Ok, know what a low side switch circuit looks like, but are the requirements the same for both a mosfet and an igbt? The reason I ask is I've seen some igbt low side circuits that have a 'Kelvin' style resistor in series with the emitter and ground. In a low side mosfet, it's my understanding that the source can't have anything between it and ground or it then becomes a high side switch, am I wrong again?

If the igbt is the same in a low side circuit as a mosfet is, how can a kelvin(4 wire resister) be used with it but not a mosfet. Please explain it to me. Thanks

 

As usual, it depends on the circuit. I think the Kelvin connection is only for measuring. If the measured voltage is small, it won't make any difference. This idea about an emitter resistor causing the transistor to become a high side switch makes no sense to me.

There was a thread about which transistor is right for this job, and it came down to how much current. Above some current, the mosfet wins.

 

A high-side switch means the switch is connected between the power supply and ground.
A MOSFET only becomes a high side switch if the resistor in series with the source is the load.
And if it's a standard N-MOSFET with a positive supply then the gate needs to be driven to 10V above the positive supply voltage to be fully turned on as a switch with a grounded load.

 

Ahh, I think I see what is bothering you.
Your kind of right. If the resistor is to large and the current to high the FET won't act as a switch. Lets say to have 10 volts to drive the gate and a 5 ohm load tied to 100 volts in the high side. No problem you put 10 volts from gate to source and the FET turns on hard. - 20 amps in the load.
Now if you put 1 ohm in the source and 10 volts on the gate the voltage at the source will rise to somewhere around the threshold voltage of the FET - say 5 volts and the FET will only be on hard enough to hold the source at 5 volts. So only 5 amps is flowing. So 25 volts across the load, 5 volts across the resistor and 70 volts across the FET. Now the FET is burning a lot of power because it is running in the linear region (which for FETs is called saturation for some perverse reason)
But if the resistor is small enough say .1 ohm the gate to source voltage will still be high enough to turn it on hard (9 volts).
That didn't come out to good, but I tried.

 

Thanks for the answers, guys. I was cleaning out my computer and came across some circuits I had saved, and two motor controls had this in them. It was to measure current in the legs of 3phase motor controls, wish now I'd saved them so I could post them.

Ron you got what I was thinking. As an old, self learner, some of this stuff is hard to understand, very hard. I got into electronics for a project that is mechanically done but still needs the electronics part to be finished. I could use already developed circuits from some of the plans out there, but they are really just basic ways of doing the project. I want something better. I've come a long way from where I started but still have a ways to go.

Is there a limit to the resistance that can be in series with source to ground in a low side switch? I only need between 1 t0 1 1/2 ohms. Can I just add a little extra voltage onto the gate drive volt to overcome source/ground resistance voltage drop? As long as it doesn't go over the maximum gate voltage?

And Ron it took me a lot to get the fet saturation, linear region thing to click.

 

A high-side switch means the switch is connected between the power supply and ground.

Question: Isn't a high-side switch connected between the power supply's positive and the load? (and then to ground)

 

Is there a limit to the resistance that can be in series with source to ground in a low side switch? I only need between 1 t0 1 1/2 ohms. Can I just add a little extra voltage onto the gate drive volt to overcome source/ground resistance voltage drop? As long as it doesn't go over the maximum gate voltage?
.

Yes you can. For example if the most you will drop across the source resistor is 5 volts, you could drive it with 15.