What might be different is current: in your case the current going into the A0 pin will be ridiculously small compared to the one flowing into the Arduino power connector.įar from being perfect, but still very simple to implement. Voltage is always the same along a wire or directly connected wires. That link return 403 FORBIDDEN error: I believe you must be logged in to see it. Is it any safer to lets say use a 10k resistor compared to a 1k?Īlso, in your drawing, what is that thing in between the two resistors? Is that voltage dividor? I thought that voltage division was just a principle and that if i used a wire between the two resistors, the wire would carry the voltage thats been divided into whatever? Yes, I realized just today that voltage division might be the perfect thing.ĭoes it matter what value of resistor I use? I mean if according to the formula, if R1 and R2 are equal, the voltage will divide by 2. You can now use an analog pin (A0 in the wiring I made) to measure the halved battery voltage by using analogRead and then use the map function to remap the value to the proper led bar scale. Please find the wiring attached: you use two resistors to reduce the voltage within 5V, in this case I divided the battery voltage by 2 so you will measure 3.7V when the battery is at its maximum (btw, fully charged two cells battery is 7.4V at maximum, not 8.4 as reported above) and 3.1 when fully discharged: discharging below that will damage the battery. Would such a thing work? What I am really confused about is whether the voltage is the same across both of them (7.4) or whether it splits to 3.7? Also if the voltage is the same across both of them, when the voltage drops to lets say 7, will there be 7V running through both jacks? Sorry if the wording is a bit confusing. You need another connection to the battery and a voltage divider with a resistor since Arduino is capable of measure max 5v.
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