I have a 12V DC input from a power adaptor. A LM7805 5V regulator takes this as an input and powers an MCU and some other digital circuitry that will eventually drive a relay through a high-side load switch.

I'm using the recommended capacitor values from the data sheet of the 7805 on both the input and output side.

Below is a simplified version of the real circuit:


simulate this circuit – Schematic created using CircuitLab

I'm worried that when I drive the relay that the sudden increase in power draw (up to 3-4 A) may cause the input voltage to the 7805 to drop and possibly causing the MCU to BOD reset. Should I add a larger electrolyte decoupling cap parallel to C2 to mitigate this, and if so, how would I go about determining how big it should be. Do I need to take some steps to prevent reverse polarisation of it by using diodes? I've never used electrolytes before (although I understand how they work).

Are there any other considerations I should make in regards to the 12V part of this circuit?

Is it worthwhile to add a capacitor between the marked +12V node and the the ground wire after the relay?

  • \$\begingroup\$ Have you calculated/measured how much it actually drops? Depending on your power source it may still be way over 5V. Also important is the duration, you won't get anything done with caps if you need it for minutes. Depending on the involved loads and impedances it can make sense to just increase C1 a lot or decouple C2 (and increase it too) from the other load. \$\endgroup\$
    – PlasmaHH
    May 30, 2017 at 15:07
  • \$\begingroup\$ What @PlasmaHH said, and you need a fly-back diode over that relay coil.... also I'm pretty concerned your micro can't drive the relay on it's own like that. A diode across the pump would be a good idea too. \$\endgroup\$
    – Trevor_G
    May 30, 2017 at 15:11
  • \$\begingroup\$ Layout is also import analog.com/en/analog-dialogue/articles/…, C2 can be much larger. Need more specs on motor and power supply, can you measure motor ESR? \$\endgroup\$
    – sstobbe
    May 30, 2017 at 15:29
  • \$\begingroup\$ @PlasmaHH I haven't chosen the motor yet. The adaptor will be able to supply 5A which is enough for the motor and MCU. So that's not an issue. I'm only concerned for the transient voltage drop due to sudden load connection. \$\endgroup\$
    – Emily L.
    May 30, 2017 at 15:37
  • \$\begingroup\$ @Trevor I said I have a high-side load switch to drive the relay. If you wonder, it's a MIC94090 with built in load discharge. \$\endgroup\$
    – Emily L.
    May 30, 2017 at 15:39

2 Answers 2


One really good way to deal with this is to simply add a diode (eg. 1N4004) in series with the input and increase the 330nF (or parallel it) with something a bit larger such as 1000uF/16V.

That way you are not trying to supply the surge to the pump, merely to supply the regulator with its input voltage and the capacitor can be much smaller for the same effect.

  • 1
    \$\begingroup\$ That is very clever, and to the point! \$\endgroup\$
    – Emily L.
    May 30, 2017 at 16:59
  • \$\begingroup\$ +1 for isolation diode, as there is also a substrate diode from the out terminal to in terminal of the 7805. So if you short the input you still clamp the output +1 diode drop. \$\endgroup\$
    – sstobbe
    May 30, 2017 at 17:03

Some thoughts about your circuit, some as mentioned in the comments.

You need a snubber diode across the relay coil. This will protect the relay driving circuit from the back EMF created when the coil is turned off.

Most logic output circuits will not directly drive a relay coil. You may need an NPN or MOSFET driver there.

If you are concerned about the motor creating transients during startup, you can isolate your 5 volt supply by placing a diode in series with the input of the regulator and then inserting a larger capacitor between the diode and regulator to ground. This technique keeps the motor from drawing current out of the capacitor, leaving the charge in the cap to prop up the 5 volt regulator during a short sag in the 12 volt supply. The size of the cap depends upon the total 5 volt load but you could start with 220 uF and experiment from there. The diode also helps slightly with offloading some of the thermal load on the regulator since it will drop ~ 0.6 volts.


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