Bottlehead Forum
Bottlehead Kits => S.E.X. Kit => Topic started by: jake111 on February 01, 2015, 08:57:02 PM
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How many watts should the bleeder resistors be for the filter caps?
Thanks, Jess
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If you use the stock value, I think you could get away 1/2 Watt parts. This assumes no other modifications to the circuit. (You need to have a bleeder on each 220uF cap in the SEX circuit)
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OK, so just two bleeders then. Thanks again.
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When connecting two resistors in series to increase resistance does the watt value change too?
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You want to use Ohm's Law here.
We will approximate the SEX voltages here, for relevance.
If we put a 270K resistor across the power supply (let's just call it 450V), let's examine how much current this bleeder resistor will draw. Ohm's Law tells us V=I*R, and we have V and R, so we have:
I=V/R, or 450/270,000, which is 1.6mA. Ohm's Law tells us that power dissipated will be about 0.75W (I*I*R, V*A, or (V*V)/R will all give this). We would recommend tripling this for reliability, which gets you to 2.25W.
If we put two 270K resistors in series, we now have a 540K resistor across the power supply.
I=V/R now gives 450/540,000, which is 0.83mA. To calculate heat in each resistor, we can use V=450/2=225, R=270,000, and I=0.00083 and any of the Ohm's Law formulas for power. I get just under 0.2W, so a 0.6W resistor is sufficient here.
So, when putting two resistors in series, you can add the power they are able to dissipate, provided you are adding the same two values. This also is informative regarding why you won't see one of our 270K 1W resistors by itself across the power supply in any of our big power amps.
-PB