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The basic Cockroft Walton circuit consists of 3 stages of diodes and capacitors creating a voltage of around 1860 from standard AC 220V/50Hz supply.

I tried creating a bug zapper with the same circuit and achieved similar voltage at output and could generate spark on touching positive and negative terminals. The actual requirement is to generate 5000V best case or at least 3000 volts from 220V AC supply without the need of a transformer. Now for this I added 3 more stages to the circuit and measured the voltage with HV probe and it was around 3700V which is decent enough. I am using 0.47uf 800 volt capacitors and 1N4007 1A diodes for this

The moment I short the positive and negative terminals it generates a good spark but when I touch again them the spark starts to fade away and the voltage at the output starts decreasing on every spark. I checked what went wrong found out diodes keep getting damaged. If we go from 3 to 4/5/6... stages, I thought diode has reverse voltage rating of 1000V, so I replaced them with 16kv 5mA HV diode, it couldn't even generate 1860 on 3 stages.

Question :

  • Why does the diode get damaged if I add one or more stages to below given 3 stage circuit?
  • How do I fix it and avoid damaging diodes if I have to use 1N4007 and generate 3000+ volts to be used as a bug zapper?

Ref video link to 1.8k fly/bug zapper: Youtube Link

This is the basic circuit for your reference:

enter image description here

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    \$\begingroup\$ It's also a human zapper because of the way it connects directly to live AC. Don't do it. \$\endgroup\$ – Andy aka Nov 2 '20 at 13:01
  • \$\begingroup\$ Can we isolate it some way and still work as bug zapper, I mean can we use small transformer since current requirement isn't big enough in this application. Please suggest ? \$\endgroup\$ – Embedded Geek Nov 2 '20 at 19:58
  • \$\begingroup\$ The example ones I've seen tend to have a large value through-hole resistor (or even several in series) on the output, both to protect the device from a short and to put some resistance between you and the power lines in the event that the capacitors fail as shorts. \$\endgroup\$ – user1850479 Nov 3 '20 at 17:48
  • \$\begingroup\$ Can you share any link of the ref circuit? \$\endgroup\$ – Embedded Geek Nov 3 '20 at 18:07
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Stop short circuiting it.

In normal operation, very little current flows through the diodes. Just enough to charge the capacitors, really.

You usually let it charge and then draw a low current from the high voltage terminal. You want a high voltage, but a low current DC source. The higher the current goes, the less the output resembles DC - the "ripple" on the output gets very bad.

When you short circuit the output, the output voltage drops to the point that all of the diodes will be forward biased (conducting) at once.

With the output shorted and all of the diodes conducting, the only thing limiting the current is that first capacitor.

You say you are using 4.7uF capacitors. That's an impedance of 677 ohms at 50 Hz.

The peak to peak voltage of 220VAC is 622 volts.

When you short circuit that thing, you are exposing all of the diodes to peak currents at about their maximum. That can't be good for them.

Do not short circuit that thing without a large resistor or a small capacitor in series with it.

I'd use a smaller capacitor for the first one (top left capacitor in your diagram.) Reduce it to maybe 1uF or less to limit the short circuit current.


What ever you do, be careful. As Andy aka mentioned in the comments, that thing is a people zapper. It is just waiting for you to touch it in the wrong place.

  • Your - terminal is connected directly to one side of the AC source. If that's the live wire of your outlet, you will get zapped with 220VAC if you touch it.
  • When the capacitors are discharged, the + and - terminals will be at 220VAC from one another - touching them will zap you.
  • The high voltage you are generating can zap you quite hard. Those 4.7uF capacitors can store enough energy to kill you if you touch + and - while it is charged.
  • If that thing is meant to really zap bugs, then enclose it in a mesh that will prevent people from sticking their fingers in and getting zapped.

That circuit is a killer. Treat it with lots of respect, and consider locking it away for safety's sake.

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  • \$\begingroup\$ All the safety is taken into consideration while making this bug zapper. Mechanical unit is ready along with the mesh. I do understand it's fatal consequences if not handle with safety and care. Can you please just draw the circuit will be more helpful ? Also I don't understand why didn't diode damaged when operating in 3 stage and why they start getting damaged as we increase the voltage from 1860V i.e 4,5,6 stages ? And why 16kv 5mA diode isn't working ? \$\endgroup\$ – Embedded Geek Nov 2 '20 at 19:51
  • \$\begingroup\$ You didn't just build it with more stages. You said you built the 6 stage multiplier with 4.7uF capacitors, but the three stage was with the 470nF capacitors. It's the current that's killing your diodes - and you get more current through the bigger capacitors. Stop short circuiting the multiplier. \$\endgroup\$ – JRE Nov 2 '20 at 20:05
  • \$\begingroup\$ I made the correction I tried 6 stages and 3 stages with same capaciotor i.e 0.47uf / 470nf 800V. So Okay but how we are gonna zap the bugs if we don't short circuit it because eventually the bug will get trapped between positive and negative terminal and will die. If we don't short circuit the multiplier how are we gonna do it ? Please tell me. \$\endgroup\$ – Embedded Geek Nov 3 '20 at 17:33
  • \$\begingroup\$ Bugs aren't short circuits. Bugs have resistance. A bug gets between the two contacts, gets zapped, and dies. Zapped bugs aren't short circuits, either. \$\endgroup\$ – JRE Nov 3 '20 at 18:11
  • \$\begingroup\$ I understand we shouldn't short it, but incase it gets shorts accidentally, diode will get damaged, can we protect them in anyway? \$\endgroup\$ – Embedded Geek Nov 6 '20 at 9:15
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Put 1M resistors in parallel with the diodes to distribute the voltage evenly up the ladder. This will avoid destroying diodes which are exceeding their reverse bias rating due to variations in voltage. Then go back to your original choice of diodes.

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  • \$\begingroup\$ Okay. Will try that. \$\endgroup\$ – Embedded Geek Nov 2 '20 at 19:59
  • \$\begingroup\$ There is a significant drop in the voltage when adding resistor in parallel to diodes. and if short circuit, the diode get damage \$\endgroup\$ – Embedded Geek Nov 6 '20 at 8:50
  • \$\begingroup\$ Do the resistors fix the problem with diodes getting damaged with 4, 5, 6, ... stages? \$\endgroup\$ – mhaselup Nov 6 '20 at 8:56
  • \$\begingroup\$ No. they get damaged eventually after 4-5 sparks. Tried with 4 stages at 2200v \$\endgroup\$ – Embedded Geek Nov 6 '20 at 9:11
  • \$\begingroup\$ Adding resistor adds voltage drop and when switched off discharges the cap. not fixing the main problem \$\endgroup\$ – Embedded Geek Nov 6 '20 at 9:14

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