> 3.3.1 | Example with Low Current Regulator IC

> The average power consumption of a wireless module is 440mA, while the peak consumption is max. 3A during the Tx burst, which can be 1 x or 2 x 577μs.
> In this example the IC delivers 500mA and covers the average power supply needs. However, it is not able to handle current peaks. Current peaks must instead be handled by the capacitor. The capacitor should be able to deliver 2.5A (3A – 500mA) for 1 x or 2 x 577μs. It should also have low ESR and high ripple current. The higher the rated voltage is the better are - as a rule - ESR and ripple current.
> The resistance in the power supply feeding line together with the ESR resistance of the capacitor is 50mOhm. 50mOhm @3A mean a 150mV voltage drop of the module. The IC is a zero resistance current source. The power supply IC delivers 0.5A, the capacitor the remaining 2.5A. The maximum voltage drop is 400mV.
> Since the Tx burst frequency is 217Hz, it can be considered as ripple current. The required capacitance can be calculated using the formula:
> C = Q/U = (I x Δt)/(ΔU)
> C = capacitance (Farad)
> Δt = discharge time (here: 577μs)
> Q = charge
> ΔU = voltage change (here: max. 250mV)
> I = current (here: 2.5A)
> With the sample values this results in:
> C = 2.5A x 577μs / 250mV = 5770μF
> The above example is a worst case scenario with bad antenna matching and strict power restriction by the regulator chip. In practice 4700…7200 μF are enough for GSM; for GPRS class 10 the values will have to be doubled.
> NOTE: A capacitor loses its capacity through ageing or low temperatures. If the product is likely to operate at low temperatures, check the datasheet for remaining capacitance at low temperatures. It is also a good idea to compensate for ageing and low temperatures by choosing a capacitor with higher values than required. Often it is easier to meet the requirements by adding two or more capacitors in parallel.
> These capacitors will dissipate power depending on ripple current and ESR; power that will additionally heat up the capacitor:
> P = I 2 x R / 8 = 39mW (RESR = 50mOhm, I = 2.5A, 1Tx)