so i keep burning out resistors for some reason, anyone know why? I'm using this data sheet http://www.robotshop.com/media/files/pdf/flexinol-technical-data.pdf
at 6in it should be 7.8ohms resistance on the .006diameter wire
which leads to equation:
(12v/.4a)-7.8ohms = R
which i get to be 22ohm resistor.
but when I connect it the resistor burns up pretty fast.
.4*.4*22 is 3.5 W, so you need a pretty big resistor to handle the power.
KeithRB:
.4*.4*22 is 3.5 W, so you need a pretty big resistor to handle the power.
ah, I was assuming it might be a watt thing. thank you.
Doublechecking:
12V/(7.8 +R) = 0.4
12V = (7.8+R).4
12V = 7.8.4 + R*.4
12v - (7.8*.4) = R *.4
(12-3.12)/.4 = R = 22.2
Power dissipated in resistor = IIR = .4*.4*22 = 3.5W.
I suggest a high wattage resistor, like a 5 W part.
Don’t worry, Hilukasz, it happens to all of us. I was in a meeting where I questioned whether a smallish 50 ohm termination resistor could handle the power required since it had -5 V on it. The reply was “It’s OK, it’s digital!”
Ha! Cargo cult engineering.
Concerning the problem at hand, a 3.3V supply would be about right for the SMA wire it seems, so
a DC-DC converter to 3.3V with 0.5A or more current handling would be more power efficient,
perhaps only 150mA needed from the 12V supply then.