i trying read square wave takes on range of 50hz 150hz. using oscope found signal high @ 6v , low @ 0v. found out hard way arduino pin 8 can handle signal 10 seconds before pin fried. (should have measured first...)
anyway, how can lower signal appropriate level without changing signal frequency or muddying up?
i thinking simple voltage divider, should use high or low value resistors? example, should use 1-ohm , 5-ohm or 100kohm , 500kohm (math may off on this...) or in between or matter?
application details:
the car sensor outputs 6v square wave. ecu of car reads directly. making display reads hz signal (by splicing it) , converts etc... display numerical value based on signal frequency. imperative original signal unmolested unreliable signal cause car run poorly @ best, lean out , fry cylinder @ worst. arduino can reliably read 5v, lower signal voltage.
the ecu has 10k pull resistor believe on input. can tolerate 1-2% change in signal, prefer not to.
anyway, how can lower signal appropriate level without changing signal frequency or muddying up?
i thinking simple voltage divider, should use high or low value resistors? example, should use 1-ohm , 5-ohm or 100kohm , 500kohm (math may off on this...) or in between or matter?
application details:
the car sensor outputs 6v square wave. ecu of car reads directly. making display reads hz signal (by splicing it) , converts etc... display numerical value based on signal frequency. imperative original signal unmolested unreliable signal cause car run poorly @ best, lean out , fry cylinder @ worst. arduino can reliably read 5v, lower signal voltage.
the ecu has 10k pull resistor believe on input. can tolerate 1-2% change in signal, prefer not to.
yes, simple voltage divider
vout = vin * r2/(r1 + r2).
5v = 6v*5k/(r1+5k)
5vr1 + 5v5k = 6v5k
5vr1 = 6v5k - 5v5k
r1 = (6v5k - 5v5k)/5v = (30000 - 25000)/5 = 1000 or 1k.
load on 6v line = 6v/6000 = 1ma
or, use 6v drive logic level n-channel mosfet connected digital input 1k pullup resistor. 6v goes high, digital pin goes low, 6v goes low, digital pin goes high (so, inverter).
load on 6v line = uas, next nothing.
vout = vin * r2/(r1 + r2).
5v = 6v*5k/(r1+5k)
5vr1 + 5v5k = 6v5k
5vr1 = 6v5k - 5v5k
r1 = (6v5k - 5v5k)/5v = (30000 - 25000)/5 = 1000 or 1k.
load on 6v line = 6v/6000 = 1ma
or, use 6v drive logic level n-channel mosfet connected digital input 1k pullup resistor. 6v goes high, digital pin goes low, 6v goes low, digital pin goes high (so, inverter).
load on 6v line = uas, next nothing.
Arduino Forum > Using Arduino > General Electronics > Reducing the voltage of a low frequency signal
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