I’m assuming that those "v"s are meters and not constant voltage sources. A 2.4v zener does NOT have exactly 2.4v across it. A 1.2 v supply will not be precisely 1.2v. Both will also vary with time and temperature. So, in practice there WILL be a dc component across the sensor - which will destroy it.
Thank you. Do you recommend me something (circuit) else? The 2.4v zener would provide about 2.4v. To keep 1.2V exactly half of the zener produced by the zener, I was considering split its voltage with 2 x 10k resistor in series and feed into an opamp in buffer configuration, so even it the 2.4 reference changes a little bit, the other side of circuit would be always zener / 2.
How I would do it is to use two digital IO pins on the processor to generate the reference square wave. Put the sensor plus a series precision resistor between them and just pull one IO pin high as the other is pulled low and then swap them. That presumes that IO pins can both source and sink IO current. Then take the junction between the two to an Analogue in pin. You get two measurements each cycle. Use a lookup table of values and interpolate between them. If you wanted more precision - add more series resistors of different values covering the range of humidity that you want to sense, going to different IO pins. So you can choose the IO pin pair that brings the centre point between sensor and resistor closest to the mid-voltage point. It’s effectively a balanced half bridge arrangement - using the precision of the resistors to determine overall measurement precision. OK it ties up several IO pins - but microcontrollers are so cheap, I’d probably just dedicate one to this sensor and that’s all that it would do.
The smd codes kind of suck. They’re used on devices where there isn’t room for the full PN. But they’re not standardized well. Are often unique per footprint, but even then, not a guarantee.
I looked up “CAZ” here: smd.yooneed.one/code4341.html and found a part that matches the footprint. Then googled around and found the LN61CC3002MR-G on lcsc.
It can be very hard to find a part on Google, or say Digikey, if it’s made by a Chinese company. LCSC can be helpful since they’re based in China.
See if you can swap the connector, those pins can be released if you gently lift the locking tabs and pull the wire away from the connector. The metal connectors from your replacement fan might not fit into the original, but it’s worth a try.
Thanks for this suggestion! Is there any chance of breaking the connector by doing this? I replied to another commenter, but to emphasize, I am only looking to modify the fan and connector as a last resort because of how rare this equipment is.
Breaking the plastic would be tough so long as you’re not hammering it. The connectors crimped to the wires it’s another issue and if you pull too hard you could get the wire to come off the connector if it isn’t soldered.
Yes you could break the connector if you’re not careful. However at the end of the day, it’s not like you can’t replace both sides of the connector with standard hardware, and at the very worst you could solder the fan to the pins and cover each wire with heat shrink.
I am trying to be non-destructive since this is a rare piece of equipment. So I would prefer to put in something entirely new in and preserve the original hardware in case something goes wrong.
One possiblity would be to cut the original cable and add a new connector to both sides. A female one to the one that goes to the board, and a male one to the one on the fan.
That way you have all the original parts still there, but you can use the one part of the cable as an adapter.
Any chance to replace the whole PSU and then going full tilt modifying the replacement? I only see one offer for 400-5494-91 (400€!!) so maybe any other in that form factor? Seems to be mostly standard PC connectors on these, but not sure from the pics I can find.
The diodes at least provide some steps - which just LEDs and resistors don’t. The dissipation in the main series resistors will be quite something, also. (You will probably need several resistors in series, if you use standard ones, with their standard voltage limits).
I think that it is the same stuff, just will propellant added in the aerosol. A small squeezy bottle with a needle is better if you have live stuff nearby - but you can always fill one from an aerosol
Assuming that the ribbon cable is standard - you could consider adding TWO IDC connectors, side by side. Then slice the cable through between them. Then add a standard extension cable to link the two. Indeed if one of the two is male, the other female - the extension can be removed if the thing is relocated to where the extension isn’t needed - or a longer one is needed.
I confess to having done this sort of mod several times, myself. It’s also quite an easy way of sticking a protocol analyser/sniffer between the two and/or modifying the data that is sent on its way. Or adding an additional sensor (even of a different type) and converting its output to something suitable.
That sounds… Difficult :p I really don’t want to slice anything to void warranty and such. I’ve been spending a lot of time trying to look for some sort of extension cord or connectors so I can make my own, but I was wondering… can I just use this? www.amazon.nl/…/B01EV70C78/ Would be so much easier 🤔
Lots of people use them for something similar. Obviously, the situation changes once the lines are power lines and not signal lines. If you can stabilise the result and add strain relief.
it worked! (at least in the sim)
thanks a lot!
I'm not sure I got all the math right, since I get 0 - 5.05V on the output, but it's not a deal breaker for me.
I attached the schematics, but not sure that images in comments federate from kbin to lemmy, so if you don't see it, try opening an original link to this comment.
ah, a little bit of thinking helped me realize that I don't need 1k in inverting loop and then I'll get 5v instead of 5.05v :) and 100k can also be replaced by 1k
No, with 0 Ohm I get a perfect 0+5v sine, just as I need. It's the other way round, it'll amplify the signal and clip off the top at about + supply voltage when you increase the resistance in the feedback loop, but it's not exactly square wave, as the bottom part of sine will remain intact. I use clipping schottky diodes to protect the arduino from voltage outside 0+5v range, this way if you increase input signal to say 20v peak to peak, you'll get more square-ish wave in 0+5v (I've attached the schematics)
You’re right of course. Two more questions if you don’t mind:
what’s the 1k || 100k doing?
if you don’t want more than 5V at the output of the OpAmp, wouldn’t it be easier to just supply the OpAmp with 5V instead of adding a shottky diode which has a forward voltage of around 0.6V? As I understand the screenshot the voltage is 5.4V at that exact point and it could theoretically raise to 5.6V.
At the moment you’re also mixing up your + input. 10V AC + 5V DC result in 7.5V input in your sim.
it does nothing 😅 I just forgot to remove it from the sim
you mean like power it with 0+5v? I think it won't work with negative input voltages then, right? and I'm using tl074 and I'm not sure that 0+5v is enough juice for it, and I use remaining 3 opamps for other stuff, so it's not really an option for me. the sim didn't have shottky diodes, so this measurements are not 100% accurate, but I used this approach to protect arduino inputs in another project, and it worked well. And yeah, I cranked up the input signal to +/-10v just to check the clipping, it'll normally stay within +/-5v
I already ordered PCB for the prototype. I hope it'll work fine. Thanks a lot for your help :)
In my opinion an oscillator always produces an AC sine wave. There is usually no need for a DC overlapped oscillator signal. The DC supply of an oscillator produces a AC sine wave relative to GND.
Where exactly did you measure a DC sine wave, relative to what, and what do you mean by “AC removes a DC component”?
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