Maker Pro
Maker Pro

basic general question

Hi,
Im a total noob to this field and was wondering if anyone could
provide input to prevent me from blowing up my ac output.
What im doing is charging a home made HV cap with the following.
I have a 1 amp 12vdc adapter, which i will be connection to a 12v
1amp neon sign power supply. output is 2kv ac @ 10ma. I will be
rectifing & doubling voltage to charge the cap however i was wondering
if this AC type conerter could possible draw too much current to burn
it out?
I have found other similar circuits however only reference to a
resistor or charging resistor was after the diode to prevent feedback
on discharge of the cap. Im not entirely sure if this was to limit
current charging as well as it wasnt clearly explained. I thought this
would be the best place to ask before try. I was thinking of adding a
1 or 2amp fuse between the ac adapter and neon supply to prevent
killing the adaptor however this wont limit charging current if i need
to. For my charging resistors i really dont know what i should put in
as i dont want to drop below the 10ma current if possible and also
dont want to drop voltage either if i can get away with it.
Any recomendations would be greatly appreciated.

for those that are curious this cap is for a pulsed hv lazor
experiment

Thanks,
Caltus
 
H

Homer J Simpson

Hi,
Im a total noob to this field and was wondering if anyone could
provide input to prevent me from blowing up my ac output.
What im doing is charging a home made HV cap with the following.
I have a 1 amp 12vdc adapter, which i will be connection to a 12v
1amp neon sign power supply. output is 2kv ac @ 10ma. I will be
rectifing & doubling voltage to charge the cap however i was wondering
if this AC type conerter could possible draw too much current to burn
it out?

First, this will kill you no problem.

How will you convert DC to AC?

Last, this will kill you no problem.
 
J

Jon Slaughter

Homer J Simpson said:
First, this will kill you no problem.

How will you convert DC to AC?

Last, this will kill you no problem.

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).

Breakdown of air is about 30kV/cm so its not like this thing will jump out
at him. Obviously if he's storing charge on the cap then he will need a safe
way to discharge it before sticking his hands in there.

http://engr.nmsu.edu/~etti/3_3/3_3h.html

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

You would have been more helpful to try to answer his questions instead of
stating the obvious.
 
J

John Fields

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).

Breakdown of air is about 30kV/cm so its not like this thing will jump out
at him. Obviously if he's storing charge on the cap then he will need a safe
way to discharge it before sticking his hands in there.

http://engr.nmsu.edu/~etti/3_3/3_3h.html

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

You would have been more helpful to try to answer his questions instead of
stating the obvious.

---
I notice that even though you were critical about Homer's not
answering the OP's questions, neither did you. Is there a reason for
that?

Also, you obviously aren't familiar with the obvious if you think
that a cap charged up to 4kV won't supply more than 10mA into a
human load.
 
H

Homer J Simpson

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).

Most people know a gun will kill them. Not as many know a sign transformer
will do this as well.

The fact that he's confused about what a transformer will transform is a red
flag that he is in the unknowing group.
In any case 10mA isn't, in general, enough to kill.

You wanna take a chance on that?
You would have been more helpful to try to answer his questions instead of
stating the obvious.

I asked him a basic question and I warned him of the danger. You'd have done
better to mind your own business since your knowledge is so slim.
 
J

Jon Slaughter

Homer J Simpson said:
Most people know a gun will kill them. Not as many know a sign transformer
will do this as well.

The fact that he's confused about what a transformer will transform is a
red flag that he is in the unknowing group.


You wanna take a chance on that?


I asked him a basic question and I warned him of the danger. You'd have
done better to mind your own business since your knowledge is so slim.


Oh!! Great come back. You asked him a question... did he ask for a question
to his question or to answer his question? If you know so much about this
shit then surely you could supply more information. Your question about how
he will convert DC to AC shows just how ignorant you are... Your trying to
act like your intelligent by misdirection. The crux of the matter is not
that how he will get AC from DC(which I assume he will do by an inverter
like everyone else) EXACTLY what he asked.

Unlike you I do not know the true answer so I did not respond. I wouldn't
want to give him bad advice. Maybe you should take that advice instead of
trying to act like you know something about something you don't.

My answer would be to add about 100kohms in series with the cap to generate
a peak instantaneous current of V/R ~= 2k/100k A = 1/50 A. While this is
over is rated max of 10mA it is only 10mA above it and will die down to 7%
of that within ~5 RC's. For his application this might be to long. He
mentions that it is for a pulsed laser but how long between each pulse does
he need. How big is the cap? In anycase he could work down from the 100kOhm
resistor to increase the charge rate. Although having a resistor of 1ohm
produces an instantaneous peak current in the secondary of about 2000 A and
that means its over 4M in the primary. I'm sure that he will definitely not
get this out. But even if it pulls just 10 amps from the 12VDC supply it
could eventually cause problems. (depending on how fast and how many times
it is pulsed) Obviously there will be a saturation point that will limit the
peak inverse current and most devices can handle a substantially higher peak
inverse current but I have no specific answers.

But unlike you have I have no practical experience with the subject so I did
not answer... You also obviously have no practical experience but answered
with nonsensical questions trying to act like you know something that you
don't(which you do a lot of).

I mean, come up... WTF is up with "How will you convert DC to AC?"??? WTF do
you think you can ask such stupid questions. You are assuming the guy is a
total idiot... and this really means you are the idiot. He did not ask
anything about that and you should assume that he as a little bit of a clue
what he is doing. Now maybe it is the case he will hook up the 12VDC supply
directly to the sign transformer and it won't work... GUESS THE **** WHAT???
YOU KNOW THEY MAKE 12VAC POWER SUPPLIES TOO? Your fucking question is moot
and just a waste of time. If he does have that problem he will post back and
say it doesn't work and then someone will solve that problem very quickly
and he will just have to run to wal-mart and by a 12VAC adapter or tear out
the rectification in his 12VDC adaptor.

Now instead of that bs you should have tried to answer his real problem but
no, since you have no clue you misdirect like a magician. A magician has no
real magical powers so they pretend. You are nothing but an intellectual
magician. I see your BS all the time in this NG and you do more harm than
good and I think its time someone pointed it out. Surely though there will
be those that will jump on me because they too are intellectual magicians
and everyone knows you guys stick together.

Also, it is true its a probably a good idea to point out that it could kill
him but if he doesn't know this already then maybe natural selection needs
to take place. Also, every time someone plugs in a device you need to warn
them they might die too. Do you know that only about 200 people a year die
from electrocution? Of those how many do you think are not electricians or
kids?

"A low-voltage (110 to 220v) 60 Hz AC traveling throughout the chest for a
fraction of a second may induce ventricular fibrillation at currents as low
as 60 to 100 mA; about 300 to 500 mA of DC are required. If the current has
a direct pathway to the heart (e.g., via a cardiac catheter or pacemaker
electrodes), much lower currents (>1 mA, AC or DC) can produce fibrillation.
Body resistance (measured in ohms/sq. cm)* is concentrated primarily in the
skin and varies directly with the skin's condition. Dry, well-keratinized,
intact skin has an average resistance of 20,000 to 30,000 ohms/sq. cm,
whereas the resistance of moist thin skin is about 500 ohms/sq. cm. "

So assuming the length of is arms to chest is ~30cm gives about
2000/(30*30k) = 2mA. Ofcourse if he's playing with the shit wet or swetting
a lot then that goes up to about 133mA.

Instead of being a moron about the dangers you should just point out that
its not safe and maybe link to a site about the dangers and some saftey
precautions. Ofcourse if he's poking around in the circuitry while its life
then maybe he deserves to get electrocuted. Its not that bad though, I've
messed around inside of life TV's and not be electrocuted. (Maybe it was a
stupid thing to do but I'm alive.)

The point is, Homer, is that if your not going to try and answer is
questions then why bother to post? I know you need to feel useful but its
just a waste. Try volunteering at something your good at like picking up
trash.
 
J

Jon Slaughter

My answer would be to add about 100kohms in series with the cap to
generate a peak instantaneous current of V/R ~= 2k/100k A = 1/50 A. While
this is over is rated max of 10mA it is only 10mA above it and will die
down to 7% of that within ~5 RC's. For his application this might be to
long. He mentions that it is for a pulsed laser but how long between each
pulse does he need. How big is the cap? In anycase he could work down
from the 100kOhm resistor to increase the charge rate. Although having a
resistor of 1ohm produces an instantaneous peak current in the secondary
of about 2000 A and that means its over 4M in the primary. I'm sure that
he will definitely not get this out. But even if it pulls just 10 amps
from the 12VDC supply it could eventually cause problems. (depending on
how fast and how many times it is pulsed) Obviously there will be a
saturation point that will limit the peak inverse current and most devices
can handle a substantially higher peak inverse current but I have no
specific answers.


This was just for a rough approximation. Obviously its an RLC circuit but
this should just result in reducing the overall peak instantaneous current
which means RC will effectively be larger. I did most of those calculations
earlier so they may be wrong. (My memory isn't so good these days). In any
case it wasn't suppose to be a detailed analysis but just an overall general
idea. The point is that increasing R increases the charge time but
decreases the maximum instantaneous current. I would imagine that its not so
important because it will be limited by the transformers and as long as the
average current isn't to high over the charge time then it shouldn't be a
problem but the average will need to include the number of charges per
second too.

I would think that if you just use it intermittently then you probably don't
need a resistor. If this is pulsed many times a second then you'll need a
more detailed analysis. (or just experiment) (also I'm assuming V = 2kVDC
but since its suppose to be 2kVAC the above numbers will be off)

In any case I'm not really trying to give any info because I have not done
this sorta stuff before and I coudl be wrong. There are many web sites that
talk about HV and stuff and you'll find more practical experience there.
Also, an important side note, which you probably know, is that the capacitor
will probably be the most dangerous thing there. You should make sure you
safely discharge the cap if you plan on messing around it. The 2000V could
span a mm or so and get you. Probably won't kill you but could hurt.

If your eager to start you might try using a 1M resistor and taking some
measurements such as charge time(which you can approximately compute
anyways) and run it for a while and see if things get hot(very doubful).
Then work your way down from there until you reach your specs. My guess,
and its only a guess, that you could probably deal with maybe even just a
few hundred ohms or even less and everything will work fine if your not
pulsing it hundreds of thousands of times a sec.

You might want to give more information on the specifics so someone that
knows about this stuff can give a more informed decision. (although maybe
all pulsed lasers follow a specific implementation). I think the two
important things are the capacitatance, pulses per second, and operation
time.

Jon
 
H

Homer J Simpson

Oh!! Great come back. You asked him a question... did he ask for a
question to his question or to answer his question?

You're still an idiot. Hooking up a 12 VDC supply to a sign transformer
isn't going to do much of anything. Asking questions is the best teaching
method.

And your assumptions could kill him. I've been working with electronics and
electricity for over 50 years and I don't fool with sign transformers or the
like without following a lot of precautions. For someone with as little
knowledge as he shows his next few steps could end his world.
 
ok, I appreciate all your concerns for safety and believe me all
precautions will be taken.
the cap size will certianly be more than 100 times more powerful
enought to kill myself but i will be doing whatever so that wont
happen. So lets not worry about that for now .
the basic charge curcuit after ac adapter and fuse is as follows
the ac current to charge the cap will be 1/2 wave rectified on + and
1/2 wave on - using 4 6kv 200ma diodes.
each ac lead will have 2 diodes connected to it ( forward and reverse
on each)
the the 2 + ends will be positive out and the 2 negative ends will
be negative out. this will evenly split the ac into a 1/2 wave
rectified positive and 1/2 wave rectified negative. 2kv ac usually
means peak + and - is 2kv so by seperating them i will have
effective 4kv diference on leads for cap charging without
multiplying.
now this is where i have questions. 1) imaging this is going
straight to a large cap. will discharging the cap cause feedback to
the 2kv transformer? I have seen others use a 1k ohm resistor past
the diodes for feedback prevention when the cap is discharged? I know
that resisters for this could vary and ohms law dictates that if i
use large enough Resistors most the current and voltage will go thru
however if a return spike came back on discharge it would filter and
prevent most or some from hurting my diodes or transformer. Thing is i
would like to prevent as much feedback as possible however do not
want to drop as much voltage. My experience comes form Whimhurst
machines and variants so i dont know alot about HV resistors. Im
stepping up from those as my projects need more current that
vandegraff. and believe me when it comes to safty, with leyden jars
theese can be just as dangerous or more so.


i would also prefer to sync charging ( or at least discharging ) as +
and - are perfectly out of phase on diferent leads.
can i achieve this by charging first to a 3 way cap ( think of
leyden jar setup with ground connection each on like how some
whimhurst machines are configured, positive - ground chain -
negative ) the discharging from them automatically?
this circuit would be the same as a Greinacher voltage doubler

thanks, any thoughts or jokes about the guy that will blow himself up
is appreciated ;)
lol
Thanks,

Caltus
 
J

Jon Slaughter

Homer J Simpson said:
You're still an idiot. Hooking up a 12 VDC supply to a sign transformer
isn't going to do much of anything. Asking questions is the best teaching
method.

Obviously you don't know how to fucking read.
And your assumptions could kill him. I've been working with electronics
and electricity for over 50 years and I don't fool with sign transformers
or the like without following a lot of precautions. For someone with as
little knowledge as he shows his next few steps could end his world.

omg! you must be fucking l33t all those years of installing ceiling fans and
you call yourself an expert. I guess you really think that just because
you've been around for 50+ years means your an expert. To bad you don't
understand that age != experience. Ofcourse one might think it is correlated
but usually arrogance and ego wins out every time.
 
J

Jon Slaughter

ok, I appreciate all your concerns for safety and believe me all
precautions will be taken.
the cap size will certianly be more than 100 times more powerful
enought to kill myself but i will be doing whatever so that wont
happen. So lets not worry about that for now .

Yeah, I agree. If your stupid enough not to be careful then your stupid
enough to die. (Ofcourse there are always accidents but if one is
irrational like homer then they will never get anything done. Sometimes you
gotta take a risk to make progress)
the basic charge curcuit after ac adapter and fuse is as follows
the ac current to charge the cap will be 1/2 wave rectified on + and
1/2 wave on - using 4 6kv 200ma diodes.
each ac lead will have 2 diodes connected to it ( forward and reverse
on each)
the the 2 + ends will be positive out and the 2 negative ends will
be negative out. this will evenly split the ac into a 1/2 wave
rectified positive and 1/2 wave rectified negative. 2kv ac usually
means peak + and - is 2kv so by seperating them i will have
effective 4kv diference on leads for cap charging without
multiplying.
now this is where i have questions. 1) imaging this is going
straight to a large cap. will discharging the cap cause feedback to
the 2kv transformer? I have seen others use a 1k ohm resistor past
the diodes for feedback prevention when the cap is discharged? I know
that resisters for this could vary and ohms law dictates that if i
use large enough Resistors most the current and voltage will go thru
however if a return spike came back on discharge it would filter and
prevent most or some from hurting my diodes or transformer. Thing is i
would like to prevent as much feedback as possible however do not
want to drop as much voltage. My experience comes form Whimhurst
machines and variants so i dont know alot about HV resistors. Im
stepping up from those as my projects need more current that
vandegraff. and believe me when it comes to safty, with leyden jars
theese can be just as dangerous or more so.

Yes, there is things such as inductive kick bad, EMF radiation, etc that can
cause problems. I cannot really tell you much because I do not know either.
There are several sites I have seen that do discuss this topic though. What
you want is some type of path to ground that the feedback can take.

For example, metal oxide varistors are used when a voltage spike on the
mains is larger than a certain amount and the mov sorta opens up and proves
a least resistive path to ground. This prevents that spike from going
through the transformer. You could use somethinglike that but not sure if
they make them for that high of a voltage. (although maybe you could use
several in series) Theres probably a better method and ofcourse theres a
whole industry built around HV stuff and they all have these very special
techniques and exotic devices for dealing with these kinds of issues. You
might look into how they handle this stuff in power distribution. (usually
though its for high currents but maybe something will work).

I would say maybe you could just ignore it for now and see but you might not
want to take the chance and ruin your components.

Maybe you can put a relay switch between the cap and diodes to completely
remove it from the circuit? The relay can be triggered to open when the cap
has reached a certain voltage. Might be a good solution. Ofcourse when it
closed there would be an inrush of current and this could effect the
contacts.
i would also prefer to sync charging ( or at least discharging ) as +
and - are perfectly out of phase on diferent leads.
can i achieve this by charging first to a 3 way cap ( think of
leyden jar setup with ground connection each on like how some
whimhurst machines are configured, positive - ground chain -
negative ) the discharging from them automatically?
this circuit would be the same as a Greinacher voltage doubler

So you want to charge up one side while discharging the other? (so you would
increase the pulse rate by a factor of 2)
thanks, any thoughts or jokes about the guy that will blow himself up
is appreciated ;)

No, I have no issues with you doing this. Its better than sitting on your
hands or twiddling your thumbs cause your too afraid. There are a lot of
weenies out there. If it wasn't for people taking risk we wouldn't be where
we are. Ofcourse its not that big a risk and you seem like you know the
risks. There are many things you can do such as wearing insulated gloves or
even a suit but this isn't really necessary if you just pay attention(which
I think you will and its not really my place(or anyones) to tell you what
you can and can't do)

Have fun.

Jon
 
J

John Fields

This was just for a rough approximation. Obviously its an RLC circuit but
this should just result in reducing the overall peak instantaneous current
which means RC will effectively be larger. I did most of those calculations
earlier so they may be wrong. (My memory isn't so good these days). In any
case it wasn't suppose to be a detailed analysis but just an overall general
idea. The point is that increasing R increases the charge time but
decreases the maximum instantaneous current. I would imagine that its not so
important because it will be limited by the transformers and as long as the
average current isn't to high over the charge time then it shouldn't be a
problem but the average will need to include the number of charges per
second too.

I would think that if you just use it intermittently then you probably don't
need a resistor. If this is pulsed many times a second then you'll need a
more detailed analysis. (or just experiment) (also I'm assuming V = 2kVDC
but since its suppose to be 2kVAC the above numbers will be off)

In any case I'm not really trying to give any info because I have not done
this sorta stuff before and I coudl be wrong. There are many web sites that
talk about HV and stuff and you'll find more practical experience there.
Also, an important side note, which you probably know, is that the capacitor
will probably be the most dangerous thing there. You should make sure you
safely discharge the cap if you plan on messing around it. The 2000V could
span a mm or so and get you. Probably won't kill you but could hurt.

If your eager to start you might try using a 1M resistor and taking some
measurements such as charge time(which you can approximately compute
anyways) and run it for a while and see if things get hot(very doubful).
Then work your way down from there until you reach your specs. My guess,
and its only a guess, that you could probably deal with maybe even just a
few hundred ohms or even less and everything will work fine if your not
pulsing it hundreds of thousands of times a sec.

You might want to give more information on the specifics so someone that
knows about this stuff can give a more informed decision. (although maybe
all pulsed lasers follow a specific implementation). I think the two
important things are the capacitatance, pulses per second, and operation
time.

---
Yes, nice post.

What we need to know in order to go forward is the value of the OP's
homemade capacitor, the output frequency of the neon PSU, and the
discharge time of the cap through the laser. From that we can tell
him what the charging impedance must not fall below in order to keep
from damaging his wall-wart and what his pulse repetition rate must
not exceed in order to get maximum power out of the laser.
 
J

John Fields

You're still an idiot. Hooking up a 12 VDC supply to a sign transformer
isn't going to do much of anything. Asking questions is the best teaching
method.
 
H

Homer J Simpson

Obviously you don't know how to fucking read.

Yet I do it so much better than you.
omg! you must be fucking l33t all those years of installing ceiling fans
and you call yourself an expert.

Think entire building climate control systems of the most complex kind, not
to mention all sorts of industrial experience, some with very unusual
equipment.
I guess you really think that just because you've been around for 50+
years means your an expert.

It does.
To bad you don't understand that age != experience.

Actually that's exactly what it means.
Ofcourse one might think it is correlated but usually arrogance and ego
wins out every time.

No, being able to fix equipment and systems that others have tried and
failed to service is more than enough proof. Passing the electrical exams
with the highest numbers of anyone is just a bonus. And of course I can also
write grammatical English.

I gave the OP good advice and tried to start helping him. You came in and
pissed all over the reply with your bone headed comments.
 
H

Homer J Simpson

ok, I appreciate all your concerns for safety and believe me all
precautions will be taken.
the cap size will certianly be more than 100 times more powerful
enought to kill myself but i will be doing whatever so that wont
happen. So lets not worry about that for now .

Do. Use one hand only and stand on an insulated surface - even a stack of
dry newspaper will do in a pinch. Try to use well insulated tools and think
the thing you are going to do through.
 
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.
So you want to charge up one side while discharging the other? (so you would
increase the pulse rate by a factor of 2)



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.

Caltus :)
 
---
Yes, nice post.

What we need to know in order to go forward is the value of the OP's
homemade capacitor, the output frequency of the neon PSU, and the
discharge time of the cap through the laser. From that we can tell
him what the charging impedance must not fall below in order to keep
from damaging his wall-wart and what his pulse repetition rate must
not exceed in order to get maximum power out of the laser.
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.

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.
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.

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
 
J

John Fields

I gave the OP good advice and tried to start helping him. You came in and
pissed all over the reply with your bone headed comments.

---
You're a liar.

All _you_ tried to do was to piss all over the OP in a
self-aggrandizing effort without offering anything other than
pejorative comments.

You obviously don't even have the capacity to envision the system he
was describing (as evidenced from your retort about 12V into the
primary of a transformer, when the OP clearly stated that the
_supply_ was 12VDC in and 2KVAC out) and yet you pretend to know
what you're talking about?

You're an utter disgrace to decent Canadians everywhere, eh?
 
L

Lord Garth

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.

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.
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.

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

Don't sweat the extra current capacity, the circuit will draw what it
requires. If your voltage is regulated to the correct value but your
circuit needs more than the available current, expect the voltage to
sag. Otherwise, you're good to go.
 
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