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'[EE]: What is the max. current allowed for battery'
2002\05\08@183212 by John Waters

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Hi All,

I've a circuit that consumes about 100mA during operation, it now uses a
wallwart as the power source, I need to convert it to a handheld device that
will run on batteries. As the microcontroller works on 5 V, I did an
experiment using a 9V battery, but it soon drained out in less than an hour
when the voltage dropped below 5V. Is my circuit comsuming too much power
than is impossible to work with battery? If so, I get to modify it to reduce
power consumption, but how low the current consumption should go before it
is considered suitable to run on battery?

Thanks in advance!

John




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2002\05\08@184438 by Jinx

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What have you got in the circuit ? Display ? Motor ?
Heater ? Laser ? Taser ? Phaser ?

You should be able to make some improvements
by shutting down or sampling major current users

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2002\05\08@185506 by Dale Botkin

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On Wed, 8 May 2002, John Waters wrote:

> I've a circuit that consumes about 100mA during operation, it now uses a
> wallwart as the power source, I need to convert it to a handheld device that
> will run on batteries. As the microcontroller works on 5 V, I did an
> experiment using a 9V battery, but it soon drained out in less than an hour
> when the voltage dropped below 5V.

If it's a PIC, chances are it will work fine on lower voltages - like
anything above 4V for some, and even lower for others.  9V batteries
typically aren't a good choice if you don't need 9V anywhere else in your
circuit.  For example, let's say you want to run just a PIC and a few
other bits from battery.  Your circuit draws 100mA total.  If you use a 9V
battery and a linear regulator, you're drawing 100+mA from the battery all
the time, which will drain the battery quickly and waste nearly half the
power in the mean time.  You can use a switchmode power supply and improve
things, or you can use (for example) 3 AA batteries to give you 3.6
(NiCad) to 4.5(alkaline)V, lose the regulator and not waste any power.
Put the PIC to sleep when you can and you'll stretch the battery even
further.

A 9V NiCad is typically good for 120mAH.  Three AA NiCads can give you
over 1000mAH.  Makes a lot of sense if you can live with 3.5V or more.
Even one or two AA's with a boost converter looks pretty good compared to
a 9V - I was surprised at how little current capacity they have.  Take a
look for example at Duracell's site - an alkaline AA cell will last 60
hours with a 43 Ohm load (34.8mA) before it gets down to 1.2V; a 9V will
last 25 hours with a 300 Ohm load (25mA) before it gets below 5V.  Big
difference!  The 9V is OK for low-current use that requires theh higher
voltage, but it's a poor choice for PIC circuits that don't need it.

> Is my circuit comsuming too much power than is impossible to work with
> battery? If so, I get to modify it to reduce power consumption, but
> how low the current consumption should go before it is considered
> suitable to run on battery?

All depends on how long you need the batteries to last, whether you can
use a slower clock and/or sleep mode, and what otehr changes you can make.
I will say three 1000 or 1500mAH NiCads will give you quite a boost from a
9V - like 9 or 10 hours between recharges even at your current rate (no
pun intended).

Dale

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2002\05\08@192748 by Jinx

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> If it's a PIC, chances are it will work fine on lower voltages -
> like anything above 4V for some, and even lower for others.
> 9V batteries typically aren't a good choice if you don't
> need 9V

Yes, you'll get much better value by using 3 or 4 NiCd/dry
cells or a 6V sealed lead (if you've room) and a low dropout
regulator (if you need regulation) - ie not a 7805 - use an
LP2950/S81250 or similar. Presently 4/9ths of your 9V is
being wasted as heat in the regulator. The other step to take
is reducing current consumption - strobing displays or even
turning them off and make them visible on request, powering
sensors only when necessary and so on

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2002\05\08@192806 by Byron A Jeff

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On Wed, May 08, 2002 at 03:31:19PM -0700, John Waters wrote:
> Hi All,
>
> I've a circuit that consumes about 100mA during operation, it now uses a
> wallwart as the power source, I need to convert it to a handheld device that
> will run on batteries. As the microcontroller works on 5 V, I did an
> experiment using a 9V battery, but it soon drained out in less than an hour
> when the voltage dropped below 5V. Is my circuit comsuming too much power
> than is impossible to work with battery? If so, I get to modify it to reduce
> power consumption, but how low the current consumption should go before it
> is considered suitable to run on battery?

There are so many variables here that my head is swirling. Let's get a handle
on some things:

1) Do you have an idea of what is sucking down 100ma of current? LED? heaters?
You're going to have to put the circuit on a power diet. Give us an overview
of what it contains and why it consumes 100ma.

2) You said 9V battery to 5V. How are you converting 9V to 5V? Linear
regulator? Switch mode? Very important because linear regulators drop voltage
by wasting power. So if you're using one nearly 1/2 of your power is wasted.

3) 9V battery. Are you married to it. 9V batteries have the absolute worst
capacity among the common battery types. What size batteries can you afford
in terms of size and space?

4) How long does this puppie have to run? 1 hr is too short. 2 weeks? 10
years?

The game is all power consumption vs. power capacity. Decrease the former,
increase the latter, and use no linear regulators, and you'll increase the
amount of time that your project will run on batteries.

BAJ

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2002\05\13@005115 by Ann & David Scott

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snip:
> 9V batteries typically aren't a good choice if you don't need
> 9V anywhere else in your circuit.

I've built an application that did need a higher voltage (the
Laipac RF modules I've just asked about) so I used a 9V.  I
powered the whole circuit directly from the 9V battery.  For
the PIC I used a 16HV540 which has built in voltage regulators.
Do these regulators have the same losses as a 7805?  My guess
is that they don't since my current consumption was low and I
calculated(?) a 5 year battery life.

The 16HV540 seems to be a unique chip.  With the built in
regulators there are also several features to lower voltages
during sleep, power the I/O ports, etc.  Are there any other
PICs or other MCUs that will take the high (9V+) supply voltages?

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2002\05\13@091824 by Eoin Ross

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I have used the 16HV540 in a design too...

I used a 9V battery as:-
a) there was a case that hard a compartment for them
b) the higher voltage gave a better volume from a Piezo alarm

With the part in sleep mode - and the circuit designed so that the PIC I/O pins supplied power to the peripherals (meant everything could be turned off) I was able to have a sleep current of 5uA.

I don't believe there are any other PICs with the regulators onchip ... it seems a shame as the HV540 is basically a 16C54 so it has very few hardware peripherals.

>>> spam_OUTandscottTakeThisOuTspamLOGANTELE.COM 05/13/02 12:38AM >>>
<snip>
For the PIC I used a 16HV540 which has built in voltage regulators.
Do these regulators have the same losses as a 7805?  My guess
is that they don't since my current consumption was low and I
calculated(?) a 5 year battery life.

The 16HV540 seems to be a unique chip.  With the built in
regulators there are also several features to lower voltages
during sleep, power the I/O ports, etc.  Are there any other
PICs or other MCUs that will take the high (9V+) supply voltages?

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