SLA battery: Connectivity detection

[jst3712]

Hi there. I am working on a basic 12V SLA battery "float" charger for an emergency power backup application. Works fine and everything, except for one little thing that's bugging me (yet it's probably very simple, but can't for the life of me come up with a solution).

I need a small "add-on" circuit that will warn me (perhaps with an LED) when/if the battery is disconnected from the charger. Possible scenarios:- Battery connection becomes loose; the fuse blows due to a short or incorrect polarity; or the 'Battery On/Off' switch is "Off") - otherwise there is no way of telling that the battery is connected or not.

I have thought about using the battery's internal resistance as a detection method, but I'm not sure how I would go about this, and its reliability.
What needs to be taken into consideration is that the battery may or may not be drawing current from the battery. Therefore, a 'current flow detector' may not work in this case.

I have attached a diagram explaining how the switch, fuse and diode are connected to the battery at the moment.

Voltage applied to battery = 13.5-13.8V
Current supplied to battery = <0.39A.
Battery = 12V SLA, 1.3Ah.

Any help on this matter will be much appreciated.

Jason.

 

[ericgibbs]

hi,

Can you tell us how the
Voltage applied to battery = 13.5-13.8V
is generated.?

In other words how is the 13.8V pre reg powered.?

 

[jst3712]

hi,

Can you tell us how the
Voltage applied to battery = 13.5-13.8V
is generated.?

In other words how is the 13.8V pre reg powered.?

Hi Eric.
I removed the rest of the circuit as I didn't think it was that important for my query. It's basically an LM317 adjustable regulator (Constant voltage) set somewhere between 13.5 and 13.8V with some very basic current limiting using a 5W resistor, with indicator LED's and the like. Do you really need to sight it?

 

[ericgibbs]

Hi Eric.
I removed the rest of the circuit as I didn't think it was that important for my query. It's basically an LM317 adjustable regulator (Constant voltage) set somewhere between 13.5 and 13.8V with some very basic current limiting using a 5W resistor, with indicator LED's and the like. Do you really need to sight it?

hi,
OK.
What you need is a comparator circuit, something like a LM393.

Using resistor dividers, connect one input of the LM393 directly to the battery and the other LM393 directly to the output of the LM317.

As soon as the LM393 detects a voltage diff of 100mV or so, it switches a buzzer.

Power the LM393 from the battery.

Do you follow OK.?

 

[ericgibbs]

hi jst,
Something like this would get the job done.

EDIT:
You could use the second half of the LM393 to set up an additional 'low battery' alarm, connected across the battery.

 

[jst3712]

hi jst,
Something like this would get the job done.

EDIT:
You could use the second half of the LM393 to set up an additional 'low battery' alarm, connected across the battery.

Thanks for taking the time in drawing up the schematic, Eric. I can see your idea, however, with due respect, something tells me it's not going to work very well...

The "What If's" that come into my head are:
- what happens if the main power supply fails due to AC failure? The buzzer would sound due to the huge voltage difference, yeah? I don't want the buzzer/LED to come on during a power failure.
- how would the comparator work if the battery is not present? (as the comparator is powered off the battery).

Also, the "Normally Connected" bit in your schematic: Do you mean commonly connected/same power circuit, by this?

Jason

 

[ericgibbs]

Thanks for taking the time in drawing up the schematic, Eric. I can see your idea, however, with due respect, something tells me it's not going to work very well...

The "What If's" that come into my head are:
- what happens if the main power supply fails due to AC failure? The buzzer would sound due to the huge voltage difference, yeah? I don't want the buzzer/LED to come on during a power failure.
- how would the comparator work if the battery is not present? (as the comparator is powered off the battery).

Also, the "Normally Connected" bit in your schematic: Do you mean commonly connected/same power circuit, by this?

Jason

hi Jason.
What would be the point of any means of detection if the battery wasn't present.

It could be configured not sound if the mains failed, but that wouldnt give what you asked for... a battery disconnection.
But it would warn you that the battery isnt being float charged.

 

[jst3712]

Ok, perhaps I haven't explained myself correctly.....

My circuit is a power supply (PSU) and float charger in one. There is a “LOAD” output terminal, and any load connected to it remains “uninterruptable”. You could say it’s like a UPS, only a 12-volt version. So when the mains power fails, the load is still powered up by the SLA battery. The PSU also keeps the SLA on float charge. I apologize for not explaining this earlier; perhaps I should have posted the whole circuit 

With my circuit (for what I'm using it for), it's very critical that that battery is *always* connected to the charger - whether it's fully charged, or partially.

If it's unintentionally disconnected from the charger (blown fuse, switch left in 'off' position, or the wires to the battery become loose), there is no warning that this is occurring, which is why I'm seeking help.

Hope this is clearer now.

 

[ericgibbs]

Ok, perhaps I haven't explained myself correctly.....

My circuit is a power supply (PSU) and float charger in one. There is a “LOAD” output terminal, and any load connected to it remains “uninterruptable”. You could say it’s like a UPS, only a 12-volt version. So when the mains power fails, the load is still powered up by the SLA battery. The PSU also keeps the SLA on float charge. I apologize for not explaining this earlier; perhaps I should have posted the whole circuit 

With my circuit (for what I'm using it for), it's very critical that that battery is *always* connected to the charger - whether it's fully charged, or partially.

If it's unintentionally disconnected from the charger (blown fuse, switch left in 'off' position, or the wires to the battery become loose), there is no warning that this is occurring, which is why I'm seeking help.

Hope this is clearer now.

hi Jason,
The circuit I posted does just that, all thats needed is a small mod to kill the buzzer when the mains fails.
Simple way would be to connect the top wire of the buzzer to the input side of the charger.

 

[jst3712]

hi Jason,
The circuit I posted does just that, all thats needed is a small mod to kill the buzzer when the mains fails.
Simple way would be to connect the top wire of the buzzer to the input side of the charger.

Thanks Eric. I shall try it out on my breadboard sometime tomorrow (hope I have a spare comparator to play with), and I'll let you know how it goes. Sorry for the confusion and all... I always seem to forget something out in my posts, grr.
I might even post the whole circuit to make it simpler to understand; we'll see.

Thanks again for your help

 

[ericgibbs]

Thanks Eric. I shall try it out on my breadboard sometime tomorrow (hope I have a spare comparator to play with), and I'll let you know how it goes. Sorry for the confusion and all... I always seem to forget something out in my posts, grr.
I might even post the whole circuit to make it simpler to understand; we'll see.

Thanks again for your help

hi,
Not knowing the pre reg circuit, something like this change should be OK.
I would consider using the second half of the LM393 as I suggested, it will warn you when running on battery power, the battery is getting loooooow.!

 

[jst3712]

hi,
Not knowing the pre reg circuit, something like this change should be OK.
I would consider using the second half of the LM393 as I suggested, it will warn you when running on battery power, the battery is getting loooooow.!

Thanks for re-posting the schematic, as it confirms my understanding of what you meant about the positive lead of the buzzer/LED.
As with the Low Battery warning suggestion, yeah I could make use of that second half, but I already have a (2-phase) voltage monitor implemented into my design that even disconnects the load as soon as it drops to a certain voltage range

Signing off now Eric, thanks once again, and I will let you know how it went as soon as I can test it out.

 

[ericgibbs]

Thanks for re-posting the schematic, as it confirms my understanding of what you meant about the positive lead of the buzzer/LED.
As with the Low Battery warning suggestion, yeah I could make use of that second half, but I already have a (2-phase) voltage monitor implemented into my design that even disconnects the load as soon as it drops to a certain voltage range

Signing off now Eric, thanks once again, and I will let you know how it went as soon as I can test it out.

hi,
Updated circuit.

 

[jst3712]

hi,
Updated circuit.

Hi Eric, I haven't had a chance to test it out as yet; been busy. I really appreciate your help.

 

[jst3712]

hi,
Updated circuit.

Hi Eric, I tested your circuit (not the second half though) and I can't get it to work properly. I am using an LM358 because I don't have an LM393/2, and I don't see much difference between the two anyway. I have tried all sorts of things. I even tried swapping the grounded 10K (for the potential divider) with an LM336 2.5V reference (as seen in various schematics) and overall I can only get the circuit to work as a Low Battery monitor, not as a "Check battery connection" monitor.
Also, what role does the R4 resistor play?

Thanks.

 

[MrAl]

Hi there,

If you use LM358's instead then you should connect a 1k resistor (or more)
in series with the transistor base (one for each circuit). The LM358 can
source and sink current, while the LM393 needs a pullup (R4). R4 is the
pullup which isnt required with the LM358.

I am assuming that the connection is broken between the battery and the
LM317 ahead of this circuit, and that the battery, when disconnected,
is still connected to the LM358's. This might be a bit of a stretch as
we dont know if the battery is going to stay connected to the LM358's when
it's disconnected from the LM317.

An improvement might be to diode OR the battery AND the LM317 so that
no matter which one is alive it will still power the circuit. Only the buzzer
gets connected to the LM317 though.
The diode OR circuit would simply be two 1N4001 diodes connected with
cathodes together, and the two cathodes power both LM358's. The
two anodes go to both the 12v battery and the LM317, one anode to
each. This keeps the LM358's alive as long as either source is alive.

To make this even a little more reliable, you can provide a separate connection
for the LM358's to the battery, or even have a good quality back up battery
just in case BOTH power sources go down. The back up battery would be
wired through a third diode, cathode to LM358 supply pins and anode to the
back up battery positive (+) terminal.

 

[ericgibbs]

Hi Eric, I tested your circuit (not the second half though) and I can't get it to work properly. I am using an LM358 because I don't have an LM393/2, and I don't see much difference between the two anyway. I have tried all sorts of things. I even tried swapping the grounded 10K (for the potential divider) with an LM336 2.5V reference (as seen in various schematics) and overall I can only get the circuit to work as a Low Battery monitor, not as a "Check battery connection" monitor.
Also, what role does the R4 resistor play?

Thanks.

hi jst.
I have some LM358's on the bench, I'll let you know the changes you have to make to the circuit.


BTW:
the LM393 or LM358 is powered from the battery under charge.
Only the top of the 10K [top left] is 'powered' by the direct output of LM317

 

[MrAl]

Hi Eric,


I made some suggestions in my previous post i think you should take a look at.
There can be some simple improvements that would definitely help a lot.

 

[ericgibbs]

Hi Eric,


I made some suggestions in my previous post i think you should take a look at.
There can be some simple improvements that would definitely help a lot.

hi Al,
The circuit has been tested using a LM393 and it works fine.
If he uses a LM358 all he has to do is remove R4 [3K9].
Again testing on the bench this morning using a LM358, the circuit works fine.

If you consider that all the OP wants to do is detect a 'poor' or open connection between the LM317 charge output and the battery on float charge and the float battery powers the OPA it worls OK.

I utilised the 2nd half of the LM393 as a simple float battery 'low' alarm.
[it might as well do something useful]

As I have the time and resources, I always try to knock up the circuits that I post so that I can debug/test, in that way the poster dosnt waste his time and money.
As you know there are so many dodgy circuits on the web that just dont/cant work and the poster/builder gets frustrated.

I have seen some of your posts and I do realise that you know what you are talking about.

 

[MrAl]

Hi Eric,


Well it's nice to meet someone with the integrity to
actually construct a circuit they are recommending.
I myself like to at least simulate the circuit but
many times i will not be able to build it, especially
if it is very complex or has parts i dont have on hand.

Anyway, i didnt mean to say that the circuit doesnt
work, only that i spotted a few things that might
improve the circuit a bit. Let me go into a tiny bit
more detail now.

First, the basic connectivity. If the battery becomes
completely disconnected, the way the circuit is drawn
it looks like the LM393 is getting it's power from
the battery so now it wont get any power and wont be
able to power the buzzer alarm. Correct me if this
interpretation is not correct.
I also thought that *maybe* a separate battery for
back up would be able to power the circuit in the event
that all the external power systems failed. This is
only if needed of course.
All of the above could be achieved with a few 5 cent
diodes, so there wont be any cost increase to speak of.

Second, when switching to LM358 we should think about
the output current to ground through the transistor
BE junction. Yes, the LM358 is protected, but at 12v
it's going to be close when it comes to the power
dissipation within the ic chip, and if both outputs
are conducting it's really going to be a toss up
just how hot the chip gets. For this reason i
suggested the 1k resistor in series with the output.
This will limit current from the ic outputs to a
very comfortable value, and also wont draw excessive
current from the battery either. The total without
resistors could be as high as 80ma, but with the
resistors it's down to 23ma, which is comfortable.
If i was building this i would probably even use
2.2k resistors, limiting current to around 12ma
max from BOTH outputs combined.

Perhaps you can do me a favor here since you already
have the circuit built up...
Maybe you could do a test where you force both op
amps to drive the transistors BE junction by forcing
their inputs to force the outputs to go 'high', without
the 1k resistors. You'd have to leave it run for a
good hour or two, then come back and feel the ic
chip to see how hot it got.
Repeat with 2.2k resistors on both outputs in series with
the base of each transistor. Compare the heating results.
Would be nice to hear about this test. It's a little hard
to simulate this kind of thing.

Thoughts?