USB 6008 output voltage is not enough?

[ivy037]

Today, we just connected the relay board to the DAQ and external 12 V

We just rely on the led's status on/off to see if the DAQ can command the board to activate the switch.

 from video 1,

https://www.dropbox.com/s/cefbolmyp4sng46/video1.mp4

you can see the LEDS DID NOT follow the DAQ whose status is reflected on our VI?

from video 2, https://www.dropbox.com/s/m67eijjdjzenypd/video2.mp4

we only connect port 1 , and it follows the computer screen indication.

Conclusion, the daq can drive up to 1 relay AT THE SAME TIME..

I contacted the manufacturer, they claimed the board can be activated by about 4V TTL signal pulse(ideally 5V),and they restrict the current under 8.5mA based on my need.

so the USB 6008 power is not enough?

need to know what's wrong quickly,thank you very much

[Joseph_Loo]

If you take a look at your specs, the USB 6008 for analog is 5 ma per output. If you use the digital I/O you have an open collector with 8.5 ma per output. Based on your specification, there is a good chance you can burn the unit up. If you are using Digital I/O, what is your pullup resistor and the voltage you supplied.

[ivy037]

this is how the manufacturer replied:

It looks like you want to use a TTL pulse from your DAQ to control my relay.  That is fine - I will have to make a design adjustment in order to do that

I will also put a 680 ohm resistor on the TTL input to make sure that the current draw on your signal never exceeds 8 mA.  To trigger the relay on my board, you will need to have your DAQ provide a short pulse every time that you want to toggle on or off.

[ivy037]

should I contact NI technical support? Because the moment the DAQ send out Voltage pulse for 200ms, the voltage is measured to be 2.5-2.9V, not 3.3V?  The manual claims it can reach 3.3V, I guess it is because of the short time period of impulse time and the DAQ needs to take time to reach 3.3V.

I really dont know the true reason 😞

[Joseph_Loo]

Do you have a 5 volt supply for the TTL output connected to each pullup resistor that is inputted to the box? The output of the USB 6008 is an open collector, thus it does not have any inherent voltage associated with the output.

[ivy037]

thanks for the reply, but what do you mean by "have a 5 volt supply for the TTL output" and "connected to each pullup resistor that is inputted to the box" ?

From the diagram, that's why I connected, that's it, nothing else.

Each relay on the board needs at least 3.5V to be triggered and the board itself is driven by external 12V power.

I originally assume that 6008 digital output has ability to send out 3.3V to the relay board TTL and if i only connect one digital port to the board, the relay on the board can be trigger withnout problem; if i connect more than one, i mean P0.0 and P0.1 is connected to the board and run the VI, ONLY one relay can be trigger, that triggered relay sometiems No.1 sometimes No.2,

is there any mistake?

[apok]

try this circuit... you need a driver from your dig output

[ivy037]

I look at the circuit diagram you posted in your old thread, my question is:

1. ULN2003 can really be directly connected to the DAQ and it will not exceed usb6008 8.5mA sink/source current ability?

2. I notice the output of ULN2003 is connected to the relay which requires 12V DC, so external load like motor cannot be directly connected to the ULN2003, correct?

anyway, thanks for reply

[apok]

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@ivy037 wrote:

I look at the circuit diagram you posted in your old thread, my question is:

1. ULN2003 can really be directly connected to the DAQ and it will not exceed usb6008 8.5mA sink/source current ability?

2. I notice the output of ULN2003 is connected to the relay which requires 12V DC, so external load like motor cannot be directly connected to the ULN2003, correct?

anyway, thanks for reply

---

1. yes, ULN is a TTL device

2. look at the ULN specs, it can drive up to 500mA on a single output....if you need more driving current, look into fet's,ssr's and high powered transistors

[ivy037]

one more question, from the data sheet , it says:

==========================================================================

The ULN2003A and ULQ2003A have a 2.7-kΩ series base resistor for each Darlington pair for operation directly with TTL or 5-V CMOS devices

===========================================================================

1. it doent mention how high the voltage should be to consider a valid TTL signal (does the datasheet assume that the TTL signal is standard 5V),

USB6008 only output 3.3V when it is connected to outside pullup resistor(load)?

2. and if it works, then 6008 send out one TTL to turn on the transtor which in turns activate my load; after that, send another TTL to shutdown my load?

BIg thanks