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The end goal is to use an N-channel mosfet (SiS176LDN) to drive as strip of LEDs with a stack voltage drip of 20 V and run at 4 A, I chose this specific mosfet because it seemed to be logic driven. It works as expected as a low side switch i.e. gate pulled to ground and to Arduino PWM and source to ground. Though, it doesn't work as expected as a high side switch i.e. gate pulled up to 3.3 V and to Arduino PWM and source to 5 V - the drain is always floating at ~3 V. I can't use a low side switch as the end goal of my project includes using 3 mosfets for 3 strips of LEDs which would introduce multiple ground planes that I don't want to deal with. Hence, high side is the only approach for me. Also, I had selected this specific N-channel as someone recommended it to me and am stuck with it right now, can't change to P-channel even if that's the better approach.

As I've been reading questions on this site, it became quite clear that the Vgs has to be higher than the source voltage. I've tried switching source to 3.3 V and gate at 5 V that doesn't work. But what I'm even more confused about is how people drive the gate at much higher voltages. On the datasheet, under maximum limits, Vds is 70 V and Vgs is +/-12V. So, if someone is switching say 50V a Vgs >50 V is impossible.

Really confused if gate voltage needs to be greater than source and if so, how and why isn't it working with a 3.3 source and 5 V gate. And if the gate voltage can be logic level i.e. 3.3 V why isn't the switching working and the mosfet always on?

EDIT:

Following the commenters solution for the high side, I implemented something similar but with much lower voltages. The 3.3V, 5V and GND all from an arduino. It should be right based on the calculations. When open Vgs = 0 and when closed, Gate voltage ~8.3 hence Vgs = 3.3V. The multimeter reads odd values tho. enter image description here

EDIT 2: I don't have access to a schematics software atm so please excuse this hand drawn schematic, I have used the schematic of the user that answered to test out the high side switch. On the left side you can see an arduino uno plugged into a laptop.
The n-channel power mosfet : SiS176LDN
and a power supply (from mean well) : which outputs 20V and 3A

Based on the answers below that talk about this working, this should switch on and off with just a 3.3v gate since Vgs is within the range that this should conduct and turn off since the gate is pulled to ground. Instead the output is always on at 20V's

enter image description here

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    \$\begingroup\$ Please a schematic diagram. A picture is worth a thousand words. There is a tool for adding schematics above the edit window when you are editing your question. \$\endgroup\$ Commented Aug 1, 2023 at 19:50
  • \$\begingroup\$ sure i'll do that \$\endgroup\$
    – roaibrain
    Commented Aug 1, 2023 at 19:57
  • \$\begingroup\$ How do I get this question re-opened? I made edits that were required. \$\endgroup\$
    – roaibrain
    Commented Aug 1, 2023 at 21:35
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    \$\begingroup\$ You're confusing Vg and Vgs. Vg is the voltage on the gate relative to some 0V ground reference point in the circuit. Vgs is the voltage from the gate to the source, and has no connection to any 0V ground reference. Vg could easily be 100V while Vgs is only 5V if the source is up at 95V relative to 0V ground. The MOSFET doesn't 'know' or 'care' what Vg is - all it 'knows' and 'cares about' is Vgs. \$\endgroup\$
    – brhans
    Commented Aug 1, 2023 at 21:52
  • \$\begingroup\$ If / when it gets enough votes, the question will be re-opened. \$\endgroup\$ Commented Aug 1, 2023 at 21:52

1 Answer 1

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But what i'm even more confused about is how people drive the gate at much higher voltages. On the datasheet, under maximum limits, Vds is 70V and Vgs is +/-12V. So, if someone is switching say 50V a Vgs >50V is impossible.

\$V_{gs}\$ is the voltage of the gate relative to the source. You are right that \$V_{gs}\$ can't be >50V for your transistor. However, it doesn't have to be. Consider the following circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

When the switch SW1 is open, the gate will be pulled down to 0V by R1, the MOSFET will be off, the load will have no current, and so the drain will be at 50V.

When the switch SW1 is closed, the gate will be pulled to 10V, the MOSFET will conduct, and the load will have whatever current Ohm's Law says it should have.

Voila, we have switched 50V with a transistor whose maximum \$V_{gs}\$ is 10V.

To drive the same load, with the same MOSFET high side, we need to lift the gate voltage, as so:

schematic

simulate this circuit

This time, the 10 V source used to activate the gate is tied to the source of the MOSFET, rather than to ground.

When the switch SW1 is open, the gate will be pulled to the same potential as the source. \$V_{GS} = 0\$. The MOSFET will not conduct. The load will get no current. The voltage at the MOSFET source will be 0.

Now, when the switch is closed, the gate will be pulled to 10V higher than the source, whatever potential that might be. \$V_{GS} = 10\$. The MOSFET will conduct. Most of the 50V supply will appear across the load. The the MOSFET source will be at some voltage slightly below 50V. Say 49.7 V. The drain will be at 50V, and the gate at 59.7 V above ground, but only 10 V above the source. That is, as stated, \$V_{GS}=10V\$.

In real life, when using an N-channel MOSFET as a high side switch, we will use a gate driver circuit to create a gate voltage higher than the source voltage. Such a circuit might involve a "bootstrap" capacitor and diode, or it might involve a small transformer, or any number of ways that, in the end, achieve the same voltage relationships shown in this example.

Now let's consider a circuit that won't work as intended!

schematic

simulate this circuit

In this circuit, the negative terminal of the gate supply is grounded. As a result, the MOSFET source will always be at ground potential. Therefore, there will never be any voltage across the load, and therefore there will never be any current through the load.

The MOSFET, will, however, turn on and off unless/until it is damaged. When SW1 is open, R1 will pull the gate to the same potential as the source. So, \$V_{gs}= 0\$ and the MOSFET will be off. When SW1 is closed, the gate will be pulled up to 10V. Since the source is always at ground potential, \$V_{GS} = 10V\$. The MOSFET will therefore be on. However, because the source is grounded by the 10V supply, the current through the MOSFET will not be limited by the load. This large current could either pull down the 50V supply to a lower voltage, or the large current could damage the MOSFET, or both. Once the MOSFET is damaged, it will no longer switch properly. Often, a MOSFET damaged by over-current will fail "short". That is, it will always be "on".

The 3.3V, 5V and GND all from an arduino. [See edit 1 of original question]

That approach will not work. The negative side of the 3.3V is connected to ground. This is precisely the problematic situation presented in the "WILL NOT WORK AS INTENDED" schematic. The source will always be at ground potential, the load will never get current, the 3.3V supply may be pulled to a lower voltage by excessive MOSFET current, and the MOSFET may be damaged.

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  • \$\begingroup\$ you put this down faster than i've made the circuits, thanks. I understand that this is a low side switching situation, which worked for me. How do I go about high side switching i.e the source connected to the 12V and the other side of the load connected to ground instead of the 12V \$\endgroup\$
    – roaibrain
    Commented Aug 1, 2023 at 20:04
  • \$\begingroup\$ The reason I can't do that is say I have three mosfets switching an led, the led have different voltage drops (didn't mention this sorry), that would mean i'd have 3 different ground's right? \$\endgroup\$
    – roaibrain
    Commented Aug 1, 2023 at 20:06
  • \$\begingroup\$ Look at my high side switching example. The ground is at the low voltage side of the load. If you had three loads, the ground (there would be only one, because, by definition it tells you what is 0 V), will connected to the low side of each load. You will need 3 gate driving circuits however. \$\endgroup\$ Commented Aug 1, 2023 at 20:16
  • \$\begingroup\$ I understand now, I used the exact same high side switch circuit but with 5V instead of 50V and 3.3V instead of 10V. The 3.3V, 5V and GND all from an arduino. It should be right based on the calculations. When open Vgs = 0 and when closed, Gate voltage ~8.3 hence Vgs = 3.3V. The multimeter reads 0V tho. Any idea? \$\endgroup\$
    – roaibrain
    Commented Aug 1, 2023 at 20:42
  • \$\begingroup\$ Show a schematic, if you want a circuit debugged. \$\endgroup\$ Commented Aug 1, 2023 at 20:44

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