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Old 8th September 2002, 05:35 PM   #1
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Default Power rating of resistors?

Hi All,

Will .6 watt resistors be acceptable for biasing this regulator?
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Old 8th September 2002, 07:01 PM   #2
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Default Resistors

Hi

Well, Try this

U = R *I
P = U*I

If you are unsure of your results, ask again. This is DIY

\Jens
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Old 11th September 2002, 08:12 AM   #3
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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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