# MOSFET Constant Current Source

Can someone explain to me how to set the IREF current with just RSET? One end of the resistor has VDD which is fixed but how do I know what the other end (gate/drain of Q1) will be??

If Vdd >> Vgsth, then the variation of Id with Vgsth will be fairly small.

If Vgsth is not well controlled, or Vdd is very low, then this circuit is too simple to deliver a settable, or even constant, current.

• So Q1 will just about turn on and remain on. Because it is diode connected, it will always be in saturation. And the voltage at the gate will be the threshold of Q1? And there is very little control over that (apart from the body)? Commented Apr 21, 2022 at 17:20
• Does this circuit not need a startup of some sort? There's no current being forced in Q1 like in a normal current mirror, so technically, the resistor is perfectly happy with 0A and so too is Q1 which will just stay off? Commented Apr 21, 2022 at 17:20
• @alayoiskgfbfqhxjiw No, it will start up fine as the resistor pulls the gate high. As it starts to conduct, the current through the MOSFET and resistor will lower the gate voltage to reach equilibrium. Commented Apr 21, 2022 at 17:50
• @Finbarr Thanks Commented Apr 21, 2022 at 17:54

Take, for example, the 2N7000 MOSFET and look at figure 1 in the data sheet: -

Because the drain is connected to the gate, $$\V_{DS} = V_{GS}\$$ hence we can draw a line on the above graph like so: -

So, given that current into the gate is zero (DC), you can estimate $$\V_{GS}\$$ (aka $$\V_{DS}\$$) based on the current into $$\R_{SET}\$$. This then means you can calculate $$\R_{SET}\$$ based on $$\V_{DD}\$$.

So, if you were looking for a drain current of 250 mA, $$\V_{DS}\$$ would be about 3.6 volts. If $$\V_{DD}\$$ is 10 volts then, $$\R_{SET}\$$ equals (10 - 3.6)/0.25 = 25.6 Ω typically.

But, remember there will be considerable variation if you bench tested several devices from different batches of MOSFET.