NI 9203

[Zilvis1972]

Hi,

I am having a problem in connection of NPN hall effect sensor 24VDC 25mA to NI9203

 (http://ie.rs-online.com/web/p/products/7659321/?cm_mmc=IE-PPC-_-google-_-3_IE_EN_BM_A_and_C_Exact-_-...

Could not get an output in Labview graph.

Any ideas or wiring suggestions.

Cheers,

Zilvinas

[Henrik_Volkers]

You have a 20mA analog current input modul, and a NPN (open collector) digital  output sensor.

You need a power supply (5-24V) and a pull up resistor.

Power the sensor, share the COM with the 9203, connect the sensor output to a AI and add a pull up resistor of

R= U_supply/20mA ~ 1.2kOhm  

(well, missed the input burden resistor .. R= U_supply/20mA -138 😉 )

from AI to the powersupply of the sensor.

The sensor output will 'shortcut' the current to ~0mA , if you want to make it ~ 4mA add another resistor of ~20 Ohm between sensor output and AI (keep the pull up between AI and U-supply).

[Zilvis1972]

Thanks Henrik,

There are 2 resistors inside 9203 already(see picture attached). Does it required additional resistor?

Regards,

Zilvinas

[Henrik_Volkers · Accepted solution]

No, there are 8 resistors with 138Ohm, each channel has one to measure the current 😉

here is a quick draft:

R3 = 1.2k limit the current to ~20mA (~17.94mA) , this resistor dissipates ~0.4W , so it should be rated at least 0.5W , better 1W  (common is 0.25W or less with SMDs 😉 )

If you (or someone with EE knowledge in your company) only has .25W resistors on stock, you can use four 1.2k resistors (two pairs parallel and these serial) ..

[Zilvis1972]

Thanks Henrik,

Will give a go with that.

Cheers,

Much appreciate

Zilvinas