I have a simple question about a depletion type jfet current transduction equation. It's from the book "ARRL handbook of radio communications".
Id=Idss(1-Vsg/Vp) ^2
I can't seem to make sense of it, first I'm not sure why its Vsg instead of Vgs. But most importantly if Vsg is the same voltage as Vp the drain current does not come out to be zero. In fact the equation can be simplified to (1/Vp -Vsg/Vp)^2 *Idss
When Vsg=Vp the drain current doesn't equal zero, in fact the smaller Vsg is the larger he drain current, I've seen other equations that make more sense like Idss(1-Vgs/Vp)^2 which makes much more sense.Could someone please help me understand.

 

The correct equation should look like this

\(\text{I_{\small{D}}=I_{\small{DSS}}\(1- \frac{V_{\small{GS}} }{V_{\small{p}}}\)^2 \ \)

 

Yeah that's what I thought. Is any one else that has read the ARRL handbook for radio communications able to explain what the equation is the way it is?

 

The equation show in ARRL has only a small error.
They swap Vsg to Vgs. And I don't see any other error.
Because for Vgs = Vp we have

Idss * (1 - 1/1)^2 = Idss*0 = 0A

 

Actually I didn't write the equation clear enough it actually goes
Idss[(1-Vsg)/Vp]^2

 

Do you have any link to this ARRL handbook about JFET?

 

Actually I didn't write the equation clear enough it actually goes
Idss[(1-Vsg)/Vp]^2

Must be wrong. V is a voltage and cannot be subtracted from "1".

 

The equation is in the lower right, I hope the text is large enough to read.

 

Well clearly they have something printing error. Try to find errata for this book.

 

Ok thank you , you have been very helpful