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# how to decrease the impedance from a battery ? Answered

if i have a battery that give me 16 M ohm and i would like to decrease the amount to a 1 ohm , if i try and put an inductor in parallel with the battery , it does decrease the amount , but the inductor and the battery get hot and may do damage , so i'm looking for a converter maybe to connect a battery to power transistor and then to an inductor or a transformer , not to increase voltage but to decrease impedance , anyone can help me through this?

thank you

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## Discussions

To decrease ESR or o=impedance, Icing mentioned using a supercapacitor, which is good for small transient loads, but for a continuous load, you will just need to use a different/bigger battery. Adding more batteries in parallel should help. Also if they are rechargible, then generally batteries have the lowest ESR when fully charged.

In fact years ago when NASA published a weekly tech ideas and developmental multi science innovations for engineers... They NASA had a small very high capacity energy but high impedance primary battery.... Which would be used to charge a lower capacity secondary, low impedance battery for occasional intermittent current draw beyond the primary ability... This was used in untended applications for months of time.... Not much different from my super-cap suggestion.

You're trying to do something intrinsically impossible

16 M ohm? Am I reading that right? I mean, where I'm from, "M" stands for "mega", also pronounced "million"

M = 10^6 = 1000000

Unless you're talking about a battery with a potential of similar magnitude, like 10^7 = 10000000 volts or so, this thing you call a "battery" is useless as a power source, and there are no impedance matching tricks that can help to extract power from it.

In the immortal words of Roy Scheider, "You're going to need a bigger boat."

Or in this case, a bigger battery.

Wait. I've got that part about the voltage wrong. What I should have said is square of the voltage should be about the same magnitude.

For example, if it were a 4000 volt battery, and you had a 16 Mohm load, the same size as the battery's 16 Mohm internal resistance. Then the total power dissipated across the internal resistance plus load, would be P = (4000V)^2/(32 Mohm) = 16 M/32M = 0.5 watts dissipated by both the internal resistance and the external load, or 0.25 W for each individually. In terms of voltage and current, the load has a voltage of 2000 V across it, and a current of 125 uA (microampere) flowing through it.

Best way to decrease a battery impedance is to put a super-capacitor across the terminals making sure the peak charge voltage does not exceed the supercap peak voltage... Depending on the capacitor farad rating you will have a period of very low impedance (milliohm range) that is very useful for motor starting.

In the meantime any time the battery is unused the capacitor is automatically recharged to provide the mini impedance you seek...