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NI 9203

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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

 

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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).

 

Greetings from Germany
Henrik

LV since v3.1

“ground” is a convenient fantasy

'˙˙˙˙uıɐƃɐ lɐıp puɐ °06 ǝuoɥd ɹnoʎ uɹnʇ ǝsɐǝld 'ʎɹɐuıƃɐɯı sı pǝlɐıp ǝʌɐɥ noʎ ɹǝqɯnu ǝɥʇ'


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Thanks Henrik,

 

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

9203.jpg

Regards,

Zilvinas

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Solution
Accepted by topic author Zilvis1972

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

 

here is a quick draft:

NPNsensorto20mA.png

 

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) ..

 

Greetings from Germany
Henrik

LV since v3.1

“ground” is a convenient fantasy

'˙˙˙˙uıɐƃɐ lɐıp puɐ °06 ǝuoɥd ɹnoʎ uɹnʇ ǝsɐǝld 'ʎɹɐuıƃɐɯı sı pǝlɐıp ǝʌɐɥ noʎ ɹǝqɯnu ǝɥʇ'


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Thanks Henrik,

 

Will give a go with that.

 

Cheers,

Much appreciate

 

Zilvinas

 

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