How much will the arbitrariness of the instrument drop when the supply voltage drops by 0.8 volts?

Question

Electronics

I have a good old screwdriver, and his battery died. I can’t find regular ones for it anymore, I decided to transfer it to lithium-ion banks (18650). However, the nominal supply voltage is 16.8 volts. Selecting tenths of a volt is a complication of the design, and then all the components will not fit into the case of the old battery. However, you can easily arrange 16 volts. How much will its arbitrariness drop if these 0.8 volts are missing?

Answer 1

almost linear to
pick the formulas - you will find the formulas for the dependence of the speed on the applied voltage.
for example https://servomotors.ru/documentation/electromechan...
0.8 / 16.8 =~ 4.7%
you can not even pay attention :) it will hide against the background of the 4s battery operating voltage range - depending on the discharge, this will be from 16.8 to 14 volts.
just take the banks, be sure to take high-current ones, otherwise they won’t be able to pull and will die quickly. well, bms is a must.

Answer 2

forum.fonarevka.ru/showthread.php?t=33409
5 parts, 500 pages

Answer 3

4.2 volts on 4 batteries will give you 16.8 volts.
But your question is meaningless, the voltage of lithium should only roughly correspond to the voltage of nickel. during operation, it one figs sags and floats.
For dummies: https://youtu.be/GIXKdcaruFk
Approximately:
12V = 3 lithium 16-18V
= 4
lithium
It is not necessary to put exactly 18650, high-current batteries of other sizes are now sold.
21700 for example or 18500, based on the size of the compartment, you can choose the size of the cans.
And yes, it is convenient to take jars with already welded petals.

Answer 4

W \u003d U ^ 2 / R
When the voltage decreases by 1.05 times, the power will drop by 1.1025 by about 10%

Answer 5

But if the screwdriver at some point works at the minimum allowable values, such a fall can lead to a stop.

Answer 6

There is a chance that performance will not only not fall, but even increase. That's exactly what happened to me after the conversion.
I think the point is that Li-Ion batteries are capable of delivering significantly more current with the same voltage drop on the battery - and, accordingly, delivering more power.