John Stumbles wrote:
On Wed, 30 Dec 2009 23:59:58 +0000, Ian Bell wrote:

John Stumbles wrote:
On Thu, 24 Dec 2009 14:43:30 +0000, Ian Bell wrote:

I am trying to select a relay for a delayed HT switch (which will also
discharge the HT when off). Most relays I can find have contacts rated
at 250VAC which translates into a peak of about 350V. However, data is
scarce on what dc voltage these relays can switch. So far I have found
only one that gives a dc current versus voltage curve and that stops at
210V dc (and 200mA) and I really want to be able to switch up to 350V at
up to 200mA. The rest just give a dc voltage at max current value.
The reason DC switching ratings are lower than AC ratings is that, when
breaking a circuit, AC limits the duration of arcing as the current
through the contacts drops to zero twice a cycle thus helping to
extinguish the arc. With DC the contacts have to open wide enough to
extinguish the arc on their own.

Indeed, I have found a very informative application note by Tyco that
explains wvery well what happens when contacts make and when they break.

You say:

The PSU is remote and the relay has an interlock to turn off the HT if
the PSU HT output lead is disconnected. I need to either disconnect the
HT or bleed it very quickly to avoid a possible shock hazard.
I assume we have the following setup in the PSU:

relay
HT+
to -------------o COM
LOAD /
o o----------- HT+ SUPPLY
NC | NO
|
-----
| R | - discharge resistor
-----
|
GROUND -----------+---------------- GROUND
Not quite, swap the LOAD and SUPPLY and it is correct - in other words
the HT supply goes to the common and is switched either to the load or
the discharge resistor because nearly all the charge storage is in the
HT supply not the load.

Could you not arrange it as above? I mean that all the above components
are in the PSU box and the node marked 'HT+ to LOAD' is the pole on the
output connector of the PSU to which the amplifier/load is connected.

All those components will be in the power supply and the connection to
the electronics made as you state.

Thus
as soon as the relay falls back to its rest position the output lead is
connected via the discharge resistor to ground (discharging any residual
capacitance in the lead electronics) and the internal HT smoothing
capacitors within the PSU can be allowed to discharge under their own
steam, or with a small, high-value bleed resistor.

Doesn't that do what you want?


Pretty much but I was trying to avoid having a bleed resistor
permanently across the supply. The PSU has a lot of stored charge and a
bleed resistor that takes only a nominal 1% of the load current would
take about 5 minutes to discharge it. That in itself is not a problem as
the PSU is in an enclosure but when I am testing it, it is a pain to
have to wait that long.

Cheers

Ian