Switching 30-VDC at 30-A in Compact Design

I would try a Solid State Relay http://en.wikipedia.org/wiki/Solid_state_relay

Try the Globalspec site: http://search.globalspec.com/ProductFinder/FindProducts?query=Solid%20State%20Relay

Or this is one I found with a quick search on the internet: http://www.futurlec.com/Relays/SSRDC100V40A.shtml

Don't know if any are small enough for your project.

Hello, thanks for the suggestion! It looks like the size of these should be sufficient, but I have some questions. I'm not really an "electronics guy", and have never worked with MOSFETs in a circuit.

From what I've read though, it seems that to use a n-MOSFET as a switch the gate voltage must exceed the source voltage by at least the threshold amount? This would mean that in order to switch a 30VDC supply using the above mentioned MOSFET, I'd have to use at least a 34VDC signal to the gate?

Alternatively, it seems that a p-MOSFET would be able to switch a 30VDC signal with a gate voltage less than the source voltage - threshold voltage? So, assuming a 4V threshold as above, I could switch the 30VDC signal using anything less than 26V -- such as the 5VDC from the card I mentioned in my first post??? What would the "body" be connected to in this case? Would this be a 4-pin/contact MOSFET, as opposed to the 3-pin IRF3710ZS?

Am I confused here, or starting to get this? Any help, pointers, or link to articles would be appreciated!

There are a few ways to approach this. An N channel FET does need several volts (at the gate relative to the source). If you can configure your circuit so the the FET is on the low side - i.e., the source is connected to ground, then the gate voltage is also relative to ground. If you have to have the FET on the high side (i.e., connected to the power rail) there are two approaches. One is to use a P-channel MOSFET as you suggest. The source connects to your power and the drain connect to the load. Tie a resistor from gate to source to pull up the gate (the gate to source voltage difference = 0) to keep if off, and pull the drain down toward ground to turn it on. Use a resistor from the gate to the pull down device so that the ON gate to source voltage is ~ -10V or so.

The other approach is to use N-channel FETs and a MOSFET gate driver chip. This uses switched capacitors to create a gate voltage 10-15V higher than the source, regardless of the voltage at the source. One I've used is the Intersil HIP4080A. You can find cheaper versions, but the data sheet will give you an idea of how they work.

But for simplicity, the p-channel solution is probably the best.

Good luck!

Thanks again for the help and the quick reply!

For whatever reason (probably since I've never worked with them before), attaching the source to ground didn't occur to me. This should be possible in the application, although it looks like there is a diode in the IRF3710ZS that would prevent current from flowing from drain to source... Am I incorrect there?

If I am, then it sounds like it might work. Sounds like I could just tie my 5V signal to the gate, attach my 30VDC power supply to the drain, and then tie the source to ground (through a resistor?) and my intended "output path"? Would that work? Or would it switch off when current started flowing, thus making the source voltage 30V and as a result higher than the gate voltage?

Here's a quick explanation of High side vs low side switches;

http://www2.tech.purdue.edu/Eet/Courses/eet257/Text%20Errata/p395.pdf

In the figures the resistor is your load - whatever you want to control current through, and the switch is the MOSFET.

Without a special gate driver, the high side configuration would be a p-channel, and the low side configuration would be n-channel.

The diode you see in the MOSFET data sheet - that conducts from source to drain - is an artifact of the way the silicon is layered in the device. It has no effect since the current flows in the opposite direction when the MOSFET is ON. It has some use as a flyback diode to suppress voltage spikes if you are switching inductive loads - like relays or motors, but you can ignore it otherwise.

This is just a guess, but it sounds like you could do this fairly easily with an op-amp circuit... They're extremely versatile with this sort of thing.