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[on:]

Any suggestions how to easily differentiate between 1.0 VAC and 1.5 VAC.  The current will be low (~340 mA).  I do not need to read the actual value, just tell the difference between the two readings.  Can I bring that AC voltage directly into the Ardunio?  If so, can I just collect data and then look for the MAX value? Any other suggestions?  Does anyone have any sample coding? 

I need to differentiate (48) signals, so plan to use the Mux Shield.  Anyone have any sample code for this?

Thank you & Hope you have a nice day!

[Reply #1 on:]

I'd suggest darlingtons array,  ULN2003A or may be ULN2001A with base R barely enough to trigger with higher voltage and not with lower. Other option is optocoupler, with couple resistors in front as voltage divider.

[Reply #2 on:]

TO read the AC voltage with an Arduino is going to require that you rectify it first - check your diodes carefully - they have approx 0.7 V drop. Bring your signal to the Arduino through a Diode with a Capacitor to ground to clean it up a bit and then to the Arduino. you will have approx 0.3 and 0.8 volts,

With a silicon Diode you have about 0.7 volts across the diode, Germanium about 0.3, but the silicon diodes are easier to find.

[Reply #3 on:]

I'd suggest darlingtons array,  ULN2003A or may be ULN2001A with base R barely enough to trigger with higher voltage and not with lower. Other option is optocoupler, with couple resistors in front as voltage divider.

Funny to me neither of those 'solutions' seem useful or possible to differentiate between two signals, one at 1vac and the other at 1.5vac? I don't think those ICs or any led based opto could be turned on with such low voltage levels?


 

[Reply #4 on:]

Amplitude voltage of 1 VAC is 1 x sqrt(2) = 1.41 V
and 1.5 VAC x sqrt(2) = 2.12 V
this is why I think Darlington or optocoupler would works.

[Reply #5 on:]

The current will be low (~340 mA)

1) 340mA is not low current it is quite high
2) you signal does not define the current the load does.

I would put the signals through  4051 multiplexer giving it a negitave refrence voltage, then at the output have only one rectifier diode and smoothing capacitor and feed it into an analogue input and measure.

[Reply #6 on:]

Amplitude voltage of 1 VAC is 1 x sqrt(2) = 1.41 V
and 1.5 VAC x sqrt(2) = 2.12 V
this is why I think Darlington or optocoupler would works.

Again I think it would be too low a voltage to reliably operate either kind of device. But opinions can very that's for sure.