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[on:]

Hi Guys,

Im making a giant scoreboard for my local rugby club. I'm using LED strips as 7-segment displays, they are drawing 300ma under 12v per segment.
I have 4 numbers so total power usage is (300 x 7) x 4 = 8,4 A

I'm using a IC for the 7 segement portion, the CD4543. according to the datasheet i can control it with 4 outputs.
My question is, as the IC cant handle the current for the display, how can i use a mosfet to handle the extra power?
I've bought a bunch of IRF830 mosfets, but i think they are too much for the job. They are rated for 500V.

Please some advice on the mosfet part (and maybe the IC-> arduino part.

As thanks i'm making an instructable in the process, for the community

Regards

Mike

[Reply #1 on:]

How are you current-limiting your LEDs? At that power level, a resistor inefficient. I recommend you use an off-the-shelf LED driver, Meanwell makes a variety of them for example. Also take a look at "Buckpucks", they are inexpensive and popular. Just make sure you get ones with an 'enable' pin, usually the ones that are dimmable have one. You'll need one driver module per string /segment.

Then just drive the 'enable' pin of your LED drivers with the CD4543 directly, no transistors required.

[Reply #2 on:]

They are just normal LED strips with built in resistors, as you can get from Ebay. I cant use led drivers, it will be to costly.

Thanks for your reply

[Reply #3 on:]

I've bought a bunch of IRF830 mosfets, but i think they are too much for the job. They are rated for 500V.

The voltage rating is the maximum rating and it's fine to use it in a lower voltage application.   Just about any MOSFET is gong to be rated for more than 12V, so usually that's one spec you don't need to worry about. 

The current rating is also a maximum.   But, you usually need to heatsink the device if you are running near the maximum.    And, you can burn-up the device with less current if you operate it in a linear (non-switching) mode by dissipating power, with current through the device and voltage across it at the same time. (Power = Voltage x Current).

[Reply #4 on:]

If the intent is to drive the MOSFET directly with a logic signal (5V), that could be an issue with the IRF830, it may not turn on fully. The datasheet specifies V_GS = 10V in the test conditions. Might be better to find a "logic level" MOSFET, typically these will specify V_GS ≤ 5V. Just as one example that I've played with a bit, check out the IRLB8743.

OTOH, since 12V is available, the IRF830 gates could be driven by the 12V through some small BJTs.