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Driving a reed reay

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#1 Jan Oberthuer

Jan Oberthuer
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Posted 23 April 2004 - 07:20 AM

We want to drive a reed relay on the DXX ports of LabJack. The choosen Relay has a internal diode. The driving voltage is 3.5 - 10V with a resistance of 380 Ohm. first question: Can I use this relay directly at the DXX port without the CB25 board (1.5kOhm resistor)? second question: Which setup is better: apply continous GND or +5V to the driving circuit. That means for me if I apply GND: if the port is enabled the relay shuts if +5V are applied continously: the relay shuts when the port is disabled (there I have the problem that the closed port perhaps not sinks the current) Regards Jan

#2 LabJack Support

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Posted 23 April 2004 - 09:50 AM

5/380 = 13 mA, so a D line should be able do control this relay directly. Even if you use the CB25 for the convenient screw terminals, you would need to install the jumper to short the 1.5k resistor. If the 1.5k resistor is not shorted, you would not be able to source or sink the 13 mA required.

The purpose of the 1.5k resistor on the CB25 is to provide extra protection for the D lines. You can use some series resistance, but not 1.5k. The minimum required current for your relay must be about 3.5/380 = 9 mA. At 5 volts, this means you can have an impedance of 5/.009 = 540 ohms. When sinking as an output-low, a D line has an impedance of about 30 ohms. 540-380-30 = 130 ohms, so I would say you can use a 100 ohm resistor in series with the D line for extra protection if needed.

I suggest you connect +5V to the + control line of the relay, and connect Dxx to the - control line of the relay. When Dxx is output-high or input, the relay will be off, when Dxx is output-low you will sink the 13 mA and the relay will turn on. Like most digital I/O, the D lines are better at sinking than sourcing.

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#3 Jan Oberthuer

Jan Oberthuer
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Posted 23 April 2004 - 09:58 AM

Thank You very much for this fast response. Regards Jan Oberthuer

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