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Can someone help me identify this part?


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I'm not sure what it is, but it appears to take 120V AC and turn it into about 18V AC. The only markings on it are 564K 400V, not very helpful.

I've taken a couple of photos and drew up a wiring diagram. In the photos, it is the large red thing.

As far as the schematic goes, R1 is a 100 ohm ceramic resistor and R2 is labeled red red black orange brown, or brown orange black red red. (I'm not sure what value it is...)

Thanks for your help!

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I'm thinking Polypropelyne capacitor? Normally they are used in that sorta thing and since the PCB has a photodiode on, it may be some light activated night light or such?

From what I can see, the capacitor is:
Rated voltage 400v
Rated capacitance 564k = 560nF or 560000pf (56 with 4 zero's)

:) best of luck!

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Thanks for the reply. That's what I kinda thought it was at first, but shouldn't it say 564KF (minor detail)? Then wouldn't that make it a 564,000F capacitor!? At 400V, that thing would be massive!! I've heard of AC caps, but I always thought they were rare, and expensive. Then I wonder where this huge voltage drop is coming from...

I didn't think about this before, but the voltage drop I am seeing could be do to the resistor. I don't know for sure, but I'm guessing a resistor on an AC circuit works the same as it does on a DC circuit. 220K ohms doesn't seem like very much though. Maybe I will ask about this in the projects and idea section, because until now, I thought the only way to lower AC voltage was with transformer...

Another thought that just popped into my frawn (oober Geek!) is what if that thing is a resistor not a cap. 564K ohms seems to be an odd number, and I've never seen a voltage rating on resistors, but I suppose they have to have them.

Any other thoughts?

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I think that most resistors (standard ones) are rated to 250v but im not 100% on that.

for a cap to be "AC" I suppose it just needs to be unpolarised so really, it can be any capacitor that isnt a polarized electrolytic, e.g. Mylar or Polystyrene etc.

To lower the AC voltage, if you put the capacitor in series with a resistor (in the sense of a potential divider with the capacitor being on the top) Due to the capacitors reactance, it will be "equivelant" to a 5684Ohm resistor at 50Hz (a capacitors reactance decreases as the frequency is increased) Therefore, if you had a resistor in series at a vaue of 5684Ohm, it would act as a potential divider of 2. The capacitor will also limit the current.

I doubt it would be a resistor, I haven't really seen brown resistors of this shape. I have seen brown resistors in circular shapes (carbon film) but never a boxy shape. I knew it was a capacitor because if you search polypropylene 400v capacitor into google, the pictures pretty much match the capacitor in your pictures :)


On that website, it shows multiple polypropylene capacitors that have the same brown colour and shap to yours :)

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That sounds reasonable enough to me, but that thing isn't in series with the resistor, it's in parallel... The main reason I wanted to know was because I thought it was some kind of weird voltage regulator or something. I thought it was weird that the AC voltage drops like that without a transformer.

Now as far as this voltage drop goes, do you know any formulas for calculating this drop based on the value of the components? Perhaps I should ask this somewhere else, the parts request forum is probably not the best place for this sort of thing. I was going to use strait AC for my project, but as it turns out, it would force me to use 180V DC, which is just too much. I was then going to use a transformer to step the voltage down, but finding the right transformer is a major pain! Then, when I saw the little motion night light plugged into the wall, I knew it had to be running on low voltage DC but it wasn't big enough to house a transformer, I took it apart to find that weird little red thing which seems to be lowering the voltage, and I thought eureka!

That's where I'm at now, thanks for your assistance. I think I will start a thread in project designs / ideas to get a little more assistance with this, unless you object....

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I think the output voltage will be dependant on the load. I can imagine the 220k resistor is there for safety reasons. When the night light is unplugged, the resistor will ensure that the 560nF capacitor is safely discharged. I couldn't find anything from a quick search on google but I can imagine adding parts together that this would be called a Series Capacitor Regulator?

I'm quite sure that this will drop the load through acting like an AC potential divider. It seems as if it would be equivelant to a High pass filter.

Example: If the capacitor has a reactance of 5.6kOhm (at 50Hz), and its driving a 100R load, due to this being the same as a potential divider I suspect the voltage across the load would be 5.95v Peak.

(Vin*R2)/(R1+R2) Since the capacitors reactance would be R1, that value would be 5.6k. R2 would be the load so in this example, that would be 100. The voltage in would be the peak voltage of 240v afaik, (240v*1.414 = 339.36v)

As you can imagine, as the resistance of the load varies, the output voltage will also vary.

I made a visual represenation of this circuit on the free applet: http://www.falstad.com/circuit/

Just click on File>Import

Then copy in the code below and click import :)

$ 1 5.0E-6 6.450009306485578 50 5.0 50
c 208 144 368 144 0 5.6E-7 -62.32904764451111
r 368 144 368 272 0 100.0
O 368 144 480 144 0
g 368 272 368 304 0
R 208 144 160 144 0 1 50.0 339.39 0.0 0.0 0.5
x 400 218 453 224 0 24 Load
x 136 97 218 103 0 24 240vAC
x 214 185 355 191 0 24 PP Capacitor
o 2 16 0 34 9.353610478917778 9.765625E-55 0 -1 out
h 3 1 0

As you can see in the waveform, the voltage across the load is 5.97v, close to what I had calculated!

Also, in your schematic the 100R resistor will just be to limit the current probably as a safety feature to keep the capacitor safe.

After having a quick search of your chip, it turns out the thing I thought was a photodiode, it actually is PIR sensor, usually used within a motion sensor. I found out that the chip operates at 5v meaning you would need to use a regulator for whatever voltage you are using for this project.

In your pictures you have shown me, I can see two Zener diodes (Possibly) so it must be an regulated supply for the chip.

I have attached a PDF file of your chip. Unfortunately its in chinese but I hope its helpful!


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