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basic general question

J

John Fields

Jan 1, 1970
0
here is the transformer i bought and i have 2 of them.

http://www.allelectronics.com/cgi-bin/item/HVS-1/480/GAS_TUBE_SIGN_POWER_SUPPLY_.html
Specifications: Input: 12 Vdc @ 1 Amp (Suggested power source CAT#
DCTX-1215). Output: 2,000 Vac @ 10 mA. Open circuit voltage: 3,000 Vac
30Khz. Short circuit current: 15 mA. UL.

---
Small inconsistency: The input power is specified as 12 watts, max,
while the output power would be 20 watts if the output voltage was
RMS.

From the output voltages shown on the PDF data sheet:

http://www.allelectronics.com/spec/HVS-1.pdf

It appears that the simple rectified DC is 3kV unloaded, which would
make it 2kV, peak, with a 10mA load. 2kV peak is 1414VRMS, so
that's 14.14 watts with a 10mA load, which is still higher than the
input so there's something gone awry somewhere. In your
application, though, it shouldn't matter. One thing to remember,
however, is that as the capacitor charges it will draw less and less
current from the high voltage supply so when it's fully charged the
supply will have risen to its full unloaded value of 3kV and the cap
will be charged to 6kV!
---
I got the recomemded 1.5 amp power supply however other findings
show that is a bit much current for this devic and i got another
adapter that outputs 1 amp , again i dont want to burn anything
out.

---
As someone else posted, the high voltage supply (the 'inverter')
will take whatever current it needs from the DC supply, so as long
as the output of the DC supply stays close to 12V when it's loaded
by the inverter you should be OK. That 1.5A part doesn't seem to be
regulated, though, which means its output might rise high enough to
hurt the inverter if the inverter doesn't load it enough, So I'd use
the 1 amp one just to be safe.
---
i got an asortment of resistors and diodes as i was not sure what to
get.
6 x 220 ohm 3 wat , 3 x 1k ohm 5 wat.
10 x .2 amp 6k diodes and a wackload lower voltage ones that prob
will not ever need now

as for the home made cap i made from 6mil polyethelene which someone
else used, surface area for diferent caps for now are ony small for
testing now and will get bigger later. ( dangerous either way im
sure )
foil surface area is about 30 cm x 5 cm for each pos and negative
plate and tightly rolled around a dowel.

---
Those two plates, with a polyethylene dielectric 0.006" thick and a
dielectric constant of 2.25 will get you a capacitor with a
capacitance of about 2nF.
---

other caps are same materials but smaller surface area . later for
good discharges i may have to go 10 or 50 times the size as the one i
mentioned. i dont expect you to do all the math as im lazy however
just want you to have a better idea what i have to start with other
then loads of pexiglass

---
OK. :)

Just for grins I ran a simulation with two different doublers you
may want to play around with:


Version 4
SHEET 1 880 680
WIRE -784 160 -864 160
WIRE -656 160 -704 160
WIRE -624 160 -656 160
WIRE -496 160 -560 160
WIRE -864 176 -864 160
WIRE -80 176 -112 176
WIRE 16 176 0 176
WIRE 112 176 80 176
WIRE 160 176 112 176
WIRE 272 176 224 176
WIRE -496 192 -496 160
WIRE -112 240 -112 176
WIRE 112 240 112 176
WIRE 272 240 272 176
WIRE -864 288 -864 256
WIRE -496 288 -496 256
WIRE -496 288 -864 288
WIRE -496 320 -496 288
WIRE -112 368 -112 320
WIRE 112 368 112 304
WIRE 112 368 -112 368
WIRE 272 368 272 304
WIRE 272 368 112 368
WIRE -112 400 -112 368
WIRE -656 416 -656 160
WIRE -624 416 -656 416
WIRE -496 416 -496 384
WIRE -496 416 -560 416
WIRE -496 448 -496 416
FLAG -112 400 0
FLAG -496 448 0
SYMBOL voltage -112 224 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(0 3000 25000)
SYMBOL diode 160 192 R270
WINDOW 0 71 28 VTop 0
WINDOW 3 65 26 VBottom 0
SYMATTR InstName D1
SYMATTR Value 1N4148
SYMBOL cap 80 160 R90
WINDOW 0 -39 33 VBottom 0
WINDOW 3 -35 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 2e-9
SYMBOL diode 128 304 R180
WINDOW 0 -38 32 Left 0
WINDOW 3 -73 -1 Left 0
SYMATTR InstName D2
SYMATTR Value 1N4148
SYMBOL cap 256 240 R0
WINDOW 0 47 33 Left 0
SYMATTR InstName C2
SYMATTR Value 2e-9
SYMBOL voltage -864 160 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value SINE(0 3000 25000)
SYMBOL diode -624 176 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D3
SYMATTR Value 1N4148
SYMBOL diode -560 400 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName D4
SYMATTR Value 1N4148
SYMBOL cap -512 192 R0
WINDOW 0 47 33 Left 0
SYMATTR InstName C3
SYMATTR Value 4e-9
SYMBOL cap -512 320 R0
WINDOW 0 47 33 Left 0
SYMATTR InstName C4
SYMATTR Value 4e-9
SYMBOL res -688 144 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R2
SYMATTR Value 100k
SYMBOL res 16 160 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R1
SYMATTR Value 100k
TEXT -42 386 Left 0 !.tran .1



The simulator, if you don't already have it, is
"LTSPICE/SWITCHERCAD III", is free, and you can download it from:

http://www.linear.com/company/software.jsp

One final thing, notice that I used 1N4148's for the HV diodes.
Through the magic of simulation that works, but don't try it at
home!
 
J

Jon Slaughter

Jan 1, 1970
0
i think ill do the throw switch like you suggested to prevent that
feed back. Im going to making my own relay switches to insure
proper insulated and safely encased . basically all my work will be
contained in pexiglass with a few redundant grounding options in
case a path burns out.

One thing you will have to worry about is when you close the switch to
recharge the cap. I will arc because of the high voltage and there will be a
huge current flow. This will probably wear down the switch. You probably
need some way to limit the current until the switch is fully closed.
Ofcourse adding any number of switches will not help cause the final one
will still have the same issue. If you were to add an inductor this would
prevent the large instantaneous current but will can cause other problems
such as oscillations and slow charging times. Something like a choke might
work but not sure. It would need to be chosen so that there is not much
overall effect on the charging of the cap except initially when the current
is highest.

Although I would imagine that the transformer itself would act as a choke to
some degree and you might not have this problem. Maybe the best thing to do
would be to experiment and actually see if there is any arcing. I doubt it
will be a big issue but could be.

this is not really necessary for the lazor project. Im looking however
for a simple solution to to sync the negative and positive pulse that
i split from the same output. the negative ac phase must match the
exact positive phase.
End result i want for this seperate project is pulsed dc that is equal
and opposite + & -. this is for hv tests regarding field polarization
effects. Non phase is usless for this test as the oppsite current
is gone or stored before the other pulse comes out. That why i was
thinking on using the doubler using the ground . just not sure on a
circuit that will pulse out from that . that curcuit " i think"
just maintains HV voltage if we dont use too much load where i want
pulses like discharges instead.
I thinking my best option might be to wrap my own transformer with 2
equal output windings. for the sync, any better suggestions? If i
wrap one winding in one direction and the next in the opposite
direction would each output winding be opposite polarity in sync? im
not 100 % sure tho i think thats how it works. of course if this works
i woudl run it on a figure 8 laminated core with input on centre
and oppsite outs on either end to prevent it from shorting across
the increased diference. in voltage. how does that sound?
Again thanks for all your input.


I'm not sure. (still a little unclear on what exactly your trying to do)

What do you mean sync? You mean fire them both at the same time? Do you have
more than one? If you just need to pulses of opposite polarity then you
don't have to do anything fancy. Just charge the caps up in parallel,
disengage them from the circuit using switches. Now you have two caps that
are isolated. Now when using them in the discharge side its just a matter of
how you hook them up that determines there relative polarity. You can set
this up using relays so that it can all be automated.


Suppose for a sec that you could only charge the caps up by "hand". That is,
you had a 2000V batter and you connected the caps up to charge them then
removed them. Now you have two caps disconnected from any circuitry. Could
you then hook them up to your discharge circuit and have it do what you
want? If so then notice that you did nothing special in charging the caps
w.r.t polarity but just "flipped" them in the discharge circuit. You can
effectively do all this with just switches and automate it.

The only draw back to this is that it requires more switches(relays or
motorized switches) and is slow compared to charging up the caps with
polarity taken into account.

Hopefully that makes some sense. I'm not sure about the transformer though.
I would say to go ahead and try it though for the experience as it probably
won't hurt anything. (might catch something on fire but thats the fun in
learning ;)

Jon
 
You Guy's are awesome!
Thank you all for your input, I greatly appreciate it!

Down the road sometime I will try to make a transformer. I will
save those questions for another post and ill do more reading and
learning first on voltage and current limiting.
thanks again !!

Caltus :)
 
J

Jon Slaughter

Jan 1, 1970
0
You Guy's are awesome!
Thank you all for your input, I greatly appreciate it!

Down the road sometime I will try to make a transformer. I will
save those questions for another post and ill do more reading and
learning first on voltage and current limiting.
thanks again !!

And please post your results so that some of us might learn something too ;)
Eventually I will probably want to try something similar and it will be
helpful to hear how it worked out for you.

Jon
 
J

John Fields

Jan 1, 1970
0
One thing you will have to worry about is when you close the switch to
recharge the cap. I will arc because of the high voltage and there will be a
huge current flow.

---
That's the wrong way to do it, I think.

The right way, with no current-limiting resistor in the charging
circuit, would be to disconnect the charged cap from the doubler and
connect it across the laser, then to disconnect the inverter from
the 12V supply. Once the cap was discharged and needed to be
recharged it would be connected to the doubler and then 12V
connected to the inverter.
 
A

Andy Dingley

Jan 1, 1970
0
Your point? A gun will kill him too if he points it at his head(first he
will need to make sure it is loaded).

How many people still manage to shoot themselves in the head / foot or
bollocks _without_ first checking that it's loaded? It's not the one
you're expecting that bites you...

In any case 10mA isn't, in general, enough to kill.

At 2kV ? That's well into the "too dangerous to **** with" region. It
might not kill you, but then people also survive being shot in the
head.

Here's a poster who doesn't spot the issue that a DC input is the
wrong starting point for a basically AC circuit, and who spells it
"lazor". Now I'm all for encouraging experimentation, but _not_ for
someone this inexperienced at multiple kV voltages.


As Homer said:
 
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