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I have a 30V power source and I need to get several different voltages from it using linear regulator. 24V for op amps, 12V for relays and 5V for MCU. They all going to be installed on a heatsink.

If I wire them in parallel 7805 regulator have to drop 25V and that means a lot of heat. if I wire them in series all of the current is going to pass through the 7824 regulator and that is going to generate a lot of heat too.

schematic

simulate this circuit – Schematic created using CircuitLab

I know linear regulators basically are not efficient but for this specific project I need a linear output.

Comparing this two configuration, which one is more efficient? series or parallel?

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  • \$\begingroup\$ Do you know roughly how much current will be consumed by each rail? \$\endgroup\$
    – TI_Lover
    Jan 6, 2020 at 19:29
  • \$\begingroup\$ @TI_Lover each of them is going to pull around ~200mA, 600-700mA total. \$\endgroup\$
    – ElectronSurf
    Jan 6, 2020 at 19:31
  • \$\begingroup\$ There are pretty low cost and extremely efficient Buck converter modules available these days which are usually as simple as linear regulators. They could certainly be a better option if you want to avoid the heat losses. \$\endgroup\$
    – Nouman Qaiser
    Jan 6, 2020 at 19:46
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    \$\begingroup\$ Or you could combine switcher modules (3v higher output) and these linear regulators to "have the best of both worlds." \$\endgroup\$
    – rdtsc
    Jan 6, 2020 at 20:18
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    \$\begingroup\$ For 3 or 4 currency units you can get switching regulator drop in replacements for a 7805. If you take heat sink size into account for the wattage you're using, size is comparable. Mine are Murata from digikey and you can also find a variety of possibly questionable or possibly fine ones on aliexpress or ebay. You can also step voltage down some other way depending on the source and loads, like if you have a bunch of 12v solenoids and just logic running at 5v, use a 30v to 12v switching converter and power your 7805 with its tiny load off of that. \$\endgroup\$
    – K H
    Jan 7, 2020 at 5:18

1 Answer 1

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They are about as inefficient either way. You have ~15mA drawn from 30V (almost half a watt) before you start drawing any current from the rails. Iq goes up slightly with voltage but it can be ignored.

If you need any substantial current from this, especially from the 5V supply, you'd be far better off using a switching supply, followed by a linear regulator if that's really necessary. Either way you'll be throwing away almost 85% of the power used on the 5V bus.

Also, the original uA7805s are only recommended for 25V input, and 7812s for 30V. So you would be exceeding the maximum recommended rating with the parallel connection, and are uncomfortably close to the absolute maximum 35V rating for 7805/7812.

Edit:

  • 200mA from 5V = +5W in the regulator(s)
  • 200mA from 12V = +3.6W in the regulator(s)

  • 200mA from 24V = +1.2W in the regulator(s)

    on top of the 0.5W or so quiescent

So you need to get rid of more than 10W of power. Very, very wasteful and you'll need a big heatsink or a smaller one and a fan.

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    \$\begingroup\$ It seems that the power loss distribution is a bit better in series. 7805 = 1.4W, 7812 = 4.8W and 7824 = 3.6W. \$\endgroup\$
    – ElectronSurf
    Jan 6, 2020 at 20:12
  • \$\begingroup\$ Using 200mA load current and 15mA self current the total heat dissipation is 10.535W in both series and parallel config. It just moves the heat from one regulator to another. With the parallel approach the regulators only need to support 0.2A output - so are perhaps smaller. The main issue will be how to get rid of the heat. In the parallel config, you might be able to use series resistors to move some of the heat from the regulators to power resistors. \$\endgroup\$
    – scorpdaddy
    Jan 6, 2020 at 20:30
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    \$\begingroup\$ Looks like a switcher for 12V, and 1.2-1.4W in each 78xx (7805 from 12V) is a reasonable compromise, and probably saves money over those heatsinks. \$\endgroup\$
    – user16324
    Jan 6, 2020 at 20:41

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