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Best design to switch a load on/off drawing ~5A

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m1tch37

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Hi,

I'm looking to design a simple little circuit to switch on and off a load that in the worst case will draw 5A at 3.3V.
The circuit needs to source the current, not sink it. The load is 192 LEDs and I want them all to have a common anode that I can control.
Furthermore, it needs to have a fast switching time, on the order of 1kHz at least, so I want to use something solid state.

I'm guessing some kind of power transistor will do, but I don't really know what design to use and how to handle heat (if that will be an issue at all.) I was looking at a 2N6488.
Also, is there any special considerations I should make when designing a PCB to handle that much current? Are there any issues I should know about?

I imagine this is quite a simple problem to someone with electronics experience, but I just don't know where to start and want to do it right.

Thanks,
Mitch
 
So you want a 5A high-side switch, switching time in the order of 500us. You could use a NMOS with appropriate high-side driver or a PMOS that can turn on with 3.3V. What voltages do you have available?
 
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Why not put the leds into strings of 3, that way you can use a 12 volt supply and you will draw approx 1.3A max.

This will be easier to manage than 3.3v @ 5A
 
Thanks cobra1, but they all need to be connected to a common anode. I know it may sound strange, but these are the constraints I am working within and I am looking for a solution to this particular problem.

Thanks dougy83, I'm pretty flexible with voltage sources. A friend of mine recommended using a MOSFET, but again I'm not really sure about how to design this.

Mitch
 
If you have a single 3.3V supply, using a logic-level gate driven PMOSFET is the easiest. e.g. a https://www.electro-tech-online.com/custompdfs/2011/06/71937.pdf should be pretty good - <0.024 ohms when on. The source pin connects to the voltage source (3.3V), the drain to the anode of the LEDs and the gate to your control circuitry - 3.3V for off, 0V for on.
 
Hi Dougy83,

Thanks for the link. That MOSFET looks neat, and the SMD package is a plus.

So do you believe if properly soldered I should have no issues with heat at 5A @ 3.3V?

Also, what does a 2W max power dissipation mean?

Thanks for your help,
Mitch
 
Leds

The little guy will be pretty toasty - A little over 100C if you give it the maximum PCBA heat-sink. There are a couple of things we should make sure of before we get to far down the line.
What is Vf of your diodes? (what color)
You say you are flexible as to supply voltage so lets revisit the idea of series/parallel. I think your objective could be met with the circuit below?
This would make the FET standard instead of logic level and lower the current requirement.
Had you planned on a resistor to limit the current through the diodes?
 

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It says it can handle 5.9A continuous in 25°C ambient if mounted on 1"x1" FR4 pcb; the sheet doesn't say the pcb has to be copper fill. The 5.9A is not @ 3V3 - it certainly can't handle that - it's when the device is turned on, in which case the on resistance of fet is ~20milliohms. 20mR x 5.9A^2 = 696mW

The 2W max diss is for 5seconds only; it's 1.1W continuous. The power dissipated depends on the channel (Drain-Source) current multiplied by the channel resistance - which as seen above is 696mW once on. The dissipation will be more during turn on and turn off, but as long as you control it with a decent signal it won't be a problem @ 1kHz (if you increase the frequency, the time spent turning on and off increases & therefore the dissipation increases also).

The thing will heat up a bit, without a heatsink it will be 90°C/W x 696mW = 62°C above ambient (e.g. 87°C at 25°C amb). The traces will act as a minimal heatsink also though.

EDIT: P = IIR, not IR
 
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Dougy,

The formula is I squared X R. So with his 5 amps and Rds on of about .03 ohms with 3.3 volt gate drive it will dissipate .75 watts. Mounted to a 1X1 copper plane gives 110C per watt. So 82.5 C rise over 25C ambient or 105.5C.
Or pretty toasty.;)
 
Thanks guys.

Yes, that is fairly hot. Would a larger transistor dissipate heat any better? I'd rather something that wouldn't get that hot.

Also, in reply to your alternative ronv, that wont work. I should elaborate on my design a little.

I am making an 8x8x8 RGB LED cube. My plan is to have 8 MOSFETs sourcing current and 8 TLC5947s sinking current. This is why the MOSFETs must be connected to all the LEDs common anodes in parallel.

By multiplexing the MOSFETs I now have a 8*196 LED matrix. Also, before you suggest I split it into eight 8*24 matrices, this will not work because it is a 3D matrix (8*8*8 [*3 for RGB]) and doing anything else would greatly complicate the lattice.

I guess my original question might have sounded odd, but when considering my application, there aren't many alternatives.

Now given all of this, it is worth mentioning the MOSFETs will only have a 1/8 duty cycle, but I am still concerned with the heat.

Also, the specs of my LEDs are:
Common-Anode
Forward Voltage: R 1.8-2.2V, G 3.0-3.4V, B 2.9-3.3V
Wavelength: R 620-625NM, G 517-523NM, B 467-473NM
Light Intensity: R3000-4000MCD, G 7000-8000MCD, B6000-7000MCD

Thanks!
Mitch
 
So you would want to use a 5 volt supply so you can turn on the Green and Blue ones.
Dougy's IC would work fine at the 1/8 duty cycle. Here are a couple more with higher current capability just in case:
IRF9204 - maybe over kill
IRL5607f - 40 C per watt.
 
Yes, the supply voltage will actually need to be higher than 3.3v, I'm sorry. Is there any advantage in using an ever higher voltage?

Also, I find no results for a IRL5607f. Do the larger transistors run cooler?
 
Wow I really do need to get my eyes checked.:eek:
How about IRL5602S.
In your circumstance there is no advantage in going to higher voltage.

In the case of say the 5602 the main advantage of the larger package is that it has better thermal conductivity to the heatsink. (40C/W vs 110C/W).
 
That's a great part if you don't mind the small package. The difference in performance is that the one you chose has an Rds on of .0085 ohms, while the one I listed has an Rds on of .042 ohms. So power in the IC is I squared X resistance. So 25 X.0085 = .21 watts vs 25 x .042 = 1.05 watts.
 
Yep, so is there any reason to go for the larger one? Is there a trade off for going to the smaller package?

Mitch
 
No, that part should be good for your application. Don't run more than 5 volts as the supply as the gate voltage is 8 volts maximum.
 
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