Power rating of resistors?
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diyAudio Member

Join Date: Apr 2002
Location: Belleville, IL.
Power rating of resistors?

Hi All,

Will .6 watt resistors be acceptable for biasing this regulator?
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Gavin

 8th September 2002, 07:01 PM #2 JensRasmussen   diyAudio Member     Join Date: Jul 2002 Location: Denmark - Jutland Resistors Hi Well, Try this U = R *I P = U*I If you are unsure of your results, ask again. This is DIY \Jens
 11th September 2002, 08:12 AM #3 TheFettler diyAudio Member   Join Date: Sep 2002 Location: south yorkshire, UK lets see if I've got this right: theres 2 stages to consider 1) the divider stage from the 1st reg to the second, which has 6.57 volt across it 2) the divider stage to set the 2nd reg , with 30-odd volts across it. taking them in order: with a 6.57v drop, and 2 resistors of 120R and 620R from V=IR , where V = 6.57 and R = (120+620) = 740R so I=V/R = 6.57 / 740 =8.8x10-3 = 0.0088 Amp or 8.8 mA P=I*I*R (or I^2*R) = (0.0088^2) * 740 = 0.058 Watt. and V = 30v ( assuming set up something like accurate) R = 270+5.6K+1K max - assume worst case (pot is 0R) so R = 270+5600 = 5870R min I = V/R = 30 / 5870 = 0.00511Amp or 5.11 mA P = I^2*R = (0.00511^2) * 5870 = 0.153 Watt. so your resistors need to be able to handle 0.058Watts and 0.153Watts respectively ;-) assuming I havent stuffed up somewhere..... ray

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