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Need math check

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charlie_r

Member
This might be a hairbrained scheme, but need to check this.

If I have a capacitor bank consisting of 20 3300uF capacitors, that should give me 66F right?

If I charge this bank to 15V that should be 7425 Joules of energy, + or - the 20% tolerance of the caps.

Since 1J = 1W/S[econds] I should be able to draw 7425W for 1S or 1W for 7425S. Am I right so far?

Ok If I have a circuit that draws 500mA or 7.5W, I should be able to power it for 990S, or 16.5 minutes, disregarding losses. Realistically I would expect it to work no longer than 5-7 min.

Have I done the math right?

BTW this is not a homework question, it is for a project I am currently working on.
 

MrAl

Well-Known Member
Most Helpful Member
1 Farad = 1000000 microfarads so i think your capacitance calculation is a little off.
Check it again.

Also, the total energy available is 1/2 C*v^2 so check that calculation again too.

You also have to consider the drop in voltage over time when loaded to see if what
you are powering will still work after some time period.
For example, a 66 F capacitor charged to 15 volts and loaded with 15 ohms will
discharge to about 5.5 volts after 990 seconds. It will also discharge to about 12v
in 12 seconds. This of course will affect the load. It certainly does not deliver
15 volts for the entire time it is discharging. If you need that kind of operation,
you'll need to use a voltage regulating boost circuit between caps and load.
With a regulating boost converter that can operate between 15v and 5v and put
out 15v for example, you can get the full 15v output for approximately 440 seconds
and then the output will drop down suddenly (assuming 66F charged to 15v again
and 15 ohms load on the output of the boost converter, and not including the
inefficiency of the boost converter). If the boost converter itself dissipates about
1 watt then the total run time will drop to about 412 seconds.
 
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charlie_r

Member
Ok thanks. So, this would still be a "fool's errand" to try to use cheap caps as a battery replacement, for a keep-alive for a couple of minutes.

It would therefore be cheaper to use a rechargeable battery bank to keep the circuit active.

That is what I needed to check.

Thanks!
 
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MrAl

Well-Known Member
Most Helpful Member
Hi again,


Well, if you have a constant current like 500ma then you can use:

dv=i*dt/C

to calculate the time dt, and doing that we get about 7.9 seconds until
the output reaches 9v at which time the circuit will stop working.
That's if you charge the 0.66F caps to 15v though, if you only charge them
up to 12v (because circuit is limited to 12v max) then you will only get
about 3.9 seconds of operation.
 
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charlie_r

Member
As stated in the OP, the bank would be 20 3300µF (25V) caps, that I would be charging to at least 15V. If I could get a run time of 2 minutes minimum I would try this. But as the [corrected] math is showing me that it won't work as desired, I will do the smart thing and abandon the idea.

Thanks for your help.
 

MrAl

Well-Known Member
Most Helpful Member
Hi again Charlie,


I just thought i would let you know that when longer run time is needed usually the
solution ends up being batteries instead of capacitors. The batteries are the
rechargeable type like NiCd or Lead Acid. NiMH are not used as much for this
kind of thing because they have specialized charging requirements.
You can also look into Li-ion cells, as three in series would give you enough
voltage and run time would probably be over two hours.
You could get a charger off the web or build one yourself (carefully of course).

Just some ideas there.
 

Warpspeed

Member
Stored energy is proportional to voltage squared. (1/2CV^2)

And at some minimum voltage, your circuit will stop working, so you cannot keep extracting power from it right down to zero volts.

Capacitors are only really suitable for that type of energy storage application if charged to truly lethal voltages to begin with, and then run all the way down to some much lower voltage.

your example of 20 x 3300uf is 0.066 Farads
Charged to 15v is 0.033 x 15 x 15 = 7.425 Joules

Suppose your circuit stops working at 5v
Energy at 5v = 0.033 x 5 x 5 = 0.825 Joules.

The energy you can actually recover as it discharges from 15v to 5v will be 7.425 - 0.825Joules

Now think about one only 3300uF capacitor charged up to 400 volts.
Energy for 0.0033 Farads.
0.00165 x 400 x 400 = 264 Joules. And you only need ONE capacitor to store that much.
 

MrAl

Well-Known Member
Most Helpful Member
Hi again,


Yes a bank of 20 caps each 3300uf is only 66,000uf, which is really not
much at all. For your constant current of 500ma then you can still use:

dv=i*dt/C

to calculate the time dt, and doing that we get about 0.79 seconds until
the output reaches 9v at which time the circuit will stop working.
That's if you charge the 0.066F caps to 15v though, if you only charge them
up to 12v (because circuit is limited to 12v max) then you will only get
about 0.39 seconds of operation.

As i was saying before, this application would probably be better off using
rechargeable batteries unless you feel like also building a voltage converter
so you can make use of the cap's energy better.
 
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