Application information
L5980
Equation 18
R4
=
B-----W----
fLC
⋅
-1--
K
⋅
R1
where K is the feed forward constant and 1/K is equals to 9.
3. Calculate C4 by placing the zero at 50 % of the output filter double pole frequency (fLC):
Equation 19
C4
=
-------------1--------------
π ⋅ R4 ⋅ fLC
4. Calculate C5 by placing the second pole at four times the system bandwidth (BW):
Equation 20
C5 = 2----π-----⋅---R-----4----⋅---C----C4----⋅4---4-----⋅---B----W--------–----1--
5. Set also the first pole at four times the system bandwidth and also the second zero at
the output filter double pole:
Equation 21
R3 = -4---------⋅------B-----R-----W----1-------–-----1-,
fLC
C3 = 2----π-----⋅---R-----3--1--⋅---4----⋅---B-----W----
The suggested maximum system bandwidth is equals to the switching frequency divided by
3.5 (FSW/3.5), anyway lower than 100 kHz if the FSW is set higher than 500 kHz.
For example with VOUT = 3.3 V, VIN = 12 V, IO = 0.7 A, L = 47 μH, COUT = 22 μF,
ESR < 1 mΩ, the type III compensation network is:
R1 = 4.99kΩ, R2 = 1.1kΩ, R3 = 120Ω, R4 = 5.6kΩ, C3 = 6.8nF, C4 = 10nF, C5 = 100pF
In Figure 12 is shown the module and phase of the open loop gain. The bandwidth is about
57 kHz and the phase margin is 45°.
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