this post was submitted on 07 Jul 2023
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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
So what should happen if you replace the 1k in the inverting loop by 0 Ohm is, you should get a 5V square wave.
Is that correct?
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:
At the moment you’re also mixing up your + input. 10V AC + 5V DC result in 7.5V input in your sim.