Newsgroups: comp.robotics
Subject: help...battery power
From: andrew.jenkins@pcohio.com (Andrew Jenkins)
Path: brunix!news.Brown.EDU!agate!usenet.ins.cwru.edu!ncoast!pcohio!andrew.jenkins
Distribution: world
Message-ID: <24.4345.2500.0NB2D631@pcohio.com>
Date: Mon, 21 Feb 94 09:32:00 -0500
Organization: PC-OHIO PCBOARD - Cleveland, OH - 216-381-3320
Lines: 70

<<  CHENGB@ECF.TORONTO.EDU to ALL   on    help...battery power  >>
CB>   I am building a self-contained robotic biped and I confront a
CB>   problem of not enough power from the battery to drive the
CB>   motors. The dc gear-motors that I use are 12.8V, and I need
CB>   to use dry cells to drive four motors.         My question
CB>   is: How many batteries do I need in order to drive the motors
CB>   half an hour? Of course the main consideration is the weight
CB>   because I don't want to put too much weight on my robot.
CB>   I am thinking to use DC-DC converter or voltage regulator to
CB>   step down the voltage by connecting several 9V batteries
CB>   together and increase the current, but I am not really sure
CB>   whether it is a suitable way to do or not.
CB>   N.B. The current of a motor can run is 0.55., 20 oz.in. at
CB>   120 RPM, and the weight of my biped is about 800g.

* A fully charged battery stores a certain amount of charge, usually 
specified in N Amp-hours, where N is a number (1, 2, 1.3, 500milli,etc). The 
spec means that you can run at a current of N amps for one hour before the 
battery is fully drained. If for instance, you have a 12 volt cell that is 
specified as a 2A-Hr battery, and draw a constant current of 0.5 amp, your 
device will be powered for a total of ~4 hours. 
(approximately 4 hours, since the voltage vs. charge characteristics of most 
batteries are such that the voltage drops off as charge is lost. Therefore, 
as the 4 hr time approaches, the battery may not have sufficient voltage to 
force a draw of 0.5 amps,from I=V/R.)

 Lithium and Alkaline batteries have a very flat characteristic curve. They 
essentially retain constant voltage until charge>>zero. They are very good 
at providing a constant voltage source, as long as the current load remains 
fairly low. I don't believe that these batteries are suitable for powering 
motors, but someone more experienced with robots may prove me wrong.

 Lead-acid (and NiCad I think) cell-voltage is more charge dependant, though 
they generally can handle much larger current draws, have significantly 
larger charge capacities and are, of course recharge-able. I noticed that 
alot of people in this conference are using a Gates X-Cell rated at 2.0V, 5 
A-Hr. It might be a little bit heavy for your application though, so maybe 
something more like a standard C or D-cell might be more appropriate.

A good source for specific battery information is direct from the 
manufacturers. 

You'll have to consider the extra battery weight in your calculations also, 
since the added weight will increase your current draw.You can find out the 
various battery weights and empirically determine the load currents 
asssociated with this added weight by adding the  appropriuate weight to 
your robot and measure the increase in load cuurent. (Start-up current is 
something to think about also)

Ohm's Law should solve the time question.

I'd like to have calculated the time involved, but you didn't specify the 
battery, the actual constant load current (or average), and whether the 800g 
weight includes batteries or not.

One last thing. Are you using PWM to power your motors? This *could* help 
you save battery charge.

AJ



AJ



AJ

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