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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

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

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

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  • \$\begingroup\$ What have you tried so far on this? \$\endgroup\$ – Spehro Pefhany May 20 '20 at 15:47
  • \$\begingroup\$ Please show all of your work and ask a more specific question. This is not Chegg, we won't do your homework for you. \$\endgroup\$ – Elliot Alderson May 20 '20 at 15:53
  • \$\begingroup\$ 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. \$\endgroup\$ – Pare Kanes May 20 '20 at 16:17
  • \$\begingroup\$ The capacitor start be charging when \$ v_s \$ is \$ v_d \$ above the output voltage and stop when it falls below this voltage. Without effort on your part we can't do homework for you. \$\endgroup\$ – Warren Hill May 20 '20 at 16:22
  • \$\begingroup\$ I edited some of the questions, sorry for ambiguous posting. \$\endgroup\$ – Pare Kanes May 20 '20 at 16:42
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[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*}

schematic

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

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    \$\begingroup\$ Since the OP is talking about 60Hz it's a sinusoidal input not DC. Also your formula is for a series resistor, not parallel. \$\endgroup\$ – Warren Hill May 20 '20 at 16:23
  • \$\begingroup\$ The formula is just to show the equation, I know he's asking about AC why the formula is not for parallel? \$\endgroup\$ – bouzaid sohaib May 20 '20 at 16:28
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    \$\begingroup\$ The capacitor charges via the diode and not via the resistor. I suggest you quickly remove this post to save face. \$\endgroup\$ – Andy aka May 20 '20 at 16:31
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    \$\begingroup\$ With your circuit when you close the switch the R you need to use is that of the switch not the one you have drawn. The capacitor will charge very quickly. \$\endgroup\$ – Warren Hill May 20 '20 at 16:31
  • \$\begingroup\$ Yes yes now I can see what you mean thank you for clarifications \$\endgroup\$ – bouzaid sohaib May 20 '20 at 16:40

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