How can I find the time at which the diode charges the capacitor every cycle(half wave rectifier)

Question

Half wave rectifier is given and given values are :

V_p = 10V, V_d = 0.7V, 1/T = 60Hz, R = 10.

I want to know when C = 0.1F, how can I find the time at which the diode charges the capacitor every cycle.

I failed to approach this problem because I'm not sure at which parameter I should find out. I'm understanding capacitor charging time is T, but I don't think T=1/60 is answer.

I tried so far on this :

$$i_{DMAX} = \frac{V_p-V_d}{R} + C\cdot V_p\cdot \frac{2\pi}{T}\sqrt{\frac{2V_R}{V_p}} = 65.02A$$

$$ dv/dt = V_P\cdot \omega \cos(\omega t)$$

---redo---

Now I find when charging time is \$ \Delta T\$

$$v_0(T-\Delta T) = V_p-V_d-V_r$$

and

$$V_r =\frac{(V_p-V_d)T}{RC}$$

$$v_0(T-\Delta T) = V_pcos(\frac{2\pi}{T}\Delta T)-V_D$$

Answer 1

[Wrong answer] the capacitor never charges completely, we often calculate the time at which the capacitor will charge to 63% of Vs

First have the charging formula:

\begin{eqnarray*} V(t)=V1(1- e^{\frac{-t}{RC}}) \end{eqnarray*}

^{simulate this circuit – Schematic created using CircuitLab}

\begin{equation} V(RC)=V1(1- e^{-1})= 0.63\times V1\\ \end{equation}

So now you have your time constant in seconds which is: RC

you still must verifiy a condition on RC, it should be smaller than the priod of your signal