Capacitor - How to work out how long they can supply a load?

[smeagol]

Hi all,

Bit of basic electronics help please.

I need to work out if it is feasible to have a reservoir capacitor to supply a circuit if the power fails, it needs to supply the circuit for a short time.

So the question is knowing the voltage applied to the capacitor, and the circuit obviously, and knowing the current drawn by the circuit. Is there a formula to work out how the voltage will decay over time. It would be nice to be able to plot a graph of voltage against time to see if the capacitor can supply the circuit for long enough before it reaches the voltage that the circuit fails to work at.

Hope that makes sense.

Cheers gavin

 

[eclectic]

See here:
http://www.kpsec.freeuk.com/capacit.htm

for an excellent introduction.

e

 

[hippy]

When I previosuly tried this I couldn't find the brain power to do the maths so simply connected the capacitor across +V and 0V, disconnected the supply, used a stop watch to time how long a LED driven by a PICAXE kept flashing.

What sort of time period are you hoping for, and how much current will be drawn ?

 

[smeagol]

Approx 200ma @ 4.1V.

I was trying to be smart, rarely works for me, maybe your way would be easier.

The issue I am trying to solve is small video cameras (search ebay for gum camera if your interested) attached to a motorbike. The cameras record to microSD cards and are normally supplied by LIon batteries, but they only last an hour max, I have hacked the cameras to supply power from the bike, and have remote control and LED indicator for operation. The camera is the size of a stick of chewing gum, hence the name gum cam, and soldering wires onto the board was a bit of a challenge Anyway the problem is that if I forget to stop recording before turning the bike off the video files are corrupted and I can't salvage them.

So what I'm trying to do is supply enough power long enough so an 8m can terminate the video.

If your interested my youtube channel is http://www.youtube.com/user/BlackBettycam?feature=mhum#p/u

 

[BeanieBots]

Is there a formula to work out how the voltage will decay over time.

Yes there is but it won't help much if you don't know the current of your circuit, how it behaves as voltage reduces or at what voltage your circuit stops working. As hippy suggests, trying it will yeald the best results.

If the current is constant (which it won't be), the decay will be linear.
If your load is resistive (which a PICAXE is not), the decay is exponential.

All the equations are available with a quick Goggle search but you would need to substitute a complex expression for either R or I so it's simply not worth the hastle. They'll give you a ball-park figure though.

EDIT:
200mA is a very high current to attempt to support with even a large (10,000uF) cap for more than a fraction of a second.
Having a large cap on the input side of a regulator will help a lot as the regulator output will hold while the input drops.

 

[eclectic]

Just rambling thoughts.
And I may have got the scenario totally wrong.

A very small UPS

1. 08M and a small LiPo / NiMh pack,
in parallel to the supply derived from 12v.
(diode protected)

2. 08M monitoring the 12 volt supply (readadc).
When the supply drops below 12v,
the small power supply runs for a short time,
while the Picaxe
"presses" the stop-recording button.


e

 

[hippy]

Approx 200ma @ 4.1V

That seems a hard push for a standard capacitor, though a supercap may do the job.

 

[smeagol]

Ok looks like I'm trying to get too much out of a capacitor, unless I put a sidecar on packed with them :-{

thanks fot the help guys.

 

[NXTreme]

Just a quick note. I wanted to see how long a 08M would run on three 1,500 uF caps. I wired it up with a piezo to beep every ten seconds. I charged the caps to 3.2V and then disconnected the batteries. It ran for ~1 hour 17 minutes before it finally quit. The capacitor voltage was at .95V when the piezo stopped beeping (i.e., the circuit shut down). I measured the whole setup up at about 30 uA, jumping up to 40 uA briefly when it woke up to beep the piezo. In other words, you probably would need a sidecar full of caps .

 

[BeanieBots]

What we don't know is how long it has to run for.
If the cap is charged to a nominal 12v and the device runs at 4.1v (no indication of how that is generated), then several seconds is quite acheivable.
Alternatively, ECs idea of a small battery (with controlled shutdown by monitoring the 12v line) would be very easy to implement.

 

[womai]

As a rough estimate (at least will tell you if it is minutes, seconds, or milliseconds): assume the supply current is constant - reasonable as long as voltage drop isn't huge. The voltage drop dV will be

dV = I * t / C

or t_max = C * dV_max / I

where I is the supply current, t the time, and C the capacitance. So if you start out at 5V when the power fails, and the device works down to let's say 3V, then dV can be 5-3=2V before the device stops operating. If the supply current is e.g. 5mA and the capacitor is 5000uF then

t_max = 5000E-6 F * 2 V / 5E-3 A = 2 secs

That result will be reasonably close to reality as long as the supply current is much larger than the self-discharge of the capacitor. So if you calculations yields something in the order of seconds or a few minutes, you can trust it. If it gives 3 months the forget the formula

Wolfgang

 

[manuka]

Smeagol: That gumstick cam of yours has produced some great videos- especially in the Cotswolds! Mmm= "Hills Angels" ? I'd similar 2 wheel excursions myself in that area ~40 years back. Ah-nostalgia...

Supercaps are NOT suited to such high current drains,as they'll only deliver small (mA level) currents. However beefier ultracaps are. These little lightweight powerhouses are billed as the storage device of the future,& may especially suit hybrid vehicles.

I've some physically small ~2.5V 25 Farad "Boostcaps" here that will deliver several Amps - for a short period of course. They're only about the size of normal power supply type ~2200µF electrolytic caps. Check the Q=VxC=Ixt maths,& you'll see that a time constant = (VxC)/I. Hence (2.5x25)/.2 = ~300 seconds until the voltage (falling exponentially) sags to the usual ~37%. To get an initial ~5 V of course you'd need to put a couple in series, but that means the available capacitance would then halve! Still- just a minute should be enough for you to run a normal camera shutdown I'd say,even if the lads have made a sudden pit stop at a rustic inn.

Naturally first run some bench top trials with such a setup !

 

[smeagol]

Supply is regulated by an LM317, output set to 4.1V. Camera will work down to approx 3.7V (need to test further to confirm this is a reliable cutoff point). Time needed is approx 2-3 seconds.

Just looked at some ultracaps at farnell, bit pricy, maybe batteries are the more cost effective solution.

 

[inglewoodpete]

Seems like you're trying to do this the hard way. Why complicate things by adding more batteries or fancy capacitors?

You have a motorcycle. With a battery (6v or 12v). Take a feed directly from the battery, in-line fuse mounted close to the battery for protection. Have the 08M sense the "ignition" side of the ignition switch. When the key is turned off, the 08M detects this and starts the shut-down process. The last thing it does is shuts itself down (eg. turns off a FET or releases a small relay).

 

[srnet]

The issue I am trying to solve is small video cameras (search ebay for gum camera if your interested) attached to a motorbike. The cameras record to microSD cards and are normally supplied by LIon batteries, but they only last an hour max, I have hacked the cameras to supply power from the bike, and have remote control and LED indicator for operation.

Very likley to have been powered by a Lipo battery.

If you have gone to the extent of hacking the camera, to run it off an external 12v supply, I would be using a simple diode switchover between the external supply and the origional Lipo battery supply.

Detect the external supply side, switch off camera when it dissapears.

But OTOH, how much power does the camera use in the first place, not a lot to worry a motorcycle battery I would have thought. So does the camera switch itself off when the memory card is full ?