Power Budgeting

Is this a valid way of looking and calculating power budgets?
I have 24 "AA" Cell NiMH cells that I want to use to power an autonomous
line seeking robot. The cells are rated at 2100 mAh.
I have several constraints. I need 500 mA at 24VDC to run the motors, and I'd like to reserve 250 mA at 12VDC for future wireless stuff.
So I ask myself how much 5V current can I then get from the battery pack if I want a 2 hour run time.
I'm planning on using switching regulators, so I come up with something like this:
2100 mAH / 2 Hours = 1000 mA current per hour 24VDC * 1000 mA = 24 W power budget
24VDC @ 500 mA / 85% efficiency of regulator = 14.12 W 12VDC @ 250 mA / 92% efficiency of regulator = 3.26 W ----- 24 W Budget - 17.38 W 6.62 W
6.62 W * 85% efficiency of regulator / 5 VDC = 1125 mA
or if I look at it this way
5VDC @ 1000m mA / 85% efficiency of regulator = 5.88 W + 17.38 W (from above) --------- 23.26 W
or 97% of the available 24W. Does that amount to a 3% safety margin?
Putting aside that I may not get a full two hours from the battery if I draw the current that fast (not sure what the internal resistance is) and I'm trusting the efficiency numbers from the spec sheets alot; can I safely believe that I have 1A at 5VDC for the rest of the system (sensors, controller and such)?
Is this a valid way to approach power budgeting of a battery pack? I'm very interested in factors I might have missed in thinking this through.
Thanks,
Kevin
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Kevin C. wrote:

[..calculation..]
It looks OK, but...

For electronics, if you can avoid it, it is better not to use the same power as your motors. The same goes for the wireless stuff.
If you can use a different source for the motor, you don't need to regulate it (=more efficient).
An other thing. For the motors, is the 500mA at 24VDC the average power or the peak power?
Good luck,
Peter
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
The 500 mA is the stall current (should be the peak as well)
500 mA at 24 volts is only 12W. This is a fairly small device and using a separate supply is impracticle.
Thanks,
Kevin

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Kevin C. wrote:

The average power used for the motor is less. It is a bit difficult to guess but it could be less then 50% depending on the task of your robot.

I don't know if it's practical and probably impossible but if you got 24 AAA cell for 500 mAh for running the motors you could use 4 or 5 AA cells with 2100 mAh for 5 volt only. When the wireless stuff arives, you can add some batteries (eg. 9volt) for it. I know, this is not what you asked but it may solve some problems and get your thoughts on an other track.
Good luck,
Peter
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Why regulate this voltage? Use a chopped current mode stepper chip, and be more efficient.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
A simple approach: Take the amount of amp hours your battery is rated at, and divide by the amount of total amperage you will draw from it. There is no need to get in to all kinds of calculations with wattages. Convert everything to Amperes. For the value to use, apply the total sum of the Amperes that will be loading the batteries.
Ahr = Battery Ahr / Load in Amps
Ahr = Ampere Hours.
When the battery is depleted to below 90% of its nominal voltage output under load, it would be considered discharged.
For an example, we can look at a standard 12 Volt gell battery rated with a nominal at 12.5 Volts. When loaded, the output will decrease over time. When the battery output reaches 11.25 Volts, it would be considered discharged.
For example if you have a battery rated at 2 Ahr, and you load it at 450 ma or 0.45 Amps, the amount of time considered for this battery to be able to supply the load would be 2 / 0.45 = 4.44 hours.
Jerry G. snipped-for-privacy@hotmail.com
--
snipped-for-privacy@glenevin.com (Kevin C.) wrote in message
news:< snipped-for-privacy@posting.google.com>...
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Jerry Greenberg wrote:

Actually, bear in mind that the AH rating is usually determined by discharging the battery at a relatively low rate over a long period of time (often 20 hours -- sometimes this is printed on the battery as 'T=xx', where xx is the discharge duration).
This means that the battery will likely NOT last as long at higher discharge rates, despite the stated AH number. The faster you'll be discharging the battery, the worse the deviation from the expected working time.
I'd plan on subtracting a good 30% or so to what you'd expect to get by naively calculating the run time in the above manner. There may actually be a better rule of thumb for doing this -- I usually just use 30% and go with that; others may have a better approach.
Hope that helps -- tAfkaks
--
(reply to mikey at swampgas dot com -- ignore the spamtrap Reply-To)

Teleoperate a roving mobile robot from the web:
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Reminds me of the joke, "For every deep complicated problem there is solution that is short, simple and wrong."
Battery amp-hour ratings are defined for a certain discharge rate (generally around 1/10 C or 1/10 their capacity in amp-hrs). At different discharge rates they can have radically different lifetimes.
That is why every battery manufacturer publishes a graph showing discharge curves drawn at different rates, and if they are really good at different temperatures as well.
--Chuck
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
The total amperage I'm drawing is the actual question (actually it was how much current do I have for 5V?) but the questions are related.
Given that the battery has to supply multiple voltages, the watt is the lowest common denominator for finding the solution.
Thanks,
Kevin
snipped-for-privacy@hotmail.com (Jerry Greenberg) wrote in message
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

This is how I'd approach this problem. I don't have the spec sheets for your batteries but there is a 2100mAhr battery available from Digikey with NiMh chemistry, its part # P017-ND. The spec sheet is here: http://www.panasonic.com/industrial/battery/oem/images/pdf/Panasonic_NiMH_HHR210AA-B.pdf
You only need 20 batteries to get 24V (1.2 * 20) (presumably you knew that) You want to draw 500mA for the motors, and another 250mA for the "wireless" at 12V. Plus some unknown current at 5V. Assuming your switchers are 85% efficient means you need (.250 * 12)/.85 watts (3.5 watts, or another 145 mA). So you've committed 645 mA. Now the battery spec sheet says that at a discharge rate of 2000mA these batteries poop out in 60 minutes. (vs 40minutes at 4000 mA and 5 hrs at 400mA) trying to interpolate between the numbers on the spec sheet I'd guess a discharge rate that gives you 2 hrs will be between 1000 and 1100mA.
I arrive at that by figuring that considering data points that the data sheet provides : 4A = ~.66 Hr 2A = ~1 Hr 1A = xx Hr .4A = ~4 Hr
Now you're already using 645 mA, so that leaves you say 355mA or 8.5 watts, * 85% gives you 7.25 watts or about 1.5Amps @ 5 volts. The numbers are only approximate because you're really converting chemical energy and the voltage across the batteries is falling during the whole process. Doing the computation in Joules (or Coulombs) would give you a more accurate number.

Depends on how much margin for error you want. It should work with batteries at 25 degrees C that are fresh.

The "best" way is to model it with energy density but your approach is ok as well.
--Chuck
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Chuck:
I don't have the spec sheet either, so it's a bit of a guessing game. I saw the panasonic sheet and am using those numbers just b/c I have nothing else to go by.
(20 x 1.2) = 24, but I'm using 24 cells to make my LOWEST possible voltage 24V even as the cells deplete. I'm also thinking that the extra 4 cells (4.8V) is my error margin.
Your calculation came up with 1.5A @ 5VDC, but the calcualtion seems similar. Makes me think I'm on the right path at least.
Thanks for all the help.
Kevin

http://www.panasonic.com/industrial/battery/oem/images/pdf/Panasonic_NiMH_HHR210AA-B.pdf
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

The actual manufacturer of your cells with have the specs if you can find them :-)

The 1.2 number is the "nominal" voltage for NiMh chemistry cells. What that means is that you should design for that voltage and be able to deal with a slightly higher voltage when they are fully charged and a slightly lower voltage when they are discharged. When fully charged, NiMh cells have a potential of 1.4 volts, when they are fully discharged they have a potential of 1.1 volts. So using 20 cell packs you should expect 28V when fully charged, and 22V when fully discharged. Going with 24 cells puts you at 33.6V fully charged and 26.4V when fully discharged. Taking the cells below 1.1V/cell risks "over discharging" them. When drained outside the cells "comfort zone" they develop dendrites (nickel "threads" between electrodes) that can kill the cell.
Further, NiMh cells get to 1.2V when they are approximately 75% discharged (check the spec sheet!) You're motors will still work at 23V and your switching regulators sure as heck can maintain regulation at 22v, so consider using 20 cells, and take the Vbatt rail across a 24v zener with anode of the zener feeding a transistor like so: Vlogic | / \ 47K / | +------> uP input | / / /| |/ Vbat ---\/\/\/-----|< |---------| 10K / \| |\ 24V Zener V | | Gnd
This works as follows:
As long as there is 24+ volts on the batteries, current flows through the zener (limited by the 1k resistor to ~2.4mA) which turns on the transistor (2n3904 or 2n2222 would work). The transistor keeps the uP input low by tying it to ground.
Now when the batteries get to 24V and below, the zener stops conducting and the transistor turns off and the uP pin goes up to a logic 1 level. This is your signal that the batteries are 75% discharged and you need to consider going to recharge.
Be aware that when you first cross this threshold the uP pin will go up and down a lot because the voltage will droop when the motors are running and return when they stop. So don't bother recharging until the "low battery" signal stays on even after you've stopped the motors. At that point you've got anywhere from 15 to 30 minutes of run time to "find a charger."
Hope this helps, --Chuck
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.