Engineering Electrical Engineering - Circuits - FET Transistor - Voltage Divider

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The discussion focuses on calculating the voltage gain (A_v) and output impedance (Z_O) of a FET transistor circuit. Participants analyze the small signal equivalent circuit and express concerns about the accuracy of the derived equations. They emphasize the importance of correctly identifying the relationships between the drain current (i_D), output current (i_O), and the resistors involved. Confusion arises regarding the treatment of the source resistor and the implications of bypassing it with a capacitor. Ultimately, the conversation highlights the need for careful analysis and verification of the circuit's parameters to arrive at the correct expressions for gain and impedance.
  • #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
 

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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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