How many watts does each resistor dissipate?
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How many watts does each resistor dissipate?
Completely dependent on the input voltage.
P = I*R = V^2/R
And blue resistors? Do you have the 3.3v version?
Maybe! That would make sense, since right now the divider isn't making any sense at all.Quote:
Originally Posted by 2k1Toaster
I just looked at the Mp3Car store image. The board is different from what I received.
What I received looks like the following, with big blue resistors in the circled locations which I'm assuming is not the default configuration. Especially since moving the jumper barely makes any difference and no matter what voltage I feed (between 12 and 14 volts) it shows roughly 4.34 - 4.44 on the 5v output side and I couldn't figure out why the range was so small.
http://www.mp3car.com/vbulletin/imag...0407b4d0d0.jpg
Yup, you have the 3.3v version... Not sure how that happened. Just clip the resistors and that should help.
Yeah weird, something must have ended up in the wrong box. It doesn't even look like the Mp3Car store sells the 3.3 version though. Hmmm. Oh well, I'll just clip the resistors.Quote:
Originally Posted by 2k1Toaster
We do not sell the 3.3v versions to anyone but a very specific couple of clients who they were special made for. That is why this is somewhat confusing.Quote:
Originally Posted by PaulF
Okay I clipped the resistors, much better as far as heat goes haha
Now whats the official equation to get the voltage for 15 volt mode. Testing with a multimeter, I came up with
volts = sensor * 2.96 + 9.63
It seems that the accuracy is good to within about +/-.1 volts. Is there a better equation?
That accuracy is actually worse than I thought. It's way off for higher voltages like 13.5+
So I'm going to assume there is a much better equation which probably isn't y=mx+b
http://a.imageshack.us/img208/3199/opto12v.jpg
y = 3E‐06x4 ‐ 0.0004x3 + 0.0183x2 ‐ 0.1488x + 0.2318
y = What the FB reads
x = Input voltage
This is for the 12v jumper setting