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Why the diodes?

For entertainment/educational purposes I'm reverse engineering a cheap fluorescent nightlight that died and I found it uses an arrangement of diodes and resistors in the input that I'm unfamiliar with. Excuse the poor ASCII schematic, hopefully the formatting holds up. The combination of D1 and R1 is in parallel with D2 and R2, of equal value wired such that current flowing in either direction will pass through one diode and one resistor. Why have the diodes there in the first place instead of just one resistor?

--------------------------------------------------------- Rest of circuit D1 R1 Mains -------||-------/\/\--------- D2 R2

The nightlight uses a very simple capacitor ballast to limit tube current, and the thing failed very quickly, it didn't even last 3 months. I suspect the poor photocontrol causing it to flicker near dusk killed it but the whole thing is not very well made.

Reply to
James Sweet
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Passing current through the diodes reduces the noise that the fluorescent light introduces to the 120v wiring.

Reply to
Rich.

Are you sure that what you think are resistors are not in fact capacitors? You might be describing a form of a voltage doubler rectifier.

Bill

Reply to
Salmon Egg

Are you sure they are resistors and not capacitors? A diode/capacitor ladder is a cheap way to make high voltage.

Reply to
gfretwell

Not sure if my other post went through, I got a news server error.

Yes, I'm positive they are not capacitors. They are 56 ohm 1W flameproof resistors and are wired exactly as shown in my crude schematic, each in series with a silicon diode with the whole thing in series with one leg of the mains. I'm well familiar with various voltage multiplier topologies and this is certainly not one of them.

I had a sneaking suspicion it may be a sort of interference filter as someone else suggested, but I'm unclear on just how this works.

I've drawn up most of the rest of the schematic of the nightlight in a CAD program just for fun, I need to finish with the photocontrol portion and then I'll post it.

Reply to
James Sweet

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