I have a microcomputer circuit that runs on 5v. I am using an LM7805 to bring the 13.8 battery power down to 5v and the 13.8 v draw is .34 amps. The LM7805 is getting very hot. The spec says it can output 1.5Amps so why is it getting hot with only 1/3 of an amp?

 

I have a microcomputer circuit that runs on 5v. I am using an LM7805 to bring the 13.8 battery power down to 5v and the 13.8 v draw is .34 amps. The LM7805 is getting very hot. The spec says it can output 1.5Amps so why is it getting hot with only 1/3 of an amp?

How big is the heatsink?

Work out how much power it dissipates (Input Voltage - Output Voltage) * Current.

 

So you have the recommended caps on input and output?
Disconnect connection to micro and test temp.

 

It doesn't have a heatsink.

13.8v - 5v = 8.8v
8.8v * .34A = 2.992 W

It has a 10uf cap on the output.

3 watts doesn't seem like a lot of power to dissipate, but it is getting hot. So, I guess the question is, how big a heatsink is needed to dissipate 3 watts?

 

A TO220 package can manage 1W without a heatsink.

 

but it is getting hot

How hot is hot?

The regulator has built-in thermal protection. An inexpensive heatsink would increase lifespan.

This one will limit junction temperature to about 100C:

It still won't be comfortable to touch the case.

 

I don't have a way to measure the temperature, but it burns my finger, so I will buy one of the heatsinks you posted.

Thank you, Dennis.

 

I don't have a way to measure the temperature, but it burns my finger, so I will buy one of the heatsinks you posted.

Thank you, Dennis.

If you can just keep your finger on it, it's 50°C. If you have to let go it's 60°C. If you lick your finger and it sizzles when you touch the device, it's 100°C.
Note: Don't try it with triacs.

 

The spec says it can output 1.5Amps so why is it getting hot with only 1/3 of an amp?

A TO-220 package has a thermal resistor to air of about 70°C per watt, so with a typical ambient of 25°C and 3W dissipation, the case temperature would be 235°C, well above its maximum rated operating temperature.

 

Would suggest a different approach; a Buck Converter.
The 7805 has to waste 8.8 volts whereas a buck converter will switch on and off rapidly. Depending on the desired output you adjust it down. In the end the buck will be OFF more than it is ON. This results in an average temperature. A rough estimate of percentage of ON time is approximately 40%. It will run a whole lot cooler too.

 

Here's six of them for $10.00 (click the "Here")
This is its operating range:

 

You could also split the power dissipation between two devices by using a pre-regulator. This might not be practical for a one-off project, but it was quite successful in an industrial application. The pre-regulator would consist of a power transistor, Zener diode, and a resistor. Suppose the pre-regulator put out 8V to accommodate the typical dropout voltage of a 7805 and it consumed the same 340 mA. Then:

\( (13.8\text { V}-8\text{ V})\times0.34\text{ A}\;=\;1.97\text{ watts} \)

Follow that with the 7805 regulator which has the following characteristics:


\( (8\text{ V}-5\text{ V})\times 0.34 \text{ A}\;=\;1.02\text{ watts} \)

We have not eliminated ANY power dissipation; we have ONLY divided it between two devices. This would allow you to choose an alternate package for the power transistor which might have superior dissipation characteristics e.g. a TO-3 package while keeping the TO-220 package for the 7805. This solution may not be appropriate in all cases, but it is a possible way to go.

 

My favourite easy-to-use regulator at the moment is the LMR38010. It takes any input up to 80V, only needs 7 external components and just works - provided that you don't forget the decoupling capacitor on the input.

 

Nobody has mentioned....if you can supply a lower voltage, the power dissipation will be less.

A 4 watt incandescent night light bulb will burn your finger.....

 

I am using TO92 78L05 with a 9 pin micro and 12vdc in put with no problem. It can take up to 30v max DC.
Ensure the 7805 has the 100nF at input AND at output otherwise it oscillates and overheats.

 

I have a microcomputer circuit that runs on 5v. I am using an LM7805 to bring the 13.8 battery power down to 5v and the 13.8 v draw is .34 amps.

What is drawing that current besides the micro?

 

It doesn't have a heatsink.
13.8v - 5v = 8.8v
8.8v * .34A = 2.992 W
It has a 10uf cap on the output.
3 watts doesn't seem like a lot of power to dissipate, but it is getting hot. So, I guess the question is, how big a heatsink is needed to dissipate 3 watts?

You didn;t confirm yet if it has the mandatory 100nf on the input & output?

 

You didn;t confirm yet if it has the mandatory 100nf on the input & output?

It has a 10uf cap on the output., nothing on the input.

 

I bought these too.
Amazon.com: EBOOT Mini MP1584EN DC-DC Buck Converter Adjustable Power Module 24V to 12V 9V 5V 3V (6) : Electronics

 

It has a 10uf cap on the output., nothing on the input.

That should be corrected for a start!
Post 3 & 15 (100nf = 0.1uf) each side.