0
\$\begingroup\$

I have a 5v 10A power supply that I am using to power my project. The project consists of a raspberry pi and 3 servos that are controlled by the pi through a PWM driver. The 3 servos and the Pi are powered from the same power supply.

The problem is that when I move all 3 servos I see the low voltage indicator pop up on the Pi. I don't have an oscilloscope to test how much current the servos are using but peak current is 600mA so the power supply should be sufficient to power everything.

Is there a way to split the incoming +5v into 2 isolated +5v rails that share a common ground?

That way I can use one rail to power the Pi and the other rail for the servos without the voltage drops triggered by the servos affecting the Pi.

I need to have a single power input that goes into the housing so I can't use two power supplies to power the Pi and servos separately.

Edit: Here is a an image of the circuit. I've marked out the important parts to make it clearer:

enter image description here

\$\endgroup\$
  • \$\begingroup\$ How are the power supply cables arranged to both Arduino and servos? How long and thick are they? \$\endgroup\$ – Harper May 27 '18 at 0:06
  • \$\begingroup\$ Try adding a large capacitor. \$\endgroup\$ – lucky bot May 27 '18 at 2:13
  • 2
    \$\begingroup\$ "isolated" but "share common ground", this doesn't make sense. You just need some decoupling capacitors. \$\endgroup\$ – Long Pham May 27 '18 at 2:46
  • \$\begingroup\$ @Harper Currently the power supply goes into a common breadboard and then jumps to both the PWM driver and raspberry pi using 22 gauge wires \$\endgroup\$ – shekit May 27 '18 at 15:55
  • \$\begingroup\$ @LongPham thanks for clearing that out, I think my understanding of it was limited. Will try adding caps. \$\endgroup\$ – shekit May 27 '18 at 16:02
4
\$\begingroup\$

Isolated (galvanically) means "no direct conduction path between the two sides" and imply "floating", which is in direct contradiction with "share a common ground". However, isolated does not imply that if there is a high current load, the voltage does not drop, which is your problem here.

So "isolated" is not what you are after.

What you are after is a more stable supply. Even if your supply is correctly rated for the total current needed, it may not be able to cope with high current transients without the voltage dropping. The servos need higher current when they start moving.

You can add a capacitor close to the Pi to overcome this. A big electrolytic could be appropriate. Its minimum value depends on the layout, peak current value and duration, but something like 1000uF will most probably fix it.

\$\endgroup\$
  • \$\begingroup\$ Thanks, I'll add a cap across power and ground near the Pi and see if it helps. There is currently a 1000uF capacitor on the PWM driver near the servos. Is that not sufficient? \$\endgroup\$ – shekit May 27 '18 at 16:00
  • \$\begingroup\$ My understanding was common ground is needed to properly control the servos from the pi. And my thinking behind separating into 2 lines was so that voltage drops only happen on the servo line, leaving the Pi unaffected. I think my understanding of it was incorrect. \$\endgroup\$ – shekit May 27 '18 at 16:09
  • \$\begingroup\$ The cap near the servo is a good thing too. It smoothes the current requested by the servos and therefore reduces the voltage drop. But the Pi needs its own reservoir too, so that it is less dependant on the remaining fluctuations of the supply voltage. This is called decoupling. And yes, your understanding of isolation is incorrect. Isolation does not reduce voltage drop. The power requested by the secondary is drawn from the primary, and this may equally result in a drop. And if the primary drops, the secondary drops too. So it does not solve the problem. \$\endgroup\$ – dim May 27 '18 at 18:12
  • \$\begingroup\$ ...Well, unless you use an isolating device that also actively regulates the output, but the isolation requirement us totally useless here. You may as well just use a non-isolating device that actively regulates the output voltage, stepping it up if necessary. Anyway, a large enough cap should solve your problem. Simple. \$\endgroup\$ – dim May 27 '18 at 18:17
  • \$\begingroup\$ Thanks, I added a 1000uf cap but the problem still persists... I updated the post with an image of the breadboard circuit and marked out the important parts. Could you please have a look and see if something is amiss? \$\endgroup\$ – shekit May 27 '18 at 19:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.