Purpose of electronic elements in a capacitive dropper circuit

Question

I have found a capacitive dropper circuit in an LED light bulb:

I have a few questions about this circuit:

• Normally, in capacitive dropper circuits, I would expect a resistor parallel to C1. I am missing that one here. How would C1 discharge in that situation?
• What is the purpose of R1? Usually I would only expect something parallel to C1, but I do not understand the purpose of R1 here.
• What is the purpose of L1? All I can think of is to prevent voltage spikes.
• What is the purpose of R2 and is this necessary? The LEDs in series already determine the total voltage that falls over the load.
• I know of light bulb brands that use the exact same capacitive dropper circuit (even with same values for the elements) for bulbs with more or less LEDs. Isn't that actually bad in the sense that more current will flow through each LED when there are less LEDs? Is there a reason why this is done? Is it just cheaper to produce or is it also on purpose so that it will fail sooner?

Answer 1

The circuit you provided is borrowed from this answer I gave to a question you posed earlier. I borrowed it from a blog post I wrote a couple of years ago about cheap LED bulbs.

In that answer, I answered pretty much all of your new questions.

• C1 is discharged through R1 and R2 through the bridge diode.
• L1 and R1 catch some of the higher voltage spikes and noise that come in through the power lines. They make a low pass filter. L1 will also slow the inrush current when C2 charges at power on.
• R2 is a discharge path for C2.
• The maximum current is set by C1 and the AC line voltage. It will deliver about 300mA into a short circuit. An LED as a load will lower that a bit because of the voltage difference. More LEDs means slightly lower current. If the manufacturer always uses LEDs rated for 300mA or more, there's no problem.
• The circuit depends on the line voltage and the capacitor to limit the current. If the line voltage is noisy (lots of high voltage spikes) then there will be high current spikes through the LEDs.
• Using the same circuit for various numbers of LEDs isn't really any worse for the LEDs. You design it for a maximum current, then use as many LEDs rated for that current as you like. More is fine because that lowers the current.
• The circuit itself is already crummy enough that LEDs will be damaged over time. There's no need to especially make it worse to damage the LEDs faster. Merely using that circuit is all the "sabotage" that the manufacturer needs to engage in.
• The circuit itself is certainly cheap to produce. There's nothing in there that's especially expensive to purchase or install. Everything in there (except the LEDs) are common parts you can buy by the zillion from any manufacturer.

It's a cheap circuit built in the cheapest way possible to maximize the manufacturer's profit without regard to the costs to the purchaser or the environment.

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The LEDs were on an aluminum PCB:

The "regulator circuit" was on a separate PCB:

Besides the crummy circuit, the wires weren't soldered or welded to the base. They weren't even properly crimped. They were simply squeezed between the metal of the base and the plastic housing.

Answer 2

Normally, in capacitive dropper circuits, I would expect a resistor parallel to C1. I am missing that one here. How would C1 discharge in that situation?

R1 is already in parallel with C1. To see how, just replace the fwd-biased diodes of the bridge with ideal connections on each cycle.

The idea behind putting R1 there instead of in parallel with C1 might be to make the layout easier.

What is the purpose of R1? Usually I would only expect something parallel to C1, but I do not understand the purpose of R1 here.

Same as above.

What is the purpose of L1? All I can think of is to prevent voltage spikes.

I'm pretty sure it's a fusible resistor, not an inductor. I can't see the point of putting an inductor there. If "33" is the number that you read on it from the colour stripes (presumably) then it must be a 33R resistor. Anyway, it's an inrush limiter and a crude, possibly a slow-acting fuse.

What is the purpose of R2 and is this necessary? The LEDs in series already determine the total voltage that falls over the load

R2 has two purposes:

• To ensure a non-zero current and therefore non-zero drop across the dropper capacitor. This is needed to prevent the output cap blow up, especially when the load goes fail open
• To discharge the output capacitor.

Is there a reason why this is done? Is it just cheaper to produce or is it also on purpose so that it will fail sooner?

The main reason is, quite possibly, that it's cheaper and less complex compared to buck (or SMPS, in general) LED drivers. Although the far-eastern manufacturers sell buck ICs for about USD 0.05 or even less than that, the availability of the chip can sometimes be of a concern. Capacitors, resistors and diodes are the most common components, so why not build an LED driver with these only? Of course, there are disadvantages such as having the brightness input-dependent i.e. if the input voltage changes then the brightness changes accordingly. But for the price tag the bulb has (or is going to have), this could be ignored.

Answer 3

Normally, in capacitive dropper circuits, I would expect a resistor parallel to C1. I am missing that one here. How would C1 discharge in that situation?

What is the purpose of R1? Usually I would only expect something parallel to C1, but I do not understand the purpose of R1 here.

Both of these have the same answer--C1 can discharge through R1 and R2, like so:

^{simulate this circuit – Schematic created using CircuitLab}

What is the purpose of L1? All I can think of is to prevent voltage spikes.

My best guess is that L1 is there to limit inrush current, and possibly also to improve power factor (though a capacitive dropper is going to have a bad power factor regardless). It will also act as a rough line filter, but this circuit doesn't really need a line filter.

What is the purpose of R2 and is this necessary? The LEDs in series already determine the total voltage that falls over the load.

R2 provides a discharge path for C1 and C2.

I know of light bulb brands that use the exact same capacitive dropper circuit (even with same values for the elements) for bulbs with more or less LEDs. Isn't that actually bad in the sense that more current will flow through each LED when there are less LEDs?

A capacitive dropper is closer to a current source than a voltage source, especially when most of the voltage is dropped across the capacitor. Slightly more current will flow through the LEDs, but not significantly more.

Is there a reason why this is done? Is it just cheaper to produce or is it also on purpose so that it will fail sooner?

My guess would be that it's probably mostly done just for simplicity's sake, so the probably-overworked engineer can get things finished and to manufacturing quickly and move on to the next thing rather than spending time recalculating values for every variant. It's also marginally cheaper, as they don't have to source more different component values. If they wanted to make the LEDs fail sooner, there are more effective ways to do that, like skimping on the heatsink (which they probably did as well).