Some cheap multimeters have 1 Megohm input impedance. Also right now I only have a 10uF capacitor I took of a broken device. I'd like to be able to read voltage quickly and frequently. That will slow down the response substantially, but that is fine for infrequent voltage measurements. The analog input has very high DC input resistance (>100 Megohm), but it also has a capacitor that must be charged.įor your high impedance voltage divider, you must add a 100 nF capacitor from the divider output to ground, in order to reduce the effective impedance of the voltage source. The article you read is giving bad advice. But I only bought 820k and 270k resistors, I can buy others, but as of now I still don't know what's going on with the wrong reads, and it would be better if I could use these I already have. That seems to me also a good combination of resistors. So I thought it was a good number, as it would just draw a few microamps. It says multimeters usually have 10MOhms impedance, and then gives an example with about 1MOhm impedance. So when I read that article, I thought the bigger resistance, the better. I wanted to interfere as little as possible with the circuit that is being read. Why such high values for your divider? Try R1 = 22k and R2 = 6.8k. a 1k2 and a 22k resistor combination will do fine for your 20V input. You have to calibrate this reference voltage once (it has a 10% tolerance) but it's stable, usually far more stable than your 5V supply, and independent from the supply voltage. The recommended input impedance for the ADC is <10k, you have over 200k (calculate the parallel resistance of the two resistors).įor overall more accurate measurements: bring down your input voltage to <1.1V, and use the internal reference. The better method is of course to lower the resistor values. The first reading sets the MUX to that input, and as it takes a moment to charge the internal cap the reading is bound to be off (a bit too high, likely, as the voltage has to drop) Then the second reading will be much more accurate as the ADC has had time to settle. Nonetheless it should work without the external capacitor, but you have to read A1 twice. 1-2 nF should be more than enough, as it's about two orders of magnitude larger than the internal one. I would use a much smaller capacitor, no need for 100 nF. I also tried including the real measured values of the resistors in the online calculator, but it doesn't seem to make much a difference. I can't see why it's reading 1.43V instead of the expected 1.26V. The resistance will decrease, the current increases, and voltage decreases. It's like connecting a big resistor in parallel with the voltage divider resistor. I discovered that there is a "special" effect when reading these voltages with my multimeter: as the multimeter at 20V range doesn't have a much different impedance than the voltage divider (while the voltage divider has 1.09Mohm, the multimeter seems to have 1Mohm, or at least a few Mohms), when I try to read the voltage across the 270kOhm resistor using the multimeter, it influences the circuit so that the serial readings start to be 1.30V at A1 pin, and the multimeter reads 1.09V.īut it seems to me that if this influence of the analog input high impedance was the problem, then the read voltage should be lower than expected, not higher.
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