Datasheet

V =50mV/div(accoupled)
OUT
I =1 A/div,
0to2 A step
OUT
Time=2ms/div
V =50mV/div(accoupled)
OUT
I =1 A/div,
0.5to1.5 A step
OUT
Time=2ms/div
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
I -OutputCurrent- A
O
Efficiency-%
V =3.3V
I
V =5V
I
0
10
20
30
40
50
60
70
80
90
100
0.001 0.01 0.1 1 10
I -OutputCurrent- A
O
Efficiency-%
V =3.3V
I
Vin=5V
OUT OUT
R C
C3 =
R3
´
OUT OUT
ref Igm
2 × c V C
R3 =
gm V V
p ¦ ´ ´
´ ´
TPS54218
SLVS974B –SEPTEMBER 2009–REVISED JULY 2013 www.ti.com
(38)
2. Place compensation zero at the pole formed by the load resistor and the output capacitor. The compensation
network’s capacitor can be calculated from Equation 39.
(39)
3. An additional pole can be added to attenuate high frequency noise. In this application, it is not necessary to
add it.
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.
TRANSIENT RESPONSE, 1 A STEP TRANSIENT RESPONSE, 2 A STEP
Figure 37. Figure 38.
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