Lithium Batteries Jump Pack Update

Hello

I just finished assembling the pack yesterday. I charged it to 12.6 volts. There are 3 packs of 10 Lithium Ion cells. 2Mah capacity cells. Did not rate the. The packs are is series for 12 volt. I charged the pack to 12.6 volts. I disconnected the negative cable from the mower battery, and used the pack. First picture is the voltage of the pack after I charged the caps. Second picture is of the pack voltage while cranking an 18HP Briggs. Third picture voltage after cranking. Fourth picture voltage with the capacitors hooked up. Fifth picture engine cranking voltage with the caps. Last picture is the voltage pack after cranking. I though the caps would stabilize the voltage, and the did,

Guido.

Will your battery pack start mower without a battery? Did you try using it on a dead battery?

Hello Geo-TH, In,

YES! I removed the negative cable as I stated on the post. better voltage while cranking with the caps. See the pics,


Guido.

Mr. Guido Sir,

Where is your initial post indicating what you are trying to do with what equipment?

Hello Texasmark1,

I guess in the archives? I don't know how to direct you there. I was however trying to get enough cells to make a 20ah pack at 12Volts. They are used cells out of the recycle bin.The pack will start/crank my lawn tractor while maintaining 10v doing it. I'm sure none of the packs have the 20ah capacity. There are 10cells each in parallel, so 20ah?. Internal resistance may change though. Do you know when is the internal resistance taken? Charged or not. I don't know. I will have to check it as well as the individual pack capacity. So far starting an 18hp on its own is pretty good. That was my goal. Surprisingly after it crancked the engine it's voltage was. 12.47 volts and it started to go higher as the engine was running taking charge. When I added the caps the voltage was 12.50 volts.,I crancked the engine the second time and the voltage was higher with the caps in the circuit. See the picture, caps are adding and stabilizing the voltage. I'm guessing the pack now may be 100 CCA?


Guido.

I would assume the internal resistance would be constant regardless. It obviously comes into play when discharging at high current and the mechanism would be things like plate size and dielectric constant-thickness sort of things, wire size.

On the smoothing effect you mentioned when using the Cs, chargers are full wave or bridge rectifying devices with no filtering, so what happens when the half sine waves come off peak? One would assume the battery just holds it but the internal resistance of the battery would cause a time lag (phase shift, not all that much since 60 Hz half sines aren't all that fast) such that the peak would/could come and go before the battery got up to it. Adding the C storing energy, when the source "Sin omega T" comes off 90 or 270 degrees, the cap continues to feed the batteries, thus raising the average value. Course that assumes the internal resistance of the C is less than the battery and much less than the source...the charger.

Otherwise I see no need for the C as they are the wrong type to have low internal impedance when you are dumping your batteries (and them too)....ceramics and micas are better for that.

I'm rambling along here. I'm retired and slowly loosing what I used to make a living. I don't want to do that so these little sessions help to keep what's left alive......I think. Grin

Hello Texasmark1,

The caps are keeping the load voltage a bit higher then the first try. Not sure how. They are DC caps in parallel one is 10 V DC the other 30 V DC. We are talking straight dc here. The phase shift got me though. You are saying then that the internal resistance stays the same regardless of state of charge? The ohmic test method is the best way I thought. Measuring ocv and load voltage will give a voltage difference. Then the voltage difference / current = resistance if I recall. But never tried an actual test. I guess that is my next step. Hey if you don't use it................


Guido.

Why would it change between charge and no charge? The R is in the construction of the device, or an increase due to decreased surface area, like a metal foil used for AC circuits where spikes blow holes in the insulating material (deliberately designed to clear the fault when overload spikes occur) thus reducing the area.

Not sure you could accurately measure it due to the capacitance effect and at that you'd need an analog meter. With c's that size, it'd take forever for the needle to return to 0 and would be gross confusion if you asked me.

Hello Texasmark1,

I do have more then one analog meter, but here is what I will simply do: Take an R.P.M. test. First with the pack alone,I did that the first time. Then I will add the caps and take the R.P.M. test again, without charging the pack. Did that too the first time, and the voltage was higher with the caps. I expect the R.P.M.'s will be higher with the caps, proving that they do add power-capacity to tbe mix,


Guido.p

Energy stored is ? Capacitance (Farads) x Volts exp2 .....volts squared, unit is watt-seconds. Consumption is energy times rate of consumption VIT (Volts, current...amps, time.....seconds). ? is used in the equation because ? of what one though would be stored is used up in the charging and discharging process.....per the owner of the equation....have no idea as to who that is.

So add that to your battery's stored energy to drive your motor. Speed is amps per second which is where your resistances come into play, with V initial distributed across all the circuit loads till dissipated.