Datasheet

50
55
60
65
70
75
80
85
90
95
100
0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3
I -OutputCurrent- A
O
V =3.3V
I
V =5V
I
Efficiency-%
0.001 0.01 0.1 1 10
I -OutputCurrent- A
O
50
55
60
65
70
75
80
85
90
95
100
Efficiency-%
V =5V
I
V =3.3V
I
OUT OUT
R C
C3 =
R3
´
OUT OUT
ref Igm
2 × c V C
R3 =
gm V V
p ¦ ´ ´
´ ´
¦
¦ ¦ ´
C
sw
= p mod
2
¦ ¦ ´ ¦
C
= p mod z mod
ESR OUT
1
z mod =
2 R C
¦
p ´ ´
TPS54318
SLVS975A –SEPTEMBER 2009–REVISED SEPTEMBER 2013 www.ti.com
(35)
(36)
(37)
The compensation design takes the following steps:
1. Set up the anticipated cross-over frequency. Use Equation 38 to calculate the compensation network’s
resistor value. In this example, the anticipated cross-over frequency (fc) is 45 kHz. The power stage gain
(gm
ps
) is 13 A/V and the error amplifier gain (gm
ea
) is 225 μA/V.
(38)
2. Place compensation zero at the pole formed by the load resistor and the output capacitor. The compensation
network’s capacitor is calculated from Equation 39.
(39)
3. An additional pole can be added to attenuate high frequency noise. In this application, adding the additional
pole is not necessary.
From the procedures above, the compensation network includes a 9.53-kΩ resistor and a 3900-pF capacitor.
APPLICATION CURVES
EFFICIENCY EFFICIENCY
vs vs
LOAD CURRENT LOAD CURRENT
Figure 35. Figure 36.
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