Calculate voltage drop across Linear MOSFET in saturation region

Considering the below schematic, how is the voltage drop across the MOSFET calculated when it is operated in the saturation region? I am trying to calculate the power dissipated by the MOSFET so that I can choose an appropriate heat sink. RDS(on) is only listed for one V(gs) in the datasheet, but in this use case, the gate voltage will be varied in order to achieve a programable constant current through the MOSFET. For example, see the datasheet for IXTK90N25L2 which is designed to be used in linear mode. The "Input Admittance" graph in Fig. 7 shows the relationship of V(gs) and I(D) but there is no V(DS) listed.

EDIT: I think I might be misunderstanding what the "Output Characteristics" graph is showing. According to this article, that graph is showing the voltage drop. But I don't understand how that graph is being read. Can anyone explain that?

Based on this article, when in the saturation region, the drain current is related to the gate-source voltage and the threshold voltage and not RDS(on). Does this mean that the RDS(on) is always the same for all V(gs) that are above the threshold voltage? If that is true then the transistor in question would only dissipate 16.2W at its rated 90A (.18V=90A x 0.002 Ohm) but then why is it rated for P(D) = 960W?

After reviewing answers on this site (ie this one) related to MOSFET power ratings, they are all referring to switching losses which does not seem to be relevant here.

The "Input Admittance" graph in Fig. 7 shows the relationship of V(gs) and I(D) but there is no V(DS) listed.

$$\V_{DS}\$$ is held constant at 10 volts as per this statement: -

The symbol $$\g_{fs}\$$ (transconductance) is called input admittance in Fig. 7 but should be called transconductance. You'll get used to it! Anyway, the point I'm making is that the table above refers to transconductance ($$\g_{fs}\$$) and therefore $$\V_{DS}\$$ will also "be held" at 10 volts for the graphs.

I think I might be misunderstanding what the "Output Characteristics" graph is showing. According to this article, that graph is showing the voltage drop. But I don't understand how that graph is being read. Can anyone explain that?

There are several gate-source voltage trajectories and when each gate-source voltage is applied, $$\V_{DS}\$$ is adjusted and drain current is recorded. The vertical line corresponding to $$\V_{DS}\$$ = 10 volts yields the admittance graph in your first question above.

Does this mean that the RDS(on) is always the same for all V(gs) that are above the threshold voltage?

RDS(on) is irrelevant for your particular application but, if you want an answer, it's the inverse-slope of the above trajectories when drain current is 0.5 amps (as per the statement in the top picture). The slop is I/V therefore the inverse is V/I which is resistance.

why is it rated for P(D) = 960W?

If you look at the safe operating area graph you can see that it is capable of handling a power of 960 watts if the case temperature is held at 25 degC (harder than you think): -

Approximate estimations: -

• red circle = 250 volts x 4 amps = 1000 watts
• purple circle = 100 volts x 10 amps = 1000 watts
• blue circle = 30 volts x 30 amps = 900 watts
• green circle = 10 volts x 100 amps = 1000 watts

Never get close to these levels in practice except for very short periods of time.

PS, I'm glad to see that you took on board some of my simplified circuit recommendations: -

But you really do need to use the RC network around the op-amp!

• oh sorry about that. I didn't realize that was the final step i should take once my question is answered. I'm new here! :-) and yes! I am definitely going to use the RC network. I just wanted to be as specific with my question as I could without being distracting with the circuit design. Thank you! – Dædalus Wheless Jun 13 at 19:56
• @DædalusWheless no problem dude. Glad to help. – Andy aka Jun 13 at 19:59

The op-amp will control the gate voltage such that the drain current is as commanded by the input voltage. The voltage across the MOSFET, assuming it is able to maintain the set current, is just:

$$\V_{CC}-I_D \cdot (1\text{m}\Omega+R_L)\$$, so the MOSFET power dissipation will be:

$$\I_D(V_{CC}-I_D \cdot (1\text{m}\Omega+R_L))\$$

If you can hold the case temperature to no more than 75°C and are willing to allow the junction temperature to rise to 150°C (not very good for reliability) the lowest curve in the SOA applies, which implies it is thermally limited at 600+ watts. Maximum DC current would be around 57A for Vds = 10V, dropping to 2.5A or so at Vds = 250V.

• Could you provide a link to where you are getting the equation from? Also, using that equation, all cases where Vcc is less that I(D) give a negative voltage drop. Is that correct? – Dædalus Wheless Jun 13 at 3:26
• You have me really confused. Fig. 4 is a temperature graph. The gate is high impedance and does not have an appreciable amount of current flowing through it. You are saying to multiply the gate voltage with the drain current? i don't understand your method here. The gate voltage and drain current are not linearly related. they are related as is given in Fig. 7. Are we looking at the same datasheet? Also, can you explain your equation that will give negative voltage drop? – Dædalus Wheless Jun 13 at 3:41
• The Rds(on) goes up when it gets hot. There's a typo in my above comment which I will correct. – Spehro Pefhany Jun 13 at 3:43
• The equation comes from looking at the parts and adding up the voltages using Ohm's law. Power dissipation of the MOSFET is voltage (Vds) * current (Id). If the current commanded is higher than can be achieved with the Vcc and Rl then the MOSFET will (assuming high enough Vgs- which depends on the drop across the 0.001 resistor, the op-amp and op-amp power supply) look like Rds(on) and will no longer be a controlled constant current source. Rds(on) with >=10V Vgs will be around 50-55mΩ at Tj = 75°C (Fig. 4) – Spehro Pefhany Jun 13 at 3:43
• We don't much care what Vgs is, the op-amp will try to drive it to whatever it needs to be to get the commanded current, which may be impossible (too low Vcc for the resistances) or unwise (outside the safe operating area). – Spehro Pefhany Jun 13 at 3:50