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I'm building a small LED driver microcontroller board and I'm looking for any advice y'all have on powering it up to 9 A (would take 3 A, if necessary) at 3 - 5 V in a small package. I've not worked with high-side switches before, so I would love any recommendations or feedback you have.

For reference, this is the size of the board (a couple of chips were added recently, so it has even less space):

enter image description here

Hardware switch

I'm currently using a (GPTS203211B) push-button switch. For the size, the highest rating I can find is up to 1 A at 30 V.

Question: If it's rated 30 V/1 A (30 W), could we assume that it would be okay at ~6 A at 5 V?

Otherwise, I feel like I should pair it with a solid-state high-side switch.

Solid-state high-side switch

Any recommendations on a good high-side switch circuit or IC for this application? The board is already tight on space, so smaller and fewer additional parts is better. Through-hole would also be okay.

I'm currently looking at this one.

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"If it's rated 30V/1A (30W), could we assume that it would be okay at ~6A at 5V?" - No, as the current would obviously exceed the maximum rating.

You can use the below circuit to switch your load: enter image description here

You'll have to look for a PMOS with appropriate current rating and a gate-source threshold-voltage below 2.5V, though.

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  • \$\begingroup\$ You probably need a threshold less than 1.5V for reliable switching at 3.3V, unless the spec sheet says it is suitable. Thresholds seem to be highly variable for many devices. \$\endgroup\$
    – PStechPaul
    Commented Feb 16, 2023 at 21:28
  • \$\begingroup\$ @PStechPaul the MOSFET I chose has a GS-threshold voltage of max 1.1V ;) But now as you mention it: even don't use logic level MOSFETs for this kind of gate driving... I figured out, "normal" ones do work quite well at those low load currents. \$\endgroup\$
    – Sim Son
    Commented Apr 20, 2023 at 15:48
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Short take: you're gonna need a bigger FET (and you found one, it should work fine.)

Why?

The switch itself behaves as a low-value resistor when it's 'on'. It sheds power in proportion the current flowing through it, not the voltage being switched.

Your assumption that because 30V @ 1A is the same power as 6V @ 5A is the same power and therefore the switch is ok is fundamentally wrong. Instead, the switch carrying 5A will be dissipating 5 times as much power internally as it would be carrying 1A. It will have 5 times the thermal rise.

The switch will burn....

What about a FET? Same thing applies.

FET on-state resistance is expressed in the datasheet as drain-source on resistance, or Rds(on). Rds(on) is determined by the size the the FET, its technology, and also the gate drive voltage.

Unsurprisingly, larger FETs driven with higher gate voltages have lower Rds(on) and can handle larger currents. Rds(on) will vary between several ohms for a small switching FET like a 2N7002 to single-digit milliohms for a large power FET.

Going back to your example, if you replaced your 1A-rated switch with a 1A rated FET and hit it with 5A, the excess current will heat it up and the FET will be quickly destroyed, just like the physical switch.

That said, the NCP45520 load switch you selected is rated for 10.5A, and has Rds(on) of about 10mOhm. Should be fine.

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