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I am designing a Capacitive power supply, I found someone who did the circuit shown in the figure. I don't know what is the usage of the 47 ohm resistor in series with the Zener diode (12V MELF package). Also what is the purpose of the 1uF ceramic capacitor in parallel with Zener diode. I didn't find any one who suggest that. In fact this product is known as the most robust one among its peers.

There is Protection Varistors and inrush resistor, but not included in the schematic.

The question is:

What is the use of this resistor and 1u capacitor? How to make power supply robust? beside the use of varistor and TVS.

enter image description here

I have edited the schematic with adding the other polarity.

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    \$\begingroup\$ That circuit does not work. Whenever you see a diode and capacitor series connected there's something wrong ;) \$\endgroup\$ – carloc Jul 27 '17 at 11:23
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    \$\begingroup\$ @carloc How is that? It is a transformer-less power supply, the capacitor to decrease the voltage and the diode for the rectification. \$\endgroup\$ – Ashraf Almubarak Jul 27 '17 at 11:28
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    \$\begingroup\$ Are you sure the circuit in your question is correct? Where did you take it from? \$\endgroup\$ – Chupacabras Jul 27 '17 at 11:41
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    \$\begingroup\$ @AshrafAlmubarak your circuit is NOT in any of those documents you mentioned. So I assume you designed that circuit. It is designed not properly. \$\endgroup\$ – Chupacabras Jul 27 '17 at 12:00
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    \$\begingroup\$ It's doubtful the Zener does anything real in this schematic.. if the output was from the top of the Zener I'd say maybe... but right now it just provides an isolated reference and perhaps a little WEAK over-voltage protection, but not a lot. However, you have decided to omit the "inrush resistor" which changes things. You can't always explain everything in isolation like that. \$\endgroup\$ – Trevor_G Jul 27 '17 at 13:36
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What is the use of this resistor and 1u capacitor?

The purpose of R2 is to limit the power dissipation of the zener diode by limiting the current that flows through it.

A zener diode produces white noise when it is operated in the reverse breakdown mode as shown in your diagram. It is likely that the designer included C3 in an attempt to bypass (filter out) this noise.

Edit: The OP asked why R2 if C1 already provides 5.6k of impedance to limit current? The impedance you quoted for C1 assumes a 50 Hz signal. Any transients on the line will have a much higher frequency and will pass through C1 largely without reactance. R2 ensures a minimum current limiting factor under these conditions. C3 may also help to bypass some of these transients.

How to make power supply robust?

This is a very subjective and application dependent question. You haven't provided enough information to adequately address this question. But I would comment that at a minimum, the circuit lacks required safety mechanisms such as a fuse.

Also be aware that this circuit has the potential to expose the user to direct line voltages. This should never be used in a situation where any access to the electrical connections is possible.

Experimenting with the circuit carries significant risks including the possibility of death by electrocution.

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  • \$\begingroup\$ The current in Zener diode is limited by the impedance of C1, which is 5.5K at 50Hz. Why additional limitation is needed? the rating of the zener is far more than this circuit could produce. For the white noise, the data sheets don't have info about its levels, but I will search more. \$\endgroup\$ – Ashraf Almubarak Jul 27 '17 at 14:08
  • \$\begingroup\$ For the second part, I have skipped the varistors and inrush resistor, as I thought it is common sense to be used. This power supply intentionally not isolated and it is not for public use. \$\endgroup\$ – Ashraf Almubarak Jul 27 '17 at 14:16
  • \$\begingroup\$ I will add to my answer to address this. \$\endgroup\$ – Glenn W9IQ Jul 27 '17 at 14:27
  • \$\begingroup\$ One last questions, I don't find explicit information in the data sheet about that transient current maximum value. \$\endgroup\$ – Ashraf Almubarak Jul 30 '17 at 5:25
  • \$\begingroup\$ Vishay, for example, lists a "surge current" for a 10 ms period. \$\endgroup\$ – Glenn W9IQ Jul 30 '17 at 14:48
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For reference and to protect against future edits, here is the circuit you show:

This can't possibly work long term. This is because C1 will get charged up, and has no way to discharge. After just a few cycles, it will get to a steady state voltage such that the voltage on its right side gets just to the point of conducting thru D1 at the positive peaks of the input voltage.

After that, every time the input peak is little higher, or the voltage on the right side of D1 a little lower, C1 will conduct. However it will also be charged up more. The next cycle a even higher input voltage will be required to allow any current to flow. After a few cycles, effectively no current flows anymore.

This is not a "power supply".

Added

Argh! You now say there is additional circuitry connected to the right side of the diode. By not showing the complete circuit, you wasted everyone's time here.

There Zeners are apparently there to act as shunt regulators for the output voltage labeled "Load". R2 is probably just to limit current thru the Zener when there are short term spikes on the input voltage.

It doesn't look like C3 does much of anything useful. It's time constant with R2 is only 47 µs. Perhaps the top of the Zener is going to yet more circuitry that you don't show, and the capacitor reduces the high frequency noise on this 12 V node.

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  • \$\begingroup\$ Thanks you, I have edited the schematic by adding the other polarity. I was more focusing on the resistor with the zener, so I forget about the other polarity effect on the operation. I did that for simplicity only. \$\endgroup\$ – Ashraf Almubarak Jul 27 '17 at 13:28

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