There's a wattage value buried in the 7805 datasheet somewhere. And remember, your circuit is drawing 1.1 watt, but your regulator is dissipating 1.54watt, and you are drawing 2.64watts from your battery total.Originally Posted by lerameur
The amount of temperature rise on the back of the 7805/heatsink/whatever, is directly related to the ability of that 'place' to dissipate heat.
If you've got something that just doesn't give up heat, it'll get hot...if it's very conductive, it'll stay cool.
So, with that being said, even 1 watt can feel very hot depending on what the heatsink is, if there is any heat conductive compound in between the 7805 and the heatsink, ambient temperature, etc.etc.
Think about it this way....take a 40 watt light bulb... It's hot right? All the way around it. It's also pretty big when compared to a 7805 with a heatsink. Now, take that same 40 watt bulb, shrink it 40 times, and apply 1/40th of the heat...sounds like it might be in that same ballpark. Obviously I don't know what the actual numbers would be, I'm generalizing quite a bit, I think it sounds reasonable.
And take another look at what I was saying earlier about how much power the regulator has to dissipate. Do you understand that much of it? (As in why the regulator dissipates 1.54watt when you power it with 12v and only draw .22amp from the regulator?) Remember, a linear regulator, such as a 7805, regulator by dumping extra voltage off as heat...that's the way they work (in simple terms anyways)...kinda like an infinetely variable power resistor that's always turning itself to keep the output voltage at a certain point.
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