# LED driving circuit, related to MOSFET V(gs)

The N-MOS used is an IRF3710. If you look at the IRF3710 datasheet, VGS is ±20 V.

However, in the picture, 24 V is added to VGS.

PWM signal: 24 VDC, 0 VDC
Is it okay to think of it this way?

What I'm curious about is that the datasheet says that VGS is 20 V or less, but in the circuit I implemented, a PWM signal of 24 V is applied.

I am wondering if there is a circuit problem.

• +/- Vgs is a "maximum" voltage ... You don't need it. 5V PWM should be enough. See figure 1 of infineon.com/dgdl/… Max Vds = ~ 100 V. Commented Feb 5 at 7:39
• Try adding a resistor in series with your LED. Commented Feb 5 at 7:41
• 24 V PWM is "not" ok ... It is "too" great for Vgs max +/-20 V ... Commented Feb 5 at 9:48

Maybe it works, as you comment, but it is not OK. 24 V is above the absolute maximum ratings.

It works because 24 V is above Vgs-th (4V) and keeps the MOSFET on, and 0 V keeps it off, while it survives. But it can work for one hour, one year or for one minute. It's out of specs.

You can see from this:

That if the drain and gate are tied together and the drain current is modest, then at most you need $$\V_{_\text{GS}}\le 4\:\text{V}\$$. There's also a transfer curve, later:

That typically at $$\V_{_\text{GS}}= 5\:\text{V}\$$ the drain current can be quite high regardless of operating temperature. Certainly, high enough for most LEDs.

You will need to use a method to limit the LED current. If you are excited by the idea of using this particular MOSFET and if your supply voltage for the LED is more than about $$\2\:\text{V}\$$ (which it is, as you say) more than the LED itself needs, then you can add a BJT like this:

simulate this circuit – Schematic created using CircuitLab

So, if the LED current is $$\20\:\text{mA}\$$ then $$\R_1\approx 10\:\text{k}\Omega\$$, $$\R_2\approx 22\:\text{k}\Omega\$$, and $$\R_{_\text{SET}}\approx 33\:\Omega\$$.

Feel free, though, to use a different value for $$\I_{_\text{LED}}\$$ that meets your needs.

$$\Q_1\$$ can be any junkbox small signal NPN BJT.

($$\R_1\$$ and $$\R_2\$$ are arranged to provide a voltage divider that keeps the NFET gate voltage within a valid/acceptable range.)

In the circuit you have 5V PWM so you have 5V on gate and 0V on source, so a Vgs of 5V, not 24V.

Also the LED is used without a resistor to limit current, but also no single LED will handle 24V so it must be a LED light fixture that works at 24V.

• Unlike the notation in the picture, A 24VDC PWM signal was given to V(gs). 1. The operation is going well. 2. However, in this case, will the MOSFET break down? I wonder. Commented Feb 5 at 7:46
• @contain If there is really 24V on gate, it exceeds the 20V maximum rating and can degrade slowly or it might suddenly damage at any moment. The FET is not meant to handle more than 20V on gate. Due to manufacturing tolerance, some FETs might handle 24V forever and some might blow up at 21V. Commented Feb 5 at 9:18