Battery life
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Posts: 46,084
Hi everyone,
I was contacted offline about this and I thought it would benefit the group
if I posted his question and my response. I think I may have miscommunicated
in my original message.
MY RESPONSE:
"I apologize. I probably should have been a little more specific with my
application. The product was a portable device with a constant load on the
power supply. The power supply was a boost converter (switching power
supply). The batteries were "AA" alkaline (or NiMH). In this type of
application, you have a big cap on the battery. As the power supply
switches, it would charge up this cap and discharge into the inductor
accordingly. This caps purpose is mainly to "help" the battery supply enough
current into the inductor. However when you first turn on the unit, there is
a large voltage drop, caused by a huge current spike while that capacitor
charges. In addition to that, the power supply itself needs a minimum Input
current. That calculation is:
(Ibatt)= [noparse][[/noparse](Vout)* (Iout)] / [noparse][[/noparse].8 *(Vin)]
Ibatt= current required from battery
Vout= output voltage of power supply
Iout= output current of power supply to load
.8= efficiency
Vin= Battery voltage
So, lets put in some givens:
Vout = 5v
Iout = 1A
@ Vin = 2.8V Ibatt = 2.23 A
@ Vin = 2V Ibatt = 3.125 A
@ Vin = 1.2V Ibatt = 5.2A
So as you can see if you need a stable voltage and current out of your
power supply, as Vin drops that power needed to maintain the voltage and
current out is provided by an increase in current load on the battery.
Now, I don't want to just give you a formula and not tell you where it came
from. Since I was using a maxim switching power supply, I used an article
they wrote to help. You can find it at
http://dbserv.maxim-ic.com/tarticle/view_article.cfm?article_id=46 .
I hope that is a little more specific and I hope this helps."
Jason Linscott
PCB Layout/Engineering Technician
ILX Lightwave
jlinscott@i...
(303) 516-8877
Original Message
From: George [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=RlpZvN5Gh2Vy7TOL0yxIcItoHKmeZmFcg6LvIr6hJ1N5omCRLxOpH50BIawXfGQWlVP--UufCYUQ3Y8cJQ]george@c...[/url
Sent: Monday, October 30, 2000 6:29 PM
To: jlinscott@i...
Subject: Battery life
Hi Jason:
I noticed your comments to Andrew Tucker on the Stamp digest. You comment
that current goes up as battery voltage goes down.
I find that curious, in that any circuit I have designed and tested shows a
decrease in current as a function of voltage.
I too, am interested in calculating, or at least estimating battery life,
and haven't done so yet with a current project. Aside from this
seemingly discrepancy in what you see as a current/voltage relationship and
my past experience, I am most interested in any means you
have devised to get to battery life.
Thank you,
George Jerome
9001 Oso Avenue Suite "C"
Chatsworth, CA 91311
(818) 407-1235
fax (818) 407-1237
Please direct all email to:
george@c...
I was contacted offline about this and I thought it would benefit the group
if I posted his question and my response. I think I may have miscommunicated
in my original message.
MY RESPONSE:
"I apologize. I probably should have been a little more specific with my
application. The product was a portable device with a constant load on the
power supply. The power supply was a boost converter (switching power
supply). The batteries were "AA" alkaline (or NiMH). In this type of
application, you have a big cap on the battery. As the power supply
switches, it would charge up this cap and discharge into the inductor
accordingly. This caps purpose is mainly to "help" the battery supply enough
current into the inductor. However when you first turn on the unit, there is
a large voltage drop, caused by a huge current spike while that capacitor
charges. In addition to that, the power supply itself needs a minimum Input
current. That calculation is:
(Ibatt)= [noparse][[/noparse](Vout)* (Iout)] / [noparse][[/noparse].8 *(Vin)]
Ibatt= current required from battery
Vout= output voltage of power supply
Iout= output current of power supply to load
.8= efficiency
Vin= Battery voltage
So, lets put in some givens:
Vout = 5v
Iout = 1A
@ Vin = 2.8V Ibatt = 2.23 A
@ Vin = 2V Ibatt = 3.125 A
@ Vin = 1.2V Ibatt = 5.2A
So as you can see if you need a stable voltage and current out of your
power supply, as Vin drops that power needed to maintain the voltage and
current out is provided by an increase in current load on the battery.
Now, I don't want to just give you a formula and not tell you where it came
from. Since I was using a maxim switching power supply, I used an article
they wrote to help. You can find it at
http://dbserv.maxim-ic.com/tarticle/view_article.cfm?article_id=46 .
I hope that is a little more specific and I hope this helps."
Jason Linscott
PCB Layout/Engineering Technician
ILX Lightwave
jlinscott@i...
(303) 516-8877
Original Message
From: George [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=RlpZvN5Gh2Vy7TOL0yxIcItoHKmeZmFcg6LvIr6hJ1N5omCRLxOpH50BIawXfGQWlVP--UufCYUQ3Y8cJQ]george@c...[/url
Sent: Monday, October 30, 2000 6:29 PM
To: jlinscott@i...
Subject: Battery life
Hi Jason:
I noticed your comments to Andrew Tucker on the Stamp digest. You comment
that current goes up as battery voltage goes down.
I find that curious, in that any circuit I have designed and tested shows a
decrease in current as a function of voltage.
I too, am interested in calculating, or at least estimating battery life,
and haven't done so yet with a current project. Aside from this
seemingly discrepancy in what you see as a current/voltage relationship and
my past experience, I am most interested in any means you
have devised to get to battery life.
Thank you,
George Jerome
9001 Oso Avenue Suite "C"
Chatsworth, CA 91311
(818) 407-1235
fax (818) 407-1237
Please direct all email to:
george@c...
Comments
> (switching power supply).
>> You comment that current goes up as battery voltage goes down.
>> I find that curious, in that any circuit I have designed
>> and tested shows a decrease in current as a function of voltage.
That is typical of switching power supplies. They use feedback, and they
provide the power (I*V) needed to operate the load. So if the load (I*V) is
constant, the supply current goes up as the supply voltage goes down.
The original question had to do with estimating the life of a 9 volt
battery, powering a stamp project. No switching supply? If the load on the
battery is a resistor or something like a motor or an led/resistor
connected directly across the battery, then the current will decrease as
the battery is depleted. I=V/R. However, if the battery supplies a
voltage regulator, which in turn supplies a load like the Stamp, then the
current from the battery will tend to be constant down to the point where
it can no longer supply the minimum voltage to the regulator. So a simple
BS2, not driving any loads, will draw about 8 milliamps constantly from the
9 volt battery, until the battery is depleted to about 5.7 volts. That
will use up practically all the 500 milliamp-hour capacity of a 9 volt
battery.
Of course the current will vary if the Stamp is turning on and off power
loads like leds , or if the Stamp is napping or sleeping. If the battery
is nearly depleted and the stamp suddenly turns on a power load, that is
likely to pull the voltage below the critical level and cause the stamp to
reset.
I hope that clarifies it a little in addition to Jason's comments about his
switching supply.
-- Tracy Allen
electronically monitored ecosystems
http://www.emesystems.com