04-07-2010 06:05 AM
Hi to all the forum, this is my first post, so excuse me if I posted in the wrong section.
I'm going to briefly clarify my situation.
I have to connect the USB-6009 DAQ with a PLC: the application consist in an analog acquisition task, started from the PLC, which wait for a signal for the completed task. The PLC work at 24V, so I found an external interface board based in the ULN2803 and relay for adapting signals [http://www.datasheetcatalog.net/it/datasheets_pdf/U/L/N/2/ULN2803.shtml].
The IC accept a TTL 5V command input, but in the datashet (and also in the figure), is specified that the activation voltage is about 3.85V@0.93mA.
Now my question arise: what is the correct configuration of the digital port on the DAQ between open-collector and push-pull? Also the open-collector maximun current is rated 0.6mA, can I damage the USB-6009 in this configuration?
Another question, from the digital input side. I'm thinking to use another relay activated from the 24V, that switch the digital input line between 0V and 5V. Is there any consideration to take? What configuration I must use?
PS: the interface board would be shipped in few days, so I don't have it at the moment.
Waiting for a reply.
Best Regards
Marco
Solved! Go to Solution.
04-08-2010 05:15 AM
Hello Marco,
I've taken a look at the USB 6009 and you're right when you say that in Open Collector the output current is 0.6 mA. That's because of the 4.7 KOhm pull up resistor. So if you need to activate the IC at 3.85 V @0.93 mA it means that you need to add an external pull up resistor in parallel to the one built in the USB 6009 board. You need a voltage drop of 1.15 V (5V-3.85V), so if you do some math you get these values:
1.15 V=0,93mA*Req
Req=1236 Ohm
Req= 4700//R
So if you do the parallel between 4700 and the external pull up resistor you can find that this resistor needs to be less than 1677 Ohm.
In this way you'll have a voltage drop of less than 1.15 V and you'll activate the IC.
As for the Digital Inputs, you just need to verify that the signal is within the specifications of the USB 6009.
Ciao,
Andrea
04-09-2010 04:03 AM
Thanks for the reply, but I have to post again to have same clarification.
The interface board was arrived yesterday morning, before your reply: with my surprise this board is equiped with the ULN2804 IC, who is quite similar to the other one; but it is rated for 6-15V CMOS input.
I have make some little test with an external power supply to feed the input line of IC: the IC activate the output relay at a minum voltage of about 3.5V. So: what is the right way to interpreter the datasheet? In any way, I'm happy with the new IC, in fact (from DS) it absorb less current compared to the ULN2803 and I am not forced to usse an external resistor.
For the input side, I haven't understood if I have to connect an external pull-up resistor (as I see from figure14 of the USB6009 manual), and how i can calculate its value?
PS: the input line was connected to another relay that switch from 0v to 5V (the relay was activated by a PLC at 24V).
Best Regards
Marco
04-09-2010 08:01 AM
Hello Marco,
good that you don't need an external resistor for the DO anymore.
As for the input, like I told you yesterday you don't need an extra resistor, you should be fine with your current setup.
Ciao,
Andrea