Electrical – High Voltage (3,000V) Capacitor Charging Circuit Needed

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.

What you are trying to achieve does not seem overly hard but it needs to be better defined than has been done here so far because the situation that you describe sounds to be different enough than the usual charge & run situation that you need to make specific provisions.

You need to identify desired pass through Imax and I_pass_through_average for a given Ah capacity. The most demanding case will be that with the largest pass_through_Ah ratio.

Usually with simultaneous 'charge and run' situations I load is less or much less than Ichg_max. In such cases it is often OK to address the battery needs and accept that the load may somewhat prolong charging times. Clearly I_load_average must be less than I_charge_averge or the battery would never charge, but you may get situations where I_load is > I_charge for significant periods, or is a substantial fraction of I_Charge_terminate, and this will affect charging results.

Usually Imax in constant current mode Icc is the same in mA as the capacity in mAh. So a 1000 mAh cell is charge at 1000 mA max, a 5Ah cell at 5 A max etc.

If you just increase charging capacity without regard to interaction with charge termination you may 'have problems'. This is because LiIon/LiPo charging typically terminates when I charge in CV mode drops to some preselected percentage of ICC = Imax and I_pass_through is seen as part of Iterminate if special steps are not taken to prevent this.

Taking the extremes of your spec ie 3A charge and 5Ah cell that is 0.6C charge (still under the usual 1C max) the usual charging terminate current will usually be in the 0.5C to 0.1C range with 0.25C being typical. So here with a 5 Ah cell Iterminate will be 2.5 A / 1.25A / 0.5A max/typ/min. If I_pass_through is say 1A mean it will swamp a 0.5A terminate current and interact significantly with a 1.25A or even 2.5A terminate current. In a dumb charger, if the C/10 rate was used for terminate the charging would almost certainly never cease, at 1.25A terminate (C/4) it may cease on load dips and at 2.5A terminate charging may be prolonged.

A LiPo cell charged at CC of C/1 will reach about 75%-85% of full capacity when CV level of 4.2V is reached. If charging is at less than C/1 (as will be the case if you charge a 10 Ah cell at 5A or C/2) then an even larger % of the full capacity has been transferred when you reach 4.2V. If you are happy to accept say 90% of full capacity then the easy choice is to hard terminate charge when 4.2V is reached and have no CV final stage. This removes the need to make end of charge determinations based on combined pass through and CV currents. Some charger IC's are available with no CV stage.

Also, as now you only need a CC stage and a voltage cutoff at 4.2V then a current limited power supply and a comparator will fill your need. Adjustment of the current limit on an off the shelf supply may suffice in a small volume research situation.