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Cezario
modified 3 years ago

Comparator - integrator - derivator amp op

4
2
411
05:19:01
The first part makes the signal saturate at the given tension source for the operational amplifier (reverting the signal in the process) That square frequency is source to an integrator, that makes the operation Vout = (-1/Ri.Cf)* ∫ (Vin dt) The extra resistor is for real case uses, being the one responsible for enabling the capacitor to release its energy from the first charge, otherwise it won't discharge This operation results in a triangle wave that is feed to a derivator amp op, which has the following principle for conversion Vout = -Cin*Rf* d(Vin)/dt Rf being feedback resistor The resistor before the capacitor is used so at high frequencies there is still impedance for the circuit entrance, since the impedance from a capacitor dissapear when frequency reach high values like a few hundred MHz You can alter the first DC source so you can have an offset for detection The alternate current source would be the entrance of any received signal If you use this circuit don't forget it causes phase shifting on the original source
published 3 years ago
592azy2circuitdude
3 years ago
Nice circuit and good explanation of the details 👍. Check out this circuit which has lots of math symbols you can use (for instance the integral sign). It may help you type the opamp equations better. https://everycircuit.com/circuit/5283024345759744
Antonio1961
3 years ago
👍

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