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Voltage Multiplier (CW) - some clarification

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yroethiel

Sun Nov 29 2009, 10:45PM

Registered Member #2488 Joined: Sat Nov 28 2009, 11:26PM
Location: UK
Posts: 8

Possible half-wave VM diagram deliberately unlabelled. Includes fuse, filter, damping resistors, final load. Does not include the circuit breaker in power supply (the coil on left). The simulator linked earlier suggests a minimum of 15 stages for the requisite 5kV. Hopefully this matches roughly what you guys suggest, and I can then start plugging values into the components...?

EDIT: .jpg uploads forbidden although seemed to work earlier. Also, no delete post function?

yroethiel

Sun Nov 29 2009, 10:50PM

Registered Member #2488 Joined: Sat Nov 28 2009, 11:26PM
Location: UK
Posts: 8

Sorry for double post. Attempting to reupload image with it now resized to <400px.

Dr. Slack

Mon Nov 30 2009, 08:13AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

You seem to have a spare capacitor across the transformer secondary? If you were intending power factor correction, then a primary capacitor would be easier, but it's not required.

You don't need 2 resistors in series with each diode, one will do.

You do need a discharge resistor across the whole stack for safety. If you make the storage leg from seriesed photoflash caps, then including balancing resistors (cheaper than zeners) will also perform the function of the discharge resistor.

Adding a photo to an existing post is a bit convoluted. Post the picture to the "Pictures for use in threads" thread in the attachments board. You must have some text in the post otherwise you get an error. Doing an edit on the resulting post will allow you to copy/paste the right incantation into the target post in the right thread. This is also the method to use if you want multiple pictures in a single post. As you are linking to rather than including the picture in a post, I'm not clear whether the 400 pix rule applies here, it gets thumbnailed automatically.

yroethiel

Wed Dec 02 2009, 05:03PM

Registered Member #2488 Joined: Sat Nov 28 2009, 11:26PM
Location: UK
Posts: 8

Thanks for the tips Dr. Slack. I've had a look through my power supply and it has a voltage regulator in it - should be OK on that front.

I did wonder about the dual resistors suggested earlier, thought this was perhaps a convenient way of keeping cost down by using lower spec components/buffering both positive/negative spikes?

Can anyone make any suggestions to literature about voltage multipliers? Went to our university library today, and in the 15 or so electronics books I scanned, neither "voltage multipliers" nor "Cockroft-Walton" turned up once. They were fairly general textbooks/engineering references. I'm not sure how useful they'll be compared to what I've read so far anyway - I unfortunately don't know anyone with practical experience I can chat/discuss this with.

Would you guys recommend sticking with the photoflash caps or just spending a few extra pounds getting some more "reliable" components...

yroethiel

Fri Dec 04 2009, 01:50AM

Registered Member #2488 Joined: Sat Nov 28 2009, 11:26PM
Location: UK
Posts: 8

Thank you everyone for the suggestions; I'm going to have to ask for a bit of help with power correction in a bit. First things first, is this set-up a step in the right direction? I haven't included resistors, main filter or an indicator light as the fundamentals have to be right first. At the moment I'm sticking with the variac to CW multiplier idea, simply because I want to get into electronics a bit more and this is a challenging and exciting (for me, :P) first step.

Input, variac off mains 0-240V. Target voltage, 5kV at 0.0001A.

Diodes 1N4007, Capacitors Metal Film 1uF 760V. (buying 0.68uF caps would be cheaper, but these are closest to the optimum capacitance for a 50Hz input. The difference is about Â£0.12 per component, not much.)

10mA fuse for safety - anyone know a good supplier for low volume?. The weird arrangement at the end (between final resistive load and ground) is for discharging capacitors after use (I'm aware of RC, would use a lower resistor value so it doesn't take forever, but will address that later).

Using the simulator found at Blaze Labs, which takes voltage drop into account and makes going through the equations a bit easier...

Input parameters: CW, Sine Wave, 240V, 5000V, 0.0001A, 50Hz, 0.

This suggests a 12-stage full-wave CW multiplier for 5200V output under full load. Ripple is very low at 0.19% and the output is just 0.52W due to the low current, matching my general requirements very nicely.

In order to achieve the 0.0001A current output, V=IR for the final load. This implies 52M Ohms. Sticking 25 x 2.2M Ohm resistors in series gives 55M Ohms. P = V^2 / R means 0.5W. The power ratings on resistors in series are additive, right? I know this is the case for parallel, but energy dissipation would be the same in equal spec series resistors? 25 x 0.5W (the rating of the 2.2M Ohm resistors) = 12.5W of power dissipation, many times more than enough. I was thinking of placing all these resistors in an aquarium tubing arrangement, much like the HV diode tutorial on the 4HV Wiki.

The components across each stage see about 480V max, so the capacitor ratings are high enough with a good margin (as are the diodes).

Putting a resistor in series with each diode (not in diagram, but as suggested for safety/power fluctuation purposes earlier) is (I think), P = V^2/R. I have a whole bunch of 0.25W resistors, so re-arranging to find R (using 0.125W as a safety net) gives 2M Ohms of these spec. This means the current across these branches will reach a maximum of (I = V/R again) 0.00024A. Are these realistic figures at all? I feel like I've done something wrong, and should work from a fixed value of I (0.0001A).

Lastly! In searching the older archives for CW posts, I found a useful Java based simulator. I don't think it's powerful enough for this application (can someone suggest a circuit simulator, preferably freeware?). I built a small CW multiplier to get a feel for what's happening, and am slightly confused by the results. According to the various voltage multiplier theory sites I've read, the polarity of the diodes in the above picture gives a negative DC output. This simulator gives me a positive output - they don't agree with each other. As far as I can see, this diode arrangement DOES give out a DC output, but the simulator has shaken my confidence in that. Is it right to do so?

File > Import > paste the following to make a quick full-wave transformer to see what I'm talking about.

$ 4 5.0E-6 15.472767971186109 49 15.0 45
r 48 224 128 224 0 100.0
R 48 224 16 224 0 1 50.0 240.0 0.0 0.0 0.5
d 464 224 432 320 1 0.805904783
d 336 128 304 224 1 0.805904783
c 336 128 256 128 0 1.0E-6 -364.48901613944963
d 304 224 256 128 1 0.805904783
d 256 128 224 224 1 0.805904783
g 192 224 192 240 0
r 656 224 560 224 0 1000000.0
c 304 224 224 224 0 1.0E-6 -427.1433901577185
d 464 224 416 128 1 0.805904783
d 384 224 336 128 1 0.805904783
w 192 224 224 224 0
w 128 224 176 128 0
d 352 320 304 224 1 0.805904783
d 432 320 384 224 1 0.805904783
d 272 320 224 224 1 0.805904783
d 304 224 272 320 1 0.805904783
d 384 224 352 320 1 0.805904783
w 128 224 192 320 0
d 416 128 384 224 1 0.805904783
g 656 224 656 256 0
w 464 224 560 224 0
c 464 224 384 224 0 1.0E-6 -320.52346727686506
c 384 224 304 224 0 1.0E-6 -364.61285857541316
c 416 128 336 128 0 1.0E-6 -320.55901617167285
c 432 320 352 320 0 1.0E-6 -320.5590161716734
c 352 320 272 320 0 1.0E-6 -364.4890161394494
c 272 320 192 320 0 1.0E-6 -213.80439248780573
c 256 128 176 128 0 1.0E-6 -213.80439248780584
o 1 64 0 35 320.0 0.1 0 -1
o 8 64 0 35 1280.0 0.00625 1 -1

Thanks for any help!

Dr. Slack

Fri Dec 04 2009, 08:14AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Still no discharge resistor shown in your diagram, tut, tut! Learn from our advice before you get bitten by a piece of kit that's been sat powered off in the closet for a couple of days, not after.

You might find that a 10mA fuse is a bit on the fragile side, even if you do wind the variac up slowly from zero to avoid inrush. The purpose of a fuse is to protect you and the components, IN4007s are tough as old boots, 30A surge IIRC, so a 1A or 3A fuse ought to be plenty small enough to prevent fire, and more robust against the inevitiable little glitches that will occur.

The components across each stage see about 480V max

You do realise that 240v mains means volts rms, and the peak is 340v don't you?

yroethiel

Fri Dec 04 2009, 10:58AM

Registered Member #2488 Joined: Sat Nov 28 2009, 11:26PM
Location: UK
Posts: 8

Thanks for quick reply :).

Regarding peak voltage across each stage, my bad - was playing around on a phone calculator and had lots of numbers, not the most organised way of working. I am aware of peak voltage versus RMS which is why I chose my capacitors on 2xPeak with a bit of a margin, the 760V (2x340 + a bit) instead of 630V from earlier.

Thanks for the tip about 1A fuse.

Regarding the discharge capacitor, I thought grounding the disconnected output through a big resistor would be OK (as badly illustrated in diagram), dumping the stored charge through that? I guess having resistors built into the circuit makes it intrinsically more safe should I forget to discharge it, so great..I'll follow the experienced :).

Regarding putting a resistor in series with each capacitor in the central arm. Proud Mary mentioned 10W 50R resistors earlier, but I suspect that was for a higher current application. I don't quite know how to spec these, whether to calculate their spec for ideal discharge times or minimal wattage/current drawn, etc.

If the resistance is low (ie. RC is kept to a minimum - both quick discharge (safety) and minimal change to VM's operation), the wattage needs to be higher - I don't have enough experience to draw from to make an educated guess at what would be a standard, cheap resistor to use where these two are balanced. Any suggestions?

[For example, using a 2.2M Ohm resistor would mean the power rating only has to be 0.25W (although more like 0.5W for safety), but the RC constant then becomes 2.2 seconds - buggering up the VM cycles. I don't know componentry well enough to think, "A cheap solution would be...". Browsing Farnell's catalogue gives items like this, but the power ratings don't seem to anywhere near P=V^2/R.]

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