4
\$\begingroup\$

Let's say I've got a device which requires 700mA at 5V and I want to power it of a capacitor for one full hour. Now if I calculate the required capacity I get: C = I*t/V = 504 Farad. Capacitors like these are quite expensive.

Does anyone here have a better idea of getting 700mAh @5V on a capacitor? Perhaps using an array of lower C capacitors? Maybe a chemical cell?

\$\endgroup\$
  • \$\begingroup\$ You want to power your 5V device for an hour, but after 12 minutes at 700mA the voltage will be down to 4V ..... ? I would take steven's advice if I were you. \$\endgroup\$ – MikeJ-UK Apr 19 '12 at 8:39
  • \$\begingroup\$ But wouldn't batteries have the exact same problem? \$\endgroup\$ – Rob Apr 19 '12 at 10:15
  • \$\begingroup\$ No, batteries have a fairly flat discharge curve to start with, followed by a much faster drop-off as they reach exhaustion. Capacitors have a linear discharge curve at constant current, or an exponential curve with a constant resistance load. \$\endgroup\$ – MikeJ-UK Apr 19 '12 at 10:19
16
\$\begingroup\$

You don't want a capacitor solution when you draw that large currents for such a long time. The word is battery. Why don't you use that? It's cheap and can store a lot more energy than a capacitor the same size.

This capacitor is "only" 50 Farad and already costs 1500 dollar. How many batteries can you buy for that? It also stores the amazing amount of 8000 J! An typical AA NiMH will store 12000 J (1.2V \$\times\$ 2800mAh \$\times\$ 3600 seconds). Less than 3 dollar.


One application where the use of high capacitance caps is justified is in a strobe flash. For my photography I have a set of studio strobe flashes rated at 500Ws. These 500J are dumped in 1/2000 second, which means that for a split second you have to supply 1MW. No battery can dump so much energy so fast.

|improve this answer|||||
\$\endgroup\$
  • 2
    \$\begingroup\$ To put the 8000J even more in perspective: a car battery can store something like 1.5MJ! \$\endgroup\$ – Federico Russo Apr 19 '12 at 12:02
  • \$\begingroup\$ there are few constrains on supercapacistors these days. \$\endgroup\$ – Standard Sandun Apr 19 '12 at 15:01
  • \$\begingroup\$ No it won't cost 1500$, only 4-5$,here is a 10F one. sparkfun.com/products/746 \$\endgroup\$ – Standard Sandun Apr 19 '12 at 15:07
  • \$\begingroup\$ I'm sure there are batteries that are 1000V 1mAh that cost $100k+ with gold plated connectors that you could use against battery argument too. :) \$\endgroup\$ – Kenny Robinson Apr 20 '12 at 0:56
  • \$\begingroup\$ @sandun - There are supercaps and supercaps. The one at Sparkfun is for low current applications. \$\endgroup\$ – stevenvh Apr 26 '12 at 10:02
8
\$\begingroup\$

To supplement the other answers, below is a graph of typical discharge characteristics for a battery and a capacitor. Of course you could choose a capacitor with a much higher capacitance to match the shallow discharge part of the battery curve but that would be even less practical than the 504F capacitor.

But even a battery will not usually be suitable for direct connection to a device which requires a regulated 5V supply since such devices usually require the supply voltage to be within tight limits (4% or 10% for example).

Discharge curves

First you need to look at the the specification of your device. What are it's input voltage limits? Remember that you will only get 4.5V from 3 X 1.5V alkaline cells, or 4.8V from 4 X 1.2V NiMH cells.

You will probably end-up needing a dc-dc converter but I don't know of a commercially available 4.5V-to-5V or 6Vt-to-5V pre-packaged solution. If you have a 12V battery pack, you could use an old in-car charger for a mobile phone. I recently dismembered an old 7V Nokia car charger and changed a resistor to get 5V for a similar reason.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ " since such devices usually require the supply voltage to be within tight limits " So are there any other options to tackle that problem except for finding a battery that has an extremely shallow discharge? Thanks for your input, I really appreciate it! \$\endgroup\$ – Rob Apr 19 '12 at 11:35
  • \$\begingroup\$ @Rob - see my edits ... \$\endgroup\$ – MikeJ-UK Apr 19 '12 at 11:54
  • \$\begingroup\$ Wait, these types batteries all have a specific input voltage too, right? \$\endgroup\$ – Rob Apr 19 '12 at 12:24
  • \$\begingroup\$ Well, yes, if you use rechargeable batteries, they will need to be charged to a specific voltage. \$\endgroup\$ – MikeJ-UK Apr 19 '12 at 12:50
  • \$\begingroup\$ Okay that's clear to me, the reason I asked that was because I forgot to mention that I was planning on powering the battery using solar cells (@~2V). So I first need to step up the voltage to the specific input voltage that is needed to charge the battery and then I need to step up (or down depends on which battery I chose) that Voltage to 5Volt in order to power the device. \$\endgroup\$ – Rob Apr 19 '12 at 13:21
7
\$\begingroup\$

Steven is right, it's too much energy to be stored on the capacitor; batteries have much higher energy density than capacitors, whose advantages are others.

But there is also another consideration.

The voltage which your capacitor will give, no matter how big, will be decreasing exponentially, so, considering that your load will be smart and continue drawing 700 mA with any voltage, you'll have a descending slope.

Your capacitor will store a charge of \$504 \cdot 5 = 2500 \$ C (impressively high!), or also 2500/3600 = 700 mAh (as your requirement says). Now let's suppose that you run it for 10 minutes, draining \$ 10 \cdot 60 \cdot 0.7 = 420 \$ C. Now the charge in the capacitor will be lower, but the capacitance will still be the same, hence \$ V= \dfrac{Q}{C} = \dfrac{2500-420}{504} = 4.13\$ V.

So after only 10 minutes, your voltage will be almost 1 V lower than specified. Probably it's not acceptable for your application, unless you use a SMPS.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ But wouldn't batteries have the exact same problem? \$\endgroup\$ – Rob Apr 19 '12 at 10:16
  • \$\begingroup\$ @Rob to a certain extent yes, but batteries have much more energy density (so they can supply for longer) and a flatter voltage curve, so it will be less critic. \$\endgroup\$ – clabacchio Apr 19 '12 at 10:32
  • \$\begingroup\$ Ah, I get it. Thanks for your efforts, I really appreciate it! \$\endgroup\$ – Rob Apr 19 '12 at 11:30
1
\$\begingroup\$

did you search supercap stocks ? they are not expensive as you think. also known as 'ultracapacitors'.

But it's charge circuit is quite complex. If you got 100$ you could do this project.

I have seen them in the sparkfun stock. There are constrains applied.

and because of that discharge curve problem , I recommend you to use a joule theft to ramp it up to a voltage like 20v like voltage and then use a SMP regulator to step it down back to 5V.

see : http://en.wikipedia.org/wiki/Joule_thief

You could do this.

[note:- please note that I'm not working for Sparkfun ]

|improve this answer|||||
\$\endgroup\$

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.