EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: MichaelW on October 14, 2013, 10:37:07 pm
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I have a bunch of laptop power supplies that I've saved for a few years and since I started working on an EE degree I've been wanting to work on some cool electronics projects, and I'm sure at some point I'd have to provide power to something that I'd rather not have to buy batteries for. So, I thought I'd research reusing those power supplies as a possibility.
For instance, say I have a laptop brick that supplies 12v and 3 amps. My current understanding is that it would provide whatever amperage the load pulls UP to 3 amps. So if I had a 12v circuit that needed 1 amp, that power supply would provide 1 amp, and not try to put 3 amps into the circuit, right?
Thanks for any help. :-)
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Exactly correct. The circuit only draws the current it needs.
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Yup, you understand it correctly - the current figure on your PSU states its capability to deliver up to this maximum current. It's a very common beginners' misconception that the voltage and current are two disparate and nonrelated things and you can adjust both of them independently given a constant load. Obviously that's not true and subject to the Ohm's law. Assuming that any load could be substituted with a resistance (very rough simplification but good enough for DC current calculations), I=U/R and you see there's no way to blow your laptop using a power brick with "too many Amps" (and that's exactly what many people seem to think).
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Thanks for the confirmation guys!
I actually had a mini-argument (friendly) with a classmate of mine about this, he was convinced that I'd get the entire 3 amps into whatever I connected it to and I tried to use the I=V/R argument with R being a resistance modeled by the load, but he wasn't swayed by that. :-)