Different voltages and amp draw
Your power supply can generate $P = I(A) x E(V) = 0.1A x 5V = 0.5 W.

Now there are two ways to lower the voltage. A linear regulator or a switching regulator.

In the case of the linear regulator, the input current (at 5V) is the same as the output current (at 3.3V) - except for a couple of mA needed for the linear regulator to work. So, at the output you have 0.1A at 3.3V, or 0.33W. The difference (0.5W - 0.33W = 0.17W) makes the regulator heat up.

The switching regulator maintains the same power (not current) - except for a little power needed internally. So we calculator 0.5W at the input, then at the output we have 0.5W too (minus about 10% loss -> about 0.45W). So, the maximum current at the output is 0.45W/3.3V = 0.136A.

If the 200mA peaks are really narrow, a capacitor might save your life... (preferably at the input where the current is lower)