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I was asked to calculate the expression for the voltage for this circuit in DC at the essential node $$v_o(t)$$

From the input voltage $$v_i(t) = A\sin(wt)$$

I did it for DC with nodal analysis over node a and ended up with $$i_D = i_R + i_C = \frac{v_o(t)}{R} + C\frac{dv_o(t)}{dt}$$ Solved the differential equation and got the function for the output voltage.

$$ u_0(t) = Ae^{-\frac{t}{RC}}$$ whenever the input voltage is less than the inner potential barrier of the diode (the diode induces an open circuit), and $$Ae^\frac{-t}{RC}+Ri_{saturation}(e^\frac{v_\gamma}{v_{TH}}) - v_\gamma$$ whenever the input voltage is greater than the inner potential barrier of the diode.

But now I'm thinking if I had an alternating voltage supplier (as I did in this schematic): how would this affect the expression I got in DC? I would assume that the voltage is the same at the node because the components are in parallel (similarly to what happens in DC), but does the Capacitance or Impedance now change the expression I got?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ There is a little problem ... Where is gone A * sin(w * t)? There are 2 "phases" in this circuit. When diode conduct (is it perfect ?) , and then ... when it is off. So 2 equations. \$\endgroup\$
    – Antonio51
    Apr 3 at 15:36
  • \$\begingroup\$ @Antonio51 you are right! Maybe I'm able to get the expression for AC relating i_D = Vi(1/R + jWC) = v0/R + C(dv0/dt)? Would it be correct to find a solution for the differential equation with sine function instead of the exponential? \$\endgroup\$
    – ludicrous
    Apr 3 at 15:48
  • \$\begingroup\$ You need to find a solution for the differential equation (use only this, do not mix complex impedances) with sine function (as long the diode is ON), then at the point where the diode switches off, another equation with initial conditions has to be solved analytically (exponential function). When this is done, as input voltage crosses (angle or time) the exponential voltage, the diode switches again ON and this repeats until it is "steady-state". \$\endgroup\$
    – Antonio51
    Apr 3 at 17:14
  • \$\begingroup\$ @Antonio51 thank you for your answer. But I'm having trouble finding the sine solution to the differential equation for which u(t) = -RC u'(t). Even if u(t) = sin(RCt + pi/2), it's derivative would be u'(t) = RCcos(RCt + pi/2) = -RCsin(RCt) and the solution wouldn't be correct. I want to find the u(t) of the output, but I'm afraid I'm making some mistake by misunderstanding the phases of the input source, the capacitor impedance and the output here \$\endgroup\$
    – ludicrous
    Apr 3 at 19:27
  • \$\begingroup\$ sin (w*t) ... not sin(RCt ...) \$\endgroup\$
    – Antonio51
    Apr 3 at 20:34

1 Answer 1

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EE&O, it was a long time ago ... easier using a simulator :-)

Here is an example Maple sheet showing the "starting" methodology ...
1- Starting, the diode is ON, capacitor not charged ...
This is when the diode is "conducting", perfect diode Ud=0.6 V (should be more 1 V in power systems).
Calculate all variables, one needs, in this "phase" time, the diode switches OFF at ... t1 (near after 5 ms)?
2- At t1, the diode is OFF
Next, the capacitor should then discharge until diode, condition: \$Uin > (Uc+Ud)\$, switches again ... at t2 ... (capacitor has a residual charge ...), etc ... (t2 between 10 ms and 15 ms)
3- repeat 1 and 2 with, now, initial conditions for Uc voltage ...

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