Analog output on NI9264 for variable voltage control

[ThomasCIGITI]

Hello everyone,

I am a bit rusty hooking up my electronics but I was wondering if it is possible to use the analog output from the NI9264 to control a variable DC device. I have a valve which is controlled by varying the voltage from 1.5 to 3V where it can draw up to 1.8A of current. 

I am trying to hook up the AO pin directly to the ADJ pin of a LM317 where the Vin is 10V and then controlling the AO thru Labview. 

With this direction connection, I don't seem to get any response from the LM317. For testing, I tried the standard setup where I connect a 200Ohm resistor between the Vout and ADJ. With this setup (the value is not hooked up yet), I seem to be able to re-adjust the voltage. Would this setup be safe to use with the valve - I am afraid of 1.5A of current running back to the NI9264.

Thanks in advance!

Thomas

[crossrulz]

Well, let's look at the spec for the 9264.  Each output can handle 4mA.  The spec for the Adjust line of the LM317 states 100uA.  So you should be able to drive this.

Make sure your grounds are tied together.  Did you measure your AO voltage with the LM317 connected?

[johnsold]

Thomas,

The LM317 has a minimum load current requirement of 3.5 mA typical and 10 mA maximum. In the typical two-resistor configuration the resistor from out to adjust is selected to draw the minimum current. If you did not have a load connected when you did your testing, the minimum current requirement probably explains the results you saw. In an application where you are driving the Adjust terminal from an outside source you need to assure that the minimum current requirement is met some other way. 

A 200 ohm resistor between output and adjust will draw 6.3 mA (1.25 V/200 ohms). That current must flow through the AO line. This exceeds the 4 mA limit that crossrulz pointed out for your device. 

For your lowest output voltage of 1.5 V a resistor of 150 ohms or less from output to ground will draw 10 mA. At 3 V that resistor will draw 20 mA and dissipate 60 mW. If the AO line goes to 10 V, the output of the regulator will be ~8.75 V and the power dissipated in the resistor will exceed 1/2 W.

Assuming that the valve is an inductive load to the regulator, you should also provide protective diodes to keep the energy stored in the inductances from damaging the regulator, the power supply, and the DAQ device.

Lynn

[ThomasCIGITI]

Ah - right - I forgot about the minimum load current which would make sense why there was no reading on the OUT pin. I guess if I hooked up the valve to the OUT pin, I would see a change as the valve would start drawing current.

I believe the valve works on a thermal effect where it is converting the current to thermal electric to mechanically expand and contract the mechanism so hopefully there is no inductive loads. The specs for it ask you to vary the voltage to the device so that I assume it will will draw more current - perhaps it functions just as a resistor to ground. If this is case, I think I can skip the loop back resistors to the ADJ pin. Does this make sense?

Thanks again!

[johnsold]

If the load is resistive, then that should work.  I like to have at least a minimum load on the board with the regulator so that the voltage does not go out of regulation if the valve gets disconnected (unplugged or wire broken, ...).

Lynn