I maked a schematic for trigger a siren 12v by a signal from Raspberry Pi (GPIO18 PWM 3.3V). All the circuit is powered by a power source of 12v 15A (see the image below).

SCHEMATIC UPDATED: enter image description here

It's works fine (although i prefer not power on by much time), but the LM7805 voltage regulator is heating up a lot! This is why I don't let the circuits run for long time.

Is it normal the LM7805 is heating up? If not, what i making wrong?

Thank you!

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    \$\begingroup\$ Measure the current being drawn by your circuit. Multiply that by the 7 V drop across the regulator. How many watts do you get? How good is your heatsink (°C/W)? Post the details in your question (not in the comments). \$\endgroup\$ – Transistor Jan 7 at 14:32
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    \$\begingroup\$ +1 on what everyone else says. Also, please draw your schematics with input from left and output to right, positive voltage going up and negative going down. \$\endgroup\$ – winny Jan 7 at 14:46
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    \$\begingroup\$ (a) What is the voltage rating of the relay coil? I guess that it's a 5V relay, but I don't see a part number or rating anywhere in the question. (b) What is the 7805 actually powering? The schematic shows it only powering the relay coil. (c) You said that the whole circuit is being powered from a 12V 15A supply, but your schematic seems to disagree and also shows a separate 5V supply for the RPi. There is no connection shown between the 7805 output and the RPi. So is the 7805 powering the RPi, and that connection is missing on the schematic? Please update the question to explain the details. \$\endgroup\$ – SamGibson Jan 7 at 15:24
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    \$\begingroup\$ Replacing the 7805 with a drop-in 5v switching regulator will solve the heat problem and drastically reduce the required supply current. \$\endgroup\$ – David Schwartz Jan 7 at 23:03
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    \$\begingroup\$ "Is it normal the LM7805 is heating up?" Yes, very. They are infamous for their low efficiency and the amount of heat produced by them. The circuit itself will be fine up to some 85(?) degrees C, then if overheated go into thermal shut down, cool off, go live again, heat up... round it goes. Typically you'd use the TO-220 pack together with standard heatsinks when these linear regulators were commonly used. Nowadays, they are mostly obsolete technology and you'd use switching regulators instead. \$\endgroup\$ – Lundin Jan 8 at 12:08

The 7805 is a linear regulator. When regulating 12V to 5V, it will be passing all the required current with a 7V drop across itself. It will thus dissipate about 140% of the power that everything on the 5V supply dissipates.

It's not unexpected, but it may require heatsinking.


Powering on raspberry pi by itself can be the source of the heat. Even if you consider 700 mA or current (recommended current for raspberry pi Model A) as an example, the power dissipated is about

(12-5)*0.7 = 4.9 W

Even if the power consumption is about 200 mA (typical for raspberry Pi Model A), the raise in temperature due to power dissipation will be high.

For a regulator with no heatsink this can drive the regulator to dangerously high temperature.

  • Provide a heatsink

  • change the regulator to a DC DC buck regulator

  • try to run the pi in minimal power mode


There is a bug in your circuit - probably the Q1 will damage after few relay ON-OFF cycles. This will be because the voltage spike occuring on Q1 collector during ON->OFF transition. You need to protect Q1 against this adding a reverse polarised diode in parallel to relay coil - anode to pin 2 of relay, cathode to pin 5. The diode can be 1N4148 or, better, any of 1N4001 to 1N4007.

Of course this is not related to heating of U1, but it is very important to remember about this always, when you are driving relay coil, motor or any other inductive load.

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    \$\begingroup\$ Q1 is also reversed in the schematic, but the sch is so badly drawn it's not so easy to see. \$\endgroup\$ – Brian Drummond Jan 7 at 15:32

So many design flaws there :(

The most obvious is wasting energy in a linear regulator (7805) instead of using a 12 volt coil rated relay. Also in most cases there is no reason to employ a precise voltage regulation just to power a relay coil. Relays are pretty tolerant to voltage swings and spikes. Should you for some reason need to get 5V from 12V and waste much less of power, there is a lot of ready-to-use step-down converter modules. The cheapest Chinese ones don't cost much more than the 7805 and perform surprisingly good.

But the most important problem is your choice of switching element. Your circuit takes no advantage from using a relay. There is no galvanic separation i.e. everything is interconnected through a common GND. The voltage being switched is actually low enough to just use a single N-MOSFET. That's it, a single element, and you can ditch all of this complicated circuitry. You will need a logic-level one to make it open with just 3.3 V. Fortunately there is a plenty of them e.g. FDN359BN, and a bigger one STB55NF06L.



I think you should supply the pin 2 of the relay with the 12V Vcc and not from the output of the LDO, you will draw less current from the LDO then it might be cooler, i don't think the raspberry is drawing too much current.

I believe the coil of the relay have a low resistor so it might draw a lot of current

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    \$\begingroup\$ That would also require changing from a 5 V coil relay to a 12 V coil relay. \$\endgroup\$ – Andrew Morton Jan 7 at 15:22
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    \$\begingroup\$ I would recommend this option than using a heat sink in a low power application \$\endgroup\$ – Delphesk Jan 7 at 15:28
  • \$\begingroup\$ Opto Mosfet can be an option to replace the relay \$\endgroup\$ – Delphesk Jan 7 at 15:33
  • \$\begingroup\$ Perfect solution for the problem about LM7805 heating up! Thank you a lot! \$\endgroup\$ – Cayo Fontana Jan 8 at 1:15

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