this post was submitted on 15 Mar 2025
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[–] brucethemoose@lemmy.world 208 points 1 day ago* (last edited 1 day ago) (16 children)

Shoudn't it be 25%?

Current is not controlled here, resistance (aka the soldering iron) and voltage are.

Power = Voltage ^ 2 / Resistance. Double the voltage, that quadruples the power. So you only want to plug in 25% of the time to get the equivalent power of 120V.

But it might not melt at double power? Maybe the extra heat helps, I can't find a resistance/temperature curve for a soldering iron...

Source: EE dropout.

[–] uneatable@lemmy.world 13 points 1 day ago (7 children)

Ok. I was acountless on lemmy for a long time, your comment made me finally register. Thanks!

So, yeah, with double the voltage you get 4x the power. But you you put 4 times the power at 50% of the time, you get only 2x the power. And the other half of the time, you get 0 power. On the average you get the same power output.

[–] ch00f@lemmy.world 23 points 1 day ago (3 children)

You double counted there.

You said 4x power 50% of the time and then said “the other half of the time.”

So you’re calculating 50% of 50% which is 25% duty cycle.

[–] uneatable@lemmy.world 16 points 1 day ago (2 children)

Oh no, I didn't. Should I draw a graph? Pop out some equations?

Let's say P is the nominal power. When I said "The other half" I meant when the solder iron is not plugged. So:

50% of the time at 4xP 50% of the time at 0...

Oh shizzzz, you're right!

[–] Riprif@lemmy.world 2 points 1 day ago

I imagine this is more or less what it felt like to be in the room at the time. A whole group of people discussing electrical theory and optimal soldering techniques and meanwhile the one guy standing there holding the actual device notices the power cord is a little loose and pushed it in another 1/8" without mentioning it because everyone is so involved in their nerdy conversation.

[–] Juvyn00b@lemmy.world 2 points 1 day ago

Oh man I was going through it in my head too...

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