# Charging 35V 470uf capacitor with coin cell

I want to charge a 50V 470uF capacitor up to 35V with the smallest battery possible. Can a coin cell do this 3000 times? Possibly a 2450 or 2477 coin cell? Also, can the capacitor be fully charged from 0V to 35V in 6 seconds.

Edit after some discussion:

How to boost from 2V-3V to 35V (acceptable range 30V-35V) with 24ma load (6 seconds recharge or better), with AAA or AAAA?

By now i'm using this scheme, PIC impose Q1 period .135 and Duty .12, up to 32V, i read with the meter 25ma charging and 12ma when fixed on the top. There is somewhere a heavy load on the battery, that goes down in few cycles and don't restore. L2 is the solenoid, S2 is the switch that discharge the capacitor into the solenoid. Led blink when fully charged.

This is the pic program i'm using: Pic Program

I uploaded the asm code i'm using: VRCON set V limit (0xa4 is near 30V-35V). When connected to 3V coin cell battery, i must increase VRCON value to 0xa6 to arrive to 30V, .35period, .12 DUty; on the meter i read a current flow of 25ma charging and 10ma while on top; i think it's not good for coin cell battery. 12F683 has function down to 2V so i need at least 3V to start with, an di'm not sure if 3V it's enough to fully activate Q1 (i'm asking why i have to double the voltage reference from pin6 to gain the same V)

With 9V battery i used this board:

With .14 Period, .3 Duty i obtain a 34V recharge about 14seconds or less, 14mA while charging, 9mA while on Top, less than 5mA when not recharging. Applied the load, 9V battery has 9,20V, after 80 cycles spaced into 2 hours the battery is 8,60V. imho it's too much. I cannot read the initial current draw with the meter, it's too fast.

Thanks to the courtesy of another user, Bobledoux:

FET drive frequency = 1,000,000/Period. So you are using 7.4Khz. That is a very low frequency for a converter. Q1 gate should be a perfect square wave, coming from the PIC PWM. Q1 drain depends on whether you are fully turning on Q1. To see that you need to be looking at the wave on an oscilloscope.

• See additions to my answer re boost converters. – Russell McMahon Mar 13 '12 at 23:28
• Would have been good to know of your PIC cct initially. Knowing this changes the nswer path. What battery "goes down in a few cycles and don't restore"? What FET drive frequency for Q1? What does waveform on Q1 gate look like ? What does waveform on Q1 drain look like. – Russell McMahon Mar 14 '12 at 1:37
• Added .asm and details. I'll try to answer all the question. I didn't wrote about pic because i was thinking about increasing Voltage out of the pic. – LukeP Mar 14 '12 at 8:24
• Russel, i send you a mail. – LukeP Mar 14 '12 at 16:38

There are two questions: the number of charging and the time of charge.

The 2450 battery has about 620mAh of energy stored. (dataheet from digikey). I'm assuming that you want to charge the capacitor with the same voltage as the battery: 3v. Therefore to fully charge a 470uF you'd need:

$Q_{cap} = C \times U$

$Q_{cap} = 470 \mu F \times 3V = 1.41 mC$

$Q_{bat} = 620mAh$

$Q_{bat} = 620mAh \times \dfrac{3600 s}{1 h} \times \dfrac{1 A}{1000 mA} = 2232 As = 2232 C$

$Cycles = \dfrac{Q_{bat}}{Q_{cap}} ~= 1582978$

Therefore the battery can charge the capacitor over 1.5 million times. If you want to "fully charge" the capacitor with 45v this number would drop to 100 thousand times, more than enough.

Remenber that I'm not assuming losses or other circuitry attached to the battery

The second question is if the capacitor can be charged under 6 seconds. We need to know how is the max currency we can get from the battery. Back to the datasheet this number is 0.2mA.

$Q_{cap} = 1.41 mC$ (from earlier calculations)

In order to feed this amount of charges in 6 seconds we would have a currency of:

$I_{cap} = \dfrac{Q_{cap}}{\Delta T} = \dfrac{1.41mC}{6 s} = 0.235mA$

As the required currency is lower than the one needed, there is no way to charge the capacitor under 6 seconds. The number from the datatheet is the "continuous standard load". It may be possible to charge the capacitor in less than 6 seconds, but it may be dangerous.

You may try another manufacturer aside the one I showed, they may have a similar battery with more "continuous standard load".

Another point to mention is that you need to consider the current drawn from the circuit you're building to charge the capacitor and subtract it from the available current from the datasheet.

• I want to fully charge the capacitor to 35V. What is the time needed to fully preserve battery life? Attached to the battery will be a drawing of 5ma, both for led and pic to decide when start charging and when discharging the capacitor through a solenoid. – LukeP Mar 13 '12 at 14:53
• @LukeP Are you sure you want to charge the capacitor to its limit? You will likely decrease the life your capacitor greatly by always using it to its maximum value. – Kellenjb Mar 13 '12 at 14:58
• Yes i know, in real board i'll use a higher rated capacitor 50V 470uf. – LukeP Mar 13 '12 at 15:06
• Am I missing something? How can one charge the capacitor to a higher voltage than the battery's voltage? – vicatcu Mar 13 '12 at 17:27
• @RMAAlmeida, in the past the number as the maximum for the battery current has very little to do with reality. We used coin cell batteries that had an absolute maximum of 1mA. We found until you pushed them past 10mA there was not a significant loss of life in the battery. At 20mA the loss becomes so large as to be unacceptable. This meant we attempted to never pass 10mA unless under extreme conditions, on batteries I suggest users measure it, especially on coin cells where I have found the values to be lies. We performed our test on thousands of batteries. – Kortuk Mar 13 '12 at 17:38

Use @RMAAlmeida 's provided battery capacity of 620 mAh.

Energy in cap = 0.5 x C x V^2 = 0.5 x 4.7x 10^-4 x 35^2 = 0.2879 Joule

Battery capacity = nominal 620 mAh x 3V = 1.8Wh.
In practice voltage will droop so say 1.5Wh (or less). 1.5 Wh = 1.5 x 3600 Ws = 5400 Ws.

Cycles at 100% efficincy ~= 5400/0.2879 =~ 18,800 cycles.

If a capacitor was charged via a resistor losses = 50%.
Being charge with a boost converter means that voltage can match Vcap at each point so losses can be lower.
Assume 80% efficiency achievable.

Cycles = 18,800 * 0.80 =~ 15,000

Assume 2.5V available to operate boost converter.
At 1 mA energy in = 1 mA x 2.5V = 2.5 mJ/s
At 80% efficiency energy into cap = 2.5 * .8 = 2 mJ/s

Cap energy from above is 288 mJ so time to charge = 288/2
= 144 seconds at 1 mA.

So charge time x Battery_mA = 144 Charge time =144/Battery_mA.

So if eg 5 mA was available or boost converter the charge time ~=144/5 = 28.8S
= say 30 seconds.
50 mA (if available) gives 3 seconds.

E&OE

Possible boost converter IC.

There are a substantial number of Asian sourced boost converter ICs which are used in mobile phone chargers and similar. Some have the capability to drive external MOSFETs or BJTs giving them arbitraily high power and voltage capabilities. In this case only voltage is an issue.

Nanjing Chipower make a nimber of useful ICs.
If you gargoyle search

Nanjing chipower pdf

you will find various devices that they make.

The CE9908 boost converter - datasheet here is liable to be useful here.
Guaranteed 0.9V start, external MOSFET drive, capable of bootstrapping the supply voltage to get higher drive once started, external feedback pin or voltage setting, SOT23-5 or SOT89-5 pkg. You want CE9908B125M. As your Vout is > Vddmax and Vin < Vdd_min_operate_usual you'll want a little extra circuitry to control Vdd but it's simple and cheap.

If operating from at least 2V and you can tolerate LT's pricing consider the superb LT1619 - datasheet here

Note that efficiencies shown are for typical or shown applications and that you can fine tune to suit your application.

Something like a PIC10F204 - datasheet here or similar processor will also do the job given at least 2V supply available.

• Russel, maybe you mixed 288mJ with 188 in the last part? – LukeP Mar 13 '12 at 16:08
• So two question: i cannot use coin cell because of recommended Continuous standard load is 0,20ma and extend too , how about – LukeP Mar 13 '12 at 16:26
• Battery_mA = 288/(2*Charge time) – LukeP Mar 13 '12 at 16:33
• Thanks yes - typo - 188 should have been 288. And mA.seconds should be 288/2 = 144. – Russell McMahon Mar 13 '12 at 16:34
• Some battery chemistries and constructions will allow much higher mAh rates than others. You should be able to find small battery packs that easily allow 50 mA discharge rates. What are your size constraints? How many cycles do you need? What is the application? – Russell McMahon Mar 13 '12 at 16:38