L5985 datasheet(24/37 Pages) STMICROELECTRONICS

24 / 37 page

L5985

Application information

24/37

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

L5985 数据表(PDF) 24 Page - STMicroelectronics

L5985 数据表(HTML) 24 Page - STMicroelectronics

Html Pages

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

数据表下载