# Limit current to a load

I'm making a little heated greenhouse. I have three 1.2 ohm 5W resistors and a 5V 1A power supply. Everything is controlled by my Arduino through a MOSFET. Now, I would just need 5W of power to keep my environment warm, so just one resistor, but by applying Ohm's law I get that it absorbs more than 4A of current, and I don't want my supply to blow up. Even if I use all of my three resistors it would absorb more than the maximum.

Can I limit the current without dissipating it in heat? Thanks.

• Sounds like a job for pulse-width-modulation - something that an Arduino does well. Commented Mar 16, 2021 at 14:41
• @glen_geek doesn't my circuit exceed 1A for every duty cycle? Is it safe? Commented Mar 16, 2021 at 14:52
• What I don't understand: You are trying to build a heater, so what's the problem if you dissipate current as heat?
– PMF
Commented Mar 16, 2021 at 20:32
• Four 5W 20 Ohm resistors in parallel will give you 5 Ohms for 1A at 5V with .25A and 1.25W through each resistor. You'll still probably need a fan or heat sink but it's much less abusive than running a 5W resistor at 5W. You could extend that to 16 1 Watt 80 Ohm resistors to get a high surface area for a fan to blow over. Maybe even 32 0.5 Watt 160 Ohm resistors. You should see the pattern. You would need a separate power supply for your fan, but if you needed lower heating power the array of resistors could be driven with PWM or at a lower voltage.
– K H
Commented Mar 17, 2021 at 6:01
• @PMF if I dissipate it in heat, it would still draw excess current from the power supply, exceeding the limit. Commented Mar 17, 2021 at 10:54

3 × 1.2 Ω in series is 3.6 Ω.

$$\ P = \frac {V^2}R = \frac {5^2}{3.6} = 6.9\ \text {W} \$$.

And the current is $$\ I = \frac V R = \frac 5 {3.6} = 1.38 \ \text A \$$.

Your PSU is only capable of 5 W or 1 A so it's not big enough.

3 × 1.2 Ω in parallel is 0.4 Ω. I'll leave you to work out the current that will draw.

• Well... I=V/R so 1.67/0.4=4.17A... That's worse, right? Commented Mar 16, 2021 at 14:48
• Correct. One more 1.2 Ω resistor will bring you to 4.8 Ω and just a touch over 1 A and would be OK. Commented Mar 16, 2021 at 15:12
• Oh okay... I understand. But I have just 3 resistors and I can't go out as there is lockdown here... I'll try with two fans as Tony suggested. Thanks! Commented Mar 16, 2021 at 15:16

You need a two simple 5V 2.5W fans to circulate the heat and air to run off the 5W supply. Any 5W resistors you use at max power rating will be 125’C and without air flow the temp range will be huge and gradient.

• Oh, that's a good suggestion... I have a fan and I was thinking of using it with an heat sink on the resistor. But this should work too... I'll try! Thanks! Commented Mar 16, 2021 at 15:13
• HTML entity &deg; may help. You also have &Omega;, &mu;, etc. Commented Mar 16, 2021 at 15:13
• Are all 2.5W dissipated into heat? I'm a bit confused here Commented Mar 16, 2021 at 15:20
• No not all power is dissipated in heat , but so let's assumed <50% heat and the rest to move air or create air pressure. Then you can sense temperature better with the forced air motion and lower latency to regulate heat. But you ought use a >5W supply. When I made a picnic box to create a -40'C air temp with -55 dry ice, I found one fan could not achieve -55'C no matter how close it was to the crushed dry ice due to the heat generated by the little muffin fan Commented Mar 16, 2021 at 15:57
• Ok, I understand. Thanks! Commented Mar 16, 2021 at 16:57