| Electronics > Beginners |
| Having issues getting an LM4040 reference working |
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| joeyjoejoe:
I'm doing some ADC, and I have an LM4040 to set the 5V reference on the arduino to something reliable. It's an LM4040BIZ-5.0, and I ordered it from DigiKey. http://www.ti.com/lit/ds/symlink/lm4040-n.pdf It's a pretty simple reference schematic that I'm trying to use, with a 460ohm resistor. This should give 2.5mA, which is more then the 1mA requirement. But I'm getting 12v at the output. One small (heh) detail, I might have put too much current through it - I'm noticing now the absolute maximum is 10/20mA. The magic smoke was never released... but I guess it's something we need to consider. Maybe I blew it... |
| floobydust:
Check your PCB pinout is correct and pin 3 is N/C or connected to pin 2 anode. 12V-5V/460R is high at over 15mA- which is the max. for the LM4040. Try the math again, it should be k ohms. Otherwise, it might be dead from an overcurrent spike. |
| joeyjoejoe:
Ahh, I see. Okay, I think it's probably safe to assume I nuked it :) |
| Zero999:
You've made an error in your calculations. I = (VIN - VREF)/R It's rated to give 5V at just 65μA, worst case, so why do you need to run it at over 15mA. How much current does the ADC need for the reference? Set the current for around 1mA and add a 100nF capacitor in parallel with the reference. R = (VIN - VREF)/I = (12 - 5)/0.001 = 7000. Use the nearest standard value. It doesn't have to be accurate, E6 20% tolerance will do, which gives 6k8. https://en.wikipedia.org/wiki/E_series_of_preferred_numbers#Lists Still, I doubt 15mA would fry it. There must be an error somewhere else. Check all the connections and the output voltage again with a 6k8 resistor. |
| joeyjoejoe:
Gave it a shot with 6k8 - no dice, still 12V. Time to find another batch of items to get from Digikey! :) |
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