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(updated due to remark of ChrisStratton).

I am wondering if it is a good idea to use two regulators in series.

What I want: using 2 12V Nema17 motors (using 12V, 1.7A each). Using a microcontroller board (STM32F103C8T6 in most cases, 5V input voltage). However, I'm not sure what to add (except DMX possibly also a nRF24L01+ 2.4 GHz transceiver), multiple LEDs (5 mA each). So assume I need max. 200 mA.

Than my power dissipation if I would use a single 7805 would be (12 V - 5 V) * 0.2 A = 1.4 W

I heard above 0.8 or 0.9 W a heatsink is adviced. However, what if I use two regulators in series:

  • 7809: power dissipation: (12 V - 9 V) * 0.2 A = 0.6 W
  • 7805: power dissipation ( 9 V - 5 V) * 0.2 A = 0.8 W

Would this work? And if so, should I place them with some distance not to heat each other?

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    \$\begingroup\$ There is also the possibility to use a buck \$\endgroup\$ – PlasmaHH Mar 26 '18 at 15:55
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    \$\begingroup\$ I second PlasmaHH. When you run into this type of problem with a linear voltage regulator, it usually means that the linear regulator is the wrong tool for the job. Use a buck regulator instead. \$\endgroup\$ – Dampmaskin Mar 26 '18 at 16:01
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    \$\begingroup\$ You could consider using one of these. Same pinout and almost the same footprint of a linear 7805, but is a DC-DC converter. \$\endgroup\$ – brhans Mar 26 '18 at 16:57
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    \$\begingroup\$ Not wanting to learn about new things is a bad reason to not use things. There are modules about the size of a to220 available if you don't want discrete products. \$\endgroup\$ – PlasmaHH Mar 26 '18 at 17:39
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    \$\begingroup\$ The STM32 does not take 5v input, so something is wrong with your question statement. Are you making the mistake of conflating a microcontroller board with a microcontroller? Also beware that running 12v steppers on only 12 volts will be disappointing - you typically want several times the ohmic coil voltage into a chopper driver (realistically, you want lower impedance motors). \$\endgroup\$ – Chris Stratton Mar 26 '18 at 17:52
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A better solution is to put a resistor in series with the 7805. The power dissipation will be shared between the regulator and the resistor. This will reduce the peak power dissipation in the regulator.

The 7805 has a minimum input voltage of 7V. If you need 200mA from a 12V source then you can use a resistor of...

(12V - 5V) / 200mA = 25 ohms.

The peak power dissipation in the regulator will occur when the voltage across the resistor is mid-way between 12V and 5V (which is 8.5V). The current in the resistor will be (12V - 8.5V) / 25 ohms = 140mA at that point.

The power dissipation in the regulator will be 140mA * (8.5V - 5V) = 0.49W.

Thw peak power dissipation in the resistor occurs at max load (200mA).

The peak resistor power is 0.2A * 0.2A * 25 ohms = 1W.

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  • \$\begingroup\$ Thanks, this is also a good idea. However, I don't have 1W resistors (but I can use two 0.5W resistors in series. Not sure if I have those values though. Also, I think I can reduce the load by powering the LEDs from 12 V directly. \$\endgroup\$ – Michel Keijzers Mar 26 '18 at 16:30
  • \$\begingroup\$ I calculated a bit (thanks to your equations), and I can use three 0.5W 10R resistors in series. \$\endgroup\$ – Michel Keijzers Mar 26 '18 at 16:56
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    \$\begingroup\$ @MichelKeijzers Design this with care. If your +12V supply is unregulated, or has AC ripple, you must use its minimum voltage (which may be below +12V). \$\endgroup\$ – glen_geek Mar 26 '18 at 18:43
  • \$\begingroup\$ @glen_geek I'm intending to use a 12V adapter from a store, not to make it myself. According to Chris I might even need more than 12 V (but for the microprocessor board, I have to take safe values probably). \$\endgroup\$ – Michel Keijzers Mar 26 '18 at 18:46

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