I am reading The Art of Electronics, Third Edition by Paul Horowitz and Winfield Hill but I feel like I am missing something when the book comes to talk about (begins to, instead) Power-supply filtering (Chapter 1.6.3_A, page 32) after a half/full wave rectifier with capacitor.
In the A subparagraph, here is what it is said :
It is easy to calculate the approximate ripple voltage, particularly if it is small compared with the DC. The load causes the capacitor to discharge somewhat between cycles (or half-cycles, for full-waves rectification). If you assume that the load current stays constant (it will, for small ripple), you have : $$ \Delta V = \frac{I}{C}\Delta t $$ Just use 1/f (or 1/2f for full-wave rectification) for \$\Delta t\$ (this estimate is a bit one the safe side, because the capacitor begins charging again in less than a half-cycle). You get $$ \Delta V = \frac{I_{Load}}{fC} $$ for half-wave $$ \Delta V = \frac{I_{Load}}{2fC} $$ for full-wave
Ok, I don't understand where that comes from, more specifically the \$I_{Load}\$ term instead of \$I_{in} - I_{Load}\$ term that I found (see below). I tried to retrieve it, I don't.
Let's take the following schematic (which is used in the book):
The KCL and KVL gives respectively : $$ I_{in} = I_C + I_{Load} = C\frac{dV_{Load}}{dt} + I_{Load} $$ $$ V_C = V_{Load} = V $$
The latter is quite useless in fact. So, if we work with the KCL equation : $$ \Delta V = \frac{I_{in} - I_{Load}}{Cf} $$ for a half-wave rectifier. And : $$ \Delta V = \frac{I_{in} - I_{Load}}{2Cf} $$ for a full-wave rectifier.
What am I doing wrong ? Why don't I find the same equation that the book gives ?
Thanks !