He was able to sell them in large quantities to truck companies and tractor companies. Now there are many failure modes in batteries such as metallic particles in acid causing leakage and self heating or boiling of electrolyte. So in a sense it rejuvenated the electrodes. The theory was the 25Khz or so 10 nS pulses (Tr) of small current resonate the sulphate crystals that coat the electrodes and cause battery life reduction. proving to me that the battery health was improved. Not only did ESR improve but specific gravity improved after full charge. We made many and I test one on a large motive power tractor at Air Canada for hauling planes. Company has long since sold its patent rights and owner made millions. The product was a little pulse circuit in parallel with LEAD ACID or ni-cad batteries and called "Solartech". It that was invented by an old neighbour of mine, and manufactured by my employer EMS firm "was Nortel then C-MAC. It can actually improve ESR of old batteries if not damaged by shorting or heat. There is another AMAZING benefit from pulse discharge. So - overall, multiple factors combine to reduce effective battery capacity and lifetime under pulse discharge. Lead acid cells tend to mechanically destroy their plates with current being a factor in rate. NimH undergoes secondary reactions which are affected by temperature and internal potentials. LiIon cells are mechanically pumped to death mechanically as Li metai is added and removed (hence the success of liFePo4 which maintains an Olivine solid cell structure when 'empty"). In both cases the cell is fully discharged but the effect on the system is exacerbated by higher rates. eg a 2Ah cell discharged at 2A will provide fewer full discharge cycles than the same cell discharged at 1A. (3) Cells last longer when discharged at lower rates relative to their amp hor rating. Where V_1A is the terminal voltage when 1A is drwan etc. The higher current will lower the terminal voltage so at 50% duty cycle the current will not be double but will be 2 x V_1A/V_2A (2) Worse - if this is feeding a switching power supply, the requirement is NOT constant mean current but constant energy rate delivered to the converter. Ie doubling current will double losses if Rinternal does not rise. (1) Cell internal impedance will almost certainly be higher at higher current but even if it is the same. Take as an example a 2000 mAh = 2Ah battery discharged at 1A continuous or at 2A with a 50% on/off duty cycle. If the energy out refers to what the load or user sees then the situation is even worse as the battery will have higher internal losses at higher current so lose more internally to deliver the equal energy externally. Even if the energy out is what the battery sees and not what the load sees, then higher discharge rates will decrease the available amp hours for equal delivered energy. If the same energy is taken by either steady or pulsing discharge then the pulses must be at higher current. The degree depends on battery chemistry, construction, past history and more but overall pulsing is almost sure to be worse. So this is the secondary effect of using pulsed power, not only is the battery dissipating more energy due to the I^2 term, it is "losing" available capacity due to elevated temperature.īoth energy losses and negative effects on the battery will usually be higher for pulse discharge than for steady DC discharge. This heat will elevate the batteries temperature, and this elevated temperature will also reduce the batteries capacity. The 4 W will generate signiciantly more heat than the 0.06 W due to the thermal resistance of the cell to ambient. Most importantly is the 0.06 W vs 4 W internal power dissipation. Time to deliver energy = 6.25 seconds (total "on" time),.Multiply this by time and you get energy (joules) lost to the resistor.Įxample: Vbatt = 10 V and Rbatt = 1 ohm. So the power loss in the resistor goes up with the square of the current.
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