LM76002 datasheet(28/59 Pages) TI2

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LM76002, LM76003

SNVSAK0A – OCTOBER 2017 – REVISED OCTOBER 2019

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In general, it is preferable to choose lower inductance in switching power supplies, because it usually

corresponds to faster transient response, smaller DCR, and reduced size for more compact designs. However,

too low of an inductance can generate too large of an inductor current ripple such that overcurrent protection at

the full load could be falsely triggered. It also generates more conduction loss because the RMS current is

slightly higher relative that with lower current ripple at the same DC current. Larger inductor current ripple also

implies larger output voltage ripple with the same output capacitors. With peak-current-mode control, it is not

recommended to have an inductor current ripple that is too small. Enough inductor current ripple improves signal-

to-noise ratio on the current comparator and makes the control loop more immune to noise.

Once the inductance is determined, the type of inductor must be selected. Ferrite designs have very low core

losses and are preferred at high switching frequencies, so design goals can concentrate on copper loss and

preventing saturation. Ferrite core material saturates hard, which means that inductance collapses abruptly when

the peak design current is exceeded. The hard saturation results in an abrupt increase in inductor ripple current

and consequent output voltage ripple. Do not allow the core to saturate.

For the design example, a standard 10-μH inductor from Wurth, Coiltronics, or Vishay can be used for the 3.3-V

output with plenty of current rating margin.

8.2.2.6 Output Capacitor Selection

The device is designed to be used with a wide variety of LC filters. TI generally recommends using as little output

capacitance as possible to keep cost and size down. Choose the output capacitor(s), COUT, with care as it

directly affects the steady-state output-voltage ripple, loop stability, and the voltage over/undershoot during load

current transients.

The output voltage ripple is essentially composed of two parts. One is caused by the inductor current ripple going

through the equivalent series resistance (ESR) of the output capacitors:

ΔVOUT-ESR = ΔiL × ESR

(21)

The other is caused by the inductor current ripple charging and discharging the output capacitors:

ΔVOUT-C = ΔiL / (8 × fSW × COUT)

(22)

The two components in the voltage ripple are not in phase, so the actual peak-to-peak ripple is smaller than the

sum of the two peaks.

Output capacitance is usually limited by transient performance specifications if the system requires tight voltage

regulation in the presence of large current steps and fast slew rates. When a fast large load transient happens,

output capacitors provide the required charge before the inductor current can slew to the appropriate level. The

initial output voltage step is equal to the load current step multiplied by the ESR. VOUT continues to droop until

the control loop response increases or decreases the inductor current to supply the load. To maintain a small

overshoot or undershoot during a transient, small ESR, and large capacitance are desired. But these also come

with higher cost and size. Thus, the motivation is to seek a fast control loop response to reduce the output

voltage deviation.

For a given input and output requirement, Equation 23 gives an approximation for an absolute minimum output

cap required:

(23)

Along with this for the same requirement, calculate the maximum ESR as per Equation 24

where

•

r = Ripple ratio of the inductor ripple current (ΔiL / IOUT)

•

ΔVO = target output voltage undershoot

•

D’ = 1 – duty cycle

•

fSW = switching frequency

•

IOUT = load current

(24)

LM76002 Datasheet(PDF) 28 Page - Texas Instruments

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