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Application information

L5963

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DocID028553 Rev 1

7.2

Output capacitors (COUT)

Output capacitors are selected to support load transients and output ripple current, as well

as to get loop stability.

The amount of voltage ripple can be calculated by the output ripple current flowing in the

Inductor:

Equation 4

Usually the first term is dominant. However, if a ceramic capacitor (which is recommended)

is adopted, the first term on the above equation can be neglected as the ESR value is very

low.

For example, in case Vout = 3.3 V, Vin = 14 V, fSW = 250 kHz, IL = 0.3 x 3 A = 0.9 A, in order

to have a

Vout = 5%* Vout =0.165 V, a 4.7 μF ceramic capacitor is needed. In case of not

negligible ESR (electrolytic or tantalum capacitors), the capacitor is chosen taking into

account its ESR value.

In the above example, if a 100 μF with ESR = 100 mΩ electrolytic capacitor is chosen, the

voltage drop on ESR dominates and the voltage ripple is 90 mV.

The output capacitor is also important to sustain the output voltage during a load transient.

In general, minimizing the ESR value and increasing the output capacitance results in a

better transient response. The ESR can be minimized by simply adding more capacitors in

parallel, or by using higher quality capacitors. If ceramic capacitors are chosen, in presence

of a fast load transient the output voltage will change by the amount.

Equation 5

Where:

Iout(max), Iout(min) refer to the worst case load in the system and Vout is the tolerance of the

regulated output voltage, 5% of Vout.

For example, Vout = 5 V, Vin = 14 V, Iout(max) =1.5 A, Iout(min) = 0.5 A, L = 22 μH

Equation 6

So two 10 μF ceramic capacitors in parallel are needed.

The output capacitor is also important for loop stability: it fixes the double LC filter pole and

the zero due to its ESR. In Section 7.5: Compensation network, it will be illustrated how to

consider its effect in the system stability.

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