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http://www.skillbank.co.uk/arduino/measure.htm

I am looking to build the circuit this guy recommends for making a better analog reference source. I found the voltage reference diode, and now I am looking for the proper resistor. I found a few 560Ω resistors with 1% rating. Now I am wondering what wattage they should be set at.

I think, the amperage for the 5V output of the Arduino is 500mA or so. Can anyone comment on that? Is there a best resistor watt rating you would choose for this circuit?

Are using watt ratings too far above necessary - bad? Or is that ok? Or stupid? Why?

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  • \$\begingroup\$ I want an accurate voltage reference so I can better measure ADC sources - such as temperature on a TMP36 - as shown to everyone doing Arduino projects... but never properly explained to newbies as to why its so inaccurate. Posted again after trying to work in this related question: electronics.stackexchange.com/questions/94406/… \$\endgroup\$
    – Tbarbe
    Sep 15, 2014 at 12:40

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If your situation is the same as that of the author of the article, the power dissipation of the resistor is (5-4.096)^2/560= 1.5mW. Any resistor you can buy is capable of that (the lowest I know of are rated at 20mW (0.3x0.15mm).

You can use higher rated resistors.. Any leaded or chip resistor will do. The tolerance is not important in this case, it is the LM4040 that determines the reference voltage, and it changes little with current.

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Power dissipated at the resistor is really minuscule. \$P=\frac{(V_{CC}-V_Z)^2}{R}\cong 1.5\, \mathrm{mW}\$. It will not cause significant heating even in small 0402 resistor. Using a bigger resistor will make no difference.

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