usb 6008 generate 5v dc voltage

[Dennis_Knutson]

You might want to start here to get an idea what DAQmx is. There are also labVIEW tutorials available. There are shipping examples for digital output in Help>Find Examples>Hardware Input and Output>DAQmx>Digital Generation. You might alwo want to make use of the DAQ Assistant on your Measurement I/O>NI-DAQmx palette.

[akhil011235]

Hi...

I am attaching a VI which I've just finished. But after running with some of the channels ON, it shows an Error Message as Follows...

"Error -200524 occurred at DAQmx Write (Digital 1D Bool NChan 1Samp 1Line).vi:1

Possible reason(s):

Measurements: Write cannot be performed, because the number of channels in the data does not match the number of channels in the task.

When writing, supply data for all channels in the task. Alternatively, modify the task to contain the same number of channels as the data written.

Number of Channels in Task: 1
Number of Channels in Data: 15

Task Name: _unnamedTask<1>"

Can any of you correct this?

Actually the objective of this VI is to generate +5V individually through the eight channels.[which can switch ON/OFF with the boolean switch.]

I also wanted to get a common 0V which I'm planning to take from the ground channel.

Please help meeeeeeeeeeeee...................

[Dennis_Knutson]

You are close. It looks like you copied the shipping example called Write Dig Chan but got the DAQmx Write wrong. Change it to 1Chan 1Samp. That's because the DAQmx Create Channel is set to 'one channel for all lines'.

[marvel]

Hello there. A newbie to the forum with a question:

I need to turn on and off  a DC heater (80-100W). I currrently have a usb-6008. Do I just need a SSR (for instance MPDCD3 or DC60MP)or do I need an external resistor as well?

I assume I need to  connect P0.0 and GND to the control pins of the SSR; then the output pins of the SSR to the heater (and short-circuit them to the dc voltage regulator)?

I can use a pci-6221 as well but I would rather save it for a more sophisticated job.

[nedkonz]

The USB-6008 should drive a typical SSR (that is, one with an internal resistor, rated "3-30VDC" ).

However, it won't source enough current to do so, so your suggested hookup won't work. Instead you have to use it in open collector mode (the default output mode) and have it sink current to ground to turn on the optocoupler in the SSR.

The specs for the digital output say that it can sink 8.5mA.

And since the usual optocouplers used in SSRs don't like more than 50 mA or so (see the MOC3023 datasheet), if they work at 30V then the resistor must be at least (30-1.5)/0.5 or about 56 ohms (it's probably about 100 ohms or so). So if you wanted to limit the input current to 8.5mA, you could figure that you'd be looking at (5-1.5)V through 100 ohms or so, so you would need (3.5/0.0085)-100 ohms or about 300 ohms in series.

Just hook up the SSR's + terminal to the 5V on the USB-6008 and the SSR's - terminal to an output port pin via a resistor of about 300 ohms, turn the output ON (i.e. port pin LOW) and measure the voltage drop to verify that you don't have any more than 2.5V across the 300 ohm resistor (or 0.0085*R).

When the pin is driven LOW the load will be turned ON.

An alternative way to choose the resistor is to measure the current through the SSR's control input when hooked up to a 5V input voltage; divide 3.5V (5.0-1.5) by the resultant current to find the internal resistance.You'll need about 400 ohms total (internal + external).

I don't recall if you set the port pin to a logical 1 or 0 to get the output pin driven LOW (i.e. load ON).

Message Edited by nedkonz on 02-12-2009 08:03 PM

[AleDa]

 

Hi to everybody.

I've a question about open collector configuration and how it is explained in the USB 6008 User Manual. At page 21 of the new version 2008, it explains how to determine the pull-up resistor in order to do not sink a current bigger than 8,5 mA.

First of all, if I understood well, when the digital output is high the current goes in the load instead when the digital output is low the current goes through the digital port to ground. So I would schematize the circuit like the image attached. If this image is correct, I think it shows much better how the circuit works than the image on the User Manual.

The method in the User Manual suggests determining the pull-up resistor when the output is high, as the position of the ammeter suggests. The Load has a voltage drop when the current pass through it, let's say 2,1 V. Thus in order to not exceed the limiting current of 8,5 mA, the equivalent resistance must be:

If I chose a pull-up resistor of 430 W, I have an equivalent resistance of:

Req=(430*4700)/(430+4700)=394>340W

that verify the relation to not exceed the limiting current of 8,5 mA  the current in fact will be 6,7 mA.

So here it's my question: but when the digital output is low, the load is short circuited by the SCR and the current will be:

 

My conclusion is also supported by measurement. I measured the current through the external pull-up resistor, connecting the ammeter between the 5V output and the pull-up resistor, and it was 10,6 A, that added to the current in the 4,7 kW it will reach the total current sink by the digital port of 11,4 mA.

Have I committed any mistake? Or is there any mistake in the USB 6008 User Manual?

According to this explanation, if I do not take into account the voltage drop 0,4 V in the SCR, I should size the pull-up resistor in order to have:

Thus the pull-up resistor, whatever load you connect, must be not less than 680 W, in fact the equivalent resistance becomes

Req=(680*4700)/(680+4700)=594>588W: