filter capacitor in series
 

Hello
I know that this may be a dumb question but I have found conflicting
info about this on a few web pages.

If one takes say 5 capacitors each with a value of 500wv @ 150uf and
connects them in series the result wold be 2500v @ 30uf. Correct?

Now if one takes another string just like the one above, and connects
both strings across the output of the rectifier parallel to each other
does this still give the 2500v but @ 60uf? or does the voltage drop?

Thanks

Gordon

You are basically correct in your assumptions Gordon.

One thing to think about though is the balancing of voltages and currents
across the string of capacitors. This is done by putting a high value
resistor in parallel with each one of the capacitors.

The method can be seen in some of the plans for high boltage power supplies
in the ARRL handbooks. If the voltages are not equally divided among the
capacitors, one will get too much voltage and it will arc over internally.

This will effectively destroy that capacitor and then the rest of them will
go bad in rapid succesion.

You are basically correct in your assumptions Gordon.

One thing to think about though is the balancing of voltages and currents
across the string of capacitors. This is done by putting a high value
resistor in parallel with each one of the capacitors.

The method can be seen in some of the plans for high boltage power supplies
in the ARRL handbooks. If the voltages are not equally divided among the
capacitors, one will get too much voltage and it will arc over internally.

This will effectively destroy that capacitor and then the rest of them will
go bad in rapid succesion.

Gordin wrote:

Hello
I know that this may be a dumb question but I have found conflicting
info about this on a few web pages.

If one takes say 5 capacitors each with a value of 500wv @ 150uf and
connects them in series the result wold be 2500v @ 30uf. Correct?

Under ideal conditions (perfect capacitors with no leakage resistance,
etc.) the effective capacitance would be 30 uf and the voltage
capability would be 2500 volts. If the original capacitors were pre
charged to 500 volts, the series connected capacitors would produce
2500 volts. As a practical matter 5 capacitors connected in series
would require ballast resistors to equalize the leakage current, and
some voltage derating to allow for incomplete equalization.

Now if one takes another string just like the one above, and connects
both strings across the output of the rectifier parallel to each other
does this still give the 2500v but @ 60uf? or does the voltage drop?

Capacitors in parallel retain their voltage rating, and their
capacitances add.

Thanks

Gordon

--
John Popelish

Gordin wrote:

Hello
I know that this may be a dumb question but I have found conflicting
info about this on a few web pages.

If one takes say 5 capacitors each with a value of 500wv @ 150uf and
connects them in series the result wold be 2500v @ 30uf. Correct?

Under ideal conditions (perfect capacitors with no leakage resistance,
etc.) the effective capacitance would be 30 uf and the voltage
capability would be 2500 volts. If the original capacitors were pre
charged to 500 volts, the series connected capacitors would produce
2500 volts. As a practical matter 5 capacitors connected in series
would require ballast resistors to equalize the leakage current, and
some voltage derating to allow for incomplete equalization.

Now if one takes another string just like the one above, and connects
both strings across the output of the rectifier parallel to each other
does this still give the 2500v but @ 60uf? or does the voltage drop?

Capacitors in parallel retain their voltage rating, and their
capacitances add.

Thanks

Gordon

--
John Popelish

On Sun, 11 Apr 2004 12:31:00 +1200, Gordin wrote:

Hello
I know that this may be a dumb question but I have found conflicting
info about this on a few web pages.

If one takes say 5 capacitors each with a value of 500wv @ 150uf and
connects them in series the result wold be 2500v @ 30uf. Correct?

Now if one takes another string just like the one above, and connects
both strings across the output of the rectifier parallel to each other
does this still give the 2500v but @ 60uf? or does the voltage drop?

Thanks

Gordon

Gordon,

That is correct with a 'but' thrown in. If the capacitor values are
different voltage will be unevenly distributed. With the typical + or
- 20% tolerance using the string at 2500 volts is living on the edge.

It might work. If it doesn't you will find out in a few minutes. You
don't want to have your face in the circuit when it fails.

On Sun, 11 Apr 2004 12:31:00 +1200, Gordin wrote:

Hello
I know that this may be a dumb question but I have found conflicting
info about this on a few web pages.

If one takes say 5 capacitors each with a value of 500wv @ 150uf and
connects them in series the result wold be 2500v @ 30uf. Correct?

Now if one takes another string just like the one above, and connects
both strings across the output of the rectifier parallel to each other
does this still give the 2500v but @ 60uf? or does the voltage drop?

Thanks

Gordon

Gordon,

That is correct with a 'but' thrown in. If the capacitor values are
different voltage will be unevenly distributed. With the typical + or
- 20% tolerance using the string at 2500 volts is living on the edge.

It might work. If it doesn't you will find out in a few minutes. You
don't want to have your face in the circuit when it fails.

"Allan Butler" wrote in message
news:ex1ec.9943$rg5.37333@attbi_s52...

This will effectively destroy that capacitor and then the rest of them
will
go bad in rapid succesion.

And you will be among the select few who have experienced the Big Bang.

73,

"PM"

"Allan Butler" wrote in message
news:ex1ec.9943$rg5.37333@attbi_s52...

This will effectively destroy that capacitor and then the rest of them
will
go bad in rapid succesion.

And you will be among the select few who have experienced the Big Bang.

73,

"PM"

Gordin wrote:

Hello
I know that this may be a dumb question but I have found conflicting
info about this on a few web pages.

If one takes say 5 capacitors each with a value of 500wv @ 150uf and
connects them in series the result wold be 2500v @ 30uf. Correct?

Now if one takes another string just like the one above, and connects
both strings across the output of the rectifier parallel to each other
does this still give the 2500v but @ 60uf? or does the voltage drop?

Thanks


Others have correctly answered your question. I'll
add something: one should always derate the voltage
that a capacitor "sees" in a circuit from the cap's
rating. Using a 500 volt cap at 500 volts is asking
for trouble. (Or using 5 of them in series at 2500
volts.) If your circuit runs at 2500 volts, add at
least one cap to the string.

And do not forget the bleeder/equalizing resistors.
You'll need 2 resistors in series per cap. I'd use
2 51K 2 watt resistors in series across each cap, to
give a very good safety margin on the resistors' power
dissipation and voltage exposure. Others might recommend
a single 100K 2 watt resistor across each cap. But that
way, each resistor "sees" about 416 volts, and dissipates
a little over 1.7 watts. With two in series, each resistor
"sees" about half the voltage - and the power dissipation
capability is doubled.

Gordon

Gordin wrote:

Hello
I know that this may be a dumb question but I have found conflicting
info about this on a few web pages.

If one takes say 5 capacitors each with a value of 500wv @ 150uf and
connects them in series the result wold be 2500v @ 30uf. Correct?

Now if one takes another string just like the one above, and connects
both strings across the output of the rectifier parallel to each other
does this still give the 2500v but @ 60uf? or does the voltage drop?

Thanks


Others have correctly answered your question. I'll
add something: one should always derate the voltage
that a capacitor "sees" in a circuit from the cap's
rating. Using a 500 volt cap at 500 volts is asking
for trouble. (Or using 5 of them in series at 2500
volts.) If your circuit runs at 2500 volts, add at
least one cap to the string.

And do not forget the bleeder/equalizing resistors.
You'll need 2 resistors in series per cap. I'd use
2 51K 2 watt resistors in series across each cap, to
give a very good safety margin on the resistors' power
dissipation and voltage exposure. Others might recommend
a single 100K 2 watt resistor across each cap. But that
way, each resistor "sees" about 416 volts, and dissipates
a little over 1.7 watts. With two in series, each resistor
"sees" about half the voltage - and the power dissipation
capability is doubled.

Gordon

Bill Turner wrote:

On Sun, 11 Apr 2004 05:25:02 GMT, wrote:

I'd use
2 51K 2 watt resistors in series across each cap, to
give a very good safety margin on the resistors' power
dissipation and voltage exposure.

__________________________________________________ _______

Also, avoid carbon composition resistors. They are notorious for
gradually changing their resistance over time. Metal oxide film (MOF)
types are much more stable over the long run. Richard, AG6K, sells them
pretty reasonable. See http://www.somis.org/

Richard's resistors are 100K and are rated at three watts. With 416
volts applied per your example, they dissipate 1.7 watts, which should
be a sufficient safety factor. The truly paranoid may connect four of
them in a series-parallel arrangement across each capacitor.

--
Bill, W6WRT
QSLs via LoTW

Thanks! Those are good resistors - I didn't know
about the site. My paranoia extends to the caps, too.
I would use 7, myself, yielding a total of 3500 volts
or about 360 across each cap and resistor. That would
also reduce the power the resistors would dissipate to
about 1.3 watts, but the reduction in cap voltage
is more important. The last one I built (for 1800V)
used 6 450 volt caps and 12 51K 2w resistors. Wish
I had known about that site when I built it - I would
have used those resistors.

Bill Turner wrote:

On Sun, 11 Apr 2004 05:25:02 GMT, wrote:

I'd use
2 51K 2 watt resistors in series across each cap, to
give a very good safety margin on the resistors' power
dissipation and voltage exposure.

__________________________________________________ _______

Also, avoid carbon composition resistors. They are notorious for
gradually changing their resistance over time. Metal oxide film (MOF)
types are much more stable over the long run. Richard, AG6K, sells them
pretty reasonable. See http://www.somis.org/

Richard's resistors are 100K and are rated at three watts. With 416
volts applied per your example, they dissipate 1.7 watts, which should
be a sufficient safety factor. The truly paranoid may connect four of
them in a series-parallel arrangement across each capacitor.

--
Bill, W6WRT
QSLs via LoTW

Thanks! Those are good resistors - I didn't know
about the site. My paranoia extends to the caps, too.
I would use 7, myself, yielding a total of 3500 volts
or about 360 across each cap and resistor. That would
also reduce the power the resistors would dissipate to
about 1.3 watts, but the reduction in cap voltage
is more important. The last one I built (for 1800V)
used 6 450 volt caps and 12 51K 2w resistors. Wish
I had known about that site when I built it - I would
have used those resistors.

Bill Turner wrote:
On Sun, 11 Apr 2004 05:25:02 GMT, wrote:


I'd use
2 51K 2 watt resistors in series across each cap, to
give a very good safety margin on the resistors' power
dissipation and voltage exposure.


__________________________________________________ _______

Also, avoid carbon composition resistors. They are notorious for
gradually changing their resistance over time. Metal oxide film (MOF)
types are much more stable over the long run. Richard, AG6K, sells them
pretty reasonable. See http://www.somis.org/

The time proven method uses wire wound power resistors, with somewhat
lower values (higher wattage). For 450 volt caps resistors in the 50k
range are common, this will require 5 watt resistors (use 10w to be
safe). The higher bleed current serves two purposes, it discharges the
capacitors when the power is shut off more quickly, and reduces the
peak voltage reached protecting the capacitors. High voltage capacitor
strings using higher value resistors can hold a charge for several
minutes, a hazard for the careless!


Richard's resistors are 100K and are rated at three watts. With 416
volts applied per your example, they dissipate 1.7 watts, which should
be a sufficient safety factor. The truly paranoid may connect four of
them in a series-parallel arrangement across each capacitor.

--
Bill, W6WRT
QSLs via LoTW

Bill Turner wrote:
On Sun, 11 Apr 2004 05:25:02 GMT, wrote:


I'd use
2 51K 2 watt resistors in series across each cap, to
give a very good safety margin on the resistors' power
dissipation and voltage exposure.


__________________________________________________ _______

Also, avoid carbon composition resistors. They are notorious for
gradually changing their resistance over time. Metal oxide film (MOF)
types are much more stable over the long run. Richard, AG6K, sells them
pretty reasonable. See http://www.somis.org/

The time proven method uses wire wound power resistors, with somewhat
lower values (higher wattage). For 450 volt caps resistors in the 50k
range are common, this will require 5 watt resistors (use 10w to be
safe). The higher bleed current serves two purposes, it discharges the
capacitors when the power is shut off more quickly, and reduces the
peak voltage reached protecting the capacitors. High voltage capacitor
strings using higher value resistors can hold a charge for several
minutes, a hazard for the careless!


Richard's resistors are 100K and are rated at three watts. With 416
volts applied per your example, they dissipate 1.7 watts, which should
be a sufficient safety factor. The truly paranoid may connect four of
them in a series-parallel arrangement across each capacitor.

--
Bill, W6WRT
QSLs via LoTW