> 3.3.2 | Example with High Current Regulator IC

> In the previous example there was a need for a large capacitor. To be able to use a smaller capacitor, the IC must deliver higher currents.
> In this example the IC can deliver 2.5A. It means that the ripple current over the capacitor is 0.5A.
> Assuming the IC is a zero resistance current source the capacitor delivers 0.5A. Even with an ESR of 300mOhm a maximum voltage drop of 150mV occurs. The required capacitance can be calculated as follows:
> C = Q / U = (I x Δt) / (ΔU)
> C = capacitance (Farad)
> Δt = discharge time (here: 577μs)
> Q = charge
> ΔU = voltage change (here: max. 400mV-150mV=250mV)
> I = current (here: 3A-2.5A=0.5A)
> With the sample values this results in:
> C = 0.5A x 577μs / 250mV = 1154μF
> In practice 470…1000 μF are enough for GSM, for GPRS class 10 at least 2200μF are recommended.