I have a 12V 270mA electric strike door lock that will only act as a load when unlocking.
The control system for this is a 5V Raspberry Pi zero

Which would be more prudent to use?
5V supply and a boost converter to supply the lock
12V supply and a buck converter to supply the Pi

My current thinking is to use the buck as the boost will try to run even with no load

  • \$\begingroup\$ Why would you think that a buck would not run with no load? \$\endgroup\$ Mar 9, 2017 at 18:40
  • \$\begingroup\$ Your last part of the question is meaningless.. at least to me. More importantly, which initial supply is easier to obtain, 12V or 5V. Decide that first. In my opinion though, the 5v logic requires a MUCH MUCH higher tolerance and stability compared to the lock so I'd concentrate on that. Then use the output from the logic to fire up a sloppy 12V booster to drive the lock. \$\endgroup\$
    – Trevor_G
    Mar 9, 2017 at 18:44
  • \$\begingroup\$ Sorry should have made clearer, the Raspberry Pi will always be running and so a load will always be provided for the buck converter. @AdamLawrence \$\endgroup\$
    – A.Rothman
    Mar 9, 2017 at 18:48
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    \$\begingroup\$ I'd suggest that you use a 12 V supply (I assume it's both AC and battery backed up) and generate the 5 V needed for the R'Pi from that voltage. It is never good form to run a highly variable current load from an MCU power supply via a boost convertor, there is much more chance of bad interactions. \$\endgroup\$ Mar 9, 2017 at 18:49
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    \$\begingroup\$ In your case it means you have to OVER-design the logic supply to handle the size and transient nature of the load of the lock. Whether its a boost converter, which it would not be for 12v-5v, is irrelevant. All power supplies have issue with suddenly increasing demand by a large factor from ambient. \$\endgroup\$
    – Trevor_G
    Mar 9, 2017 at 19:53

1 Answer 1


@DanielTork asked why it's better to use a 12 V supply and a down convertor for the 5 V supply than a 5 V supply and a boost convertor to generate 12 V for the door strike.

By example.....
If you use a readily available 5 V Walwart to power the raspberry Pi there would appear no problems with that. The 'Pi Zero likely draws about 100 mA (depends what your are running and how tight the loops) plus another 100 mA for a WiFi adapter. ...Total 200 mA

To create the 12 V @ 270 mA for the striker you need a DC-DC convertor (with an efficiency of about 80%) and it probably has an operating frequency of 50 -100 kHz.
This means that the average current at 5 V will be about 340 mA.

Total average current from the 5 V supply will range from 200 mA (just the 'Pi) to 540 mA ('Pi and striker).
However, the DC-DC convertor does not draw a nice average current. Because of the duty cycle involved and related to the frequency it will draw current pulses that may be 2-3 times the average current. So when the striker is activated the current pulses drawn from the 5 V supply may approach an Amp.

To help ameliorate the current pulses you can increase the capacitance on the output of the 5 V supply, and it will certainly help. But you can expect that the 5 V supply will dip. If you are an experienced Raspberry Pi user, you will already know that it's very easy to drop the 'Pi into it's reset zone (about 4.8 V). You would probably use a 5 V 1 - 1.5 A supply in this case, hoping to keep the 5 V constant under all load conditions. IMO bad Ju-Ju.

Now let's configure the supplies the other way; and buy a 12 V supply that will provide the required current for the project.
What the average current is depends on how you generate the 5 V for the Raspberry Pi.

  1. Use a linear regulator for the 5 V. So the average current at 12 volts is 270 + 200 = 470 mA. So the 12 V supply is probably a 700 - 1000 mA rating.
    Now if the striker is turned on (a sudden load) the 12 V may dip (maybe by as much as 0.5 - 1 V). However the linear regulator has a drop out voltage probably down at 7 V, so it's output won't be affected at all by the 12 V load dip.
    You do have to deal with the dissipation in the linear regulator. It'll be about 1.4 Watts.

  2. You could use a switching regulator for the 5 V. The average current now will be less. The 200 mA required for the 'Pi results in only about 80 - 100 mA average from the 12 V supply. This means the 12 V supply only needs to supply on average about 270 + 100 = 400 mA. In all probability you'd get a 700 - 1000 mA power supply again, but you could perhaps get away with a 500 mA unit since the striker is very intermittent operation.
    Power dissipation in the 12 - 5 Buck convertor is very low, perhaps only 70 mW or so.

So there is the choices to be made.
Either solution could be ok, but I'd just feel much more confident using the 12 V supply.

  • \$\begingroup\$ So because of those pulses, the boost can overload and damage the power supply and then you can get voltage dipping, too. \$\endgroup\$ Mar 10, 2017 at 5:45
  • \$\begingroup\$ The pulses would be unlikely to damage the power supply. The real danger is that you might potentially see 'Pi resets/brownouts if the 5 V supply dips. \$\endgroup\$ Mar 10, 2017 at 5:58
  • \$\begingroup\$ Cheers for the thorough explanation, I will use a buck converter @JackCreasey \$\endgroup\$
    – A.Rothman
    Mar 10, 2017 at 8:16
  • \$\begingroup\$ @JackCreasey One more question, won't running the boost in continuous mode solve this dipping problem? \$\endgroup\$ Mar 11, 2017 at 7:19
  • \$\begingroup\$ @DanielTork. Not necessarily. Since the output current is close to zero when the striker is off the Boost convertor will be at minimum PWM duty cycle (it will have some output capacitor to control ripple). When the striker is turned on the convertor has to charge the output capacitor and supply the striker current. Depending on the PWM frequency and the striker inductance you will see an input current demand larger than the striker DC current needs. ....this assumes the OP is using a fail secure striker of course. \$\endgroup\$ Mar 11, 2017 at 17:05

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