# super capacitor capacity?

#### Divedeep

Feb 2, 2014
57
Hi.

I am working on a project and have had an idea to improve it. There are 2 parts to this improvement.

First I would like to replace the battery in the device for a super capacitor. I have seen 10F , 2.7v capacitors available.

The device itself runs on around 2-5uA when asleep. When awake draw is aroun 15-30mA. The device is awake for 1 second every 15 minutes.

My question is how much power can the capacitor hold and how long could it power my device?

The second part i would like to have a go at is charging the device by wireless energy transfer.
I am wondering if it is possible to transfer enough energy to keep the capacitor charged so power is available when required.

The transmitter for the wireless power transfer will be around 3-4 meters from the device to be powered.

Thanks

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
The device itself runs on around 2-5uA when asleep. When awake draw is aroun 15-30mA. The device is awake for 1 second every 15 minutes.
OK, let's do some calculations.

I'm going to assume that your circuit will be powered directly from the supercap at 2.7V and can tolerate a supply voltage as low as 2.5V.

1 second every 15 minutes is a duty cycle of 1/(15*60) = 1/900 = 0.1111%. That percentage of 30 mA is 33 µA. Plus 5 µA continuous is 38 µA mean current drain.

If the supercap is fully charged at 2.7V and can discharge to 2.5V, that's a dV of (2.7 - 2.5) = 0.2V. You can calculate the amount of time a 10F supercap can power that circuit for, using the formula dT = C dV / I
= 10 * 0.2 / 38e-6
= 52,600 seconds
= 14.6 hours.

That's assuming worst case current consumption.

If I've made any incorrect assumptions, you'll need to correct them and rerun the calculations.
The second part i would like to have a go at is charging the device by wireless energy transfer.I am wondering if it is possible to transfer enough energy to keep the capacitor charged so power is available when required.
Mean power consumption is voltage multiplied by mean current, which is 2.7 * 38e-6 which is about 100 µW.
The transmitter for the wireless power transfer will be around 3-4 meters from the device to be powered.
This is outside my experience. Is it possible to transfer enough power, magnetically I assume, to produce 100 µW at the receiver at a distance of 3~4 metres? I can't answer that question.

If you have a continuous charging design, you don't need such a big capacitor; you just need one to hold the charge for 15 minutes or a bit longer. C = I dT / dV = 38e-6 * 15*60 / 0.2 = 0.17 farads.

Can you use some other method of getting the power into the device? Light, perhaps?

#### Divedeep

Feb 2, 2014
57
Wow. I didnt realise that a supercap would run the device for that long. In that case im thinking maybe 0.3F would be suitable. This would last about 30 minutes ??

The device is directly powered from 2.7v and can run down to 2.1v i should have mentioned this before.

I plan to use a continuous charge design for charging.

So i guess its potentially all dooable then..

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
All doable with the possible exception of transferring 100 µW over a 3~4 metre distance.

#### Divedeep

Feb 2, 2014
57
I am looking into the wireless energy transfer. If it sounds like it may work i will go with it and see if i can get it to work.

#### Divedeep

Feb 2, 2014
57
I have found a 0.22F, 2.5v capacitor that is physicall small enough for my application. Would the 2.5v be enough to power the device under the same conditions?

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
You're in a better position to know that than I am. But I will repeat my calculations in post #2 with the new values:

C = 0.22F
dV = (2.5 - 2.1) = 0.4V

dT = C dV / I

= 0.22 * 0.4 / 38e-6
= ~2300 seconds
= ~40 minutes.

#### Divedeep

Feb 2, 2014
57
Thank you.

This is a bit new to me as i havnt used a capacitor to powet a device before. It seems they work pretty much like a battery but over a shorter period. I also guess that they have a power curve much like that of a battery??

The difference i can see with a capacitor is they can be drained to 0v.

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
No, they don't have a power curve like a battery. If you mean a discharge curve, well, the discharge curve of a capacitor into a constant load current is a straight line. That's what the formula dV/dT = I/C means: for a certain capacitance, dV/dT (the rate of change of voltage - charging or discharging) is proportional to the current flowing. The more current you feed into a capacitor, the more quickly the voltage across it will increase; the more current you draw from it, the more quickly the voltage across it will decrease.

#### Arouse1973

Dec 18, 2013
5,178
Hello Divedeep.
All I would like to add to Kris's great reply is depending on the manufacture of the super capacitor you may have to consider leakage current. Some of the capacitor might have a higher leakage current than what your circuit draws on average. Also charging current will need to be considered which will be very high with a low source impedance and could be several hundred mA for a short amount of time. Example: to charge the capacitor from 0.5V to 2.5V with a 2.7V supply and a 10R series resistor will take approx. 5 seconds and draw a peak current of about 220mA.

The other option of powering over 3m - 4m, do you want to charge the capacitor or just have the circuit running from the wireless power?

#### Divedeep

Feb 2, 2014
57
The idea with the wireless charge is to ve constantly trickle charging the capacitor. The capacitor will then deliver the higher current when required.

The capacitor will have maximum draw every 15 minutes. So the charger needs to top the capacitor back up within the next 15 minute slot.

Hope that makes sense.

Heres the capacitor im thinking of using.

#### Arouse1973

Dec 18, 2013
5,178
Yep Ok I'll have a look at something for you.

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
Good point about leakage current Adam. I just did a quick sample of five types available from Digikey and here are the leakage current specifications for a 1F part:

Nichicon UW: 500 µA maximum - far too high!
Cooper Bussman PowerStor B series: 4 µA nominal, after 72 hours
Elna DZ 2.7V: 200 µA maximum, after 24 hours
Maxwell HC: 6 µA maximum
Taiyo Yuden PAS: not specified!

It looks like the Maxwell HC series is the best option. The Cooper Bussman parts only specify a nominal leakage current figure; there is no maximum figure stated.

#### Divedeep

Feb 2, 2014
57
Thanks rajeshr i will take a look.

#### Arouse1973

Dec 18, 2013
5,178
He is something rough I had a go at working out. It is for two un-tuned coils separated by 3m. Shows the amount of current needed to induce a certain voltage in the secondary coil. This is approximate and would work better if the two circuits were tuned.
Thanks

#### Attachments

• COIL.PNG
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#### Divedeep

Feb 2, 2014
57
Hi Adam. Thanks for that calculation. I dont quite understand all of it. Hopefully you could clarify.

Why are you making calculations based on 20v is this what the receiver will collect or the voltage transmitted.
If it is on the receiver side i assume i will have to use buck converter to feed the capacitor??

#### Arouse1973

Dec 18, 2013
5,178
Because the TX coil will be loaded by the circuit and it is not regulated I just added 20V to give a bit of head room for the current. It might be lower. The example just gives you an idea of what is needed to receive 20V on a coil 3m away. It is not the absolute values for your circuit. I could work it out but it would take longer.

#### BobK

Jan 5, 2010
7,682
On the other hand, at 38uA average current draw, a CR2032 would operate for about 9 months, so why go to such extremes as supercapacitors and wireless charging?

Bob

#### Arouse1973

Dec 18, 2013
5,178
Maybe because he want's to have a play with electronics.

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