I find that derivations of formulas are a lot more useful to me than just recitations of formulas. I'm looking at a self-biased JFET circuit, and I'm interested in deriving the quiescent state of the circuit under DC conditions. Neglecting all AC components of the circuit, I have a very large resistor, \$R_g\$, between gate and ground, and another resistor, \$R_s\$, from source to ground. Drain is connected to \$V_p\$, my power supply.


simulate this circuit – Schematic created using CircuitLab

I believe this is a fairly standard JFET circuit. I understand how to analyze the circuit. What I'm interested in is the derivation of \$I_{d}\$. This is given to me as \$I_{d}=I_{dss} \left(1 - \frac{V_{gs}}{|V_p|)} \right) ^2\$. None of the textbooks I've got on my shelf go into the derivation, nor did my google searches. The closest I got was from a textbook that said that the JFET is a square-law component, and this relationship is intrinsic to the component. I don't buy it. Transconductance is intrinsic to the component. Current flow isn't. Can anyone show me how \$I_{d}\$ is derived?

Edit: Sorry, I was asking for the wrong parameter and gave the wrong formula, as well. Question is updated.

Edit: Some discussion from Alfred Centauri pointed me in a good direction. I've done a bit more work that I'll bring up here. It's mostly turned into a math problem, now.

Essentially, I want to derive \$I_d\$ based on intrinsic properties of components. Transconductance is a property of JFETs, so I started from there.

Knowing that \$g_m = \frac{dI_d(V_{gs})}{dV_{gs}}=\frac{2I_{dss}}{|V_p|} \left(1-\frac{V_{gs}}{|V_p|} \right)\$, I can rearrange and work some calculus, as follows.

$$ g_m = \frac{dI_d(V_{gs})}{dV_{gs}} \\ dI_d(V_{gs}) = g_mdV_{gs} \\ dI_d(V_{gs}) = \frac{2I_{dss}}{|V_p|} \left(1-\frac{V_{gs}}{|V_p|} \right) dV_{gs} \\ \int dI_d(V_{gs}) = \frac{2I_{dss}}{|V_p|} \int dV_{gs} - \frac{2I_{dss}}{|V_p|^2}\int V_{gs}dV_{gs} \\ I_d(V_{gs}) = \frac{2I_{dss}V_{gs}}{|V_p|} - \frac{I_{dss}V_{gs}^2}{|V_p|^2} + C $$

We can find \$C\$ as we know \$I_d = I_{dss}\$ at \$V_{gs} = 0\$

$$ I_d(0) = \frac{2I_{dss}(0)}{|V_p|} - \frac{I_{dss}(0)^2}{|V_p|^2} + C \\ I_{dss} = C $$


$$ I_d(V_{gs}) = \frac{2I_{dss}V_{gs}}{|V_p|} - \frac{I_{dss}V_{gs}^2}{|V_p|^2} + I_{dss} \\ I_d(V_{gs}) = I_{dss}\left(1 + \frac{2V_{gs}}{|V_p|} - \frac{V_{gs}^2}{|V_p|^2} \right) \\ $$

This is really close. My target is \$I_{d}=I_{dss} \left(1 - \frac{V_{gs}}{|V_p|)} \right) ^2\$. Think I can switch the signs of the terms by rolling the negative factor into my absolute value denominators?


1 Answer 1


Edit: For a derivation of the JFET equation from physical principles, there are plenty of online resources. Google: "JFET device physics" and, for example, find this. The derivation is not trivial.

I'm not sure where you're getting your information but \$I_{DSS}\$ isn't given by the equation you've written. The following is from Marshall Leach's JFET notes:

In the saturation region of operation, the JFET total drain current is given by:

\$i_D = \beta (v_{GS} - V_{TO})^2\$ for \$v_{GS}> V_{TO} \$

Here, \$V_{TO}\$, the threshold or pinch-off voltage, is negative.

This can be written as:

\$i_D = I_{DSS}(1 - \dfrac{v_{GS}}{V_{TO}})^2\$

From the above, it is clear that \$i_D = I_{DSS} \$ when \$v_{GS} = 0 \$; it is the drain current when the gate and source have the same voltage.

enter image description here

Solving for \$I_{DSS} \$ gives:

\$I_{DSS} = \beta V_{TO}^2\$

  • \$\begingroup\$ Sorry, that's my mistake. My fault for writing questions on here in the middle of class. I'll update the question, since it's way wrong. \$\endgroup\$ Commented Sep 17, 2013 at 16:40
  • \$\begingroup\$ Thanks for the edit. This is good. It's helping me narrow down on what question I'm trying to ask. \$\endgroup\$ Commented Sep 17, 2013 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.