I should have looked at the picture more closely! I've looked and can't see nything obvious - 4 volts AC at no more then 0.5A or so seems hard to find.
Try it on 4.5v of batteries with a silicon diode in series (and a 500R series resistor if you want oo be cautious).
My instinct is that the first thing the thing does is rectify the AC to DC before it reaches the LEDs. Although 4v is low enough that most LEDs can tolerate that much reverse voltage.
LEDs that change colour autonmously have been around for a while.
That might prove difficult as I've tried to explain to a lecturer who was suprised I didn't keep 50 ohm resistors as a main stock item.
Resistors usually come in the E series so look for 470 ohms rather than 500, not that it'll make much differnce, but I'd try 330 and 220 at least. I think 470 might be to high as that would only allow about 11ma for a typical single LED.
It might not it may just use the +ve cycle to run the LED's as that's a cheap way of doing it.
But if teh foward voltage is high that'll mean more current will flow and that is more likely to damaged the LEDs with over heating especially if run from DC at 100% duty cycle.
Not really the LEDs don'y do this by themselves they have an inbiuld chip that does this for them. I've used these and then run on 5V DC they will probbaly work on 4V DC but haven't tried them.
The thing is as on that other thread, why is it AC? Unless its using the mains for timing, then surely a dc supply will still work.
I'm sure one could be made fairly simply from a normal small mains transformer of the right type.
When dealing with AC though, you need to be careful to see what the load is and what the actual peak to peak levels are not just the mean as measured with a meter.
And they do! I bought the 4.5v AC to DC adapter from CPC: PW02731 from the 'other' thread before I discovered the original failed unit was AC to AC and so I plugged the lights into it on the 4.5v setting (they wouldn't run at 3v).
Polarity doesn't appear to matter, and they're running a little brighter than they did, the colours are less varied at any one time, but they work.
Correct. If it's AC then the 'polarity' is changing 50 times a second. ;-)
4.5 instead of 4V?
Hmmm ... ?
Just that they could be being overdriven slightly so ('a little brighter' ) so they may not last as long?
Because you are running AC there isn't much that you could do easily to lower the output voltage to 4V (AC), especially if the load varies a lot (no lights on, all lights on etc).
You could try a series resistor (as mentioned elsewhere) but that would really only affect the 'lights all on' condition.
You might be able to 'clamp' the voltage with a pair of back to back (3.4V?) zener diodes (to clamp the peak voltage(s)), with reverse biased rectifier diodes in series (to stop the zener conducting on the forward biased half cycle) and a series resistor between transformer and diodes (to limit the current) but you will need a real electronics design engineer to help you on that one. ;-)
Ok, so it would be even easier to drop the voltage a bit so that you don't see the extra brightness / less change in colour and may even prevent them from dying early.
Basically, if you were to fit a silicone rectifier diode in series with one of the leads to the lights, it will drop about .6V and may make it more like it was originally?
4.5V is 25% higher than it was designed to run and that can be quite a bit where LEDs that may already be driven reasonably hard are concerned.
If the original adaptor was only 1VA then at 4V that is only about
250mA so a 1A diode should be ok (even at the price you could wire a couple in parallel)?
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If the (DIN was it) plug is re-wireable you could even put the diode(s) in there?
Don't forget that 4VAC is 5.66V peak. Assuming tha there is a silicon diode in the lights themselves, the actual resulting DC voltage might be as high as 5V if there is a fair sized electrolytic smoothing the output and allowing for the typical
0.7V drop across the diode.
With 4.5VDC fed to the same diode, the same forward drop will still be experienced, leaving only 3.8V on the output of the diode.
Checking back, I note that the OP says that 'Polarity doesn't appear to matter' which implies a bridge rectifier. This will have twice the diode forward voltage drop so 4.25VDC maximum from a 4VAC input or 3.1V on the bridge output from a 4.5VDC input.
On that basis, I wouldn't bother to do anything further.
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