Cheap/easy/efficient way to switch 5A without a 5A switch?

I've also created a schematic for how I think this is all supposed to be wired up:

Is the switch supposed to just go directly to ground like that, or should there be a resistor?
And is 1M a good choice for the pullup on the gate, or would a lower value be fine / better? Could I use a 10K or would that drain my battery?

  1. P-channel vs N-channel. P-channel, only the MOSFET has power live at its terminal, vs the whole circuit being 'live' waiting for a ground connection to be made.

  2. Power: power dissipated by the MOSFET is I^2R, or 5A * 5A * 0.020ohm with the MOSFET I suggested, or 0.5W.
    Voltage loss across the MOSFET Vds will be V=I
    R = 5A*.02 = 5*.02 = 0.1V.

Smaller resistor would be okay, current into the gate will be minimal.

Smaller resistor would be okay, current into the gate will be minimal.

I take it this means you think my schematic is okay as is then?
How do I calculate how much current will flow into the gate?

only the MOSFET has power live at its terminal, vs the whole circuit being 'live' waiting for a ground connection to be made.

I don't understand the concept of the circuit being "live" when power isn't applied to it. How does it make a difference if +12v line is attached to the regulator, or 0v? It's all relative potential, isn't it?

If you look at sheet 2 of the part you found,
you'll see this
Gate-Source Leakage IGSS VDS = 0 V, VGS = ± 8 V ± 100 nA
So not much current is going to flow into the gate.
When the gate switches on/off tho, you want it to do it quickly so you don't have 5A trying to flow while the part is transitioning from off to on.
A lower value resistor to pull the gate high quickly to turn off quickly is thus better.

The circuit being "live" is a safety thing - do you turn off your high power electric circuits by switching the ground off? Or do you take away the 120V source instead?

The circuit being "live" is a safety thing - do you turn off your high power electric circuits by switching the ground off? Or do you take away the 120V source instead?

Take away the 120V source, but I don't really understand why. First of all, it's AC, so current flows both ways. And second, I know that in a DC circuit electrons really flow from the - to the +, and the holes the other way, and that what charging a battery does is create a surplus of electrons on one side and a shortage of electrons on the other. So if you were to touch either side, assuming you were electrically neutral there ought to be an equal chance of electrons either flowing from you or to you.

I think there's something going on with AC that I don't fully grasp though. I know AC is grounded to earth somehow. I don't know how that works with the current flowing back and forth, but something tells me it's bad to be at ground potential and touch the "live" wire.

Back on the subject of mosfets though...

If you look at sheet 2 of the part you found, you'll see this
Gate-Source Leakage IGSS VDS = 0 V, VGS = ± 8 V ± 100 nA
So not much current is going to flow into the gate.

That part I linked to...

Would that even be okay to use in my circuit if my battery were 12-16v?

The reason I ask is it says under absolute maximum ratings: "VGS ± 8v"
And if I'm not mistaken, if I put 16v in I'll have Vg = 0v, Vs = 16v, and Vgs =-16v, which would appear to exceed that rating

On a related note, if you were looking for a mosfet on Digikey how would you determine what voltage the gate can handle? They don't seem to have a parameter specifically for that. The nearest they have is Vgs(th) (Max) @ Id, but the ratings for that are all really low, like 1.5V @ 250µA, and when I look at the datasheets for the corresponding parts, they can clearly handle far more voltage on the gate than that.

When the gate switches on/off tho, you want it to do it quickly so you don't have 5A trying to flow while the part is transitioning from off to on.

But if the mosfet's resistance is high when it is off, and low when it is on, and the voltage is constant, then doesn't that just mean while it is turning on less current will be able to flow? Why is it bad if it take 1 ms to turn on fully instead of a ns?

A lower value resistor to pull the gate high quickly to turn off quickly is thus better.

Do you have any specific recommendation? I don't know how to calculate how fast it will turn on with different resistances, or how to take this:
Gate-Source Leakage IGSS VDS = 0 V, VGS = ± 8 V ± 100 nA

...and calculate how much current will flow through the gate at different resistances.

Hmm. I'm thinking:
Gate-Source Leakage IGSS VDS = 0 V, VGS = ± 8 V ± 100 nA

Is a maximum regardless of whatever pullup resistor I choose.

I'm also thinking that what I should worry about, besides how fast the resistor I choose can pull up the mosfet, is the leakage from +12v to ground via the resistor when I have my switch closed. Which would be around 1.2mA with a 10K resistor and a 12v supply.

But that's when the device is powered up. The 100nA would be what it draws when it's off.

Now I need to know why starting the mosfet up slow is bad, and how to calculate how fast a particular resistor will pull it up, and how fast I really need to pull it up.

If you're planning to source from 16V, than a part with absolute max of 12V is underrated.

You've got a 16V, 5A source. If you supply power to your circuit, and turn the circuit on/off by connecing the ground, then your circuit is live any time power is applied. And if you happen to touch something inside, then you are liable to act as the ground.
If you turn the 16V on/off tho, then the circuit is not live when power is removed and you can tough anything except the MOSFET and not have to worry.

Same with an AC powered circuit - you have a hot wire, usually Black in the US, a return, usually white, and sometimes ground, usually green.
Return connects to ground back at the circuit breaker box - and black is hot. So if you switch off the hot wire, the rest of the circuit is not active. But if you turn it off by switching the return, then the circuit is live and looking for a ground path - and you don't want that to be you.

Back to digikey
I did a search for p-channel mosfet, single fet, in-stock, with a Vds of 20V and Ids of 6A to provide some margin, then browsed the results for a part with low Rds - and guess what came up:

Or you could save 21 cents and use a part with twice the resistance

Power dissipated: 5A x 5A x 0.02ohm = 0.5W
vs 5 x 5 x 0.41 = 1.025W

I think I'd go with the cooler running part.

You want it to switch fast. If you look at the data plots, you can see that all the best numbers occur when Vgs >4.5V - lowestRds, lowest capacitance, etc.
If you slowly go from off to the on-state, you sit in the higher Rds region longer and generate more heat.

Yes, you'll need a transistor to hold the part in the on-state - a simple NPN, wired as open collector, the collector will go up to 16V when it is not driven by an arduino pin to let the MOSFET gate get pulled high to turn off the voltage.

You keep mentioning that transistor. I'm not sure if you're telling me I have to use it even if I'm using a switch, or whether you're telling me to use it because you think I want to be able to turn the device off with the Atmega.

If it is the latter, letting the Atmega switch itself off is a neat idea, but:

  1. I don't need that functionality.

  2. It'll make the circuit even more complicated than it already is. It took me all day to barely understand the mosfet, I'd rather not throw another type of transistor into the mix.

  3. Half the reason I'm adding the mosfet is to allow the end user to choose whatever power switch they'd like. Allowing them to use a momentary button to switch the board on would be nice, but the device has to be able to be turned off again by the same switch, and the only way I can see to do that is to use a DPDT switch and have the Arduino sense when the switch is pressed again. This would limit the user's choice of switches to DPDT switches which are often more expensive, and it would make the wiring scheme too complicated for folks who don't know much about electronics. It would also use up one of the IO pins on the Atmega, and I'd rather use those for other purposes.

So if I can just pull the gate up to 16V with a 10K resistor, and switch it to ground (without another transistor) to turn it on, then that is what I'd like to do.
If I need the other transistor because I want to run at 16V then I would be fine with dropping the voltage somewhat. I'd prefer 12V minimum but even 10V would be acceptable. I don't really intend the board to be run off 16V. I do expect though a user might want to run it off a 12V source, which I would assume might be slightly higher than 12V. And I will probably recommend they run it off 6V-7.2V.

But I keep saying 16V because I don't yet understand how the voltage ratings relate to what the mosfet can handle at the gate. The voltage rating isn't simply say, 10V. It's Vgs = +-10v. If there's a voltage drop in there somewhere then maybe 16V at the gate and 0V at the source is okay. Or maybe 0V at the gate is always okay cause that's just ground, and anything up to Vs is okay because Vs-Vg = 0.

I think the way I've got my schematic laid out now is fine (I changed the 1M resistor to a 10K) but I don't really understand it completely, and while I could just toss that mosfet you selected in there and be fairly confident it would work, I'd really like to know how you know it will work. Then maybe I can decide later that 16V is a bit excessive and I can choose a less expensive 10V or something.

Ok, just go with a simple switch that grounds the gate & keeps it grounded until released.

Vgate can go from Vdrain to Vsource.

Go with a part that is rated higher than you need, will provide some margin for voltage spikes, etc.

Crossroads,

Thanks for all the help so far.

I have some more questions about that MOSFET you found:

The datasheet says the absolute maximum rating of Vgs is +-12v. Does that mean if I am simply switching the gate to ground to turn the MOSFET on, that I am limited to a maximum of 12v for my battery?

If so, what should I derate that to? 10V?

On page 4 of the datasheet, at the bottom, what does the graph "Safe Operating Area, Junction-to-Ambient" mean? It worries me because it seems to indicate that the chip can only handle 60mA of current continuously at Vds = 10V. (When I say continuously I mean driving it at that level for an hour or more. At the top of the the datasheet they seem to use "continuous" to mean 10 seconds.)

On page 5 though, there's another graph "Current Derating" which lessens my concern. According to that, it would seem the chip can handle 6A as long as the case temp doesn't rise above 125 degrees.

Okay,
In Reply #20 you indicated a need to work at higher voltage.
In #21 I suggested you go with a part with higher ratings.

Search Digikey for a part rated at 24V, 30V, I am sure you can find something with higher ratings that will suit your needs.
You also seem to be able to follow the charts, if you have a question on another part I can look at that part for you.

I'm not asking you to find a higher rated part for me, I just want to be sure that I'm interpreting the datasheet properly.

In regards to following the charts, like I said, I'm not sure if I'm interpreting those properly either because one chart seems to conflict with another, and 0.06A seems awfully low, so I'm not sure what that first chart is supposed to represent.

I understand, and I think you are interpreting correctly.
The 0.06 vs 6.0 does seem a little off, but moot since you need >16V and the part only seems good to 12V.

I don't absolutely need 16V. I'd just prefer to get there since my other components can handle it and it would make my circuit more robust.

The thing about this circuit I'm designing is it's for replica prop builders who don't know anything about electronics. Which means they won't know enough to select a power switch which can handle 5A, and they will want to do bad things like drive everything off a single battery, and they might be using an external amp which requires 12V and the battery they're using puts out 14V fully charged.

And though I can't protect them from every bad thing they might try to do to the circuit, I'm trying to make it as robust as is reasonable, within my space/cost constraints (the board is the size of a credit card) so they don't keep blowing chips up.

Also, though I don't expect they will even use half the power the board can handle 95% of the time, I'm still trying to design it to handle a full 5A 100% of the time just in case.

Few MOSFETs I found while looking for one with greater current capability. I seem to be having much greater success by looking for those with a higher wattage rating than by trying to find ones rated for specific voltages. Though with the leadless packages the wattage rating seems to be way off what seems to be indicated by the graphs. Perhaps these have high internal resistance.

Unless otherwise specified, the below can handle Vgs = +-20V:

This one is $0.55 and rated 55W, but based on Figure 11 on page 4, it seems it can only handle 5A continuous @ 10V input, and this drops to 3A @ 20V input:

This one is $1.20 and rated 81W, and based on figure 3 on page 4, it seems it can handle 5A @ 15V continuous:

This one is $1.71 and rated 120W, but can only handle Vgs = +-15V. Though based on Figure 11 on page 4, it should be able to handle 5A @ 20V, and around 8A @ 12V.

Starting to get into fairly large and pricey ones now though, and once again I'm questioning if this is really the best way to do things. I decided to give another look at voltage regulators to see if there wasn't one which could be switched on and off, and I did find this:

http://www.ti.com/lit/ds/symlink/uc385-2.pdf

I'm not certain how exactly it's used, or how much current it would draw from the switch, or what voltage you'd need to apply to VB to turn it on, but it looks like you might be able to switch the VB pin to turn it on and off. It also is a 5A regulator, but unfortunately the quantity available online is only around 100 anywhere, so it's not really a safe part to go with. But it give me hope that maybe there is another part out there like this. Maybe if I widen my search to adjustable regulators I'll have more luck finding one. The extra resistors would be worth it to avoid having a huge expensive mosfet. I suppose a buck regulator also become a possibility if it's gonna cost me twice as much with the mosfet anyway.

I think I've found my part!

http://micrel.com/_PDF/mic29150.pdf

The MIC29502 adjustable voltage regulator. $4.50, handles up to 5A, and has an enable pin which I can pull high with Vin or 5v logic. Only 20uA needed to enable it.

It's more than twice as expensive as the other 5A regulator I was going to use, but roughly the same price as a separate regulator and a mosfet capable of handing that much current. Also effectively half the size of that two part solution, and now I only have to design one simpler package in Eagle. It's also widely available, which the fixed 5V version of the same regulator was not.

Shame I won't be able to put all that stuff I learned about mosfets to use, but this is clearly the better solution for my particular application. Thanks again for all your help with that though CrossRoads. I'm sure the information will come in handy in some future circuit.

[edit]

I just noticed something else about this regulator, or rather, something I'd failed to notice with the one I was going to use. This regulator has an ultra low dropout... just 0.37v. The original vreg I was going to use with the mosfet had a dropout of 1.3v. That means with the old regulator I couldn't have run my circuit off a 6v supply if I wished; whereas with this regulator, I can. That's a nice perk.

I've also just done some thermal calculations. At first I was concerned because the datasheet seemed to indicate the package could only dissipate around 1W, but then I realized the example was using the worst case TO-252 package which for some reason has a thermal resistance of 56C/W. All the rest of the packages are more like 2C/W, which if my calculations are correct means they can dissipate 25W without a heat sink (other than the PCB).

And based on my calculations, with a 5V output:
12v in 5A out = 35W
12v in 3A out = 21W
7.2v (NiMh, so actually 8.4v max) in 5A out = 17W
6v in 5A out = 5W

Not the 5A at 16v I was hoping for, but not unreasonable. It's highly doubtful anyone will run the device at 5A constantly. It's more likely it will be run at 33%-50% power most of the time. So even 16v in 5A out, which is 55W should still be relatively safe, and I don't expect anyone will run the device at that high a voltage.

Looks like I'm good to go with this part.

How much to go up one more level & get some margin?
With an absolute max of +60V Vin, these can handle 16V. Notes say Max continous supply is 26V.
If you go with a 5-pin TO-247 package you can even bolt a heat sink on if you wanted to.
Dropout voltage drops down to 300mA at 5A too.

Or mount it offboard and let the chassis of a metal box be the heatsink.

MC29750 series - 7.5A