First time posting here :) Hope someone can help.

I have a wireless Yale home alarm system and the external siren is chewing through batteries in no time and is expensive to keep replacing them. I came to the conclusion that if I were to run it of a DC mains power adapter it would be hell of a lot cheaper to run. The external siren uses 4 x D cell batteries (1.5v each) The batteries are in series so this is 6v. With the batteries in series, I used a multimeter to check how much amps/current it is using when the siren is activated. It seems to be consuming about 450 ma when the siren is fully activated and consuming up to 220 ma in standby mode with just its warning LEDs flashing intermittently every 5 seconds.

I had a 6v DC 3500ma mains adapter lying about so thought this would be perfect to power it as it will only draw 450 out of the 3500 milliamps that the adapter supplies and should be sufficient enough. I hardwired the adapter to the black and red terminals on the siren and made sure the polarity is correct. When powered up, the siren seems to be getting some sort of power as it may make 1 beep or the leds might flash once on power up, but that is all. If I do a siren system test with the batteries inserted, it works perfectly with a continuous siren alarm and flashing LEDs. If I do a siren system when connected to the DC power adapter, the LEDs don’t flash but the siren will only let out one burst of sound for like a second. Like one loud squelch and that’s it. I’m totally confused why it won’t work on DC mains power when the voltage and the current output of the adapter seem to be similar to the battery power. Any ideas or feedback are welcome folks! Many thanks in advance :)

Please see YouTube link below to the Yale Siren I’m using:

https://youtu.be/CUIDHVbGK2o enter image description here


closed as off-topic by Leon Heller, Andy aka, Dave Tweed Mar 5 '18 at 13:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Questions on the use of electronic devices are off-topic as this site is intended specifically for questions on electronics design." – Leon Heller, Andy aka, Dave Tweed
If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ To the people voting to close: Perhaps it is benifficial to add a reason why you are voting to close. This way the OP can perhaps correct what you think is wrong with the question, instead of just seeing the question closed without any response. \$\endgroup\$ – Joren Vaes Mar 5 '18 at 11:46
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    \$\begingroup\$ How confident are you in the measurements you made? Is it possible that the siring is taking a very high pulse load for very short timeperiods (far higher than what your supply can handle), resulting in the supply going into a protection mode or such? \$\endgroup\$ – Joren Vaes Mar 5 '18 at 11:49
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    \$\begingroup\$ I understand it’s a consumer product, but I’m asking on grounds as I’m modifying it. Power reading of adapter is about 6.2v when connected and turned on. I presume the power pulses as it would rapidly increase when it activates and must fluctuate slightly with the flashing of the LED’s \$\endgroup\$ – Jimmy Mar 5 '18 at 12:05
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    \$\begingroup\$ @Jimmy you are not modifying the electronics - if you were folk here would expect some attempt at a schematic to be shown in your question. You have a consumer thing and you are having problems when you attach it to another consumer thing. Those things don't have schematics (that I can see) and people here are not magicians with endless time on their hands. \$\endgroup\$ – Andy aka Mar 5 '18 at 12:35
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    \$\begingroup\$ 220 ma in standby mode seems very excessive. It may be that there is a fault with the unit \$\endgroup\$ – Dirk Bruere Mar 5 '18 at 13:01

It could be that the initial surge of turning on the alarm is causing the voltage to drop too low. Many electronics devices contain "brownout detector" logic that resets the system if the voltage drops below some threshold. Alkaline cells have a pretty good pulse current capability, perhaps better than your external supply.

Energizer has a good overview of battery internal resistance on their website. Internal resistance is measured by measuring the voltage change due to a step change in current a R = dV/dI, as shown below.

Internal Resistance Measurment

Energizer says that the typical internal resistance of a new cell is about 150 to 300 milliohms. At 3.5A, the current rating of your supply, 4 new cells would drop the voltage only about 2V, assuming 0.15 ohms internal resistance. That would still leave the voltage at the low end of the operable range for an alkaline cell.

You can try putting a large electrolytic capacitor in parallel with the battery terminals. That could perhaps smooth out the initial surge and allow the external supply to work.

You could also take the device apart some more to determine what it is using for the siren and look up the peak current requirements, to see if they are within the capabilities of your supply.

If your meter has a min/max capability, you could try measuring the minimum voltage in the battery and external supply cases and see if the external supply is dropping lower when the siren turns on.

One final thought, 220ma in standby mode seems very high for any modern low power electronics device. Perhaps the alarm is defective.


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