A string is required not for wattage but for voltage rating of
resistors.
YES, resistors do have a voltage rating independent of its dissipation.
It is best not to exceed 300 or so volts per resistor if they are 1/4 or
1/2 watt and probably not more than 500 volts per resistor for 1 and 2 watt
resistors. If you want to go beyond that check the manufacturer's specs.
73
Bill wa4lav
At 06:15 PM 3/22/02 -0800, Radio WC6W wrote:
>Hi David,
> If you want to the meter to read 1500V full scale with a 1.5M series
>R you'll need a meter with a 1 milliamp movement.
>
> The 10 R's in series is a bit of an overdo for the multiplier string,
>unless that's all you had handy. The dissipation in the R is only 1.5 W
>(max. at 1500V) though, unless you have an HV rated R, two or three 2W
>parts in series is a good solution.
>
>73 & Top of the wee hours to you,
> Marv WC6W
>
>
>On Sat, 23 Mar 2002 00:02:01 +0000 David & Wendy Dodds
><gm4wll@talisman41.freeserve.co.uk> writes:
> > Hi.
> >
> > I hope you big guns will excuse a really basic novice question?
> >
> > I've just finished building my first HT PSU. (Very proud of myself.)
> > It produces 1150/1350v at 400mA. Not big by the standards of this
> > list, but big enough for a 2c39 amp on 23cm.
> >
> > The beast is working very nicely, with one exception, which is the
> > HT voltmeter. It barely reads at all, when the Avo meter is twanging
> > off the end-stop! The PSU is based on the N6CA design and I've
> > followed his practice of hanging the meter off a string of resistors
> > attached to the rectifier output. As I am using a 100mA FSD meter I
> > calculated the multiplier resistors as needing to total 1M5, so I
> > used 10 2W 150k resistors.
> >
> > Can anyone tell me if I've fouled up the arithmetic? (we medieval
> > history graduates aren't too hot at this theoretical stuff!!)
> >
> > My thinking was as follows...
> >
> > R(multiplier) = voltage / I(fsd) - R(meter)
> >
> > therefore R(multiplier) = 1500v / 100mA - 0.7 Ohms
> >
> > therefore R(multiplier) = 1M5
> >
> > Power rating was calculated as follows....
> >
> > Power rating = I(fsd) squared x R(multiplier)
> >
> > therefore Power rating = 100mA squared x 1M5
> >
> > therefore Power rating = 1.5 W
> >
> > Am I talking nonsense here? The formulae are lifted from the RSGB
> > Rad Com handbook, but maybe I'm not using them right....
> >
> > 73 de David GM4WLL
> >
> >
> > Please reply to:
> > gm4wll@talisman41.freeserve.co.uk (Wendy)
> > gm4wll@qsl.net (David)
> >
>
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