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Let's say you have a standard capacitive dropper power supply like the following (from Wikipedia):

enter image description here

I understand that these work well for driving LEDs since they act like a constant current source. I was wondering if the design could be (safely) simplified further by removing the zener and decoupling capacitor - the LEDs don't care what voltage they "see" as they're current-limited anyway, and human eyes cannot detect 120Hz flicker.

I mentioned this to a friend of mine and he pointed out that the mains is subject to voltage transients from things like lightning and inductive loads switching off, and that C2 would shunt the majority of these transients away, protecting the LEDs - that seems to make sense, I suppose C2 + the input impedance of the circuit would behave like an RC filter during a transient?

If that's the case, what purpose does the zener serve if voltage regulation isn't important, and why do so many commercial designs seem to include one? Simulating a similar circuit with a 50us/1kV spike (a la IEC 61000-4-5), the LEDs only see a small increase in voltage and current - C2 does limit the transient significantly...but if C2 were gone, is a sub-millisecond overcurrent even that bad anyway?

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  • \$\begingroup\$ Addendum - I am aware of the dangers posed by the lack of galvanic isolation with this design - this is just a hypothetical. \$\endgroup\$
    – flashbang
    Jan 6 '21 at 4:19
  • \$\begingroup\$ This does not answer your question, but for safety, and assuming one is in a country with a Live/Neutral/Ground mains power system, I would place R1, C1 and R2 all on the Line side of the power. That way, if there should be a ground fault, it won't set my components on fire. \$\endgroup\$ Jan 6 '21 at 4:57
  • \$\begingroup\$ In designs I've seen before there was a fuse on the line side along with C1 and R2. R1 was on the neutral. With that setup there seems like a low risk of fire due to the fuse, no? Even better I have seen some that use a fusible resistor for R1 instead so I would assume that could be placed on either conductor safely (although it's probably better to fuse the hot I suppose). \$\endgroup\$
    – flashbang
    Jan 6 '21 at 5:09
  • \$\begingroup\$ The fault condition that I am concerned about is that some part of your circuit becomes grounded. This creates a parallel pathway for current to return to the mains, through neutral and through ground. Any protective circuitry on the neutral side, including R1 if it were on the neutral side, or a fuse if it were on the neutral side, could thus be potentially bypassed. It is always better to fuse the hot side, or at least I can't think of an occasion where it might be advantageous to do it the other way. \$\endgroup\$ Jan 6 '21 at 5:20
  • \$\begingroup\$ Good points. In this design R1 is not technically a protective element—it's just for inrush limiting—although I suppose it does sort of become a factor in protection during a transient. \$\endgroup\$
    – flashbang
    Jan 6 '21 at 5:26
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I understand that these work well for driving LEDs since they act like a constant current source. I was wondering if the design could be (safely) simplified further by removing the zener and decoupling capacitor

There are plenty of cheap LED lights just like this. They're all unusable for anything except indicator lights because they flicker at 120Hz though.

In this design R1 is not technically a protective element—it's just for inrush limiting

R1 should be a fusible resistor, which is cheaper in mass production than a resistor and a fuse. These resistors are designed to fail open just like a fuse.

R2 should be specified to withstand the voltage, which probably means several resistors in series.

mains is subject to voltage transients from things like lightning and inductive loads switching off, and that C2 would shunt the majority of these transients away, protecting the LEDs - that seems to make sense, I suppose C2 + the input impedance of the circuit would behave like an RC filter during a transient?

If the transient is fast enough, that depends mostly on C2's ESR. LEDs can also tolerate surprisingly high non-repetitive peak currents.

If that's the case, what purpose does the zener serve if voltage regulation isn't important

For LED lights, I don't see any purpose for it.

When a component absorbs a transient, the transient energy is converted into heat. So its ability to absorb a transient's energy comes from thermal mass of the part of the component where dissipation occurs. A tiny semiconductor junction can absorb much less energy than a bulky voltage-dependent resistor, for example. Unless the zener is much bulkier than the LEDs and rated for high current, it won't have much more transient handling capability.

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  • \$\begingroup\$ Good to know about LEDs handling surge currents, thanks! I did some further simulations with C1 shorted and the LED array failed open and realized the zener comes into play blowing the fuse and protecting C2 from overvoltage, respectively. I've heard fusible resistors are pretty terrible at their job, and reading the datasheet for one seems to confirm this—e.g. a 1W resistor wouldn't actually blow unless it got to 30W or more, which even a dead short on the rectifier + C1 failing short wouldn't generate. Wouldn't a fuse be a safer choice here? \$\endgroup\$
    – flashbang
    Jan 6 '21 at 15:43
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    \$\begingroup\$ Fusible resistors have a lot more thermal mass than a fuse, so they are very slow to blow open. But hey, it costs one cent less! Note I'm not sure the zener would survive long enough to blow the fuse... \$\endgroup\$
    – bobflux
    Jan 6 '21 at 21:32
  • \$\begingroup\$ In a crude simulation, the zener had to handle (IIRC) ~100W for less than 1/4 of a second. I can probably check the datasheet to confirm this, but I'd imagine a 1.3W zener could probably survive that, no? \$\endgroup\$
    – flashbang
    Jan 6 '21 at 21:46
  • \$\begingroup\$ check page 4 of this datasheet for pulse power... but 1/4s looks like too much. \$\endgroup\$
    – bobflux
    Jan 6 '21 at 23:30
  • \$\begingroup\$ My mistake - I completely misremembered. Here's a simulated circuit driving a 32V LED COB at 10mA (3.3k resistor) with a 200mA fuse: tinyurl.com/y3l2oejh closing the switch to short the X2, the zener in that datasheet would experience 2-3 65W pulses before the fuse blows. At 1/120=8.3ms, that seems to be comfortably under the .1kW peak pulse power given in that graph, correct? \$\endgroup\$
    – flashbang
    Jan 7 '21 at 0:18
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I believe I have figured out the answer to my question - at least most of it. If the X2 capacitor fails short, the load alone may not draw enough current to blow a fuse - although it may draw enough for itself or R1 to become a fire hazard. The zener clamps enough current to blow the line fuse quickly.

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  • \$\begingroup\$ Hmm. Maybe. I tend to think that the regulation, if it is used with LEDs, is more to reduce the maximum current experienced by the LEDs, and perhaps by so doing, allow them to have a higher average current. But I'm not convinced of that either, because the cost of a slightly more robust LED may be less than the cost of the "zener" and cap. Now if this supply were used with a string of LEDs, the cost of the zener and cap may be less than the cost of multiple more robust LEDs. \$\endgroup\$ Jan 6 '21 at 6:56
  • \$\begingroup\$ Yeah - I'm thinking LED COBs mostly rather than individual LEDs. The zener would also prevent C2 from exploding due to overvoltage if the LEDs go open-circuit. \$\endgroup\$
    – flashbang
    Jan 6 '21 at 7:39
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The zener is there to provide regulation, since it is a circuit for a capacitive 5V regulated power supply, not a circuit optimized for driving LEDs.

Further down in the article, a circuit that drives LEDs is analyzed, which does not have all the parts as the circuit above.

So for LEDs you can simplify the power supply even further.

But saying that human eyes can't detect the 120 (or 100) Hz flicker of LEDs that are run without a capacitor, that is simply wrong thing to assume. Many people do, some are more sensitive to see it than others, even if you don't see it at all.

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  • \$\begingroup\$ I just picked the image from Wikipedia as an example and because I figured I'd be okay to re-post it here. If you look up schematics for them, or read application notes for designing one, nearly all seem to say to use a zener and decoupling capacitor, but don't really go into the "why" beyond voltage stability—they certainly don't touch on safety, transient suppression, or anything else, which are the things that I was curious about. \$\endgroup\$
    – flashbang
    Jan 6 '21 at 7:57
  • \$\begingroup\$ Well it would be nice if you said which application notes. If it is the appnote from Microchip, you can be pretty sure it is made to support people powering up Microchip products like microcontrollers which do need regulated supplies. \$\endgroup\$
    – Justme
    Jan 6 '21 at 8:12

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