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I want to get 12V and 5V from a 24V power source. 12V drives a MOSFET H bridge and a cooling fan. 5V is for a microcontroller. I have already damaged a microcontroller due to a failed 7805 regulator.

Here are two arrangements of 78XX regulators to get required voltages.

I want to know which design is the better.

Which design improves the durability of 7805? It would be really helpful if you can provide some ideas for improvements.

enter image description here

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    \$\begingroup\$ The question which design is better can't be answered because you must know the currents going into 12V and 5V loads. If you do know them, you can calculate how much power is dissipated by each regulator and check if it is witihin limits for each regulator. \$\endgroup\$
    – Justme
    Commented Aug 15, 2022 at 5:15
  • \$\begingroup\$ @Justme Required current is below 1A. So, it is in limits. But my doubt is the input voltage for 7805. Does 12V, instead of 24V improve anything??? \$\endgroup\$
    – shan
    Commented Aug 15, 2022 at 5:21
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    \$\begingroup\$ You can't pull anywhere near 1A or the regulators heat up and burn up. Just say how much current is needed and then the power dissipation tells you if it is or is not possible to do either of the circuits. It will help if you give exact model numbers of the regulators, including manufacturer and case type, or TO220 is assumed. \$\endgroup\$
    – Justme
    Commented Aug 15, 2022 at 5:24
  • \$\begingroup\$ @Justme That's a news to me coz datasheet tells they can provide up to 1A. Anyway, from 12V H bridge will pull around 800mA and the microcontroller is Atmega328. I have no idea how much current it pulls and all it does is drive the H bridge \$\endgroup\$
    – shan
    Commented Aug 15, 2022 at 5:30
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    \$\begingroup\$ The datasheet also tells you the thermal limitations. Eg: for 12V at 1A from 24V you need to waste 12W. Without a heatsink the 7812 will probably reach 600C! With a 2C/W heatsink it will be around 24C above ambient temperature. Your heatsinks will be sizeable. The datasheet also says you need capacitors but your schematic does not show them. Nowadays you can use buck regulators that are much more efficient and wont need heatsinking or minimal at best. \$\endgroup\$
    – Kartman
    Commented Aug 15, 2022 at 5:42

2 Answers 2

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The power dissipated in a linear regulator is the difference between the input and output voltage times the current. So just to use 1 A as an example, the top 7805 would dissipate \$(24 V - 5 V) \times 1 A = 19 W \$ and the bottom one would be \$(12 V - 5 V) \times 1 A = 7 W \$

The problem then is that the other 12 W would end up being added to the dissipation of the 12 V regulator. The power to drop from 24 V to 5 V has to get dissipated somewhere, by going with the second schematic you're just shifting some of it from one regulator to the other.

From your comments it appears that you're already at 9.6 W on the 12 V regulator, so add whatever current the 5 V regulator needs times 12 V and that will give you the additional power the 12 V one will have to handle.

You can find how much current you actually need at each voltage and calculate the dissipation for both cases to see if it's possible to stay within the specified ratings of the regulators. Alternatively you could use switching regulators which will be much more efficient and won't have this \$E \times I\$ loss.

Note that the current for each regulator includes its quiescent current. In addition to the current drawn by the load there is also a small current to run the regulator and that contributes to its power dissipation. It's usually small enough to be ignored but you should take a look at it anyway.

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  • \$\begingroup\$ @GodJinyo Thanks for the explanation. I got it now Although it is not advisable, you confirmed that the second design is good for the 7805 regulator. If I use Buck converter by replacing 7812, is the rest OK? \$\endgroup\$
    – shan
    Commented Aug 15, 2022 at 5:59
  • \$\begingroup\$ If you're going from 24 V it would be best to use buck regulators for both. Do you need to use 24 V? \$\endgroup\$
    – GodJihyo
    Commented Aug 15, 2022 at 6:21
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What you suggest is not practical by any means. You are not hitting the current limit, but the regulator must dissipate the wasted power as heat and you are likely hitting a temperature limit.

If you need 0.8A of 12V out, it means the regulator must drop 24V to 12V at 0.8A which is 9.6 watts. That's completely unreasonable amount of power to dissipate in a linear regulator.

In comparison, the dropping 24V to 5V at 500mA needs to dissipate same amount of power as heat.

If you connect the 5V regulator after the 12V regulator, it will reduce the power dissipation of the 5V regulator, but it will increase the power dissipation of the 12V regulator.

So basically, either the 24V input voltage is too large for the linear regulators, or the linear regulators are a wrong components for the job and switch mode regulators should be used instead.

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