Resistors for high-powered LED circuit

Hi I’m planning a project which includes a couple of circuits for bright LEDs.
One of them draws 350mA with a forward voltage of 2.2V and the other draws 700mA with a forward voltage of 2.4V. I’m planning on powering them via separate 5V from USB cables.
My calculations say I require 6.8Ohms for the first circuit and 5.6 for the second, both rated at 1W.
Is there a way I can achieve this with components from Core Electronics? I can’t seem to find resistors rated above 0.5W on the site. Could I use potentiometers? When a potentiometer is decribed as 200Ohms does that mean 0-200Ohms? Also the potentiometers on the site are only rated at 0.1W.
I would rather get something from Core so as to avoid a separate shipping fee.

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Hi Andrew. Don’t know where your numbers come from. My calcs as follows
5V minus 2.2V = 2.8. To drop this @ 0.35A requires 8 ohm @ 1W. nearest preferred value 8.2 ohm
5V minus 2.4V = 2.6. To drop this @ 0.7A requires 3,7 ohm @ 2W. Nearest preferred value 3.9 ohm.
I think at these currents you would be pushing the USB ports quite a bit. I would tend to use a separate 5V supply or a powered hub with this capability.
Re potentiometers. Yes a pot described as 200 ohm would be a 0 - 200 ohm resistor when connected as a rheostat… Be absolutely aware that the quoted power handling capability is rated over THE ENTIRE LENGTH of the resistive element. Derating must be considered depending on the amount of resistive element used. Not a good idea using carbon pots with DC current. They tend to get very noisy and unreliable. Wire wound is the way to go when used as a rheostat to control DC currents. I think the only higher wattage pots commonly available would be wire wound anyway.
Cheers Bob Rayner

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Andrew. Follow on
Have you considered using dedicated LED driver modules. I haven’t personally had anything to do with them but I think they are constant current devices and are selected according to the current required. You then supply a high enough voltage with the required current capability and away you go (I think).
Good luck BobR

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When two resistors are connected in parallel then their overall power rating is increased. So two 0.5W resistors in parallel could dissipate 1W. (Power Resistors).
So your application could use two 0.5W resistors in parallel with each one having a resistance of twice the total.

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I will not buy into multiple resistor pros and cons but I will say this: Allow a bit of “fudge factor” even as much as 100%. That is if 1 watt dissipation is required allow for up to 2 and so on. What is sometimes not considered is derating for proximity effect, if 2 resistors are connected in parallel and are close to each other each one will contribute to the heating of the other so a derating factor applies. This means that a theoretical capability of 1 watt for 2 off 0.5W resistors does not apply. This could be significant if they are mounted close together. I cannot quote actual numbers but AWA in their “Standard Practice Manual” (3 large volumes, I wish I had a copy now) published formulas and derating curves and graphs for this effect. I have found unless into serious design I only have to be aware of this effect and if sailing a bit close to the wind go up a bit. For instance in the above example I would tend to use 2W instead of 1 and 3W instead of 2. Hope this helps.
In Andrew’s case using LED drivers gets around the problem of wasteful hot resistors. Core have such devices. Single units SKU COM-13716 or triple (3 individual on single board) SKU COM-13705.
Cheers BobR

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Hi All,

Fantastic to see everyone helping each other out!

While the failsafe (original) USB standard is 500mA you can definitely get that much current from a USB power supply without much trouble. It’s not uncommon to see 2.4A USB supplies and USB-C can (after some handshaking) do up to 100W at up to 32V.

Robert is on the money with the calculations, and with using multiple resistors to share the load. If doing this it’s generally best to do it with resistors in series so you can guarantee the current, else there’s a fair chance you’ll still end up with most of the current through one and then you’ll just get a cascade failure.

Definitely better to go with an LED driver, but resistors are cheap and even though you’re wasting as more energy as heat than you’re using to drive the LEDs, a whole lot of nothing is still nothing.

If you do want to go the resistor route, I’d suggest grabbing some power resistors that will just be well over specced for what you need:

Just be mindful that these can get hot!

Support | Core Electronics

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Agreed series is better so same current flows through both resistors. In both cases the same derating consideration applies if the resistors are mounted side by side due to mutual heating effects.
Good advice re using power resistors. Better safe than burnt offerings.
Cheers Bob

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Thanks much appreciated help. I’m going to try the femtobucks LED drivers first and see how I go.