Electrical Engineering - Circuits - FET Transistor - Voltage Divider

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Discussion Overview

The discussion revolves around finding the voltage gain \( A_{v} \) and output impedance \( Z_{O} \) of a FET transistor circuit, specifically focusing on the small signal equivalent circuit and its implications. Participants explore theoretical aspects, mathematical derivations, and potential configurations related to FETs in electrical engineering.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the correctness of their small signal equivalent circuit and seeks help in deriving the output voltage.
  • Another participant confirms the validity of the equivalent circuit and suggests a method to find the output impedance by injecting a voltage and recording the current.
  • Several participants discuss the output impedance \( Z_{O} \) and propose formulas, with some suggesting it should be calculated as \( Z_{O} = r_{d} || R_{D} \).
  • A participant points out a potential issue with the source resistor not being AC bypassed, suggesting a modification to the circuit configuration.
  • There is a debate on whether the configuration is equivalent to a self-bias JFET configuration, with differing expressions for \( Z_{O} \) being proposed.
  • Some participants express uncertainty about the values needed for calculations, particularly \( V_{GS} \), and how to derive them from the given information.
  • Discussion includes the limits of expressions for \( Z_{O} \) as \( r_{d} \) approaches infinity, with participants analyzing the implications of these limits on the correctness of their derived formulas.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating \( Z_{O} \) and the implications of circuit configurations, indicating that multiple competing views remain. There is no consensus on the final expressions for voltage gain or output impedance.

Contextual Notes

Participants highlight limitations in their understanding of certain parameters, such as \( V_{GS} \), and the dependence of their calculations on assumptions about circuit configurations and bypassing components. The discussion reflects ongoing refinements and corrections to earlier claims without reaching definitive conclusions.

  • #31
GreenPrint said:
So then V_{GS} is about -.974 V?


The equation looks like a pain in the but to solve algebraically so i graphed it.

Why is it a pain? It's a simple quadratic in V_S.
 
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  • #32
Oh you I think that was for the previous problem. Apparently he uses

\frac{R_{D}}{1 + g_{m}R_{s} + \frac{R_{D} + R_{S}}{r_{d}}} is this wrong?

This is the reason why I think it's wrong.

Oh I mean it's just a lot of math so I just plugged it into my calculator and graphed it.

However in my book for a Self-Bias Configuration with Rs bypassed they get this

Z_{O} = \frac{1 + g_{m}R_{S} + \frac{R_{S}}{r+{d}}}{1 + g_{m}R_{S} + \frac{R_{S} + R_{D}}{r_{d}}}R_{D}

and a self bias configuration with Rs bypassed has the same exact small signal equivalent circuit as the one in this problem. The only difference is that RG is RTh in this problem

but the electrician already said how it was wrong because when you take the limit as rd goes to infinity you don't get RD

weird I'm learning a lot from this lol
 
Last edited:
  • #33
He doesn't show how he got the original expression but he does this see attached.

when i solved for VGS i got -0.974 V and gm = 5.403 mS

but this

Z_{O} = \frac{1 + g_{m}R_{S} + \frac{R_{S}}{r+{d}}}{1 + g_{m}R_{S} + \frac{R_{S} + R_{D}}{r_{d}}}R_{D}

gives me 1730.826
 

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  • #34
GreenPrint said:
Oh you I think that was for the previous problem. Apparently he uses

\frac{R_{D}}{1 + g_{m}R_{s} + \frac{R_{D} + R_{S}}{r_{d}}} is this wrong?

This is the reason why I think it's wrong.

Oh I mean it's just a lot of math so I just plugged it into my calculator and graphed it.

However in my book for a Self-Bias Configuration with Rs bypassed they get this

Z_{O} = \frac{1 + g_{m}R_{S} + \frac{R_{S}}{r+{d}}}{1 + g_{m}R_{S} + \frac{R_{S} + R_{D}}{r_{d}}}R_{D}

and a self bias configuration with Rs bypassed has the same exact small signal equivalent circuit as the one in this problem. The only difference is that RG is RTh in this problem

but the electrician already said how it was wrong because when you take the limit as rd goes to infinity you don't get RD

weird I'm learning a lot from this lol

What "it" (above, in red) are you referring to?

I did not say this one is wrong:

Z_{O} = \frac{1 + g_{m}R_{S} + \frac{R_{S}}{r+{d}}}{1 + g_{m}R_{S} + \frac{R_{S} + R_{D}}{r_{d}}}R_{D}

I said this one is wrong:

\frac{R_{D}}{1 + g_{m}R_{s} + \frac{R_{D} + R_{S}}{r_{d}}}
 
  • #35
GreenPrint said:
He doesn't show how he got the original expression but he does this see attached.

when i solved for VGS i got -0.974 V and gm = 5.403 mS

but this

Z_{O} = \frac{1 + g_{m}R_{S} + \frac{R_{S}}{r+{d}}}{1 + g_{m}R_{S} + \frac{R_{S} + R_{D}}{r_{d}}}R_{D}

gives me 1730.826

What values did you use for Rs, rd, and RD?

When I use these values, I get:

attachment.php?attachmentid=63757&d=1383882848.png
 

Attachments

  • Eval3.png
    Eval3.png
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  • #36
Ya I think I punched it into my calculator wrong. I retyped it into my calculator and got the answer you got. The second one that my professor used is what is wrong. That doesn't surprise me lol. I have learned a lot from this and was able to solve the rest of the problem.
 

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