- #1

JJBladester

Gold Member

- 286

- 2

## Homework Statement

Derive the ripple voltage of a full-wave rectifier with a capacitor-input filter.

## Homework Equations

Where [itex]V_{r(pp)}[/itex] is the peak-to-peak ripple voltage and [itex]V_{DC}[/itex] is the dc (average) value of the filter's output voltage.

And [itex]V_{p(rect)}[/itex] is the unfiltered peak rectified voltage.

## The Attempt at a Solution

[tex]v_{C}=V_{p(rect)}e^{-t/R_LC}[/tex]

[itex]t_{dis}\approx T[/itex] when [itex]v_C[/itex] reaches its minimum value.

[tex]v_{C(min)}=V_{p(rect)}e^{-T/R_LC}[/tex]

**Since [itex]RC> > T[/itex], [itex]T/R_LC[/itex] becomes much less than 1 and [itex]e^{-T/R_LC}[/itex] approaches 1 and can be expressed as**

[tex]e^{-T/R_LC}\approx 1-\frac{T}{R_LC}[/tex]

[tex]e^{-T/R_LC}\approx 1-\frac{T}{R_LC}[/tex]

Therefore,

[tex]v_{C(min)}=V_{p(rect)}\left ( 1-\frac{T}{R_LC} \right )[/tex]

[tex]V_{r(pp)}=V_{p(rect)}-V_{C(min)}=V_{p(rect)}-V_{p(rect)}+\frac{V_{p(rect)}T}{R_LC}=\frac{V_{p(rect)}T}{R_LC}=\left ( \frac{1}{fR_LC} \right )V_{p(rect)}[/tex]

My issue is with the approximation that I bolded above. If [itex]e^0[/itex] approaches 1, then how does the expression [itex]e^{-T/R_LC}[/itex] approach [itex]1-\frac{T}{R_LC}[/itex]?