Simple MOSFET question

[Roff]

You can switch a load which is in the source circuit of an NMOSFET if you raise the gate voltage to 5 volts (for a logic FET) above the voltage on the drain. In your case, that would be 10 volts.
If you want loads in the source and drain circuits, you must calculate what the voltage will be at the source and drain (they will be almost identical if the transistor is on), and then raise the gate at least 5 volts above that calculated value. Always remember that the gate has to be at least 5 volts (in this example) above the source for the transistor to be on, and the voltage across the source load will decrease the voltage between gate and source.

 

[rackley]

For NMOSFETs, the gate must be 5 volts (or whatever the spec says) more positive than the source in order for the FET to be fully on. If you tie the gate to the drain (as in the schematic I last posted), the transistor will be off when the drain (and gate) are at zero volts (actually, any voltage below the threshold). In fact, this is negative feedback, and the drain voltage will be dependent on the individual transistor's threshold voltage, and on the load. Threshold voltage varies from one part number to another, and even varies considerably in those with the same part number.

LOL, ok, now you're confusing me. I didn't see you changed the gate power in the last schematic, I thought you were simply re-asking the same question you posted in the one above that. The LED direction and resistor are correct, the gate is power is not. The gate is tied directly to +5v, so I don't think there is any negative feedback going on.. It should just be Vgs, so the gate is +5v relative to the source, so it's on.

 

[rackley]

You can switch a load which is in the source circuit of an NMOSFET if you raise the gate voltage to 5 volts (for a logic FET) above the voltage on the drain. In your case, that would be 10 volts.
If you want loads in the source and drain circuits, you must calculate what the voltage will be at the source and drain (they will be almost identical if the transistor is on), and then raise the gate at least 5 volts above that calculated value. Always remember that the gate has to be at least 5 volts (in this example) above the source for the transistor to be on, and the voltage across the source load will decrease the voltage between gate and source.

OK I think I see what might be going on. Since I'm using an N-channel and I'm switching a load in the source and applying +5v, it's only allowing 5v-(turn-on threshold)volts to get through? So if the turn-on threshold for a particular FET is .5 volts, it would allow 4.5v though.

Is that correct?

 

[audioguru]

The threshold voltage of a Mosfet is the gate-source voltage that it is barely turned on and has a very low current through its load.
The logic-level Mosfets you are looking at need a gate voltage 5V higher than the source voltage to be fully turned on.

 

[rackley]

Note: Revised #'s with non-toasted FET.

Ok I went back and did some testing now that I understand the difference between source and drain loads. What I tested was with no load at all, just an open drain and a voltmeter to measure the potential between the +12v and +5v rails and the drain.

I tried driving the gate at two different voltage levels, from both a +5v and +12v source with a 1k gate resistor. Connecting either directly to gate resulted in the solder melting off the leads. I tried testing with an 87 ohm resistor but it burned my finger :-/ In the end I used a 1k gate resistor to limit the current.

There figures are all with source connected directly to ground.

Voltages were:

Actual +5v voltage level: 4.99v
Actual +12v voltage level: 13.78v

+5v to gate through 1k resistor:
Voltage at gate: 4.98v
Voltmeter w/red lead connected to +12v rail, black to Drain: 13.78v
Voltmeter w/red lead connected to +5v rail, black to Drain: 4.98v

+12v to gate through 1k resistor:
Voltage at gate: 13.78v
Voltmeter w/red lead connected to +12v rail, black to Drain: 13.78v
Voltmeter w/red lead connected to +5v rail, black to Drain: 4.99v

This FET, STD16NF06L, is supposed to have a Vgs threshold of 1v minimum, no typ or max rated.

 

[audioguru]

Your Mosfet is broken.
The gate of a Mosfet is an insulator that draws no current. Yours drew enough current to cook it and to burn your finger. If it got hot enough to melt solder on its leads then it is destroyed even more by the excessive heat.

 

[rackley]

Edit.. it's amazing what a non-toasted FET will do! Revised the numbers in the previous post.

OK, off to go find P-channel fets or redesign some of my circuits..

THANK YOU SO MUCH FOR YOUR HELP.. I was pulling my hair out for a while there! I'll likely be back with more questions though

 

[philba]

a suggestion - why don't you get a simulator like ltspice and play with MOSFETs. You won't burn out anything other than your eyeballs from late-night sessions and you can see all the various currents and voltages. it also allows you to make changes and try "what if" scenarios very quickly. Once you get a good feel for the circuit, then try it in a breadboard.

though, I have to say, letting smoke out is a time-honored tradition and you can't truely understand MOSFETs until you have a TO220 outline burned into a finger!

 

[Roff]

Edit.. it's amazing what a non-toasted FET will do! Revised the numbers in the previous post.

OK, off to go find P-channel fets or redesign some of my circuits..

THANK YOU SO MUCH FOR YOUR HELP.. I was pulling my hair out for a while there! I'll likely be back with more questions though

Why don't you stick to N-channel, and just put the load in the drain circuit?