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|8th September 2002, 06:35 PM||#1|
Join Date: Apr 2002
Location: Belleville, IL.
Power rating of resistors?
Will .6 watt resistors be acceptable for biasing this regulator?
|8th September 2002, 08:01 PM||#2|
Well, Try this
U = R *I
P = U*I
If you are unsure of your results, ask again. This is DIY
|11th September 2002, 09:12 AM||#3|
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.
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.....
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