I have a load driven at 24VDC that draws very high current for a short period of time. The following picture is the output of a current amplifier, it reads 2A/V, meaning I have to deal with roughly a 18A pulse that lasts for 50ms.
So far I have calculated the energy of this pulse like this:
Charge spent: 18A @ 50ms, using Q=I.T -> Q = 18*0.05 = 0.9 Coulombs.
Energy drawn: E = 0.5*V*Q = 0.5*24*0.9 = 10.8 Joules.
The 24V is generated by a boost converter, let's suppose I want the output capacitors of this converter to be able to provide all the energy:
E=0.5*C*V² -> C = 2*E / V² -> C = 2*10.8/576 = 37.5mF, or 37,500uF.
That is a humongous ammount of capacitance to install @ 24V. This leads to the question: What percentage of the pulse energy do I need to store on my output capacitors in order not to stress my BOOST converter too much?
I know it might have to do with the impedance between the output capacitor and the load itself, and also that I don't need output caps to be able to provide ALL the energy for the pulse, as the converter should be able to deal with some ammount of load transients.
EDIT#1 @ Brian & Peufeu: I'm considering 0.9F as the correct answer for my previous calculations, using C=dV/dt, from this point on. The boost circuit that generates the 24V is as follows:
Upstream this converter is a 4S3P - 12 x 18650 cell Li-Ion battery, more than capable of providing the current needed. Cabling is also cut to minimum lenght and twisted to minimize media impedance. At the time of design I wasn't sure of what the peak currents on the system would be, so I think I might have chosen my inductor poorly already... here are the specs:
AFAIC, my switch and diode are far beyond capable of withstanding the aforementioned scenario.