I'm using the arduino to measure [45~60] volt DC. I made voltage devider with capacitor parallel to R2
R1= 20k, R2= 1k, C=100uf 16v
so this theoretically should able the arduino to measure up to 100 volt (equal to 5 volt on the analog input).
but what is happening that the read values are unstable and goes up and down about 1 volt randomly.
I think the problem that the arduino doesn't draw enough current to measure the value.
is there a specific and definite current should i give to arduino to works right ?
or there is another problem ?
please note that I measured [20~30] volt perfectly with voltage devider
R1=10k R2=1k C=100uf 16v.
What is the purpose of the capacitor and what have you got the reference voltage set to ?
As to the current, then why not measure it ?
Don't use that capacitor.
It is possible to use a capacitor, but use a small 100nF. The 100nF capacitors don't leak.
But don't use a capacitor just yet. Make it work first.
For safety, it is better to use higher values of the resistors.
Rule of thumb: Keep the circuit impedance at the analog input maximum 10kΩ.
For example: R1 = 20k, R2 = 20k, is okay, the combined impedance is 10kΩ.
However, if both R1 and R2 are 100k, it will still work, you will notice almost no difference.
I suggest: R1 = 220k, R2 = 10k
For better accuracy, you can use an internal reference voltage. You have to change the voltage divider of course.
Suppose you choose the internal voltage of 1.1V of a Arduino Uno.
Then: R1 = 1M, R2 = 10k
To answer your question: The Arduino does not draw any current with its analog input. The ADC requires a little charge, but if the circuit impedance is 10k or lower, then the 10 bits are still accurate.
The 10 bit settling time of your enormous 100µF capacitor will be 0.7s, so unless you are waiting nearly a second between analog reads you will simply be seeing this cap in the process of charging.
100nF is plenty, in fact 10nF is probably enough. And I second the advice about using higher values of resistance - 10k in the lower part of the divider is a good value to choose. Then you'll only drain 0.5mA from the high voltage you are measuring rather than 5mA as you are now. The existing 20k resistor will be getting quite hot you'll find.
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