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I need an electromagnet so I made one out of an old microwave transformer. (something like this: https://www.instructables.com/Microwave-Electromagnet/)

To power this electromagnet I have a DC power supply with a voltage limit of 30V and a current limit of 10A. I measured the electrical resistance of the electromagnet coil to be about 0.46 Ohms.

The issue is that when I power the electromagnet I can only get around 46W of power into it because when my DC power supply is set at 4.6V it is already hitting the current limit of 10A obviously, as 4.6V / 0.46 Ohms = 10A.

My question is is there any way to modify this circuit so I could get higher power to the electromagnet, somewhere in the neighborhood of the maximum 300W the power supply can output? or is this not possible with a 10A power supply.

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    \$\begingroup\$ Rewinding the electromagnet with more turns of thinner wire is one way. \$\endgroup\$
    – user16324
    Commented Aug 6, 2021 at 16:29
  • \$\begingroup\$ A switching DC-DC converter, but at that point just get a power supply for it. \$\endgroup\$
    – vir
    Commented Aug 6, 2021 at 16:30
  • \$\begingroup\$ Current is what gives your electromagnet its pulling force. You're supply is already at its current limit. You could take advantage of its un-used voltage headroom by adding a big DC-to-DC converter (of a "buck" type). You will lose perhaps 30W of the 300W in heat. 270W into 0.46 ohms should result in 24A. \$\endgroup\$
    – glen_geek
    Commented Aug 6, 2021 at 16:32
  • \$\begingroup\$ I will look into a DC-DC step down buck for this. I see the cheap ones on amazon are rated up to a theoretical max of 20A and a recommended of 15A so that's at least a step in the right direction. I'll probably just get one and toss a fan on the heatsink and see if it works for a bit. Thanks everyone \$\endgroup\$
    – Mitchell M
    Commented Aug 6, 2021 at 16:54
  • \$\begingroup\$ You won't be able to touch it anymore. It will be smoking hot! \$\endgroup\$
    – Jeroen3
    Commented Aug 6, 2021 at 18:51

3 Answers 3

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As you said, the DC resistance of the coil is 0.46 Ω. If you want to apply 300 W, then you can use Ohm's Law (and the power law) to determine the other variables:

$$I = \sqrt{\frac{P}{R}}$$

This would give you a current of 25.5 A. This is more than double the limit of your power supply, so you would need a more capable unit.

The voltage would be:

$$E = \sqrt{P R}$$

Or 11.7 volts.

Although your power supply may be capable of 300 watts, you've hit one of its limits (10 A). In order to apply 300 W to the coil, you need to increase the voltage to 11.7, but the current would continue to increase beyond the power supply's limit of 10 A.

Also consider the wire gauge of the coil. If you were to supply enough power to reach 25.5 A through a fine gauge wire its temperature will rise quickly. You want to make sure the coil is rated for the power intended.

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Most likely you are saturating the core with 10Adc.

If you apply 2V and record the current ramp dI/dt in both directions of polarity, then you may see the slope droop 10% at the knee of the saturation point for L max.

The video only used a 1.5V battery.

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Just use a car battery as your power supply. Although not the target amperage your aiming for, it's still stronger than what you have and has worked great for me.

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