How to intentionally overload almost any value resistor with a DC power supply?

Question

When attaching a resistor to my power supply (which is a Extech 80W 3-in-1 Switching DC Power Supply, capable of 36V 5A) and the amps are set to let’s say 5A at 16V the voltage stays the same but the amps will shoot down to something like .11A, not even close to half an amp. I know this is because of Ohm's law but how can I exceed and force the resistor to take in what the power supply wants to put out? Of course it's the same situation on my weaker power supply as well. Do I have to toggle something on the PSU itself? I've seen at least 5 videos of others frying resistors, but they never show the readings on the PSU's display. I know it's possible with my own PSU, I'm just clueless as to how to configure it. I've tried it on a 4 band 120 ohm and a 5 band 1k resistor.

(I'm not a completely destructive mad man, I just have some useless components that no one wants. I want to stress test them out of curiosity instead of throwing them away.)

Answer 1

You can never "force" a current higher than 0.11A through a 120 ohm resistance while maintaining the voltage at 16V. You can either set the voltage or the current, not both. Your power supply is (within limits) a voltage supply, so it sets the voltage. If you want to increase the current, you need to either increase the voltage, or decrease the resistance.

When you set the "Amps" on your power supply, what you are really setting is the limit. This means that the power supply will supply up to 5A, but the actual current is determined by the voltage.

In order to "blow up" the resistor, you would have to exceed its power rating (by quite a bit). These are probably 1/4 watt resistors, so you would need to give them something like 5W if you want to see interesting things happen. For the 120 Ohm one, you would need 25V (P=I*V, V=I*R, P=(V^2)/R, V=sqrt(P*R)=sqrt(5*120)=24.5V).

Answer 2

When you set your power supply to 16 volts and 5 Amps, the supply will maintain 16 volts while delivering UP TO 5 Amps. The actual current delivered will be determined by the resistance of the load.

If the supply delivered 0.11 amp at 16 volts, by Ohm's Law (V = I x R), the load resistance was about 145 Ohms, and the power dissipated in the load was about 1.76 watts. A 1/4 Watt resistor would quickly go up in smoke, but a 5 watt resistor would just get fairly warm.

A 3.2 Ohm resistor would draw 5 Amps from your 16 volt supply. If you connect a 1.6 Ohm resistor to your 16 volt 5 amp supply, the supply would reduce its voltage to about 8 volts, in order to limit the current to 5 amps.

Answer 3

To fry a resistor you need to put into it more power than its rating.

The power (W) dissipated in a resistor (R) with voltage (V) across it is...

W = V^2 / R

Note that the power supply current limit is nowhere in that equation. The current limit setting on the power supply is just a max allowed value. The logic is like this.

if V / R < current_limit then the supply will output I = V / R Amps.
otherwise the supply will output I = current_limit Amps (and the voltage will drop to make it so).

You can see from the power equation, that since we are dividing by R, lower values of resistance yield more power for the same Voltage. For example...

(10V)^2 / 1K ohms = 0.1W (current is 10mA)
(10V)^2 / 100 ohms = 1.0W (current is 100mA)
(10V)^2 / 10 ohms = 10W (current is 1A)
(10V)^2 / 1 ohms = 100W (current is 10A)

So suppose you had a 1/4W resistor and you set the supply to 10V. If you wanted to find the resistor value that would be at 1/4W you would write...

1/4W = (10V)^2 / R

And then solve for R...

R = (10V)^2 / (1/4W) = 400 ohms

To burn up the resistor you might want to set the wattage to 4X that, so make it 100 ohms instead of 400.