JJBladester
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Homework Statement
Derive the ripple voltage of a full-wave rectifier with a capacitor-input filter.
Homework Equations
Where V_{r(pp)} is the peak-to-peak ripple voltage and V_{DC} is the dc (average) value of the filter's output voltage.
And V_{p(rect)} is the unfiltered peak rectified voltage.
The Attempt at a Solution
v_{C}=V_{p(rect)}e^{-t/R_LC}
t_{dis}\approx T when v_C reaches its minimum value.
v_{C(min)}=V_{p(rect)}e^{-T/R_LC}
Since RC> > T, T/R_LC becomes much less than 1 and e^{-T/R_LC} approaches 1 and can be expressed as
e^{-T/R_LC}\approx 1-\frac{T}{R_LC}
Therefore,
v_{C(min)}=V_{p(rect)}\left ( 1-\frac{T}{R_LC} \right )
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)}
My issue is with the approximation that I bolded above. If e^0 approaches 1, then how does the expression e^{-T/R_LC} approach 1-\frac{T}{R_LC}?