Originally posted by dehughes
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The drop across the diodes is about 2,4 V ( 0,8 V each, the voltage drop across a diode increases together with current because of its differential resistance ), you obviously need the resistor to drop the same voltage, and the sum of the heaters' current, if memory serves me well,is 2,1 A, so you'll need the resistor to be R = 2,4V/2,1A=1,15 Ohm.
As to the resistor's rated power, 2,4V*2,1A=5,04W, so a 5W resistor will work at its limit, thus heating considerably, so I'd rather put in a "10 Watter" for peace of mind. Since 1,15 Ohm is not a standard value, you could use two 2,2 Ohm 5 W resistors in parallel, ( BTW, this will put you a little closer to 6,3V ).
HTH
Best regards
Bob
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