Discussion:
USB-6008 Digital Amperage
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RL
2006-11-24 23:40:08 UTC
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I am trying to control 4 solenoid valves by writing to four lines on the Digital Output of USB 6008.  Though
the +5V has sufficient capacity to activate all the valves (280 mAmps)
each line can only send out .77 mAmps.  The Digital lines Out do not
appear to have sufficient power to trigger a solid state relay which
only requires 9 mAmps.  Are the Digital Outputs only good for
powering LEDs?  The specification sheet does not give any information
about this and it is a bad weekend to get information from NI.  Has
anyone figured what the problem is?Raymond
Ed W
2006-11-27 18:40:14 UTC
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Hi Raymond,

 

The USB-6008 is not capable of sourcing that much current on its own.  You will need to add external circuitry in the form of a pull-up resistor to get current values of up to 8.5mA.

 

The following Knowledge Base article discusses this issue.

 

<a href="http://digital.ni.com/public.nsf/websearch/9CDCBF3ADD0DB6288625700700656F4E?OpenDocument" target="_blank"> Why Does the USB-6008 Digital Output Line Voltage Drop Below 5V When I Connect to an External Device?</a>

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I hope this helps!

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Ed W.
AnalogKid2DigitalMan
2006-11-29 20:40:11 UTC
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Ray:
Make sure you put reverse biased protection diodes across the valve coils so you don't fry your DAQ or PC.
AnalogKid2DigitalMan
2006-11-29 22:10:10 UTC
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Ray:
It's not really about the opto feeding back power into the digital out line. Rather it is the use of the 5V from the DAQ card as a source of power for the coils. When you turn off the coils via the transistor side of the opto, the magnetic field in the coil collapses. This generates a high voltage spike of short duration (10's-100 V) of opposite polarity that needs to go somewhere- typically back into the power source. Once in the power source, it can go anywhere. Such spikes can damage 'low voltage' devices- DAQ, optotransistor, PC. So, I suggest the use of a standard 1N4004 diode across each coil with the banded end connected to the 5V side of the coil. The other lead of the diode to the opposite side of the coil. The diode will turn on and conduct when the reverse spike voltage reaches about 1V. The diode effectively becomes a short and absorbs the spike.
Some people never do this and get lucky, other's aren't so lucky. It really depends on the coil inductance, currents, lead lengths, power supply integrity, etc, etc.
Hope this helps, I was probably too verbose. Let me know if you need more info.
-AK2DM
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RL
2006-12-04 17:40:12 UTC
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Cedric78

That looks like a software issue unless your unable to write to the Digital Port using MAX.&nbsp; <a href="../view_profile?user.id=48749" class="auth_text" style="font-weight: bold;" target="_blank"> gromar</a>
started a thread brought out other people working with Delphi.&nbsp; I
did not consider the Philips because I my pneumatic solenoids only
require 5 volts.

Raymond

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