 # What is the max of a 7805 Voltage Regulator to 5v output?

Hello i looked around on some search pages for the 7805 Voltage Regulator and i found some places that say 9v is the max or up to 15v max input. my input source i have is 12v 800ma would that be okay and enough to run it or would i burn out the 7805 being to much power through it?

Get the correct datasheet for the make of regulator you have and it will tell you the max/min values. For my ST make 7805 the max voltage is 35V but the higher the input voltage the greater the heat produced in the regulator as it drops the voltage by essentially converting it to heat.

Riva thank you very much i did try to look at the datasheet got a little confused . That is why i came here to ask so Thank you for that information the last question i have for this is. my project only takes 364ma of current. at 5v. sense there is 12v coming in would i still need to use a heat sink sense there will be plastic around it be in a inclose small box that the 7805 will be near the edge of the plastic close to the wall?

my project only takes 364ma of current. at 5v.

So, with a 12V input, you'll be burning about 2.5W. That's a lot. Get a cheap switch-mode supply.

Awol yes i would do that but all radio shacks around my area are now gone there was 2 as i was thinking about it i do have some 7805 and caps with me.i need to rig something up for test into i can order one online.

but you are correct i will order one next week.

It's all about power. A 7805 is a 'linear' regulator which means that the power which your load DOESN'T use must be dissipated as heat in the regulator.

Power is measured in Watts. Watts are easy to calculate if you know current and voltage. For this problem, you need to start with the current. Measure or estimate the current that your 5v device will use. In this case 800mA. For the 7805 to supply 800mA at the output, it must have 800mA at the input. It is almost perfectly efficient with the current.

So what is the voltage? The output is 5v and the input is 12v, so that means the voltage difference across the 7805 is 7v. 7v times 0.8A equals 5.6W. Your regulator is pumping out 5.6W of heat. Where does this heat go? It will probably need a heatsink.

Heatsinks are measured by their thermal resistance. How hot does the heatsink get when you put one watt of power into it? Some heatsinks might be 20 degrees C per watt. 5.6 times 20 is 112 degrees. That's really hot. Most electronics might survive this temperature but it will have a short lifetime. A larger heatsink might have a thermal resistance of 11 degrees per watt. This gives 61.6 degrees above ambient temperature, so about 80 degrees in a normal indoors environment. Too hot to touch your finger on it but this might be okay for the 7805.

Then you should look at the datasheet for the actual 7805 you have. It will have specifications for the maximum die temperature or maximum case temperature. This will help you work out if it will survive the temperatures involved.

Hello Morgan thank you very much for that information this project box that the power supply will go into haves a fan is it to bring air in and air vent holes to let the hot hair out but you are right it will need a headsink i have a large one that will Fit in the project box and enough room for air as well. thank you

In this case 800mA

Where did that figure come from? I read 364mA

the power supply is 12v 800ma i read it. the Power from my project reads 364ma at 5v