L5985 datasheet(24/37 Pages) STMICROELECTRONICS

24 / 37 page

L5985

Application information

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Equation 22

Where fESR is the ESR zero:

Equation 23

and Vs is the saw-tooth amplitude. The voltage feed forward keeps the ratio Vs/Vin constant.

Calculate C4 by placing the zero one decade below the output filter double pole:

Equation 24

Then calculate C3 in order to place the second pole at four times the system bandwidth

(BW):

Equation 25

For example with VOUT = 3.3 V, VIN = 12 V, IO = 2 A, L = 15 μH, COUT = 330 μF,

ESR = 50 m

Ω, the type II compensation network is:

Equation 26

In Figure 15 is shown the module and phase of the open loop gain. The bandwidth is about

37 kHz and the phase margin is 46°.

ESR

------------

⎝⎠

⎛⎞

ESR

------------

--------- R

⋅⋅

⋅

ESR

π ESR C

OUT

⋅⋅

--------------------------------------------

π R

⋅⋅

-------------------------------

π R

4BW

⋅

–

⋅⋅⋅

--------------------------------------------------------------

1.1k

249

10k

68nF

68pF

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