Calculate needed capacitance

Question

I have this application which requires 18VDC as well as 12VDC. To get 12V I have designed a buck converter circuit which can supply up to 2A.

The problem:

The 18VDC (3.33A) supply may suffer from shorts which may last up to 50ms. This has a negative effect on the 12V line. The voltage drops here as well.

What I want to achieve:

If the ingoing 18VDC supply experiences a 50ms short, I want to keep the input voltage on the buck converter above 13V to prevent the 12V from dipping.

My proposed solution:

I want to use a single diode and a capacitor to bridge the shortcircuit gaps. The idea is to keep to the voltage above 13V during 50ms on the ingoing side of the buck converter.

(My other alternative is to use a bigger PSU and limit the short current to prevent a voltage drops in the first place)

My question.

I want to know, if using the capacitor is possible and what the mininum capacitance should be.

I used the following constant current discharge formula I got from Mouser's. t ＝ ｛C × （V0－V1)｝ ／ I

With

t  = 0.05s
V0 = 17.3V  (18V - V diode)
V1 = 13V
I  = 1.5A (estimated current draw on the ingoing side of the buck converter, I believe that the current draw is increased when the voltage drops)

I calculated that C would be atleast a whopping 57F. I am however not entirely sure if this calculation is correct.

Answer 1

If the voltage drops from 17.3 volts to 13 volts then, \$\frac{dv}{dt}\$ = 86 volts per second based on a 50 ms drop-out. Next re-arrange the standard capacitor formula: -

$$I = C\frac{dv}{dt}$$

$$C = I\frac{dt}{dv}$$

And plug in the numbers: -

$$C = 1.5\times\dfrac{1}{86}$$

Hence, the capacitance needed is 17,442 μF (call it 18 mF). Of course if the buck converter has a low-drop-out voltage and, you can live with the 12 volts dropping as low as 10 volts, the hold-up capacitor can be smaller.

And, if instead of a buck regulator you used a buck-boost regulator, you should be able to get it even lower.

But, bear in mind the assumption about the 1.5 amps. This will only be true at the instant the drop-out occurs because it will rise towards 2 amps as the input supply to the buck converter drops towards 12 volts.

I recommend that you simulate to double-check things.