step out source current

[LabVIEW_vikas]

I am using the NI PXI 7340 and the UMI 7764 and want to connect my 3rd party driver and a stepper motor with Opto-coupler 6N137 which need at least 100mA current for that i want to connect a 47 ohm in series with anode of 6N137 to GND.

I want to know how much current7340 can drive at the STEP out and DIR ports from the UMI 7764 are connected to my 3rd party driver.
In 7340 user manual given specification are

Stepper outputs
Maximum pulse rate ........................4 MHz (full, half, and microstep)
Minimum pulse width......................120 ns at 4 MHz
Step output mode.............................Step and direction or CW/CCW
Voltage range.................................. 0 to 5 V
Output low voltage .................. <0.6 V at 64 mA sink
Output high voltage .................Open collector with built-in
3.3 kΩ pull-up to +5 V

vikas

[Eric_R]

Hi Vikas

Its unusual for a drive to specific a current for its step and direction input. Normally these are controlled by low current digital TTL pulses that output step and direction. Can you verify for me that the step actually needs 100 ma of current, also please link the data sheet for your drive.

Thank You

Eric Reid

Applications Engineer 

[Buechsenschuetz]

It is not very unusual that a motor drive unit uses optocouplers as input devices. Drive units usually work with rather high voltages internally (24 to 48VDC) and using optocouplers is the best method to ensure safe operation of the control device (the NI systems) in case of breakdown of internal components.

But - 100mA drive current is way to high for the 6N137. The data sheet says that you should operate it with 15mA forward current maximum (10mA are recommened for safe operation), 100mA will surely destroy the optocoupler.

When calculating the series resistor connected to +5V supply you have to take into account 1) the low output voltage of the NI system, 2) the forward voltage of the LED of the optocoupler. 1) is rated at 0.4V, 2) rated at 1.5V (10mA drive current). Let's assume a total of 1.7V to be sure. This voltage has to be subtracted from the supply voltage, and the resistor is calculated for 10mA at the necessary  voltage drop across the resistor. In this case, the voltage drop is 3.3V, and a resistor of 330 ohms will result in a drive current of 10mA. 

The NI systems can handle up to 64mA in sink mode so everything is safe with 5VDC supply voltage and a resistor of 330 ohms.

Where did you get this 100mA value from? I have never seen any optocoupler requiring more than 20mA input current.

6N137 datasheet:

http://www.datasheetcatalog.net/de/datasheets_pdf/6/N/1/3/6N137.shtml 

[LabVIEW_vikas]

Dear Eric and Dear Buechsens

thanks for your valuable commenets... you both are write actually this is my mistake and that time i m unable to understand datasheet properly.

setp out terminal is sink up to 64 mA and 6N137 is drive my 15mA and i am onterface it properly and sussefully with my drive.

Onece again thanks to both of you 

 Vikas

[Buechsenschuetz]

Thank you for your feedback. Anyhow, I have to correct some basic maths errors in my posting.

The resistor is determined as follows: R = (Vs - UfLED - Uol) /  I 

Vs = supply voltage

UfLED = forward voltage of input LED

Uol = output voltage low level 

I = LED current

so 

R = (5.0V - 1.5V - 0.6V) / 10mA = 290 ohms. A 300 or 270 ohms resistor will work properly. The 330 ohms resistor I suggested earlier will also work properly but reduces LED current to 8 or 9 mA, and according to the data sheet a higher value is recommended to compensate for LED aging effects.