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4hv.org :: Forums :: High Voltage
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Voltage doubler question

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TACAMO
Sun Nov 06 2011, 01:03AM Print
TACAMO Registered Member #2794 Joined: Sat Apr 10 2010, 05:13AM
Location:
Posts: 12
Hi, this might be a stupid question!! Do the components of a voltage multiplier need to be rated for the input or the output, or do the ratings need to increase from the input to output?
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IamSmooth
Sun Nov 06 2011, 04:45PM
IamSmooth Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
The caps need to be rated for 20% more than the maximum voltage across the terminals. Each stage will see a portion of the final output. I have a doubler and each of the two caps is rated for less than the final value
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Wastrel
Sun Nov 06 2011, 05:30PM
Wastrel Registered Member #4095 Joined: Thu Sept 15 2011, 03:19PM
Location: England.
Posts: 122
Making the assumption the question is about a half wave CW multiplier, the first capacitor must be bipolar and rated for the peak applied input voltage. All other caps and all diodes must be rated for twice the peak applied input voltage. Then apply whatever safety margin people recommend, eg the 20%, which may depend on your application and the quality of the capacitors used. So for example if the input was 1KV AC and 20% fudge the first cap would need to be 1 * SQRT(2) * 1.2 = 1.7Kv and everything else would be 3.4Kv minimum.

Other multiplier designs vary. It's certainly not a stupid question, circuit diagrams are common in books and on the internet and very few give voltage ratings.

There are some advantages to using identically rated capacitors, voltage and capacity, for the whole ladder such as behavior during a fault but this is not the most efficient design for power.
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TACAMO
Wed Nov 09 2011, 11:13AM
TACAMO Registered Member #2794 Joined: Sat Apr 10 2010, 05:13AM
Location:
Posts: 12
All right, I was going to build this quadrupler(pic). Does it remain 20% no matter how many stages? I was planning on adding 4 more stages to the quadrupler. The input would be 7500vac. I have 16 18kv rated doorknob capacitors no diodes yet.
1320837087 2794 FT127784 Vq
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IamSmooth
Thu Nov 10 2011, 12:45PM
IamSmooth Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
20% for all stages. For the multiplier you will need a rating of at least 7500 x 1.7 = 12.75kvdc

Personally, if i was building this I would have every stage made from the same capacitor with the same ratings as Alexander mentioned.
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Xray
Thu Nov 10 2011, 05:38PM
Xray Registered Member #3429 Joined: Sun Nov 21 2010, 02:04AM
Location: Minnesota, USA
Posts: 288
TACAMO wrote ...

All right, I was going to build this quadrupler(pic). Does it remain 20% no matter how many stages? I was planning on adding 4 more stages to the quadrupler. The input would be 7500vac. I have 16 18kv rated doorknob capacitors no diodes yet.
1320837087 2794 FT127784 Vq


The CW multiplier is a very common circuit that's used in Dental X-ray heads (I repair them for a living) and based on my own experience, the part that fails more often than others is the very first capacitor (labeled C1 in your diagram) in the chain. I think it's because it "sees" every + and - peak of the ac waveform coming out of the transformer. All of the other caps see only one peak every N cycles, depending on how far down in the chain (or ladder) it sits. So that first cap should not only be rated for the required voltage, but it needs to be able to handle the current also. In some CW's that I've seen, the first cap in the chain is physically larger than the other caps of the same capacitance and voltage rating. I can only assume that it's larger in order to handle more current than the other caps need to handle.
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