Answer #1Originally posted by ashbreeze96@Mar 2 2006, 01:14 AM
can anyone help me with converting power from a 9.6v 700mah battery pack to 8v 200mah
Firstly the "mha" rating is a function of the cells in the battery. As mentioned, it is the amount of current that can be supplied by the battery in one hour before being unuseable. So in your situation and ignoring the change in voltage if you were to only draw 200ma then your 700mah battery should last 3.5 hours. Obviously changing the voltage would change this time. This figure is the maximum so there is no real minimum.Originally posted by Papabravo@Mar 2 2006, 04:46 PM
I can take a 9.6V source and with a three terminal voltage regulator I can make an 8V output.
The "mah" is a unit of capacity and it is computed by multiplying current in milliamperes by time in hours. What it means is that you can get some number of millamps for some number of hours before the battery voltage falls to an unuseable level. This unuseable level will be different for each application.
You original request asked me to convert a high capacity battery pack(700mah) to a lower capacity battery pack(200mah). I don't know of any way to do that, nor can I figure out why you would want to do that.
The point at which the 8V regulator will drop below 8 Volts is determined by the Drop Out Voltage. This may be as low as 300 millivolts. So the 9.6V Battery pack would need to discharge to 8.3 Volts before the output would fall below 8 Volts.
If the 9.6V battery pack is made up of idividual cells @ 1.2V each as many NiCad or NiMH battery packs are, you might be bold enough to remove one cell dropping the output from 9.6V to 8.4V. Warning: you can seriously injure yourself messing around with battery packs if you're not careful and you don't know what you're dealing with. Things can get hot in a hurry, and when a battery gets hot, bad things can happen. You don't want the nasty chemistry inside the battery to get out. It's sorta like letting the magic smoke out of a chip, only much more unpleasant.
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