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Battery powered device question

dxpwny

Jul 9, 2011
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I have a small sealed device that has a USB interface and runs off an internal rechargable battery when not connected to a USB port.

I'd like to add an external 9V battery to extend it's life before a recharge is needed. I attached a 9 V battery to a 78L05, and then the 5 V output to the USB connector. Made it run nearly twice as long (it uses only about 27 mA).

Only afterwards did I take the time to find out that the device will run with as little as 3.8 V applied. So, by using the 78L05, I was 'wasting' some of the batteries power - the 78L05 would 'shut off' once the battery voltage dropped below about 5V.

So, I'm wondering what would be the most efficient method of attaching a common 9V battery to the USB connector and allow the device to be able to run as long as possible.

I thought of connecting the 9V battery to a LM317 set to output about 3.8V. To make things as efficient as possible, I'd have to make the adjustment resistors a high value to lessen the 'wasted' current ... right ?

Also thought of just using a resistor / zener diode with voltage as close to 3.8 V as I can find. In that case I'd want to use a low value resistor to keep its voltage drop as low as possible ... right ?

Which approach would likely be more efficient ... or is there another possibility I am not seeing ?
 

Colin Mitchell

Aug 31, 2014
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Using a regulator will WASTE just as much energy, not matter how you configure it.
Just get 3 x AAA cells and connect it directly to the device, making sure you have the polarity correct. You can add a diode in one lead if you are not sure about the polarity and then remove it. The device may need more than 3.8v as it has an internal regulator and may need 5v.
 

davenn

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I thought of connecting the 9V battery to a LM317 set to output about 3.8V. To make things as efficient as possible, I'd have to make the adjustment resistors a high value to lessen the 'wasted' current ... right ?

no, that's wrong ... the resistor values set the voltage not the current
and the result would be just as wasteful and useless as the 7805

Also thought of just using a resistor / zener diode with voltage as close to 3.8 V as I can find. In that case I'd want to use a low value resistor to keep its voltage drop as low as possible ... right ?

again, wrong, The zener sets the voltage and there is some current limiting caused by the resistor
the problem with the zener method is that the as the load current varies, the voltage from the zener/resistor
combo will also vary

With your very low current required around 27mA and as long as that doesn't vary significantly, you will
probably get away with this method

cheers
Dave
 

dxpwny

Jul 9, 2011
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Using a regulator will WASTE just as much energy, not matter how you configure it.
Just get 3 x AAA cells and connect it directly to the device, making sure you have the polarity correct. You can add a diode in one lead if you are not sure about the polarity and then remove it. The device may need more than 3.8v as it has an internal regulator and may need 5v.


I applied 3.8 volts and it fully charged.

Also should have added that the smaller the better.

3 AAAs would only give me about 3.6 to 4.5 V and about 800 mAH. I found 9V batteries that can give me more then that.
 

davenn

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a 9V battery may give more voltage but its current capability is so much less only ~ 300 mA
 

Colin Mitchell

Aug 31, 2014
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"a 9V battery may give more voltage but its current capability is so much less only ~ 300 mA"

It's not the mA but the mAHr that is important.
 

davenn

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both

but initially, if the device requires 1A and the battery can only supply 350mA, then it is irrelevant
what the mA/h rating is ... it aint going to work!!
 

Colin Mitchell

Aug 31, 2014
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"but initially, if the device requires 1A "

We are not talking about device requirement. We are talking about the ability of one battery charging another battery. Keep to the requirement.

"300 mA" means nothing. It is an instantaneous value. We are talking about capacity - mA-Hr.
 

davenn

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"but initially, if the device requires 1A "

We are not talking about device requirement. We are talking about the ability of one battery charging another battery. Keep to the requirement.

"300 mA" means nothing. It is an instantaneous value. We are talking about capacity - mA-Hr.

total garbage ... he's looking for external power NOT external charging
 

Colin Mitchell

Aug 31, 2014
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Sorry, you are right.
I was put off with his remark:
"I applied 3.8 volts and it was fully charged."

The only problem is this:
Any external battery is just going to constantly charge the internal battery and the charging circuit will consume some of the energy.
You are not going to be able to do this successfully.
 

davenn

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all cool :)

it is tricky when things are not initially explained and then its easy for the topic to wander off-course and all sorts of misunderstandings to occur :rolleyes:

as in this case
 
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