L7805 destroyed - why?

Question

I used a L7805 to power a Raspberry PI and it worked for about 10 seconds then started to smoke and die. L7805 is rated for 1.5A and 35V DC input. Power supply was 24V DC (was actually two plugin 12V DC 1A adapters wired in series) that was metering 30V DC without a load (will obviously drop close to 24V DC with a load). 24V DC is industrial controls standard. I didn't put the caps on the input and output which I now see in the datasheets. Would omitting these caps cause the meltdown?

Because the input is 24V DC and output is 5V DC there would be a lot of heat generated. Would it be better to use a switching power circuit? Datasheet says heatsink is needed for anything over 1A. The PI uses less than 1A without any devices connected which I didn't have any connected.

What else could I do to make the 7805 more robust (other than adding the caps of course)?

Thanks

Answer 1

Linear regulators (like the 7805 and similar parts) have a very simple characteristic.

Power dissipated = \$ (V_{in}-V_{out}) \cdot I_{out} \$ (plus \$ V_{in} \cdot I_{q} \$ where \$ I_{q} \$ is about 5 mA for the 7805).

That means that even if we have a 10 V average input (about the lowest that is practical if unregulated power is used) the dissipation at 1 A out will be 5.05 W, which requires a fairly large heatsink or a smaller heatsink and a fan.

You must satisfy all constraints on the datasheet simultaneously, not just the ones that happen to attract your attention. The absolute maximum input voltage is 35 V, and you should make sure never to even get close to that. There is a maximum output current, and there is a maximum power dissipation. If you dissipate too much power for the heatsink etc. the chip gets too hot and the lifetime is compromised, sometimes dramatically.

The current version of Raspberry Pi 3 uses a lot of current, as much as 730 mA plus whatever is plugged into those USB ports. That's why we generally use a 2.5 A wall wart.

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TL;DR: The L7805 is totally unsuited for this application. If you have an industrial application (and still want to use a Pi) you can buy a DIN rail-mounted supply.

However, the Pi is not a hardened industrial computer, so you may have other fascinating discoveries to come.

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Edit:

There are drop-in replacements for the L7805 which use a switching regulator. They would be suitable for 24 VDC input (not 35 VDC, not 24 VAC rectified/filtered or not). They do not require additional heat sinking and some may be adequate for your output current.

Here is one from Murata capable of 1.5 A. They are actually quite inexpensive- probably less expensive than a 7805 + heatsink:

Answer 2

Your 7805 has to drop say 19 V if your effective input is 24VDC .Even if you took 100mA you would be burning 1.9watts and your finger if you touched it .You must use a big heatsink because your input /output differential is large .This is why Buck convertors are so popular.

Answer 3

In my opinion you are doing it wrong in 3 ways.

• You have two 12V transformers in series, since this does not sound like a streamlined / series production you may as well sacrifice some pennies and put a bridge + capacitor(s) at the 12V junction. You'll have a third wire that provides a much more efficient (less heat dissipated) and apt voltage to feed to the 7805.

• Second: even with 12V you could still have heat dissipation issues. There are TO-3 versions to help with that.

• Third, even a TO-3 7805 sounds quite insufficient for a fully expanded (USB and other stuff) Raspberry. You might want to check alternatives out. In example: LM323 (TO-3, 5V, max 3A). Here is a link to a LM323 datasheet.

Or, you could skip the issue altogether and use a switching DC-DC converter.