well because i still need all of these components. Its drawing way too much. 1.1 a out of 1amp max.
what about putting 2 Lm7805 in parallel?
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well because i still need all of these components. Its drawing way too much. 1.1 a out of 1amp max.
what about putting 2 Lm7805 in parallel?
Last edited by lerameur; - 2nd February 2007 at 02:28.
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dual supply
Honestly Ken,Originally Posted by lerameur
I do not know if that works or not, That is a linear IC and I am not too sure how they will interact. I think a better solution would be to use a bigger regulator IC, or use 2 or more 7805s as seperate supplies using a common ground. Your pic can sink current from the secondary supply, and your display could also run from it. Note: check the data sheet for maximum current the pic can handle in total, so as not to exceed.
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You just said you were drawing .22amp...
Which one is it? 1.1 a or .22 a ?
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sorry 1.1watt
These babies can draw 1amp, why cant I draw more then .22 without being very hot ?
I dont see the wattage value on the spec sheet of this chip,
Last edited by lerameur; - 2nd February 2007 at 03:02.
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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.
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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well it dissipates 1.54watt because it can take 1watt , so 2.54 watt total from the battery. I guess I need this part: L78S05CV for now
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You could 'preregulate' the 12v input with either another regulator ( 12v -> 7809 -> 7805), maybe a few diodes in series, etc.etc. Knock the voltage down a bit before doing the final regulation to 5v. That'll cut down on your heat.
What package is this 7805 that you have right now? TO-92? TO-220?
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it is TO-220,
by the way I once saw a sheet with all the packages types on the digi key web site, I cant seem to find it anymore. Would you have a similar sheet?
But instead of using 7809 and 7805, I think I will use a 2amp or maybe a 3 amp regulator it wont cost more.
k
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