+5V Regulator 7805 Problem

I have an Arduino project I have been working on. The Arduino controls multiple sensors that need to run off 24V. I only have room for one power supply, so I thought I would use a 24V power supply and then use the 7805 Voltage Regulator to turn the 24V to 5V for the Arduino. When I connect the 7805 in my circuit I get 5V from the output, but it also changes the 24V on the other side of the 7805 to 5V. I must have something wrong. My circuit looks like the picture I attached. Here is the data sheet as well (http://www.datasheetcatalog.org/datasheets/228/390068_DS.pdf). Any help would be great! Image (http://yfrog.com/637805p)

What kind of power supply is it? If you have something like a laptop charger it might vary in voltage depending on the circuit it is connected to.

Have you tried measuring from 24V+ to ground [u]without[/u] going through the 7805? If you measure 24V there then you should get the same voltage assuming you connected it properly.

Here is the power supply, it’s a regulated power supply.


Outputs 24V very smooth. I think I had the regulator in backwards, but I had another one. Tried it and it gets very hot very fast. Is this usual?

When the 7805's output is not connected the the Arduino the regulator works great and doesn't get hot at all. When connected the voltage drops to about 2V which isn't enough to run the Arduino and the regulator gets very hot. The load of the Arduino shouldn't be too much for this regulator as I have seen other people use them. I am connecting the 5V output to the Arduino's Vin pin and the ground is connected to the ground pin right next to it.

@eSietsema, it probably isn't a good idea to connect 24v to the Arduino's regulator. It doesn't have enough heat sinking to handle the voltage drop. If you have regulated 5V it should go into the 5V pin of the Arduino, not Vin. Vin goes through the regulator.

@mxc1090, Assuming you have the 7805 connected correctly, it is probably going to be getting very hot. Linear regulators drop the voltage difference from their input to their output. Multiple that difference by the amount of current being drawn and you get the amount of power the part has to dissipate.

24v - 5v = 19v.

Let's just say the Arduino draws 100mA. So,

100mA * 19v = 1.9W

You are dissipating almost 2W of power through that 7805. Without a heat sink, you probably can't dissipate more than 250-300mW safely.

I think I had the regulator in backwards, but I had another one. Tried it and it gets very hot very fast. Is this usual?

Yes. You are dumping a [u]lot[/u] of power as heat (which is what a linear regulator does) with a 24V input.

Use an appropriate heat sink on the regulator, heat sink compound between regulator and sink, ensure air flow around the sink, etc.

You may want to consider a switching regulator instead. They're more efficient.


If I step up to the 7812 which is a 12V regulator how big of a heat sink would I need? If you could list a specific type or model that would be great! Thanks

I think you should look into switching step down regulator to take that 24v down to 5v. http://www.pololu.com/catalog/category/84 Will get a nice 5v for the arduino without making a ton of heat.

how big of a heat sink would I need?

It depends on how much current you draw. For a discussion about this see:- http://www.thebox.myzen.co.uk/Tutorial/Power.html Along with:- http://www.thebox.myzen.co.uk/Tutorial/Power_Examples.html