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colinb

Hall effect Limit switch

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I would like to use Optek opb-850 hall effect switches to act as limit switches connected to an Omron CQM1 Id211 input module. What is the best way to do this?

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Optek OPB-850 are not hall effect switches. They are photo-transistors. You could use them on CQM1-ID211 but you will need an external 2 VDC supply. See attached datasheet. Have you bought the sensors yet? OPB850.pdf

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2V? Wonder where did this come from... @colinb well... this is only a sensing component, not a complete sensor. it is meant for embedded use, not for interfacing to PLC so to make it work you MUST add other components and size them. in other words you cannot connect it directly to a PLC, you must at least do some bare bone design around it. in fact using it with a PLC is still problematic, because commonly used 24VDC is actually too much for this component (absolute max Vce for this device is 24V). but ID211 can also work with lower voltages, as low as 12V or .... we can add more complex circuit around this device to make it work for other voltages too. how many external components are to be added and actual circuit design will depend on what your specs for completed sensor are. to turn it into common industrial sensor takes a bit more work than to just make it work with particular product (in this case ID211). to make a simplest connection.... first thing is to power the emitter. for this we just need Kirchoffs Laws: Vfd for emitter is about 1.2V. If supply voltage is 12V then we need series resistor to drop Vr=12V-1.2V=10.8V normally If=20mA, and since resistor is in series with LED, currents are same Ir=If. So from Ohms Law I=V/R or in this case R=Vr/If=10.8V/0.020A = 540Ohm. such value is not common but 560 Ohm value is. Resistor also need to be able to dissipate at least Pr=Vr*If = 10.8V * 0.020A = 0.216mW so this need to be at least 1/4W resistor, preferably 1/3 or 1/2W. also be very careful with LED polarity, if the voltage is reversed, 12V will kill it (it can not handle more than 5V when reverse biased which is typical for LEDs). normally you would: 1. bring +12V to one side of 560 Ohm reistor, 2. connect other side of resistor to pin2 (anode) of the Optek device 3. connect pin4 of optek to 0VDC if using 24V supply, then Vr=24V-1.2V; R=Vr/If=22.8V/0.02A=1140 Ohm so nearest common value is 1.2k and it has to be over 1/2W. sanity check: Pd=Vf * If = 1.2V * 0.02A = 24mW which is less than max allowed 75mW for input diode so this part is ok. second part is to complete receiver side circuit. according to ID211 datasheet: http://industrial.om...sheet-CQM1H.pdf input current is 10mA which is within max 20mA that Optek device can handle. so: 1. connect Optek pin3 to +12VDC 2. connect pin1 directly to one of ID211 inputs. the pototransistor handles only 100mW max. if using 24V instead of 12V, you will push the Optek to the beaking point. it may work at least for some time. above is the simplest possible connection and it has its limitations. this is an electronic component costing only $2-3, while industrial sensor is easily 50x more.

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Thanks for the info, it may be simpler to buy prox switches from ebay, but i will have another look what to do.

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