rechargable batteries for uninterruptible power supply

Hello,

for the uninterruptible power supply, I want to have as a matter of safety, I think about some combination of these

Idea is actually given in the picture; short version:
Normal: 1) use power from 230V converted voltage, and 2) charge batteries, so that... 3) they are fully charged in the case of a problem with power supply.
Problem (no power): 1) use the batteries' power, 2) and (as a bonus) Arduino should measure voltage after charger (=between charger and batteries), and then inform Arduino with AnalogRead)

Is that a good idea?
And can I directly use the given power by the charger without harming the batteries. I want the batteries to be fully charged while having power from wall-wart. To achieve that, (I think), the battery charging process would have to be "faster" than the amps taken by Arduino.

Is that suitable, what about feasibility?

Thank you

Entirely feasible. You might want to think about what sort of battery chemistry is well suited to spending months and months with a float charge. VRLA -- lead-acid gel type batteries -- are normally used in things like fire alarms and UPSs and such.

I did some work on something similar that you can read about here: Power monitor for small instrument package « Projects

Thanks for the quick answer.
I don't know much about batteries' chemistry. Is it true, that the current won't even flow through the battery when they are not used (normal state)? And in the case of a problem, the batteries just need to hold ~3-6h. There are no high expectations to the life duration of the batteries but they should work at all.

This looks suitable (?)
It is Pb (lead), also "gel" is named in the title.

And, how would I connect the components practically? Can I charge the battery (link above) with 230V AC? And then directly attach it to Arduino (?)

Yes, that is a 12V gelled electrolyte valve-regulated lead acid (VRLA) battery.
It will happily sit for years with 13.6V float voltage.

So you might have something like:

mains 230V --> 15V xformer or wall-wart --> 13.6V regulated float voltage --> battery --> Arduino power input.

The float voltage regulation can be done with a LM317 or a 7805 or something like that.

The float voltage is the output voltage? 13.6V should be good(?):

http://arduino.cc/en/Main/ArduinoBoardUno:
Input Voltage (recommended) 7-12V
Input Voltage (limits) 6-20V

gardner:
mains 230V --> 15V xformer or wall-wart --> 13.6V regulated float voltage --> battery --> Arduino power input.

Sorry,.. What is the input for this recharable battery (I am still referring to the ebay auction)? So I need to transform it first, I think you mean 15VDC.
The red arrow: where does this happen? Is this a special object for creating regulated voltage, I think. So the battery does not get 12V input, not 15V input, but 13.6V, where do you know? :roll_eyes:

gardner:
The float voltage regulation can be done with a LM317 or a 7805 or something like that.

This is for getting constant voltage, is that really neccessary?

Thank you for the help, I didn't think that there are additional things to consider. There should be a device including the neccessary additional components.

Whatever your standby batteries are, you will need to regulate the charge voltage and current you feed into them. For a VRLA battery the float voltage will be 13.6V or so. For a pack of NiCd or NiMH batteries, it will be something else, that I'm not familiar with offhand. In either case you will have to contemplate some electronics to manage the maintenance charge of your battery pack. If you try to charge anything to the arbitrary voltage coming from a wall-wart, it will either, not work, ruin the battery or cause the whole thing to explode.

There are plenty of sources on float charging a gel cell VRLA. Here's a couple of things I found right off:

http://www.qsl.net/wb3gck/charger.htm

I may have some ideas for modest loads, but I'd need more information. How much current needs to be supplied to the Arduino? I know an Arduino alone requires around 40mA, so I am really asking what additional loads may be involved. If there is an estimate of the duration of the average power outage and how often they occur, that is very useful as well.

How can I measure current drawn?
I think it is

power source --- (A) --- Arduino

whereby (A) is the ampermeter connected in series

I don't know if I am able to do this with my old volt/ampere-meter because it has very thick cables and my 230VAC converting power source has the Arduino power source (2.1mm) output and for + pole it has to reach the center of the socket...
I will try and post results.

Yes that is the correct way to connect an ammeter.

Small cheap multimeter might be a wise investment - and the little ones are pretty cheap these days and probably perform better than an old AVO.

Unfortunately, the ampermeter I have at home doesn't work at all. So, I guess, however, the current drawn is not very high. Have connected a 433MHz transmitter 3 LEDs with each 220R and reading some sensor values (photoresistor ,. ...) and that's it.
So can you call this modest load? I don't know.
Would be very happy with an uninterruptible power supply :slight_smile:

My project was quite similar: a parallax PING, 434MHz transmitter @ 5V, Arduino Pro 5V, an op-amp and a temperature sensor. Average current draw is about 25mA, but the radio wants another 25 or so during transmitting and the PING takes a good 50mA to do its thing. Both the radio and the ultrasonic sensor, in my case, are only active once in a while, so the main draw I have is the Arduino.

I'd guess your setup will be in the 20mA to 30mA range with the radio off, and 50mA or so whilst transmitting, from the loads you describe.

Ok, thank you, that is not very much I guess, even if I don’t know much about these units.
And I really want to simplify the battery with the uninterruptible power supply. Jack Christensen speaks about “ideas for modest loads”, I would really like to hear them.
AND I really like the idea, basically, charging a rechargable battery and using it, and then use the power from the charged battery, so please stick to that if it’s possible.

Yeah that’s probably not that much current, depending on the transmitter. Here is what I ended up going with for a project that draws maybe 100mA at most. The MCP1640 can deliver about 350mA max. I decided against rechargeable batteries because power outages here are fairly infrequent and I could achieve what I needed with ordinary AA alkaline cells (2 or 3). This reduces circuit complexity, since charging is not needed. In the case of an extended outage, I do want to power the circuit for 4-8 hours at least and this will easily do it. The circuit does check for a power outage and takes some power-conserving steps like turning off a seven-segment LED display most of the time. Probably replacing the batteries every 2-3 years is sufficient if no significant outages occur. C4 in the schematic is probably not necessary and in fact should probably be eliminated.

For me it is very important to charging the batteries. There are also AA /AAA rechargable ones?
Or do you think it is a problem when they get empty while power is supplies from the wall wart?

karlok:
For me it is very important to charging the batteries. There are also AA /AAA rechargable ones?
Or do you think it is a problem when they get empty while power is supplies from the wall wart?

My initial thought was to use rechargeable cells too. As I thought more about it, I didn't see much advantage, the charging circuit just adds complexity. I realized that plain old AA alkaline cells are probably a better route for me. In my circuit, no current is drawn from the AA cells unless the mains power fails. Normally all power is provided by a 5V wall wart. AA alkaline cells are advertised as having seven year shelf life, so while I would replace them after a significant power outage, otherwise I plan on just changing them maybe every 2-3 years, just to be conservative.

Rechargeable cells would make more sense if significant power outages were more frequent, but that's not my scenario. We might get a few short outages per year, a minute or less. Every few years we might get a longer one.

Why is it so important to you to use rechargeables?

Hello.

[quote author=Jack Christensen link=topic=127544.msg963612#msg963612 date=1350701733]
In my circuit, no current is drawn from the AA cells unless the mains power fails. Normally all power is provided by a 5V wall wart. AA alkaline cells are advertised as having seven year shelf life, so while I would replace them after a significant power outage, otherwise I plan on just changing them maybe every 2-3 years, just to be conservative.[/quote]
Ok, that is very good - especially the first sentence. I think, if I put them in dry area, this will decrease the self-drain (self-discharge).

However, I have to admit that your circuit is very complex for me.

I am thinking of something like this (red=9VDC, blue=GND)

I have heard, relatively long ago, something with diodes so that electricity can only flow in one direction...

I am really clueless now... And I really want to do it the way you told: "no current is drawn from the AA cells unless the mains power fails"

The picture you drew is in fact the way my circuit works, it just uses less than 9V. It’s actually quite simple, but a little explanation is probably in order. It’s basically just a boost converter (MCP1640) that will boost a voltage less than 5V up to 5V. J1 is input from a 5V regulated wall wart. It’s important that it not be more than 5V, so it should be a regulated (usually a switching-type) wall wart. TB1 is a terminal block where the batteries are connected, either 2xAA or 3xAA alkaline cells. It is important that the battery voltage (TB1) be less than the wall-wart voltage (J1).

Diodes work as you say, current can only flow in one direction. When the mains are supplying power, the voltage at J1 is greater than the battery voltage at TB1. This causes D2 to be reverse-biased and prevents current from flowing out of the battery. D1 is forward-biased so current flows through D1 to the boost circuit. When the mains drop, then D2 is forward-biased and current flows from the battery to the boost circuit. D1 is then reverse-biased which prevents any current from flowing back into the wall-wart.

All diodes drop some amount of voltage, I’m using Schottky diodes to minimize the voltage (and hence power) loss, and also to guarantee that the input voltage to the boost converter is less than 5V (for the output to be regulated, the input voltage must be < the output voltage).

The rest of the circuit is straight from the datasheet. If there are no power outages, the AA cells should basically last as long as their shelf life.

This is not clear for me, that means, always the higher power supply is taken?
I think, that would result that the sum of each voltage is supplied then, that might be too much.
And what is the sense of the diodes, then?

Some diodes like this,

I am thinking of something like this now

karlok:
This is not clear for me, that means, always the higher power supply is taken?
I think, that would result that the sum of each voltage is supplied then, that might be too much.
And what is the sense of the diodes, then?

Correct. Higher of the two. Not the sum. Either one or the other. Output from the boost converter is always 5V.

The diodes, as I explained, cause the higher of the two power sources to be selected, and at the same time, prevent current from flowing back into the lower of the two. In addition, the small forward voltage drop guarantees the input to the boost converter is never more than 5V. This is a requirement of the MCP1640, it will not regulate if VIN > VOUT.

Some diodes like this,

Probably would work, but are overkill. 5A or even 3A is not needed. As shown by the schematic, I’m using 1N5818, these are very common and can handle 1A.

I am thinking of something like this now

That is not going to work exactly the same, there is a subtlety there. 6 fresh AA cells can supply more than 9.5V. So assuming the typical 9V regulated wall-wart delivers less than that, then the battery will supply the circuit initially, even with the wall wart connected. It will continue to do so until the battery is partly exhausted, and the battery voltage drops below that of the wall wart. This does not occur with a 5V wall wart and 3xAA cells.