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Re: Bleed Resistor for Homemade/Large Caps - THE FULL DESIGN NOTE S



Original poster: "Jim Lux by way of Terry Fritz <twftesla-at-qwest-dot-net>" <jimlux-at-earthlink-dot-net>

At 12:37 PM 10/29/2002 -0700, you wrote:
>Original poster: "Jonathan Peakall by way of Terry Fritz 
><twftesla-at-qwest-dot-net>" <jpeakall-at-madlabs.info>
>
>In series, the resistors current rating will be the value of the lowest
>resistor in the series and the resistance values add up. In parallel, the
>current rating is added, and the resistance value added. So in this case,
>each resistor must be rated for 5 watts.


Uhhh... but, the power dissipated in any resistor in a series is I^2 R... 
the power dissipation is proportional to the resistance.  If a given 
resistor is 10% of the total string resistance, then it will dissipate 10% 
of the total power.

143 Meg -at- 5 Watts... Say you string up 14 10Meg 2W resistors... the string 
is 140Meg, 28W.. If the entire string is dissipating 5 Watts, each resistor 
0.36 watts, which is actually a nice margin.. (you don't really want to run 
your resistors at their rated dissipation, do you? )....  An advantage of 
2W resistors in this sort of application is that their surface length is 
greater, so the flashover potential is less.

> >
> > 143Mohm -at- 5 watts...
> >
> > Now how does one make such a resistor, since
> > I'm sure that one of that specification does
> > not exist in a single package...
> >
> > Would it be safe to series fifteen 10M resistors
> > rated for 1/2 watt each? Or would each of the
> > resistors have to be rated -at- 5 watts?