# Derivation of I_d of a self-biased JFET

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 $V_{gs} = 0$

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

Therefore:

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

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?

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.

Solving for $I_{DSS}$ gives:

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

• 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. – Jason_L_Bens Sep 17 '13 at 16:40
• Thanks for the edit. This is good. It's helping me narrow down on what question I'm trying to ask. – Jason_L_Bens Sep 17 '13 at 18:19