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Here is all the circuitry I have.

enter image description here

I only found thisdata sheet which may be for this anemometer: http://www.synchrodan.hu/letoltes/kriwan/kulsoerzekelok/szel/N292_INT10K_69000248_e.pdf

What I don't get is is it pulse output or current output? There is current source sign and a pulse sign for the same output. I'm confused.

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It will output a pulsed current with 4mA low current and 18mA high current (+- 1mA).

The frequency will determine the velocity as can be seen in the graph in the datasheet:

frequency and velocity

The high pulse will have a duration of 100µs +-20µs, so it's more like a PWM output than a rectangular frequency.

Edit:

To answer some of the follow up questions in the comments:

To read the signals you would probably use a resistor in the output to convert the current output to a voltage output, which can then be read by some input hardware. The total resistance is limited by the sensor and is given as 600\$\Omega\$, which includes the wire resistance.

Why would someone use a current pulse output?

Because an anemometer is typically placed in a remote location the wire length to the sensor can introduce a significant resistance. If you would use a voltage output, the resistance will distort the signals amplitude, which might be okay for a pulse output. But a current output is also more robust against capacitive coupling.

How can you get 0V to 8V for your pulse input?

A simple solution might be this (I'm not sure this will actually work, there were some glitches in the simulation, maybe someone can comment on that):

schematic

simulate this circuit – Schematic created using CircuitLab

Simulation result

But you should be aware, that the shunt resistor is quite close to the maximum allowed resistance in the output of the sensor. You also loose the information if the sensor is dead as you won't have the live zero signal present behind the capacitor.

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  • \$\begingroup\$ I dont understand I have a DAQ which can only read analog voltage or count pulses. How should I take the output. You say pwm, do you mean duty cycle is changing? \$\endgroup\$ – user16307 Feb 18 '15 at 21:35
  • \$\begingroup\$ Well if you can only read voltages, place a resistor in the output to convert the current into an according voltage. You can place up to 600\$\Omega\$ in the output, which would result in a 2.4V low signal and 10.8V high signal. \$\endgroup\$ – Arsenal Feb 18 '15 at 21:43
  • \$\begingroup\$ As for the PWM part, the pulse has a fixed width, but the frequency at which the pulses occur changes, so the off time varies. You will have a higher duty cycle at higher frequencies. \$\endgroup\$ – Arsenal Feb 18 '15 at 21:45
  • \$\begingroup\$ thanks is it possible to obtain the pulses as OV to 8V for instance? I ask because thats how pulses detected with my system. \$\endgroup\$ – user16307 Feb 18 '15 at 21:45
  • \$\begingroup\$ btw I never saw a current pulse output before why would on earth people would use it? voltage pulse output is used in general. \$\endgroup\$ – user16307 Feb 18 '15 at 21:47
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It is current pulse output. Look at the "signal output" section. The current has a form of pulses.

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  • \$\begingroup\$ How can I read the ouput? \$\endgroup\$ – user16307 Feb 18 '15 at 21:37
  • \$\begingroup\$ I guess some shunt resistor with voltage measurement on it will do.. \$\endgroup\$ – Eugene Sh. Feb 18 '15 at 21:38

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