square to sine gen

243

03:37:50

converts square wave to a sine wave. The square signal is passed into the cascading RC network, where the time consant (Tau) is 20K*2uF equalling 40 (ms). Because the period of the square wave is 250ms and half of the period is 125ms, that means the capacitors can charge up to about approx 95% of the input voltage due to half of the square wave period being more than 3 times the RC time constant (ie. 0.5*T = 125 ms > 3*R*C =120 ms). After the small sinusoidal is produced, it is then fed into a voltage follower (buffer) and then into a negative feedback Op-Amp configuration with a variable mulltiplier (potentiometer).

published 8 years ago

2ctiby

8 years ago

I think that perhaps you have got the basic setup a bit wrong here (correct me if I am wrong) ... here is my take on it: ... http://everycircuit.com/circuit/6392997007851520

ZRoussel

8 years ago

thanks! i appreciate the help, i completely forgot about differentiator/ integrator Op-Amp configurations. Now i see why my RC constaqnt methodology was backwards. All in all, this is a great learning experince for everyone (myself included).

2ctiby

8 years ago

@ZRoussel ... You're welcome, that's a kind and appreciated reply... With the Maths Integral Calculus part, you could use the now found value of the triangle wave and use it instead of the square wave input. The R and C values remain the same (as seen) for the second integrator, which is just a resistor and cap. That second integrator converts the triangle in to a sine and you will see that the Maths result is the same Vout value as seen at the final bottom sine output ... providing you remember to half that 125mV for this formula now....then halve that final pp to get the seen peak value. As a matter of interest... using a tau value of about x10 higher than the half-duration is reasonable to obtain a decent sine shape from the triangle, but with lower amplitude.

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