Quote:
Originally Posted by bandersen
Right and P is also I2 X R. So if B+ is drawing 320mA, then you'll have (0.32 * 0.32) * 39 = 4 watts out of that resistor so a 5W should be OK, but maybe 7W for a little overhead. Or use a smaller 5W resistor like 33 or 27 ohms.
|
He's 20V low, and the 39 ohms is dropping that. My original calculations called for a ~13 ohm resistor. (
see post 27, page 2) Take three 39 ohm resistors, parallel them, and place them in the circuit. That'll give you the proper (lower) voltage drop, which should give you the proper B+, and plenty of overhead on the wattage. .33A X 13 = 4.29W, and you've got (3) 5 watters in parallel.