From: Winfield Hill
Subject: Re: Simple adjustable 0 - 400V regulator, with flaw
Date: 18 Jan 2003 11:20:32 -0800
Organization: Rowland Institute
X-Newsreader: Direct Read News 4.20
> Winfield Hill wrote...
>> Thomas wrote...
>>> Winfield Hill wrote:
>>>>. +480V ___ HIP5600
>>>>. ---| |---+---+-----------(O) HV out
>>>>. |___| | |
>>>>. | | 1.5k
>>>>. | | | adjust, 0 - 400V
>>>>. '---- | --+----o->o--------,
>>>>. | o--, |
>>>>. 10uF === 300V | 500k
>>>>. 450V | fixed | pot
>>>>. | 380k |
>>>> ... As a design quiz, tell us what the problem is, and
>>>> find a solution that doesn't significantly affect any
>>>> of the circuit's functionality. ...
>>> If the switch is flipped, and the pot is in zero position, the
>>> 1.5 K resistor gets to see 300 volts, maybe even 470+ volts if
>>> the switch is open for a moment.
>>> That's 60 watts, and nearly half a joule. Probably too much for
>>> whatever little R is there.
>> Actually, the 1500-ohm resistor has never minded the short power
>> surge. Amazingly, neither did the '5600. However a 5.1V zener
>> was placed across this resistor in a design refinement; I should
>> have shown it.
>> > And the 200 mA will be too much for the pot...
>> BINGO, that's part of the answer. If the pot is turned down
>> when the switch is thrown, a small part of the pot's resistance
>> path has to absorb all the 10uF capacitor's energy. Bamm!
>> OK, what's the solution?
> A cap across the pot?
Nope, see other post, keep thinking.
> BTW, how many pot (hehe) did you smoke before you found out?
Hehe yourself! Only one, it's a good practise to figure out why
a part failed, if possible, rather than simply replacing it.
As it happens, I had specified a 2W pot, but my technician had
installed a 0.5W unit and I didn't notice when I first tested
the (one-off prototype) instrument. However, I'm not sure a
2W pot couldn't be made to fail at the right setting. Consider,
most settings are OK. Set high and the current flow isn'd too
severe. Set very low (right at zero) and the 1.5k takes up the
energy instead. So there's a setting range, probably near 5 to
15V, where only 1 to 4% of the pot's resistance pathway gets
most of the energy. BAMMO!
Well, I've given enough hints here to reveal one of the possible
solutions (the poor one).