Operation of a simple transistor capacitance multiplier
The capacitance multiplication effect is related to how the capacitor discharges. Consider the following two circuits:



simulate this circuit &ndash; Schematic created using CircuitLab
When SW1 opens, C1 discharges through D1 and R1, with an initial current of \$\frac{10 V - 0.65 V}{100 \Omega} = 93.5 mA\$. This current discharges C1 within a certain amount of time, roughly half a second.
With the second circuit, when SW2 opens, the voltage across R2 and the current through it are the same as R1. However, because the transistor has gain — let's say that the β is 99 — 99% of that current comes from the collector, and only 1% comes from the base. Since the current flowing out of C2 is only \$\frac{1}{100}\$ the current flowing out of C1, it takes 100 times as long to discharge, on the order of 50 seconds. From the point of view of R2, it seems to be connected to a capacitor that is 100× the actual value.
All of this assumes that the collector of Q1 always has a higher voltage than its base. If the collector voltage drops too low because of excessive input ripple, then C2 will also dicharge through the base-collector junction, ruining the &quot;capacitance multiplication&quot; effect.