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USB-C connectors can support 100W, which is a lot for such a small connector. So, I wanted to investigate how that works. It would be funny to investigate whether even more power could be supported.

USB-C is specified for supporting 5A at 20V, that’s make 100W. For the transport of large currents, the contact resistance must be small. I have measured the contact resistance myself and it turns out to be 40mΩ per contact and that is exactly according to the specifications (link). The power lines GND and Vbus, each use 4 contacts in parallel, which brings the resistance to 10mΩ.

The power loss at 5A can be calculated with

P = 2 * I2 * R

P = 2 x 5A2 * 10mΩ = 0.5W

### USB-C tested on supporting 2000W

With currents above 5A, the connector will become a bit warmer, and a larger voltage could also not be a big problem. So, I tested the USB-C connector with a 2000W heater at 230VAC. There was no smoke or anything like that coming out of the USB-C connector, it supported the 2000W well. The current in this case is 8.7A and the power loss in the connector is about 1.5W.

You could also connect all 12 contacts in parallel for better supporting high currents and you'll end up with a power loss of just 0.5W at 8.7A, the same as with the specified 5A current.

P = 2 x 8.7A2 * (40mΩ/12) = 0.5W.

### Using USB-C as a 25A power connector

If you connect all 24 contacts in parallel, the total contact resistance becomes 40mΩ/24 = 1.7mΩ. If you accept a power loss of 1W, that is just 2 x the 0.5W discussed above, then the maximum allowable current is √(1W/1.7mΩ) = 25A.