4 ms·
That connector (C13) is rated for 15 amps. That's 180W at 12V.
by vvv5 2y ago
That connector (C13) is rated for 15 amps. That's 180W at 12V.
- wruza 2y agoSo heat depends on current, not on power. My research: https://www.reddit.com/r/ElectricalEngineering/comments/15xfet7/soo_what_actually_generates_heat/ https://www.reddit.com/r/ElectricalEngineering/comments/15xf...
- cwillu 2y agoAnd the card requires up to 600 watts, which is 50 amps if the supply is 12 volts.
- hgomersall 2y agoNo it depends on power, but the power dissipated by the cable, not the power through the cable. The power dissipated is i^2 * r, where r is the resistance of the cable and i, crucially, is the current through the cable which depends on the power it's supplying (which with a resistive load, in this case, is i * 12v).
- gorjusborg 2y agoYes, but it seems the connectors, not the entire cable are too high resistance. Using a larger diameter wire would drop resistance in the cable, but if it has to go through the same connector, it will likely still get hot. Also, NVDA might be telling the truth about poorly seated connectors, that could raise resistance and heat significantly. That could also be handwaving away a business decision to move forward with a design with too little margin.
- dcrazy 2y agoI’m not sure what you’re trying to suggest here… the PSU is also connected to the wall with a C13 connector, and is able to supply 600W at 12VDC to the card.
- vvv5 2y agoYes, but the voltage on the wall side of the psu is not 12V it's 120-240V. And 600W at 120V requires only 5A of current. It's easier to handle more power at higher voltages because heat produced depends on current and not on voltage. If you take that puny C13 connector and move it to the 12V side with its 50A it would just melt.