Do I really need a large smoothing capacitor?
The equation is simple enough:

$$\Delta V = \frac{\Delta Q}{C} = \frac{I \Delta t}{C}$$

Solve for C and plug in the numbers:

$$C = \frac{I \Delta t}{\Delta V} = \frac{1 \text{A} \cdot 10 \text{ms}}{0.5 \text{V}} = 20 \text{mF}$$

If that's the amount of droop you want to have at that current, then you need the 20&nbsp;mF capacitor.

Supplies that use smaller capacitors are:


using less current
allowing more droop (more headroom on the input side of a voltage regulator)
operating at a higher frequency (60 Hz vs. 50 Hz)
overly optimistic(?)