Security light powered from mains adaptor instead of batteries.
I wish to power an outside security light, for an elderly neighbour, from a mains adaptor instead of batteries. After adding a DC jack & soldering the 2 wires, when powered the light stays on all the time. How do I overcome this?
By figuring out what you have done wrong. Without giving us any information, you are not likely to get a better answer than this.
OK, I will make a guess. Did you measure the *actual* terminal voltage of your adapter, when not under load? Not the quoted rating. Did you get the sign right?
The PIR part is powered from 6 volts (all 4 cells) but the LED part is powered by 3 volts (only 2 cells), or something like that?
Owain
It would seem that you've added the external power supply in the wrong place.
Not likely IME.
I wonder if the OP has tried putting the batteries back. If it works then, he probably hasn't broken anything. If it behaves the same, then he might have done. Or he could have disturbed the timer and/or sensitivity pots. Maybe he used an AC adapter, who knows?
The following are total speculation, however.....
The correct on and off setting is usually a fine balance, and often requires a few repeated tweaks of the adjustment control to get it right. If the mains adapter supply voltage is not quite the same as what the batteries provide, the setting of the adjustment might be different
- and in fact it might actually be outside the range of the setting.
Also, it might depend on whether the mains adapter output voltage is stabilised. If it's not, its output voltage will vary considerably between when the light is on and when it is off. This could make it impossible to set the on/off correctly.
On the subject of power supplies, what do the various symbols mean? I know that a wavy line is AC, but what about a continuous line with a broken line next to it? Is it raw DC, smoothed DC - are there different degrees of smoothing - or stabilised, smoothed DC?
The three supply types (with different symbols) are:
Unregulated DC A 5V adapter puts out 7V at no loading. At rated load (say 2A), it just happens to put out 5V The DC level is thus, "wavey gravy".
Regulated DC A 5V adapter always puts out exactly 5V
Alternating Current AC Sine waves
We can draw what is in the adapter too.
___ ______ \||/ /||\ AC adapter, sine waves, symmetric around 0V \||/ Main advantage is reduced shock hazard. ___/||\______ You could run an incandescent light bulb off this.
___ _----- Rectifier -----+---- (+) Unregulated DC 7V @ 0 amps <=== surprise!!! \||/ | 5V @ 2 amps <=== stamped on unit /||\ --- \||/ C --- Unregulated supplies make ___/||\_ | the motor on a label printer to work. ---------------------+---- (GND) The label printer actually needs that 7V value. Adapter is stamped as being a 5V unit. Liars.
___ _----- Rectifier -----+---- 7805 --- (+) Regulated DC 5V @ 0 amps ( Constant \||/ | reg 5V @ 1 amps DC Voltage ) /||\ --- \||/ C --- Many digital electronics devices ___/||\_ | use this type. Phone charger. ---------------------+------------- (GND) Laptop 19V 3A adapters. Many different adapters with different constant V are available for purchase.
Some regulated DC adapters have switchable output, and the user must be careful to select the correct switch DC voltage setting *before* plugging it in!
The barrel connector can either be "Center (+)" or it can be "Center (-)". Be careful not to mix the two types. Verify the two devices have the same barrel connector polarity.
On adaptaplug equipped gear, make sure the adaptaplug is keyed, as it is easy to rotate the unkeyed ones 180 degrees, and turn a barrel (+) situation into a barrel (-) and ruin you phone or digital device. Adaptaplugs are evil.
Paul
I think we would need to know the details. I'm assuming there is a pir somewhere? Brian
Solar cell that charged up the original batteries? If so the solar cell also acted as daylight switch.
Many lights with a timer come on initially for the set time when first powered on.
Some security lights can be made to be permanently on by switching on - off - on withing 30 seconds.
In answer to all, there are only 2 wires from the batteries to the electronics. All I have done is extended them to the power supply ensuring correct polarity. First I chose a power supply with the correct voltage output. Then used a basic analogue meter to measure the output & checked polarity (I know there is a symbol for that on the label). There are no settings adjustment on the light just an on/off switch. Photos here -
In message <sfdamr$8u7$ snipped-for-privacy@dont-email.me, wasbit snipped-for-privacy@hotmail.com writes
OK - back to square one.
*Disconnect the power supply, and re-fit the batteries. *Does it work correctly again? If not, you've got a problem! *If it works OK, measure the battery voltage. With the four AAs or AAAs, it should be around 6V. *Remove the batteries, and reconnect the power supply. *Measure the supply voltage. If it's not very similar to 6V, that's possibly your problem.I guess that the unit is designed to correctly work from when the batteries are new (around 1.6V each, so a total of 6.4V), to when they're getting rather flat (around 1.2V, so a total of 4.8V). However, if the power supply varies between when the light is on and off (assuming it WOULD turn off), or it isn't very smooth, that could complicate the issue.
How can you tell without an oscillocope? Is it smoothed? Is it regulated? Does it have adequate current capacity?
If the unit still works with batteries,the answer to at least one of those questions is probably 'no'.
Many multimeters (even the cheap ones), on their AC voltage range, will read the AC ripple superimposed on a DC supply (ie it ignores the presence of the DC).
A simple measurement of the off-load and on-load voltages should tell you whether it's regulated - although some switchmode power supplies might require a minimum load current before they start to regulate properly.
Does it have adequate current capacity?
With the batteries inserted, it should be possible to measure the current the unit requires by slipping the tips of the meter's probes in between the unit's contact and the end of the battery. I'd be surprised if it's more than 100mA.
Indeed.
According to an image search, that unit has a motion sensor as well as a dark sensor. Do you realise this? If not, perhaps you're moving about too much. I saw a review saying the sensor is extremely sensitive.
It might be worth adding a suitable capacitor to the output of the power supply in case it gives out voltage spikes that upset the sensitive circuitry.
Assuming that is behind the white dome, it could probably be masked off with black PVC tape (or better, aluminium tape). As others have said, you need to confirm that it still works with batteries.
Might just be easier to fit a solar powered one.
I saw this in a schematic yesterday for one.
There was a CDS cell and a 1 megohm resistor in the circuit, for light detection. As well as the usual stuff for the PIR sensor. The CDS feeds a pin "INH" for Inhibit.
Thank you for all your help & apologies for not providing the basic information.
I'm aware that these lights work for a number of seconds when power is applied, before they go off. Security light has a light sensor & PIR behind a fresnel dome.
Batteries & DC adaptor both test at just over 6 volts. My meter only has 0.5, 100 & 500 mA settings. Testing the batteries the needle goes off the scale. Testing the adaptor = 150mA
I had thought of a solar powered security light but sunlight is blocked by a tall hedge.
I can only test the light at night as I can't find a way to both simulate dark & provide movement to activate the sensor.
Replaced batteries & the light works as it should. Now I have to wait until tonight to retest with the power supply.
Can you provide more information about the power supply?
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