How to generate the maximal amount of heat given a power supply and a resistor ?


I am currently ordering some components in order to generate some heat.

I have several scavenged power supplies (all around 60W / 80 W, in the 14V/20V range).

Now I would like to short that power supply with one power resistor in order to generate heat. This is the goal of the circuit.

Can you tell me if my calculations are correct, and which variable I need to adjust in order to make the best use of those components ? Here is a reference for one kind of component that I intent to buy:

SODIAL(R) 50 W 8 Ohms 5% Resistance de puissance bobinee vetue d'aluminium Ton or

P 50 W W = V*A
-> I 2.5 A

Power Supply
18 V
60 W
Maximal current 3.33333333333333 A

Voltage drop across resistor
Ohm’s Law: V = R * I
20 V ideally if 50W of heat produced

With this power supply:
current: I 2.25 A
Power used by resistor: P = RII 40.5 W practically
81 % of its capacity

Does it mean that a (8Ω; 50W) power resistor really needs to be powered with a 20V DC power supply in order to optimize the heat loss in the resistor ( vs.. at other places in the circuit that I would not prefer to, because practically, I want to attach that power resistor to a heat sink and dissipate that heat with a fan)é-plaqué-aluminium-résistances-ohms/dp/B0087ZCPJY/ref=sr_1_4?s=hi&ie=UTF8&qid=1458992190&sr=1-4&keywords=25w+12ohm

Two parallel (6ohm) on your 18volt supply = 54watt (3A).
Practically, with some wire losses (volt drop), you might end up with 50watt.
Make sure you add some seagullshit (thermal paste) between resistors and heatsink.

The main limitation of your 18V, 60W power supply is the power rating.
To get maximum power out of this supply, you can calculate the required resistance by using R = V2/P = 5.4Ω

Calculate to see what voltage is needed, V=IR = 18V (same as supply rating).
Calculate what current is drawn trough the load, I=V/R= 3.333A (same as supply rating).

OK R = V2/P = 5.4Ω

Does this characteristic have a name ?

Just Ohms Law and Power.

OK R = V2/P = 5.4Ω

Does this characteristic have a name ?

No... It's just algebra. All of this stuff can be derived from-
Ohm's Law: I = V/R
And the basic power formula: P = V x I

I learned the "Ohm's Law Triangle" when I was in high school and I knew some algebra, but I didn't "get" that it was simply algebra until one of my classmates pointed it out to me.

BTW - About 90% of the time, I find myself calculating power as P = V2/R. That's because I usually know the voltage and resistance. (But, you didn't want to solve for power you wanted to solve for resistance.)

If heat is your goal, use nichrome wire. It's the same thing as a resistor, but with resistive wire, you can shape it to where it is the most effective.


First you need to check if the power rating of your power supply is the Maximum rating or the Continuous Rating.

Use the CONTINUOUS Rating for your calcs, also check that you may need ventilation for the supply.

What is the application, are you heating an oven, 3D printer base plate?

Do you need distributed heat or pinpoint heat.

Thanks.. Tom... :slight_smile: