Accenta 8 Mini alarm system - power failure & dead rkp
- Thread starter ddma
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There's no need to go messing with the mains voltage. Remove the duff battery and test the voltage on the battery leads. If it is around 13.8 volts then that is good. If it is way over 20 odd volts then the voltage regulator has gone and you need a new panel. If it is way under then again, the power supply has shot it and you need a new panel.
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C
chaindaisy
Hello DDMA,
It appears that I am wrong in my several paragraphs to yourself and Joe 90 in his single sentence is correct.
Please accept my humble apologies for misleading you.
Best Wishes for the future,
Chain Daisy.
It appears that I am wrong in my several paragraphs to yourself and Joe 90 in his single sentence is correct.
Please accept my humble apologies for misleading you.
Best Wishes for the future,
Chain Daisy.
I can report as follows...
When measured with a multimeter the following readings were taken without the battery connected but with the panel powered from the mains. Note: The panel is protected by a 3 amp fuse but did not blow on panel failure.
1. Voltage from transformer onto PCB = 17.5 V AC
2. Voltage at battery terminals on PCB = 7.7 V DC
3. Voltage at auxiliary terminals 13V & 0V on PCB = 13.98 V DC
Now, I'm a mechanical not an electrical engineer but straight away I'm seeing big problems here as the battery itself is only holding 8 volts.
Furthermore, if you look at the photos you should see some tell tale signs of an electrical component failure (unless this is normal burning by a capacitor or whatever is).
On a positive note though, I did power up the panel after clearing NVM back to default again and the RKP and alarm system is now all working, but obviously without a backup battery or any means of charging it!
My assessment suggests that the battery did not cause this or is itself at fault so maybe the battery isn't always the obvious culprit. Maybe?
Guys, please put me out of my misery - am I into forking out for a new panel installation? Perhaps I could do the job myself and save a few quid. In the meantime lets hope we don't have any power cuts in my area!!
When measured with a multimeter the following readings were taken without the battery connected but with the panel powered from the mains. Note: The panel is protected by a 3 amp fuse but did not blow on panel failure.
1. Voltage from transformer onto PCB = 17.5 V AC
2. Voltage at battery terminals on PCB = 7.7 V DC
3. Voltage at auxiliary terminals 13V & 0V on PCB = 13.98 V DC
Now, I'm a mechanical not an electrical engineer but straight away I'm seeing big problems here as the battery itself is only holding 8 volts.
Furthermore, if you look at the photos you should see some tell tale signs of an electrical component failure (unless this is normal burning by a capacitor or whatever is).
On a positive note though, I did power up the panel after clearing NVM back to default again and the RKP and alarm system is now all working, but obviously without a backup battery or any means of charging it!
My assessment suggests that the battery did not cause this or is itself at fault so maybe the battery isn't always the obvious culprit. Maybe?
Guys, please put me out of my misery - am I into forking out for a new panel installation? Perhaps I could do the job myself and save a few quid. In the meantime lets hope we don't have any power cuts in my area!!
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Bernard will tell you exactly what the resistor is for, but I'm pretty sure it's a current limiting resistor protecting the voltage regulator.
When the battery went short circuit the voltage regulator was asked for more and more power - until it gave up.
Replace the panel, it's not hard to do.
When the battery went short circuit the voltage regulator was asked for more and more power - until it gave up.
Replace the panel, it's not hard to do.
Without the circuit diagram it would be a calculated guess as to what that burnt resistor does. It is too far burnt to be able to read the colour code to find the value.
But even if the resistor value is known the amount of charring of the PCB would require special care to repair the board.
It is very likely the copper tracks to the resistor will have detached from the PCB and will have to be replaced before a new resistor can be fitted.
Also it is very likely that the carbonised board will create un-wanted connections between components.
But even if the resistor value is known the amount of charring of the PCB would require special care to repair the board.
It is very likely the copper tracks to the resistor will have detached from the PCB and will have to be replaced before a new resistor can be fitted.
Also it is very likely that the carbonised board will create un-wanted connections between components.
C
chaindaisy
Hello ddma,
If a clr (current limiter) were not there, in normal working circumstances the battery may receive too much recharge. Nominally, the battery requires about 500 milliamps of constant charge. When a battery deteriorates and is left inside and connected to the power supply, it trys to pull down more and more current to compensate for the deterioration. Eventually, the clr goes high value (it is unusual for a resistor to go low value) to the extent that it 'burns out'. Now before Bernard "I get it all out of a book" green reads up on the subject and comes back with some answer designed for sympathetic scientists, it should be noted that the above explanation is purely for the benefit of the layman, and of course for Joe 90, who has homework to do before playing on his X.box.
This is not the end of the world, Any zener, voltage regulator, diode, capacitor etc. are not usually affected and the resistor can be replaced by taking it to your local technical college for the electronics class to replace. It's nice to offer them a fiver for doing the job. The resistor costs about two or three pence as opposed to a new panel. Where I work, the engineers change these resistors on site, if called to a one off job.
The battery voltage is obviously low, All other readings are correct DDMA. So well done for carrying out the tests. I shall leave it to the 'famous five' to resolve your problem.
Have you got all this now Joe, since I look forward to seeing an exact copy in my next post.
It is reasonable to suggest to you DDMA that whether you decide to have the resistor replaced or spend quids on a new panel, in either case you will need to take the panel off the wall, so do you dump it or take it to the local tech ? Your call.
If a clr (current limiter) were not there, in normal working circumstances the battery may receive too much recharge. Nominally, the battery requires about 500 milliamps of constant charge. When a battery deteriorates and is left inside and connected to the power supply, it trys to pull down more and more current to compensate for the deterioration. Eventually, the clr goes high value (it is unusual for a resistor to go low value) to the extent that it 'burns out'. Now before Bernard "I get it all out of a book" green reads up on the subject and comes back with some answer designed for sympathetic scientists, it should be noted that the above explanation is purely for the benefit of the layman, and of course for Joe 90, who has homework to do before playing on his X.box.
This is not the end of the world, Any zener, voltage regulator, diode, capacitor etc. are not usually affected and the resistor can be replaced by taking it to your local technical college for the electronics class to replace. It's nice to offer them a fiver for doing the job. The resistor costs about two or three pence as opposed to a new panel. Where I work, the engineers change these resistors on site, if called to a one off job.
The battery voltage is obviously low, All other readings are correct DDMA. So well done for carrying out the tests. I shall leave it to the 'famous five' to resolve your problem.
Have you got all this now Joe, since I look forward to seeing an exact copy in my next post.
It is reasonable to suggest to you DDMA that whether you decide to have the resistor replaced or spend quids on a new panel, in either case you will need to take the panel off the wall, so do you dump it or take it to the local tech ? Your call.
C
chaindaisy
Hi DDMA,
The resistor is typically about 470 ohms, the pcb tracks may or may not require a solder dash. Oh glorious prophet of doom perhaps if you actually did some alarm installation work it would become clearer to you. Not the end of the world by any stretch of the imagination DDMA, if everyone threw away control panels because of a three pence resistor there would be none left to go around.
Take care mate.
The resistor is typically about 470 ohms, the pcb tracks may or may not require a solder dash. Oh glorious prophet of doom perhaps if you actually did some alarm installation work it would become clearer to you. Not the end of the world by any stretch of the imagination DDMA, if everyone threw away control panels because of a three pence resistor there would be none left to go around.
Take care mate.
Yes Daisy pain,. I do read books. But I have a bit of experience and the ability to think out of the box when problem solving.
Sometimes I read the books I wrote during my very sucessful 45 year long career in specifying system function, designing hardware and writing software. Systems for control and automation. safety monitoring, communications, bespoke data collection among others.
It also included several years of fault diagnosis and urgent repair of marine and industrial equipment without access to manuals or spart parts.
Edit .. The books I wrote were technical manuals, user manuals and other engineering texts.
C
chaindaisy
Hi DDMA,
Need to add a few lines for the benefit of that great writer and author of several books. The value, or thereabouts, of a current limiter is about 470 ohms. DDMA, there are coloured bands around this 'hi value' resistor. Do you think it might be possible, when you have time, to let us know what they are ?
Thanks.
Need to add a few lines for the benefit of that great writer and author of several books. The value, or thereabouts, of a current limiter is about 470 ohms. DDMA, there are coloured bands around this 'hi value' resistor. Do you think it might be possible, when you have time, to let us know what they are ?
Thanks.
500 mA through 470 ohms is 235000 millivolts or 235 volts
Heat dissipation will be 470 x 0.5 x 0.5 = 117.5 watts
Time for a re-think on the value of the resistor ?
Maybe you are thinking of the current limiting resistor in series with the current setting capacitor in a direct on mains power supply as found in mains operated PIRs and similar devices.
Edit, I left out the 117.5
On closer inspection. If you read the colour code then you are un-believable. Be honest and admit you made a mistake.
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