50KV multiplier from 5KV neon sign transformer

S

sinebar

Guest
I need to make a multiplier to take the 5KV from my neon sign
transformer to 50KV. Can someone tell me what voltage rating the
diodes and capacitors should have? I know they don't have to have a
rating of 50KV but what would be a safe rating? It's been a while
since I've made one of these things. BTW my neon sign transformer is
a 15KV 30ma but I don't want to take it that high. I want to variac
it to 5KV and then use the voltage multiplier.
 
sinebar wrote:

I need to make a multiplier to take the 5KV from my neon sign
transformer to 50KV. Can someone tell me what voltage rating the
diodes and capacitors should have? I know they don't have to have a
rating of 50KV but what would be a safe rating? It's been a while
since I've made one of these things. BTW my neon sign transformer is
a 15KV 30ma but I don't want to take it that high. I want to variac
it to 5KV and then use the voltage multiplier.
They need to handle the 5kV. Because on a multiplier, each stage is
Click to expand...
getting bummped with an offset of 5KV per diode.
 
In article <Ny5Zo.66683$sx4.23544@newsfe02.iad>, Jamie wrote:
sinebar wrote:

I need to make a multiplier to take the 5KV from my neon sign
transformer to 50KV. Can someone tell me what voltage rating the
diodes and capacitors should have? I know they don't have to have a
rating of 50KV but what would be a safe rating? It's been a while
since I've made one of these things. BTW my neon sign transformer is
a 15KV 30ma but I don't want to take it that high. I want to variac
it to 5KV and then use the voltage multiplier.
They need to handle the 5kV. Because on a multiplier, each stage is
getting bummped with an offset of 5KV per diode.
Click to expand...
Multiply by both factors of 2 and square root of 2, plus safety factor.

All diodes and most capacitors involved will by-design experience
voltage of transformer output RMS voltage times twice the square root
of 2.

That means 14.2 KV, 15 KV rating with awfuly small room for error...

Unless yu want to "Variac"-down this arrangement to 5 KV peak rather
than RMS (3.5 KV RMS, resulting from 28V input to a 15 KV transformer),
in which case you are mostly pushing 100% of the ratings of 10KV
voltage-rated parts.
--
- Don Klipstein (don@misty.com)
 
The microwave ovens also use a voltage dividing resitor across each diode to
insure voltage division, evenly.

None of the diodes will be rated high enough for the total voltage and if a
few leaks the last ones will get zapped without some voltage division
insurance. I believe a string of about 10megOhm resistors was used in the
last one I saw.



"Don Klipstein" <don@manx.misty.com> wrote in message
news:slrnija9tu.e7m.don@manx.misty.com...
Multiply by both factors of 2 and square root of 2, plus safety factor.

All diodes and most capacitors involved will by-design experience
voltage of transformer output RMS voltage times twice the square root
of 2.

That means 14.2 KV, 15 KV rating with awfuly small room for error...

Unless yu want to "Variac"-down this arrangement to 5 KV peak rather
than RMS (3.5 KV RMS, resulting from 28V input to a 15 KV transformer),
in which case you are mostly pushing 100% of the ratings of 10KV
voltage-rated parts



In article <Ny5Zo.66683$sx4.23544@newsfe02.iad>, Jamie wrote:
sinebar wrote:
I need to make a multiplier to take the 5KV from my neon sign
transformer to 50KV. Can someone tell me what voltage rating the
diodes and capacitors should have? I know they don't have to have a
rating of 50KV but what would be a safe rating? It's been a while
since I've made one of these things. BTW my neon sign transformer is
a 15KV 30ma but I don't want to take it that high. I want to variac
it to 5KV and then use the voltage multiplier.
They need to handle the 5kV. Because on a multiplier, each stage is
getting bummped with an offset of 5KV per diode.
 
On Jan 18, 11:20 pm, "Josepi" <J...@easynews.calm> wrote:
The microwave ovens also use a voltage dividing resitor across each diode to
insure voltage division, evenly.

None of the diodes will be rated high enough for the total voltage and if a
few leaks the last ones will get zapped without some voltage division
insurance. I believe a string of about 10megOhm resistors was used in the
last one I saw.

"Don Klipstein" <d...@manx.misty.com> wrote in message

news:slrnija9tu.e7m.don@manx.misty.com...
  Multiply by both factors of 2 and square root of 2, plus safety factor.

  All diodes and most capacitors involved will by-design experience
voltage of transformer output RMS voltage times twice the square root
of 2.

  That means 14.2 KV, 15 KV rating with awfuly small room for error...

  Unless yu want to "Variac"-down this arrangement to 5 KV peak rather
than RMS (3.5 KV RMS, resulting from 28V input to a 15 KV transformer),
in which case you are mostly pushing 100% of the ratings of 10KV
voltage-rated parts

In article <Ny5Zo.66683$sx4.23...@newsfe02.iad>, Jamie wrote:
sinebar wrote:

I need to make a multiplier to take the 5KV from my neon sign
transformer to 50KV.  Can someone tell me what voltage rating the
diodes and capacitors should have?  I know they don't have to have a
rating of 50KV but what would be a safe rating?  It's been a while
since I've made one of these things.  BTW my neon sign transformer is
a 15KV 30ma but I don't want to take it that high.  I want to variac
it to 5KV and then use the voltage multiplier.
They need to handle the 5kV. Because on a multiplier, each stage is
getting bummped with an offset of 5KV per diode.
Click to expand...
Well I did some investigating and came up this design: I will use 6
20KV
doorknob capacitors and 6 HV03-12 12KV PIV High Voltage Diodes in a
typical voltage multiplier configuration. With 6KV input I should get
about
51KV according to Instructables.com. Here's their calculations on
the website

To calculate the expected voltage at the output with a given input and
number of stages, plug the numbers into this formula:
Eout = (2 x Ein) x S x 1.414
Eout is the Output Voltage, Ein is the Input Voltage, and S is the
number of stages in your design. I used a 6,000 VAC Oil Burner
Transformer for my input and built 3
stages.
Eout = (2 x 6000) x 3 x 1.414
Eout = 12000 x 3 x 1.414
Eout = 36000 x 1.414
Eout = 50,904 volts

please give opinions and advice.
 
On Jan 19, 6:08 pm, sinebar <sinebar3...@gmail.com> wrote:
On Jan 18, 11:20 pm, "Josepi" <J...@easynews.calm> wrote:





The microwave ovens also use a voltage dividing resitor across each diode to
insure voltage division, evenly.

None of the diodes will be rated high enough for the total voltage and if a
few leaks the last ones will get zapped without some voltage division
insurance. I believe a string of about 10megOhm resistors was used in the
last one I saw.

"Don Klipstein" <d...@manx.misty.com> wrote in message

news:slrnija9tu.e7m.don@manx.misty.com...
  Multiply by both factors of 2 and square root of 2, plus safety factor.

  All diodes and most capacitors involved will by-design experience
voltage of transformer output RMS voltage times twice the square root
of 2.

  That means 14.2 KV, 15 KV rating with awfuly small room for error....

  Unless yu want to "Variac"-down this arrangement to 5 KV peak rather
than RMS (3.5 KV RMS, resulting from 28V input to a 15 KV transformer),
in which case you are mostly pushing 100% of the ratings of 10KV
voltage-rated parts

In article <Ny5Zo.66683$sx4.23...@newsfe02.iad>, Jamie wrote:
sinebar wrote:

I need to make a multiplier to take the 5KV from my neon sign
transformer to 50KV.  Can someone tell me what voltage rating the
diodes and capacitors should have?  I know they don't have to have a
rating of 50KV but what would be a safe rating?  It's been a while
since I've made one of these things.  BTW my neon sign transformer is
a 15KV 30ma but I don't want to take it that high.  I want to variac
it to 5KV and then use the voltage multiplier.
They need to handle the 5kV. Because on a multiplier, each stage is
getting bummped with an offset of 5KV per diode.

Well I did some investigating and came up this design:  I will use 6
20KV
doorknob capacitors and 6 HV03-12 12KV PIV High Voltage Diodes in a
typical voltage multiplier configuration.  With 6KV input I should get
about
51KV according to Instructables.com.   Here's their calculations on
the website

To calculate the expected voltage at the output with a given input and
number of stages, plug the numbers into this formula:
Eout = (2 x Ein) x S x 1.414
Eout is the Output Voltage, Ein is the Input Voltage, and S is the
number of stages in your design. I used a 6,000 VAC Oil Burner
Transformer for my input and built 3
stages.
Eout = (2 x 6000) x 3 x 1.414
Eout = 12000 x 3 x 1.414
Eout = 36000 x 1.414
Eout = 50,904 volts

please give opinions and advice.- Hide quoted text -

- Show quoted text -
Click to expand...
What value capacitors in pf or uf? At 60 Hz, the capacitors have to
be pretty large in value if you are expecting to draw any current at
the 50 kV level. What are you going to do with it except shock the
life out of whatever touches it?
 
On Jan 20, 4:35 pm, "hr(bob) hofm...@att.net" <hrhofm...@att.net>
wrote:
On Jan 19, 6:08 pm, sinebar <sinebar3...@gmail.com> wrote:





On Jan 18, 11:20 pm, "Josepi" <J...@easynews.calm> wrote:

The microwave ovens also use a voltage dividing resitor across each diode to
insure voltage division, evenly.

None of the diodes will be rated high enough for the total voltage and if a
few leaks the last ones will get zapped without some voltage division
insurance. I believe a string of about 10megOhm resistors was used in the
last one I saw.

"Don Klipstein" <d...@manx.misty.com> wrote in message

news:slrnija9tu.e7m.don@manx.misty.com...
  Multiply by both factors of 2 and square root of 2, plus safety factor.

  All diodes and most capacitors involved will by-design experience
voltage of transformer output RMS voltage times twice the square root
of 2.

  That means 14.2 KV, 15 KV rating with awfuly small room for error....

  Unless yu want to "Variac"-down this arrangement to 5 KV peak rather
than RMS (3.5 KV RMS, resulting from 28V input to a 15 KV transformer),
in which case you are mostly pushing 100% of the ratings of 10KV
voltage-rated parts

In article <Ny5Zo.66683$sx4.23...@newsfe02.iad>, Jamie wrote:
sinebar wrote:

I need to make a multiplier to take the 5KV from my neon sign
transformer to 50KV.  Can someone tell me what voltage rating the
diodes and capacitors should have?  I know they don't have to have a
rating of 50KV but what would be a safe rating?  It's been a while
since I've made one of these things.  BTW my neon sign transformer is
a 15KV 30ma but I don't want to take it that high.  I want to variac
it to 5KV and then use the voltage multiplier.
They need to handle the 5kV. Because on a multiplier, each stage is
getting bummped with an offset of 5KV per diode.

Well I did some investigating and came up this design:  I will use 6
20KV
doorknob capacitors and 6 HV03-12 12KV PIV High Voltage Diodes in a
typical voltage multiplier configuration.  With 6KV input I should get
about
51KV according to Instructables.com.   Here's their calculations on
the website

To calculate the expected voltage at the output with a given input and
number of stages, plug the numbers into this formula:
Eout = (2 x Ein) x S x 1.414
Eout is the Output Voltage, Ein is the Input Voltage, and S is the
number of stages in your design. I used a 6,000 VAC Oil Burner
Transformer for my input and built 3
stages.
Eout = (2 x 6000) x 3 x 1.414
Eout = 12000 x 3 x 1.414
Eout = 36000 x 1.414
Eout = 50,904 volts

please give opinions and advice.- Hide quoted text -

- Show quoted text -

What value capacitors in pf or uf?  At 60 Hz, the capacitors have to
be pretty large in value if you are expecting to draw any current at
the 50 kV level.  What are you going to do with it except shock the
life out of whatever touches it?- Hide quoted text -

- Show quoted text -
Click to expand...
The caps are 470 pf. I'm going to use the multiplier to charge a 50KV
10 nF capacitor
for a flashtube pumped dye laser.
 
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