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4hv.org :: Forums :: High Voltage
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Cascade Multiplier Diode - Maximum Speed??

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si2030
Sun Jul 06 2008, 07:29AM Print
si2030 Registered Member #1571 Joined: Wed Jul 02 2008, 03:26AM
Location: Bendigo Victoria Australia
Posts: 44
Hi All,

I only found out about this site about a week ago and this is my first post (Newbie). I am not an Engineer (wish I was - this would be easier) so its taken a lot longer to get my head around this field than I would have liked.. and I am no where near where I like to be.

I have some HV diodes for a cascade multiplier.

My questions..

1) How fast can I run them?? That is what would a safe maximum frequency rating be? They have a Trr of 0.4u seconds.

2) I am ganging them together to produce a combined higher reverse voltage rating. Will this have detrimental effect on speed?

Hope someone can assist.

Regards

Simon
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Sulaiman
Sun Jul 06 2008, 09:11AM
Sulaiman Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141
1) With a Trr of 0.4 us you can operate at several hundred kHz, maybe even 1 MHz...

BUT (approximately)

the charging current is up to C x 2 x PI x F x V
which may exceed the current ratings of the diodes at high frequency.
e.g. C=1nF, F=1 MHz, V=5 kVrms ..... I=10 Arms !
even at 100 kHz the initial charging current would be around 1 A rms.

2) You will have to put a resistor across each diode to share reverse voltage
and a capacitor across each diode to share dv/dt.
OR
if the diodes are from the same manufacturing run
you could use each diode at half it's rated voltage
to ensure that no diode receives more than it's rated reverse voltage.

DO NOT forget to have a resistor from the top of the multiplier if you think there's a chance of sparks/arcs
(there always is !)
the resistor is to limit current 'spikes' that would destroy the diodes.

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si2030
Sun Jul 06 2008, 09:39AM
si2030 Registered Member #1571 Joined: Wed Jul 02 2008, 03:26AM
Location: Bendigo Victoria Australia
Posts: 44
Hi Sulaiman

Thanks for the prompt reply!

Interested in the value for the resistor or how to calculate it?

By the way the diodes are rated for 100mA.

Regards

Simon
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Firefox
Mon Jul 07 2008, 02:45AM
Firefox Registered Member #1389 Joined: Thu Mar 13 2008, 12:50AM
Location: Pittsburgh, PA
Posts: 346
The current limiting diode is simple. V/R=I. V is the ideal voltage of the multiplier, and R is the limiting resistor. Also, CW cascades are usually only good for a few milliamps without a huge amount of ripple voltage, so a few hundred kilohms at least.
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