Powering Sensor from constant current source (24 Volt IO) - Page 1
Forum section: electronics / projects

author: robint91
author_rank: Contributor
timestamp: February 03, 2020, 10:35:30 pm
content: I'm making a special 2-wire sensor that can be power by the parasitic current that can flow in EN 61131-2 type 3 input. More info about this interface you can find here,https://download.beckhoff.com/download/document/Application_Notes/DK9222-0909-0008.pdfThe standard define at which voltage level a 24 Volt digital input should trigger.The idea is to connect my sensor in between the +24Volt line and a type 1/3 digital input.My sensor can consume max 2 mA and by modulating the input voltage I can send an 0 or 1.So the summery is the following,The digital input terminal can supply 2 mA minimum and the minimum voltage is 5 Volt. So I have 10mW of power to use for my processing and sensors. It the idle (0) state I have around 40mW.So I have 10mW when I put a logical 0 and 40mW when I put a logical 1 on the lineNow I have a CCS811B which is a nice TVOC/eCO2 sensor. In it's low power mode it consumes around 1.3 mW average.I checked the current consumption and it does consume for 1.5 s around 25 mA. This happens every 60 s, which that averages to the above number.Now my problem is how to bridge the power gap. For 1.5 s I need to supply at least 50 mW.Any one an idea?I was thinking about a small SMT supercap but the internal resistance of those is just way to big to be usefull. Small battery is not really an option because of the limited cycles...Currently I have a TPS62175 generating 1.8 V from the 5 to 24 V supply.

author: sam[PS]
author_rank: Contributor
timestamp: February 28, 2020, 06:15:05 am
content: What about storing much higher voltage on a capacitor (than would not need to be supper) That would make the circuit much more complex but could work if the bom cost and/or size is not too much an issue. You just need to design two high efficiency smps, one to boost the voltage to charge the cap then a down converter for the uptime. Of course top notch efficiency would be paramount so probably bigger self than you might want. That's just an idea i haven't done any math on it so it might just be a stupid idea.Just my 2cts...

author: Zero999
author_rank: Super Contributor
timestamp: February 28, 2020, 11:57:20 am
content: Quote from: robint91 on February 03, 2020, 10:35:30 pmI don't see why you need a super capacitor.If you use a low drop-out regulator to get 1.8V from 5V, then a volt sag of 3V, assuming a dropout voltage of <200mV, would only require a 12.5mF capacitor, which could be charged from a suitable 24V to 5V voltage regulator. A had a quick look in Farnel and found a 15000µF, 6.3V, which should do the job.https://uk.farnell.com/panasonic/eca0jhg153/cap-15000-f-6-3v-20/dp/1848531

author: jbb
author_rank: Super Contributor
timestamp: February 28, 2020, 05:45:07 pm
content: You mention:“The digital input terminal can supply 2 mA minimum and the minimum voltage is 5 Volt.”I think that’s 5V drop across the *PLC input*, so you might have more voltage available than you think.In any case, something on the order of 20,000uF might do well. Remember to consider tolerance, ageing and minimum operating temperature.I also see that the sensor supply voltage is 1.8V absolute minimum, so you probably need to leave margin (maybe 2.2V?). Also, the 25mA figure is typical; what is the maximum?