I have connected an ESP32 to a 3v3 voltage regulator as seen in the picture below.
Essentially: Unregulated voltage —-[4.7uF]—LDO—[4.7uF]–[0.1uF]–
Then at the ESP32 at the VIN i have [10uF]–[0.1uF]—ESP3V3 [Where all the caps are connected between LDO and ground].
The LDO datasheet https://www.digikey.com/en/datasheets/diodes-incorporated/diodes-incorporated-ap7361c
After connecting it the ESP during wifi boot detects a brownout and constantly resets.
The brown out can be seen below. The duration is aprox 1ms
And the voltage drop is around 1V
At the input side of regulator this is the brownout 
- How do i calculate the capacitors needed to avoid the brownout.
- How would you go around this problem?
I think the problem is that the capacitors i used are simply too slow (have big internal resistance?) to provide the current needed fast enough.
Best Answer
Total energy is measured in watt-seconds (Ws or W-s). Your brownout is approx 4 ms (3.36 ms in one trace), and lets say the transient current is 500 mA. The total energy you need to make up is (3.3 x 0.5 x 0.004) = 0.0066 w-s. Not much.
Regulator output (3.3 V) plus headroom (0.32 V) = Vin(min) - 3.62 V. Lets round that up to 4.0 V for margin.
Total energy (in w-s) in a capacitor is 1/2 C V^2. During the transient, the input capacitor discharges from its nominal value to the minimum regulator input value. If the nominal Vin is 5.0 V (for example), then removing 0.0066 w-s of energy from the capacitor discharges it from V1 to V2 (5.0 V to 4.0 V).
0.0066 = 1/2 x C x (5.0^2) - 1/2 x C x (4.0^2)
Moving the pieces around:
C (in farads) = (0.0066 x 2) / (25 - 16) = 1466 uF
Note: This example has a lot of worst-case assumptions and margin. It is for a complete blackout, where the input voltage source goes to 0 V for 4.0 ms. A shorter transient period, higher input voltage, and lower transient current all work to reduce the size of the holdup capacitor.
In this example, the resulting capacitor size is relatively high because the allowed voltage drop is so small. Increasing the nominal input voltage by only one volt decreases the capacitor size by over 50%.