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I am hoping someone can help me understand something. I'm an engineering student trying to analyze this circuit, but am finding myself not understanding and going in circles somewhat. I've searched as much as I can but am still stuck.

The circuit is a relay. I did not design it, only analyzing it. The only part of the analysis I am getting hung up on is the voltage across the mosfet (Vds). I feel like I have a good understanding of mosfets and their operating regions. I understand this mosfet is a switching mosfet and must be operating in the linear region. I can confirm this when I measure the voltage in simulation software and get Vds << Vgs - Vth. For the load, I am using a 24V DC LED strip light. The lights dissipate ~3.36W, so I = 3.36/24 = 140mA and Rl = 171 omhms.

Here is my hangup: I want to calculate Vds to determine the current through the mosfet, as well as determine the operating region. The source voltage is 0V, but how can I calculate Vd? I don't know the current through the mosfet, because that is determined by Vds, and I don't know Vd because it depends on the current through the load. I feel like I am missing something.

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The typical IRF540 \$I_D\$-\$V_{ds}\$ relationship is shown in this chart from the IRF540 datasheet:

enter image description here

This shows that with 10 V \$V_{gs}\$, which is about what you could expect to be applied with your circuit, the typical IRF540 will be in triode operation (what we often call "fully switched") with up to about 100 A load.

That means you can essentially treat the MOSFET channel as a resistor when it's in the on state and conducting less than 100 A. The datasheet also guarantees a channel resistance of no more than 77 mΩ with 10 V \$V_{gs}\$.

Now the current through your load will be

$$I_{L} = \frac{24\ {\rm V}}{R_L + 77\ {\rm m\Omega}}$$

again, provided the current works out to less than about 100 A.

If the circuit is well-chosen for your load, it will mean you get darn near 24 V across the load.

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  • \$\begingroup\$ Great explanation. On my simulation, I get ~11.6V for Vgs when the switch is closed. But I am used to needing to know Vds to be able to tell if I can treat the fet as a resistor (that RDS(on) value). But the datasheet only mention that Vgs should be about 10V for this to happen. It's very nice for me that this is the case, but why is this so? Is this just because the properties of this particular fet? I'm at the point where I understand the textbook for the most part, but getting the info from the datasheet is still a practice. \$\endgroup\$
    – usinjin
    Dec 9, 2018 at 0:31
  • \$\begingroup\$ @usinjin, sorry, for some reason I was thinking the gate was driven with 5 V. \$\endgroup\$
    – The Photon
    Dec 9, 2018 at 0:38
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    \$\begingroup\$ So wait. Essentially, from the datasheet, we're not determining how much current can go through the mosfet based on Vds. We're determining Vds based on the amount of current we are feeding through it! Am I on track with this thinking? I was not aware I could think about it like this. \$\endgroup\$
    – usinjin
    Dec 9, 2018 at 1:56
  • \$\begingroup\$ Yes, in a switching circuit like this we usually just make sure Vgs is "high enough" and then treat the channel as a resistor. \$\endgroup\$
    – The Photon
    Dec 9, 2018 at 2:59
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    \$\begingroup\$ @usinjin Yes, in almost all switching circumstances, it's the load that determines the current, and the switching component is turned 'fully' on, and the voltage across it is determined by its residual resistance, and any other parasitics. Trouble comes when we don't turn it on hard enough, its voltage rises and the switch dissipation increases dramatically. Trouble also comes if we don't switch on and off fast enough, and it spends too long in an intermediate region, dissipating high power, check SOA graphs for the handling of high power pulses. \$\endgroup\$
    – Neil_UK
    Dec 9, 2018 at 7:17

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