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I have made this circuit. It works fine with normal Zener diode, however when I replaced it with the LM7812 it works fine but as time passes the regulator keeps getting very hot although the load is just 5mm LED. (I replaced it with a regulator because I need it to be more stable during current fluctuations.) Why is the LM7812 getting hot in this circuit?

Note: The circuit is fully covered inside a plastic enclosure and it would be just used to drive an op-amp without any human interference.

edit: the input voltage for the LM7812 is 60Vdc at 25C

edit 2:tried using LM7824 the voltage just dropped dramatically to 3.4V and the 120ohm started a magic smoke

it would be used in an unstable power area with current fluctuations enter image description here

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  • \$\begingroup\$ What's the input voltage to regulator, and how much is the LED current? Does the LED have a resistor or is it just connected directly to 12V? Or is it some sort of 12V LED module? Please add in more info about your circuit. \$\endgroup\$
    – Justme
    Commented Jun 21, 2023 at 22:30
  • \$\begingroup\$ sorry for the lack of information, the input voltage to the regulator as I measured is 60V, yes it exceeds the datasheet specs for that I limited it with 330ohm resistor. yes the led has a 680ohm resistor connected in series and it draws maximum of 20ma \$\endgroup\$
    – Hazardous Voltage
    Commented Jun 21, 2023 at 22:48
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    \$\begingroup\$ Please edit your question with the information about the input voltage to the regulator. It's a Stackexchange thing -- the questions and answers should be stand-alone, without anyone having to dig through the comments to understand them. So when you have additional information or corrections -- edit the question. \$\endgroup\$
    – TimWescott
    Commented Jun 21, 2023 at 22:55
  • \$\begingroup\$ @HazardousV I'm getting the idea that you seem stuck on supplying from a 60 V supply -- that you intend on keeping it for whatever reasons but that from it you'd like 12 V, as well. It sounds as though you are only driving an LED from all this? Is all this work just to drive an LED at 20 mA? Or do you have other plans once the LED is working as you expect? (Note: 20 mA at 60 V is 1.2 watts. Only a tiny portion of that will be in the LED. Most of it elsewhere unless you go towards switching supply.) \$\endgroup\$
    – periblepsis
    Commented Jun 22, 2023 at 0:21
  • \$\begingroup\$ @periblepsis I had stated what is the use of it in the question above, it would drive an op amp(lm358) \$\endgroup\$
    – Hazardous Voltage
    Commented Jun 22, 2023 at 2:51

4 Answers 4

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The LM7812 is rated for 40V on the input. At some point above that, it will start drawing excessive current.

I can't say for sure, but I assume that what's happening with your circuit is that the regulator is starting to draw that excessive current at around 60V, and is, basically, acting like a zener diode.

60V on the regulator implies over 200V across that 330 ohm resistor; that implies around 600mA through the resistor and regulator.

600mA at 60V is around 36 Watts -- which would make for a very hot regulator indeed.

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C6 pushes current into C5, without a load the voltage will go up until LM7812 over voltages and dies. You are seeing that happening. I added a Zener across C5 to eat up excess current. It will get hot/worm because it must take all the power that is not sent out to the load. A LM78xx needs an input capacitor or it might oscillate. You could use 3 to 5, 5.1V Zeners in series. I have used 3V white LEDs as a Zener.
Reducing the value of C6 will reduce the power.

enter image description here

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  • \$\begingroup\$ I get your point, but would it be better if we used a resistor in series with the Zener diode so it gets less warm because I need it somehow to be stable and works for a long time? I will update the design and put the input capacitor and use a feedback diode too. \$\endgroup\$
    – Hazardous Voltage
    Commented Jun 22, 2023 at 2:54
  • \$\begingroup\$ @HazardousVoltage if you put a resistor on series with the zener, then the zener would be less effective at preventing the LM7812 from being destroyed by over voltage. \$\endgroup\$
    – brhans
    Commented Jun 22, 2023 at 3:58
  • \$\begingroup\$ do you have an idea how should I calculate the resistor for the Zener? \$\endgroup\$
    – Hazardous Voltage
    Commented Jun 22, 2023 at 17:58
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The LM7812 datasheet indicates that a 0.33\$\mu\$F capacitoris required at the input for stability if located an appreciable distance from the input filter. R7 essentially isolates the regulator from the input filter C5. The regulator is oscillating and getting hot.

Add a 0.33V\$\mu\$F capacitor from the regulator input to ground as close the the regulator as possible.

The output capacitor is required to be only 0.1\$\mu\$F. It can be higher to improve transient response to load changes. So the 10\$\mu\$F may be ok.

Addition from comments: The input to the regulator of 60V will damage the regulator. The 330 ohm resistor will not solve the problem.

There will be an approximately 58V drop accros the regulator, which at 20mA is 1.16 watts of power dissipation. The regulator will still get hot.

You must reduce the input voltage to a reasonable value before the 7812. OR. You can use a switcing regulator to reduce the 60V to 12V.

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  • \$\begingroup\$ I would try it now \$\endgroup\$
    – Hazardous Voltage
    Commented Jun 21, 2023 at 22:40
  • \$\begingroup\$ I tried it, it still getting hot unfortunately Note: the input voltage to the regulator is 60Vdc for that I connected the 330ohm resistor \$\endgroup\$
    – Hazardous Voltage
    Commented Jun 21, 2023 at 22:46
  • \$\begingroup\$ See the addition in the answer. \$\endgroup\$
    – RussellH
    Commented Jun 21, 2023 at 22:52
  • \$\begingroup\$ I could use a Zener diode to solve it but would it withstand he fluctuations of the input current? \$\endgroup\$
    – Hazardous Voltage
    Commented Jun 21, 2023 at 22:54
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Say the LED and regulator draw 25mA, 330 ohms will only drop 60V to about 52V.

That's still over maximum input voltage limit.

Even if input was 40V and 25mA load, it will need to dissipate 0.7W.

Assuming a generic TO220 case with abput 70°C/W thermal resistance, 0.7W makes it heats up by about 50 degrees C from ambient, so assuming 25 degrees C room temperature, the regulator will be at 75 degrees C.

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  • \$\begingroup\$ then can I use diode Zener to drop the voltage to 15V and then put it to a regulator the problem that this circuit would be used in unstable electricity place so I need it to survive as long as possible \$\endgroup\$
    – Hazardous Voltage
    Commented Jun 21, 2023 at 22:59
  • \$\begingroup\$ @HazardousVoltage The zener diode will dissipate heat, around 45*0.02 = 0.9 W. For a small zener, it will get hot unless you mount it on a heat sink. A switching regulator would be a better solution than a linear regulator. \$\endgroup\$
    – qrk
    Commented Jun 21, 2023 at 23:09
  • \$\begingroup\$ it would be better to keep it simple so is it possible if I could step the voltage down by changing the capacitor or other solution? \$\endgroup\$
    – Hazardous Voltage
    Commented Jun 21, 2023 at 23:12

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