Eugene's DIY Den

So I'm thinking this low measurement could be due to voltage drop on the connecting leads (which could amount to a few tenths of a volt at that current). The panel will only measure voltage at the board during a load test.. Next step is to check voltage at the actual terminals of the battery.

if the float charge is only 22 mA, it could take more than 24 to fully charge the battery, depending on its initial charge state. It is a 7 Ah battery.

I've got 5 PIRs (older ones so current consumption might be higher than newer ones). Also 6 shock sensors, a wired bell and two wired keypads. That's 14 devices (and there's also the panel itself that has to be powered), so if they're pulled a few tens of milliamps, it could add up. I'd have to check all the specs to see what the current consumption is.

So I left it for 3 days and performed a longer test this time. After ten minutes pulling 0.55 A, voltage stabilised at 11.9 V. (it was at 11.9 V after around 2 minutes).

I'll see if I can find some charge/discharge curves. Presumably if the battery was sitting on a shelf in a store, it may have self-discharged somewhat. Not sure what the self-discharge rate is for lead acid, but 12.9 V would have indicated that it was fully charged, but maybe not. I guess there's some variability. The panel is trickle charging it at 22 mA. That figure was 38 mA for the 6 year old battery (which I thought was a Yuasa, but it was an Interlogix. I presume these are just badged and there are a limited number of battery manufacturers)

I’m after putting a new 12 V, 7 AH lead acid battery into my alarm panel. Off load voltage was 12.9 V at the battery terminals before fitting. A fully charged 12 V battery should read around 12.8 V. When I did a load test, with the panel and sensors pulling 0.55 A, the voltage eventually dropped to 12.0 V before I stopped the test after a few minutes (the voltage fell faster at first). Float charge voltage for a lead acid battery (and from a panel) is typically 13.8 V. So just wondering since the battery isn't charged in a three state mode by a panel like a normal charger would charge it (constant current, constant voltage and then trickle maintenance charge), how long typically does it take a new battery to fully charge? I've seen figures of 24 to 48 hours mentioned.

Battery not at fault, it was the piezo element. However HKC say the battery should be good for 5 years, and now it's 8 years. However I tested it, and it can still power the sounder for the 15 mins set by the jumper.

I thought something like that may have caused it. Blew the plastic horn out with an air compressor, but it didn't make any difference. There was some corrosion around where the leads were soldered onto the element (it's just a flat disc with a metal plate on one side and a metal coating on the other). Cleaning that didn't do anything either. So I reckon the crystal just cracks and can't vibrate properly.

That's the one I salvaged the piezo element from, which is working fine now in the HKC one. So does anyone know whether 120 V AC is a standard voltage for these?

120 V AC. Presumably that's RMS , because peak voltage is nearly 250 to 300 V. This is the waveform on a scope. Anyway I had a stroke of luck. My old sounder has the exact same piezo transducer so I'm going to use that. Hopefully it's the same voltage, but it looks totally identical. The HKC sounder is taking nearly 200 mA when sounding after I replaced the piezo element, which tallies with the 250 mA peak they give in the spec. Also the 6 V battery is still working fine. After 15 mins of alarming, voltage fell to 5.75 V.

Yes, that worked. The next thing is to open the transducer to see if anything can be fixed, probably not. Also I'll put a scope on the sounder output on the board to see what has happened. A cheap meter shows 77 V instead of 120 V, but that's possibly because the meter has a limited bandwidth. Alternatively one of the drive transistors/MOSFETS may have blown.

I'm thinking I need to connect 0V from the power supply to SAB Hold, because that's actually the negative power supply to the sounder. Then connect 0 V to to External Bell -, because that seems to be Trig - on the schematic.

I've taken my sounder off the wall and I'm checking it on my workbench. It's powered from a desk power supply set to 12 V, connected to the external bell input terminals. I would have thought it would activate with the back tamper open, but it didn't. I closed the back tamper for 4 seconds and opened it and it still didn't activate and tried the same with the battery connected. Shouldn't all of these tamper conditions and also open circuit holds activate the unit? If a burglar cuts the cable, shouldn't the 6 V battery activate it, or is it only edge triggered (i.e when the tampers were initially closed when the sounder was powered down and then opened that triggers it? ) In this case, how can I generate a false alarm? Which terminals do I need to connect?

There are two problems I think. When the panel is powered down and the battery disconnected, the sounder is barely audible, suggesting the battery is at fault. With everything powered up, the sounder volume randomly increases and decreases and has a "scratchy" sound, like when the volume control on a radio is bad. So like you suggested previously, possibly the piezo element at fault too. I'm going to open it up today.

The time and date also reset on my wired keypad. So either they don't have a non-volatile memory to hold time settings or a lithium cell in the keypad is dead.