Mosfet's Vgs and rds

If I use n-channel MOSFET (IRF3205) for high side switch, I need a gate driver for that MOSFET to turn on. For high switch the mosfets gate require higher voltage than it vcc. I have to switch 50 positive supplies, so I have to give more than 50 V for the gate to turn it on. http://www.irf.com/product-info/datasheets/data/irf3205.pdf

But in datasheet Vgs= ± 20, what that vgs exactly means.

• I tried to correct you capitalization and English, but it’s still not clear what you mean. Can you elaborate on the “50 positive supplies”? Jul 15 '18 at 13:43
• It may be easier to use NPN drive with collector to Pch FET with pullup and series R to limit Vgs to 15V or 1/3 of 50V for Vgs then no 10V supply is needed. Jul 15 '18 at 14:46

It's easier to understand with a drawing: simulate this circuit – Schematic created using CircuitLab

The way you propose to use the NMOS as a high side switch is correct, you have to take the gate voltage above the supply voltage.

Let's use Vctrl = 10 V to open the NMOS and the supply is 50 V, that means there must be 60 V at the gate. Note that that is 60 V referenced to ground.

That -20 V / + 20 V limitation of the NMOS is for the NMOS itself, for its Vgs. That Vgs is the voltage across the Gate and Source of the NMOS. Since the source of the NMOS is at 50 V when the NMOS is switched on, the Vgs will be 10 V, not 60 V. So this condition is allowed and OK.

But watch out! If you now want to switch the NMOS off you might have a problem. Let's say you make Vctrl = 0 V then the gate will be at 50 V. That will not turn off the NMOS will as the voltage at the source will drop until it opens the so much that the NMOS will still conduct somewhat. It will work as a source follower. Avoid that ! What needs to be done is to make the Vgs of the NMOS itself equal to zero.

In order to do that you need to take the Gate all the way down to 0 Volt so ground. That means 0 Volt referenced to ground, just making Vgs = 0 is not enough as explained above. For this we would need to make Vctrl = -50 V !

• I believe naming Vgs that voltage source is quite misleading, Vgs usually means voltage between gate and source. Jul 15 '18 at 11:48
• Your statement about turn-off is incorrect. If you make Vgs =0 by setting the voltage to zero, it will work fine (by definition). If you do it by disconnecting the 10 volt source so the gate floats, then you have problems. Jul 15 '18 at 12:56
• @WhatRoughBeast Fair point, will update to make more clear, I named the source Vgs and that's confusing. Jul 15 '18 at 13:17
• That is by pulling up the gate- mosfet turn on and by pulling down to gnd- mosfet truns off. Right?
– Bud
Jul 16 '18 at 5:23
• @Nihal yes Vgate_on = 60 V and Vgate_off = 0 V. Jul 16 '18 at 12:14

That i have to switch 50 positive supply, so i have to give more than 50v for the gate to turn it on.

The absolute value of gate voltage is irrelevant - it is the difference in voltage between gate and source that will cause a MOSFET problems. With a high side N channel MOSFET with 50 volts on the drain and (say) 60 volts on the gate, the source will be pretty hard-connected to the drain and it will be at 50 volts (minus a small volt drop due to Rds(on).

But you have to be careful; when you activate the gate from a 0 volt level to 60 volts, you have to do so in such a way so as not to leave the "source" voltage behind should it be connected to a high capacitance load. In other words you MUST place a 15 volt (for instance) zener diode between gate and source to clamp the gate drive voltage to no-more than 15 volts. You also have to consider the scenario when turning off the MOSFET - capacitance at the source could keep the source at a high voltage (50 volts) and if you whip the gate down to 0 volts then you get more than -20 volts between gate and source.

So back to back zener diodes are needed OR use a proper gate driver IC.