Datasheet

Iripple
ILpeak = Iout +
2
æ ö
´ -
´
ç ÷
´ ´ ¦
è ø
2
2
1 Vo (Vinmax Vo)
ILrms = Io +
12 Vinmax L1 sw
-
´
´ ¦
Vinmax Vout Vout
Iripple =
L1 Vinmax sw
-
´
´ ´ ¦
Vinmax Vout Vout
L1 =
Io Kind Vinmax sw
TPS5432
www.ti.com SLVSB89A –MARCH 2012–REVISED OCTOBER 2012
The current flowing through the inductor is the inductor ripple current plus the output current. During power up,
faults or transient load conditions, the inductor current can increase above the calculated peak inductor current
level calculated above. In transient conditions, the inductor current can increase up to the switch current limit of
the device. For this reason, the most conservative approach is to specify an inductor with a saturation current
rating equal to or greater than the switch current limit rather than the peak inductor current.
(5)
vertical spacer
(6)
vertical spacer
(7)
vertical spacer
(8)
OUTPUT CAPACITOR
There are three primary considerations for selecting the value of the output capacitor. The output capacitor
determines the modulator pole, the output voltage ripple, and how the regulator responds to a large change in
load current. The output capacitance needs to be selected based on the more stringent of these three criteria.
The desired response to a large change in the load current is the first criteria. The output capacitor needs to
supply the load with current when the regulator can not. This situation would occur if there are desired hold-up
times for the regulator where the output capacitor must hold the output voltage above a certain level for a
specified amount of time after the input power is removed. The regulator is temporarily not able to supply
sufficient output current if there is a large, fast increase in the current needs of the load such as a transition from
no load to full load. The regulator usually needs two or more clock cycles for the control loop to see the change
in load current and output voltage and adjust the duty cycle to react to the change. The output capacitor must be
sized to supply the extra current to the load until the control loop responds to the load change. The output
capacitance must be large enough to supply the difference in current for 2 clock cycles while only allowing a
tolerable amount of drop in the output voltage. Equation 9 shows the minimum output capacitance necessary to
accomplish this.
For this example, the transient load response is specified as a 6% change in Vout for a load step from 0.75 A
(25% load) to 2.25 A (75% load). For this example, ΔIout = 2.25 A - 0.75 A = 1.5 A and ΔVout= 0.06 × 1.8 =
0.108 V. Using these numbers gives a minimum capacitance of 39.7 μF. This value does not take the ESR of the
output capacitor into account in the output voltage change. For ceramic capacitors, the ESR is usually small
enough to ignore in this calculation.
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