mosfet gate threshold voltage

Threshold means starting to turn on..

Whichever you like subject to chip power dissipation and die temperature limitations. Google for Motorola's AN1040 which goes into some detail about heatsinking.

Define 50% full on !

It all depends. Cheerfully either.

As ever the devil is in the detail which you will have to learn I'm afraid.

Graham

** Go look up the word " threshold " in a dictionary - fool. ** Why do you keep posting questions you already know the answer to ?

Troll.

** For you - it would rather blow up than be mauled by an idiot.

..... Phil

Starting to turn on, the datesheet will say what current, usually something like a few hundred uA

as long as you can get rid of the heat generated it doesn't matter

what do you mean by 50% on?

as long as you stay within the limits of the datasheet things should be ok

-Lasse

Here you have to define what ON and 100% ON mean. The gate threshold voltage is the gate voltage that will just start to make an appreciable current flow through the MOSFET. Increasing the gate voltage will cause a progressively larger current to flow between source and drain until the current reaches a level limited by the load.

Example: Vth = 4V, P.S. = 10V, Rload = 1 ohm. As you increase the gate voltage above 4V, the drain current will increase, but that current will cause a voltage drop in the load resistor until Id = 10A. When that level is reached, the whole

10V is dropped across the load resistor and drain current can no longer increase even if you keep on increasing the gate voltage. That's when it is said to be fully (or 100%) ON.

You can set the current at any value between 0 and 10A by controlling the gate voltage.

However, if the load resistance is high, a gate voltage of 4V or very slightly above it may be anough to turn the transistor fully on. If the load resistor is, say, 1000 ohms, the maximum current possible is 10V/1000ohms = 10 mA. And that level of current may already be reached at or slightly above Vth.

In the above example with a 1-ohm load, when the current is somewhere between 0 and 10A, that part of the supply voltage that is not dropped across the load resistor exists across the transistor. Now the transistor is absorbing both voltage and current. Voltage*current = Power. And that power heats the transistor.

Suppose you set the gate voltage so that Id = 4A. That will cause a drop of 4V across the load resistor, leaving 6V across the transistor. 6V*4A = 24Watts. Now you have 24W of power heating the transistor, and that will quickly burn it out unless you provide adequate cooling.

Thanks - you've been most helpful

Thanks!

1/2 ID_max

Thank you very much for the helpful replies.

Michael