# Sine Wave (input) - Rectification?

Help!

I am trying to read in wind speed from an anampmeter, into my Mega 2560. The output from the anamometer is a con stant frequency and 2.5V peak (not peak to peak). However as the anamometer rotates it passes a read switch which forces the wave low (0V). I must count how many times the anamometer passes the read switch.

How would i count this?

I was thinking about using AC-DC rectification (4 diodes and a capacitor - but this sound like allot of effort because i must have this done for tuesday and i dont have a capacitor). After the rectification i simply use a innterupt to generate whenever there is a falling edge. But the rectified signal will be bouncy giving errornous falling edges???

Does anyone know of a way in which i can do this.

Heres is a simplfied example of the output

can you not just connect to the reed switch and forget the sinusoidal signal?

No, this is the output from a bought anamometer.To be honest i dont even know whgere the sine wave comes from? the anamometer takes a 2.5V DC input, and im testing in a lab where the parts arnt even moving.

(Davis standard anamometer/wind vane combination)

You can use a differentiate circuit to detect the sharp edge from the reed switch. But that involves passing the signal through a capacitor.

Alternatively you can read the signal from the analogue pin, track the value and when you get a difference between two successive readings that is bigger than some threshold you know you have the reed switch event.

aha
I wonder if you are looking at the output from the reed switch
when open - you're picking up noise

try connecting an ohm-meter to it and see what it has to say

I wonder if you are looking at the output from the reed switch when open - you're picking up noise

Ill try that in a bit, but can noise produce a constant frequency, and amplitude sine wave?

Alternatively you can read the signal from the analogue pin, track the value and when you get a difference between two successive readings that is bigger than some threshold you know you have the reed switch event.

I only know how to generate interrupts from a input. I have other stuff going on in my program so i cannot be evaluating only every sweep through of the code as i might miss a count.

You can use a differentiate circuit to detect the sharp edge from the reed switch. But that involves passing the signal through a capacitor.

I think i will have to try this. Do you know of what value capacitor i should use?

if it's 50Hz or 60Hz (depending on your location) it could just be mains hum

try connecting an ohm-meter to it and see what it has to say

Its has a 515 ohm, but how will this make a difference?

if it's 50Hz or 60Hz (depending on your location) it could just be mains hum

It looks to be approximatly 50Hz. Im not sure what mains hum is, how can i remove it. (I assume it has something to do with the power sorce, available at my country. Im from south africa so 50Hz would be right- i think. But it has no external power input other than directly from the arduino's vcc port)

if it is mains hum, then a simple capacitor across the contacts should reduce it
mains hum is just induced current in long wires that are aywhere near the mains

I had expected you to report 0 ohms (short circuit) when the reed is closed, and very high when open

The sort of circuit I imaged for you would be:
a) connect one side to ground
b) connect the other side to +5 volts (or whatever your 'duino is using), via say 10k resistor
c) connect any pin to the resistor (obviously not the end that's connected to 5 volts

as your anemometer turns I would expect to see HIGH, changing to LOW as the reed closes

I'm assuming that there is no power to the device - purely passive?

But it has no external power input other than directly from the arduino's vcc port)

Yes it sounds like just mains pickup to me.
It is probably a very high impedance and will disappear with a small load.
I would use a 10K resistor from the input to ground and a 1K resistor from the input to +5V.

via say 10k resistor

Are you suggesting a pull down resistor?

I had expected you to report 0 ohms (short circuit) when the reed is closed, and very high when open

0 ohms closed , 515 ohms open;

The sort of circuit I imaged for you would be:

The anamometer has these 4 conductors:
Supply voltage on the yellow conductor. - Arduino 5V
Analog ground is on the red conductor. - Arduino ground
Speed signal is on the black conductor. - this is the output im plugging to my Arduino input port, i have plugged it in directly but im going to add a resistor now and get back to you.
Direction signal is on the green conductor. - dont worry about direction that works

if it is mains hum, then a simple capacitor across the contacts should reduce it mains hum is just induced current in long wires that are aywhere near the mains

This sounds interesting. I really think this could be it, since the cable from the anamometer is about 15m long and its wound in a coil at the moment sitting next to my arduino. You say a capacitor across the contacts, which two contacts?

Yes it sounds like just mains pickup to me.
It is probably a very high impedance and will disappear with a small load.
I would use a 10K resistor from the input to ground and a 1K resistor from the input to +5V.

So can i just connect a 10kohm between the black connection to the input port and a 1kohm between supply 5V and the red conductor and this should remove the sine wave? ill try this thanks.

Ok i connected the resistors as instructed. When i run my interrupt code it never generates any interuupts. So i checked it using this analogue read in example:

``````void setup() {
Serial.begin(9600);
}

void loop() {
Serial.println(sensorValue, DEC);
}
``````

here is the results when the anamometer is held fast in a position not closing the reed switch:

53
1023
0
0
0
44
1023
0
0
0
24
1023
0
0
0
13
1023
51
0
0
98
1023
0
0
0
85
1023
0
0
0
66
1023
0
0
0
50
1023
0
0

and more of the same.

When the reed switch is held closed i get a continous stream of zeros.

What is going on!!!

0 is obviously closed
not sure why you don't just get high numbers when open
that's what the resistor is meant to do, pull the line high
did you try Grumpy_Mike's suggestion - two resistors?

not sure why your interrupt code doesn't work
a) how have you wired it
b) show us the code

oh and do you have a link to the anemometer?
might tell us what we should be doing!

Here is the official datasheet

http://www.davisnet.com/product_documents/weather/spec_sheets/6410_Spec_Rev_E.pdf

I connected the output to a occilascope again and with the resistors added it is no longer a sine wave but now a square wave. I dont know if this means anything.

What about the capacitor? what value one should i use and how should i connect it.

I have a 10k resistor between the black conductor and the A0 port. I have a 1k resistor between the Arduino 5V and the yellow conductor. The red is connected directly to ground and the green hangs free atm.

Please look at previous post for color code.

square wave is good
do you get one pulse per revolution?

no, if i got 1 pulse per revolution it would be easy =(

Its a continous square wave with constant frequency and amplitude untill the reed connects and it drops to a flat zero. Like my picture in the first post with a square wave instead of a sine wave. (one flat 0V per revolution)

I cant even count the rising edges of a signal with no reed switch closes and subtract it from the signal with reed switch closes because the reed switch may be closed for varying ammounts of time.

Should i not just construct a rectifier? will a single capacitor be enough? I have four diodes but im not too clued up on that kind of thing. can i use any diodes?