Error 50405 using DAQ USB 6008 and LabVIEW 8.6.1

[Starkboy]

Hi there,

I am a student with very little knowledge of LabVIEW. I am trying to use the program to generate a digital output signal from a USB 6008 DAQ. The specific purpose is to turn a relay on and off. My VI can turn on all the necessary relays, but whenever I turn them off, I receive an error 50405 message saying specifically "No transfer is in progress because the transfer was aborted by the client. The operation could not be completed as specified." The error 50405 occurs at DAQmx Write (Digital 1D Bool NChan 1Samp 1Line).vi:10.

I appreciate any feedback. Thank you!

[Eric-J]

Have you tried resetting your device?

[Starkboy]

Yes, I have reset the device as well as self tested it. Unfortunately that has not resolved the problem. I have tried changing the USB ports and cable as well but have not had any success.

[Eric-J]

How about your LabVIEW code?  Have you written your own?  If so, I would recommend using an example VI from the NI Example Finder and see if that runs with your device correctly.  You can find this by opening LabVIEW and clicking Help and selecting Find Examples.

[Starkboy]

I have written my own. I will search the examples and see if that helps. Thanks for the suggestion.

[AnalogKid2DigitalMan]

Do you have any electronics between the 6008 and your relay?

If it is a DC relay, do you have a reverse biased protection diode across the relay coil? If not, it could be the inductive spike glicthing your DAQ card when you turn the relay off. Without one, it may eventually damage your DAQ card and the PC.

-AK2DM

[Starkboy]

I do have some electronics, but I was unaware of an inductive spike.

I did not design the circuitry between the 6008 and the relay, but I can explain it to the best of my knowledge. It is an amplifier which steps up the DAQ signal to turn on the 12 VDC relay. It contains a 1kohm resistor, a 4001 diode, and an NPN transistor. The line out of the DAQ goes into the 1 kohm resistor, which enters the N side of the transistor. The 4001 diode is connected to the P terminal of the transistor and the output of the circuit goes to the relay.

Do you have any more info on the inductive spike that I could read about? 

[AnalogKid2DigitalMan]

Sounds like you may have it. Here is a schematic:

http://www.acroname.com/robotics/info/articles/drivers/rly_intf.jpg

Google 'flyback' to learn more. Basically when the relay is energized you have created a magnetic field from current passing through the relay coil. When power is removed from the coil, the field collapses and generates a higher voltage potential within the coil. This inductive spike can damage solid state components. The reverse biased diode across the coil safely conducts and shunts the high voltage spikehrough it, thus protecting other devices in the circuit.

-AK2DM

[apok]

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@Starkboy wrote:

I do have some electronics, but I was unaware of an inductive spike.

I did not design the circuitry between the 6008 and the relay, but I can explain it to the best of my knowledge. It is an amplifier which steps up the DAQ signal to turn on the 12 VDC relay. It contains a 1kohm resistor, a 4001 diode, and an NPN transistor. The line out of the DAQ goes into the 1 kohm resistor, which enters the N side of the transistor. The 4001 diode is connected to the P terminal of the transistor and the output of the circuit goes to the relay.

Do you have any more info on the inductive spike that I could read about? 

---

you mean the 1K is connected to the BASE and the COLLECTOR is connected in parallel with the ANODE side of the 4001 and low side of the relay with the EMITTER connected to gnd.... your signal must provide a potential signal higher than gnd plus the diode drop of the emiiter

please post your vi

[Starkboy]

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@apok wrote:

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you mean the 1K is connected to the BASE and the COLLECTOR is connected in parallel with the ANODE side of the 4001 and low side of the relay with the EMITTER connected to gnd.... your signal must provide a potential signal higher than gnd plus the diode drop of the emiiter

please post your vi

---

As you described it, yes that is exactly how it is connected. Should my diode be connected to a ground as well? Looking at the schematic posted earlier it looks like the diode is in parallel with the relay. Is there a mistake in the circuit layout? Should it be constructed differently? It "grounds" to a 12VDC, could that be the issue? Maybe I should ground the diode to something else?

Excuse my lack of knowledge, this is not my area of discipline.

Will post my VI as soon as I get back to my shop space.