Diodes AP3595 Manuel d'utilisateur Page 16

  • Télécharger
  • Ajouter à mon manuel
  • Imprimer
  • Page
    / 23
  • Table des matières
  • MARQUE LIVRES
  • Noté. / 5. Basé sur avis des utilisateurs
Vue de la page 15
AP3595
Document number: DS36749 Rev. 1 - 2
16 of 23
www.diodes.com
January 2014
© Diodes Incorporated
AP3595
A Product Line of
Diodes Incorporated
Application Information (Cont.)
Calculate the C2 by the equation:
75.022
1
2
LC
fR
C
4. Set the pole at the ESR zero frequency f
ESR
:
ESRP
ff
1
Calculate the C1 by the following equation:
1222
2
1
ESR
fCR
C
C
5. Set the second pole f
P2
at the half of the switching frequency and also set the second zero f
Z2
at the output LC filter double pole f
LC
. The
compensation gain should not exceed the error amplifier open loop gain. Check the compensation gain at f
P2
with the capabilities of the error
amplifier.
LCZ
SWP
ff
ff
2
2
5.0
Combine the two equations will get the following component calculations:
1
2
1
3
LC
SW
f
f
R
R
SW
fR
C
3
1
3
Figure 10.Converter Gain and Frequency
14. Output Inductor Selection
The duty cycle (D) of a buck converter is the function of the input voltage and output voltage. Once an output voltage is fixed, it can be written as:
INOUT
VVD /
For two-phase converter, the inductor value (L) determines the sum of the two inductor ripple current, ΔI
P-P
, and affects the load transient
response. Higher inductor value reduces the output capacitors’ ripple current and induces lower output ripple voltage. The ripple current can be
approximated by:
IN
OUT
SW
OUTIN
PP
V
V
Lf
VV
I
2
Where f
SW
is the switching frequency of the regulator.
Vue de la page 15
1 2 ... 11 12 13 14 15 16 17 18 19 20 21 22 23

Commentaires sur ces manuels

Pas de commentaire