Rechargeable battery question

[Higgs1]

Hello,

I have this small consumer electronic device that is meant to be plugged into the wall at all times. It's one of those things with a little power adapter on it with the little circle plug thing. The input for the adapter is for a U.S. house and the output says: "12VDC 1000mA". This device is too valuable to me to require it to be plugged in at all times. What I would like to do is very simple to explain: I want it so that I can have it plugged in and be on, then unplug it and it will still be on, and then I can plug it back in at a later time and it will still be on, and between all of that I don't have to do anything special to the device. This is a extremely common concept: laptops (and netbooks, tablets, etc...), smart phones, digital cameras, mp3 players, sat navs, cell phones, those wireless landline phones, etc. But the battery pack also needs to have a switch because I don't want this thing to be always on until the battery runs out.

The battery life isn't that important. Needs to be long enough so I can sit with my laptop I'm using right now, with my electronic device, and allow me to unplug it so I can reposition myself, my laptop, and my device any place else in my small two story house. It would be even better if I could take this it with me on a five minute drive to my friends house a mile away and have it be on for the entire five minutes.

What would be even better - if the battery were to fit in this small empty space inside the device, but that's kinda stretching it because it's not that big. If not I'll just tape it to the top of the device (where it will block a cooling vent so maybe not) or just have it be a separate box with a wire coming out of it that will actually plug into the device.

I've already been to two forums, and asked a bunch of people I thought knew a lot about electronics and electrical things but now I'm really super confused and have very little idea of where to start. Lets say I know very little about how electricity works. I have a two toys to play with: a digital multi meter and a soldering iron. I also have a oscilloscope but I don't think I need that. Right now, in my mind, putting this together shouldn't be that difficult but, without going into too much detail, according to everyone I had asked it is beyond difficult, extremely costly, impossible, unsafe, hazardous, or illegal. My question is, is this possible, and if so, where do I start (aside from posting this question)?

Edit (Sat Jan 14): There is one thing I forgot to mention: there's at least one setting on my device that has the word "power" in it's description, so that might affect the amps.

 

[mister_rf]

In theory it does not seem to be so complicated , all you need some ‘’batteries’’ and a dedicated circuit to recharge the ‘’batteries’’. In practice there’s a minor problem related to the charging process, as usually standard chargers require minimum 2-3V above nominal batteries voltages (let’s say 15V in total) for proper charging the ‘’12V batteries’’ and now there’s no possibility to get this extra amount from the same 12V power adapter.
If the device can work with lower voltage (low up to 10V) , this may help this situation.

 

[Higgs1]

Thank you, I think you have been the most helpful person so far.

I did some research, and didn't find anything on my exact model or any model particularly close, which is annoying me, but for somewhat similar models I found out that at least one person has gotten away with using an adapter rated at 7.5 volts. Some people have said they believe the device to contain a voltage regulator, which sounds amazing because I'm thinking I can now get away with a lot of different voltages if this is true. I looked at some modding forums for the device and I have seen the number 5 volts when discussing what the internal voltage is. However, this device contains an antenna, so is it possible that a lower voltage might affect it in some way? If it only affects signal quality than it's probably fine.

But anyway, I kinda think I should not rely solely on information for other models. How do I test if my device can run on 10 volts?

The first thing I do is pull out my digital multimeter, set it in volts dc mode in the range of 20 volts and connect the leads to the little circle plug of the adapter that came with the device and it reads 17.36 volts. At first I though I was reading it wrong, but then I remembered something else from the some place I may have gone to, and multiplied it by 2^-.5 and got 12ish volts so it's all good.

Although I have a million power adapters laying around the house I can't find one that's rated at or above 1 amp or above 9 volts. However - I found this other little device that has a knob on it and has two electrical leads. When I attach my DMM and turn the knob my DMM reads 1.25 volts to 16 volts.

Umm... so 4 hours later... hopefully edaboard doesn't log me out.

The device I have with the leads and the knob that controls the voltage is now called device B. The thing I want to attach a battery to is device A.

So what I did was 1) disassemble device A and 2) construct a plug out of paper, tape, and aluminium foil. steps 3-whatever) I plugged the power adapter that came with device A into device A and normal operation started in 10ish seconds. I then measured the voltage across the power pins on device A: 14.43 volts. After unplugging device A, I turned the knob on device B so that the reading of my DMM matched: 14.43 volts. I then proceeded and plugged my paper and tin foil plug into device A and wired it to device B. After about 10ish seconds it started normal operation. I unplugged device A and then adjusted device B to make my DMM read 10.03 volts. After plugging it in the same thing happened - after 10ish seconds it started normal operation.

My conclusion: my device can be run at a lower voltage, just not exactly sure what voltage. 4.4 of some unit lower is probably good enough it sounds like to me.

 

[Pjdd]

Hard to be sure what will work reliably as we don't have any details about how your gizmo works. But here are a few things to chew on.

Your adapter output reads about 17V when it's marked as a 12V adapter. This means that it provides an unregulated output. That 17V is the no-load voltage and will drop to a lower voltage when it's loaded. It also indicates that the gizmo probably has a built-in regulator inside as most modern circuits need a regulated power supply, which in turn implies that the exact voltage of the raw input doesn't matter much as long as it's a couple of volts higher than the internal regulated voltage. What we don't know is how low we can go and still have the unit operating as it should. You've already established that it works down to 10V.

Another unknown factor is the amount of current drawn by the device. The 1A rating of the adaptor gives some indication but not a precise one because such ratings are usually round-figure ratings and are usually higher than the actual load current drawn by the device it supplies.

OK, so far I've really been thinking aloud. Now for some clarification from your side:
Any particular reason why you didn't go lower than 10V with your adjustable adapter? It will help find a practicable solution if you could find out how low you can go before the gizmo starts sulking.

Would you be satisfied with using dry (non-rechargeable) batteries as the backup and replace them from time to time? Using non-rechargeables could greatly simplify things.

 

[BradtheRad]

You are using a sensible approach. It's crucial to make sure the battery supply will be compatible with everything else.

From your tests, it appears you need a battery pack whose output ranges between 10.1 and 14.4 V.

Since your power adapter is rated 1A, I am supposing your device A needs about 1/2A.

A string of 8 AA rechargeable batteries should provide 11.6 V when freshly charged. They may (repeat, may) power your device A for five minutes while staying above 10.1V. Rechargeable AA's now are coming out with 2.5 amp-hour capacity.

You can recharge your battery pack conveniently, while connected to device A, if you install the right kind of switch and the right diode and the right resistance. All this will be tricky and you'll have to do some testing.

You can start out with 8 AA to see if that will work. If not try 9 AA (fully charged 13V). If AA are insufficient then you will need to step up to something bigger (C batteries).

If you use alkalines then you must not subject them to a higher voltage than their own value. Not even slightly higher. I've tried it and had too many of them leak chemicals.

You'll hook up a 'Y' connection of some kind. One branch brings wall power. The other branch brings battery power.

There are several questions:

(1) whether to open device A and rework circuitry, or else break up the wires which carry wall power so you can attach the Y device externally.

(2) whether you want to select power source via manual switch, or else use a jack with internal automatic switching,

(3) what to house the batteries in, and whether they will be inside device A, attached to the outside, or at the end of a length of cable.

If you can find space within the housing then you might wish to use the most consumer friendly method. It's typical to use coaxial plugs and jacks. One model of jack contains a switch which automatically disconnects internal batteries when you plug in the wall adapter. It's this way with my laptop computer.

If you use a different option then you'll need to obtain (or make your own) things such as a battery pack holder, junction box, additional plugs and/or jacks, possibly heat shrink tubing.

 

[Higgs1]

Any particular reason why you didn't go lower than 10V with your adjustable adapter? It will help find a practicable solution if you could find out how low you can go before the gizmo starts sulking.

Yeah, I wasn't sure if I was going to break it, but I'll try lower voltages if it'll help.

Would you be satisfied with using dry (non-rechargeable) batteries as the backup and replace them from time to time? Using non-rechargeables could greatly simplify things.

Not really. I'd kinda like the rechargeable batteries.

Your adapter output reads about 17V when it's marked as a 12V adapter.

Are you sure the reading on my DMM and the voltage rating on the power adapter are the same unit of voltage? I remember something vague about a root-mean-square function from school, and if I divide 17 by the root mean square value for a sine function I get 12.

Another unknown factor is the amount of current drawn by the device. The 1A rating of the adaptor gives some indication but not a precise one because such ratings are usually round-figure ratings and are usually higher than the actual load current drawn by the device it supplies.

Sounds like another test that I should do. I'll post the results of my tests probably as an edit to this post or something.

There is one important thing I did forget to mention: there's at least one setting on my device that has the word "power" in it's description, so the amps probably varies but likely not by much.

 

[Pjdd]

Yeah, I wasn't sure if I was going to break it, but I'll try lower voltages if it'll help.

That and the load current will help decide what type of backup battery will be suitable.

Not really. I'd kinda like the rechargeable batteries.

OK. Then you have to accept that you have to have some way of charging the battery. This means either incorporating a charger in the backup system or moving the battery back and forth between the gizmo and a separate charger.

Are you sure the reading on my DMM and the voltage rating on the power adapter are the same unit of voltage? I remember something vague about a root-mean-square function from school, and if I divide 17 by the root mean square value for a sine function I get 12.

17V is the peak value of 12V RMS. The filter capacitor in the adapter charges up to the peak voltage when it's unloaded, and the voltage drops when it's loaded. By how much it varies between loaded and unloaded conditions depends on a number of factors like the load current, filter cap and transformer characteristics.

Sounds like another test that I should do. I'll post the results of my tests probably as an edit to this post or something.

There is one important thing I did forget to mention: there's at least one setting on my device that has the word "power" in it's description, so the amps probably varies but likely not by much.

See above.

 

[Higgs1]

Either my computer is messing with me or the last two posts of this thread have mysteriously vanished, including one that described my test results. If my computer is messing with me this post can be ignored, but otherwise I'll try remembering what the results were:

At 3.7 volts my device does not work properly and gives me some sort of indication there's a problem. At 3.8 volts it works fine.
At 12 volts my device uses 0.31 amps, and at 3.8 volts it uses 1.00 amp.

I currently have that power setting I was talking about set to 84 out of 251 milliwatts. I believe it was originally set to 75mw.

Also when I set my power supply (device B, I forgot what it was called, plus it doesn't have any words written on it) to 3.8 volts my DMM measured it at 155+°F (my DMM has a temperature setting).

You can recharge your battery pack conveniently, while connected to device A, if you install the right kind of switch and the right diode and the right resistance. All this will be tricky and you'll have to do some testing.

Yes, I believe this is what I want. I'm fine with doing any kind of testing.

whether to open device A and rework circuitry, or else break up the wires which carry wall power so you can attach the Y device externally.

I'd rather not mess with my device, especially since theres only 4 components that aren't surface mounted.

whether you want to select power source via manual switch, or else use a jack with internal automatic switching,

I'd prefer automatic switching

what to house the batteries in, and whether they will be inside device A, attached to the outside, or at the end of a length of cable.

On second though I probably can't fit the batteries inside the device. 8 AAs won't fit and a C battery also won't fit. I can probably construct my own battery pack casing out of something and some electrical tape. I also have heat shrink tubing available.

 

[BradtheRad]

I posted a reply after seeing your result of 1A current draw at 3.8V. It's gone now. Earlier today the board was not operating. It's likely that the most recent posts were lost.

It's risky to apply undervoltage. Appparently somewhere you had components carrying 3x their intended amount. Hopefully your device will continue to operate properly at 12V.

I did something similar to a digital camera I once had. Applied supply voltages below 5V to see what range it would operate on. Afterward I started noticing it would operate for less and less time on the rechargeable batteries. Eventually it would not turn on even with fresh batteries. I can't help thinking my experiment caused components to start degrading.

---------- Post added at 14:22 ---------- Previous post was at 13:17 ----------

I'd prefer automatic switching

On second though I probably can't fit the batteries inside the device. 8 AAs won't fit and a C battery also won't fit. I can probably construct my own battery pack casing out of something and some electrical tape. I also have heat shrink tubing available.

The following describes what you might do if you find it is feasible to charge the batteries directly from your DC adapter. It's a lot of work now but it will save you trouble later. You won't need to switch between batteries and wall power (either manually or automatically).

If you can determine the right number of rechargeable cells to use, then you might be able to hook up the battery pack directly to the internal circuit board. Two wires would do it. These will run into your device through any available opening. You will solder them to the same points where adapter power runs. You may need to install 2 diodes in order to avoid overcharging the batteries. You'll also add a diode between batteries and circuitry.

What is the proper number of cells to use? I have seen 1.38 V on my rechargeables after being charged and sitting a few hours. By that token you should use whatever number of cells are found to match the incoming adapter volts. Reason: to halt battery charging when they reach 1.38V each. (Up to 1.42 V might be okay.) Example, you measure adapter V at 15V maximum. With diode installed makes 14.3 V. You would use 10 rechargeables. Or if you wish to use only 9 then you'll have to add a couple more diodes.

This will be one of those sessions of testing that's required. Success will allow you to unplug wall power from your device and have it continue operating on batteries automatically, uninterrupted.

On the other hand, you spoke about installing a switch to select battery power or wall power. Have you tested whether your device will continue to operate without power for a few milliseconds? Glitch free? Also your switch will need to be low profile. It will need to resist being moved by accidental touching.

For obvious reasons it will be wise to attach the battery pack firmly to your device. A long tube shape is more suitable than a cubical block shape.

The above is viable because you are able to open the housing.

However if you do not want to run wires inside your device then you must tap into the DC lines from the wall adapter. It will involve soldering wires to a plug, soldering more wires to a jack. Perhaps using a junction box. Etc.

And it seemed like it ought to be a simple project to add battery power, didn't it?

 

[Higgs1]

This is amazing thank you, outside of this forum I've only been able to get responses telling me this is impossible.

On the other hand, you spoke about installing a switch to select battery power or wall power.

My original post may have been confusing, and I apologize, but I just need a switch to turn the device off so it doesn't run on battery forever. If I'm not mistaken, the switch can just go inbetween one of the wires running from the battery pack to wherever I decide to connect it to the power on the device.

However if you do not want to run wires inside your device then you must tap into the DC lines from the wall adapter. It will involve soldering wires to a plug, soldering more wires to a jack.

Now that I think about I may buy a jack and plug. The website for a local electronics store is selling jacks for $1.00 and plugs for $1.20 of various sizes. Plus I already have a soldering iron and soldier.

Is the following diagram I made correct?

The thing labelled device in the diagram represents the device internals, as I have decided where the parts will be placed physically.
Sp all I need to do is get some rechargeable batteries and diodes whose voltages add up (or at least very close) to the output voltage of the power adapter? And I can find the voltage of a battery by charging it normally and then measuring the voltage with my DMM, correct?

Three more questions: 1) Is it safe to leave this plugged in or is there a risk of overcharging? 2)will the device work if the batteries are not charged but it's plugged into the wall? 3) would it be possible to extend the battery life by putting more batteries in parallel?

 

[ucsam]

You wont get any output if u implement above schematic though you battery will charge. Put the switch across diode

In J connector put the input from transformer and J2 to the output to your loadView attachment hjk.bmp

 

[BradtheRad]

Sp all I need to do is get some rechargeable batteries and diodes whose voltages add up (or at least very close) to the output voltage of the power adapter? And I can find the voltage of a battery by charging it normally and then measuring the voltage with my DMM, correct?

Your schematic will work with the the addition of components (in green) as shown.

Three more questions: 1) Is it safe to leave this plugged in or is there a risk of overcharging? 2)will the device work if the batteries are not charged but it's plugged into the wall? 3) would it be possible to extend the battery life by putting more batteries in parallel?

1.

It's safe to continue sending a few mA through AA rechargeables after they are full. I would not go over 50 mA with AA NIMH (C/50). Nicads can endure more trickle charge.

Since you are willing to experiment, you should find out what happens in various conditions.

2.

This is the reason for adding the resistor inline with the diodes. It limits charge rate. A 1A power supply should be adequate to power both simultaneously. However if you find the device does not get sufficient voltage, then try increasing the resistor to the battery pack. This will send more current to the device.

Discharged batteries rated 2500 mAH will take overnight to charge if you feed them 100mA. If they are discharged to 8 V when you begin charging, a 33 ohm, 2W, resistor will get hot but will not burn up.

Rechargeables should not be discharged to zero V. If you charge them starting at zero V, the 2W resistor will pass a much higher current and will fry.

This is an area where experimentation is needed.

3.

Parallel battery banks will work in theory. However then you have a hard time knowing if one or more cells are deficient within a string.

======

As you can see the project gets rather involved. The above method is what I would use if I were to consider that it's $20 worth of batteries. That would motivate me to do tests to see they are not being stressed.