I have a simple application where a 6V, 2A DC power supply is driving 4 hobbyist-grade servos. In most cases this is adequate, but there are cases (when all servos are suddenly loaded) when I think the power draw will exceed 2A for a short period of time.

It was suggested to me that I should use a capacitor between my power source and the servos in order to handle this kind of transient load. Unfortunately the suggestor didn't know how this would actually be implemented. I tried the University of Google, but mostly came up with videos of giant capacitors being used to dramatically explode things.

Could someone point me in the right direction, or give me a simple circuit example of how I would do this. Is it as simple as wiring a capacitor onto the positive lead?

What calculations should I make to determine the appropriate capactitor size? For example, if I wanted to sustain a peak of 3A for 5 seconds.

  • \$\begingroup\$ If you have a 2A fuse on your power supply this will still blow the fuse. \$\endgroup\$
    – Kortuk
    Nov 11, 2011 at 23:59
  • 1
    \$\begingroup\$ Then what approach should be taken? How do I charge the capacitor from the power supply, and then use the capacitor to power the circuit? \$\endgroup\$ Nov 12, 2011 at 0:08
  • \$\begingroup\$ I would suggest you get two power supplies, for single small pulses to higher current you can cope with a choke and capacitor, but that will not deal with turning on more motors then your supply has power, that will deal with a moment of noise current. I would suggest multiple supplies above all. Someone may have a better solution. Someone with time will probably also give you full schematics and explanations of how to use the capacitor so you get something useful out of your question. \$\endgroup\$
    – Kortuk
    Nov 12, 2011 at 0:21
  • \$\begingroup\$ @geometrikal - Your answer was OK and added to the discussion. I'd suggest that you reinstate it. I've added a dummy version as an answer - if you want to you can copy that and I'll delete mine. OR just leave undeleted if you wish. \$\endgroup\$
    – Russell McMahon
    Nov 12, 2011 at 6:19

3 Answers 3


Subset summary:

I = excess current to be provided.
T = time to provide this extra current.
V = acceptable drop in voltage during this period.

C = capacitance in Farad to meet this requirement.

  • C = I x T / V

In theory, and close enough to be useful in real applications:

  • One Farad will drop in voltage by one volt in one second with a 1 Ampere load.

    Scale as required.

The results are not encouraging :-(.

(1) Providing a capacitor to do everything

For over current of I ampere, droop of V volt over time T seconds (or part thereof) Capacitor C required is, as above)

  • C = I x T / V <- Cap for given VIT

    ie more current requires more capacitance.
    More holdup time requires bigger capacitance.
    More acceptable Voltage droop = less capacitance.

or droop given CIT is, simply rearranging

  • Vdroop - (T x I) / C

or time a Cap C will hold up given C I V, simply rearranging =

  • Time = T = V x C / I

So eg for 1 amp overload for 1 second and 2 volt droop

C = I x T / V = 1 x 1 x/2 = 0.5 Farad = Um.

Supercaps may save you as long as required peak current can be supported.


A Supercap (SC) solution looks almost viable.

These 3F, 2.5V supercaps are availale ex stock from Digikey for $1.86/10 and under 85 cents in manufacturing volume.Prices

For the 3F, 2.7V unit the acceptable 1 second discharge rate to 1/2 Vrated is 3.3A. Internal resistance is under 80 milliohms allowing about 0.25V drop due to ESR at 3A.

Two in series gives 1.5F and 5.4V Vmax. 3 in series gives 1 Farad, 8.1V Vmax, same 3A discharge and 0.75V drop due to ESR at 3A.

This would work well for surges in the tenths of a secnd range. For the specified wort case 3A, 5 seconds requirement perhaps 15 Farad is needed.

The same family 10F, 2.7V $3/10, 26 milliohm looks good. 10A allowed discharge. Two in series drooping from 5.4 to 5 volts at 3A gives

T = V x C / I = 0.4 x 5 / 3 = 0.666 seconds.  

Getting there.

(2) IF the droop causes system reset etc and one wishes to avoid this (as one usually does :-) ) an often useful solution is to provide a sub supply for the electronics with cap that hold them up over the dropout period.

eg electronics need say 50 mA. Holdup time desired = say 3 seconds (!). Acceptable droop = 2V say.
From above

  • C = I x T / V = 0.05 x 3 / 2
    = 0.075 Farad
    = 75,000 uF
    = 75 mF (milliFarad)

This is large by most standards but doable. A 100,000 uF supercap is reasonably small. Here the 3 second holdup is "the killer". For a more typical say 0.2S dropout the required cap is

75,000 uF x 0.2/3 = 5000 uF = very doable.

(3) A small holdup battery for the electronics can be useful for obvious reasons.

(4) Boost converter: In a commercial design where 4 x C non rechargeable batteries were used, to provide 5V, 3V3 and motor drive battery (exercise equipment controller) end of life Vbattery got well below needed 5V during end of battery life and much much below when motors operated. (The primary design was not mine). I added a boost converter based on a 74C14 hex Schmitt CMOS inverter package to provide 5V to the electronics at all times plus 3V3 regulated to the microcontroller. Quiescent current of boost converter and 2 x LDO regs and electroncs under 100 uA.

E&OE - may have got something on wrong side somewhere there, easily done. If so, somebody will tell me about it :-).


Query: It has been (quite understandably) suggested that

  • I am not sure you are answering the users main question.

    To stop from overloading a power supply it does not seem feasible.

    It is not a case of power supply cutout, it is a case of wanting to allow higher current for short periods(on the order of 5 or more seconds).

    This seems like a case of needing another power supply


I believe that I am addressing the question completely, as asked, BUT I am also addressing what I believe is liable to be the larger question as well.

Consequently, there seem to be tangents and irrelevant material here.
I have addressed points unasked as well as points asked based both on my own experiences in closely analogous applications and also on general expectations.

The issues are

  • "What if demand exceeds supply" and

  • "What if supply falls below demand".

These are one and the same in practice but may have different causes.

Note that my answer (1) specifically says

  • "For over current of I ampere"

and his question was

  • " ... but there are cases (when all servos are suddenly loaded) when I think the power draw will exceed 2A for a short period of time.

ie dealing with overcurrent is exactly what he is asking.
BUT overcurrent is caused by overload and, when the "cost" of trying to deal with overcurrent is seen (0.5 Farad caps or whatever) then the perspective may well turn to "what can we do to ride out this overload differently". The next most obvious "solution" is to accept the hit on motor performance, let the supply rail fall BUT maintain a local supply to keep the eectronics sane. Another solution which I didn't bother addresssing is to deloa the system by eg slowing servo rates when all are on at once. Whether this is acceptable depends on the application.

The reason that we can TRY to address the short term overcurrent situation is that the supply has spare capacity most of the time and this is used to charge the caps prior to the surge event. The caps do not magically manufacture extra current, just save up spare current for arainy day.

To supply current the capacitor MUST lose voltage so I specify the acceptable limit for that too. I think you'll find that if you couch his requirement in numbers and then plug them into my formulae that his question as-asked will be answered.

Re on geometrikal post.

  • But it is not a case of 6V*3A*5s. You need enough capacitance to stop the output from sagging low enough to cause the output of the power supply to need to host more current. It is really just not going to happen in a good way.

What happens depends very much on the original supply characteristics.
Imagine an LM350 was being used. Datasheet here. This is essentially an LM317 on steroids. Good for about 3A in most conditions and 4.5a IN MANY, deep-ending on application. 3A guaranteed. Fig 2 shows that it is good for 4.5A for a Vin-Vout differential of 5 to 15V depending on other issues. It can be run up near its current limit with good regulation. If being run at 3A and if the drop across it is not too high and it is well heatsunk it will not be hot and intermittent peaks of 4.5A will be provided. Do this too often and the temperature will rise and figs 1,4,5 and a few things unshown will affect how it behaves. First off Vout will start to droop on peaks and a capacitor on the output will help it serve the load. Increasing drOop and longer peaks and the capacitor will be called on to do more. If the IC decided to completely cut out for a moment (which it is unlikely to ever do) as long as T x I / C does not exceed the voltage droop which is acceptable the capacitor will do the whole job. Restore Iout to 3A and the capacitor will recharge until next time.

  • 1
    \$\begingroup\$ @Kortuk - A problem is that while I am addresing his question completely, as asked, I am also addressing what I believe is liable to be the larger question as well, so some of it is as yet unasked. || I think you'll find that if you make up an example of his requirement and then convert his requirement to numbers and then plug them into my formulae that his question as-asked will be answered - please see addition to text. ((But, as Karl Sagan was wont to say, "I MAY be wrong" :-) ). \$\endgroup\$
    – Russell McMahon
    Nov 12, 2011 at 4:33
  • \$\begingroup\$ Careful with supercaps. Make sure they can ripple out that much current. And do you mean Carl Sagan, the astronomer? \$\endgroup\$ Nov 12, 2011 at 4:50
  • \$\begingroup\$ @Kortuk - it looks like flagging up messages to people is case sensitive. please see additions to answer.(edit: Hmmm Maybe not). \$\endgroup\$
    – Russell McMahon
    Nov 12, 2011 at 4:51
  • \$\begingroup\$ Thanks Russel, As a beginner I appreciate very much the "big picture" answer. I am learning a lot, and I'm sure other people are also being helped. \$\endgroup\$ Nov 12, 2011 at 5:13
  • \$\begingroup\$ @RussellMcMahon, I thought the overall answer was important, as I said in my comments on the question but when I read your answer and tried to imagine knowing nothing I could see thinking a 5mF cap might fix it. You have them both there, but I think the post could be better organized, but that can always be said about almost anything. \$\endgroup\$
    – Kortuk
    Nov 12, 2011 at 5:19

It turns out there are several products that do this for RC receivers. They're typically specified for eliminating brownouts or undervoltages due to high current conditions such as a servo locking for a short period of time.

This is a representative unit. The vendor carries several variations with different storage capacities.

The TURNIGY capacity set helps prevent "brown outs" resetting your Rx if your servo amp draw spikes or you have an Rx glitch. It will also help reduced the load on your ESCs BEC and reduce the likelihood of glitching. Simply plug it in to any spare channel on your receiver.

Operating Voltage : 3.2V - 11.1V (1s ~ 3s LiPo)
Capacitor voltage: 15v
Storage Capacity: 783333uf
Data on a 3A load spike typically seen when large retracts jam:
Supply 6v with a voltage drop to 4.7v over 0.88sec
Supply 6v with a voltage drop to 3.0v over 3.0sec (3.0v minimum voltage of the OrangeRx 6ch)



I found this sheet to calculate the voltage drop Sur it is theorical but gives a good idea:



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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