Author Topic: circuitry battery  (Read 2483 times)

0 Members and 1 Guest are viewing this topic.

Offline RebelgiumTopic starter

  • Supreme Robot
  • *****
  • Posts: 637
  • Helpful? 0
  • It's called the future ... We like it here
circuitry battery
« on: August 28, 2007, 12:43:34 PM »
Just a quick question, What kind of capacity do you use for the battery that feeds the circuitry.
So this battery will only supply power for all the electronics, no motors and servos.

I'd like a a fairly long battery life. And the electronics will be : 2x dual Hbridge, PIC board (the brain), 2x sharp IR, 4 IR proximity sensors, and more for in the future.
So there has to be enough capacity to still be enough in the future (when there'll be more electronics added).
To relax after some hard work on robotics: A very fun free online text based MMORPG

Offline dunk

  • Expert Roboticist
  • Supreme Robot
  • *****
  • Posts: 1,086
  • Helpful? 21
    • dunk's robot
Re: circuitry battery
« Reply #1 on: August 28, 2007, 01:38:40 PM »
hi Rebelgium,
so the data sheets of the various components should be able to tell you how many milliamp (mA) each component draws.
add them all up.

now your batteries datasheet should tell you how many milliamp hours (mA/h) it is good for.

divide the mA/h figure by the total mA draw of the circuitry and you'll get the number of hours battery life you can expect.

this of course assumes that the voltage of the components is the same as the voltage of the battery pack.
if you add voltage regulators into the mix things get more complicated.

linear voltage regulators will convert the excess power into heat.
generally speaking, if you are using a linear regulator you can expect the current draw from the batteries to remain the same as the current draw of the circuitry.
this means that if you are using a linear regulator to drop the voltage of your battery you are wasting all of this power.

switching voltage regulators on the other hand draw only the power that is required from the battery.
you need to use the P=IV (Power = Current * Voltage) formula where the power coming out of the battery will be the same as the power being used by the circuitry minus a percentage for the inefficiency of the regulator. (typically 10 to 20%)

for more info on regulators.)


Offline zamboniman60

  • Full Member
  • ***
  • Posts: 55
  • Helpful? 0
Re: circuitry battery
« Reply #2 on: August 28, 2007, 11:24:42 PM »
If you don't want to pay for a rechargeable battery for the IC and low-energy electronics, you could use 2 CR123A batteries in series with a 5 volt (Important! low-drop-out!) regulator.

Offline Joe

  • Full Member
  • ***
  • Posts: 54
  • Helpful? 0
Re: circuitry battery
« Reply #3 on: August 30, 2007, 02:23:25 PM »
Yes, it pretty much depends on your voltage regulator. You will probably never need more than 1A for logic circuitry but if you ever do you can always put another regulator in parallel. You could get by with 7805 linear regulator and a "9-Volt" but you would have to change batteries pretty often. Or you could use a 7.2V "AA" or "AAA" NiMH battery pack for long life at 1A capacity (but increased weight). If you use a LM2940 linear regulator you would only need a battery 1V above (6V) your 5V supply for a 1A of max capacity, but slightly more waisted "idle" current. As for the switching regulators, at least the ones I know of, they are only more efficient if you are stepping down from a much higher voltage (like 12V or more).


Get Your Ad Here