanted to thank you again for your help. You were correct. I was unable to get more than a 3.2 gain from my "real" circuit. I think I may have had a bad capacitor as well, but I was able to finish the lab and I ended up getting a 10.5 / 10 for my grade because I had completed the calculations. I guess most students didn't get them completed.

Anyway... thanks again!
 
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