TPS54218_12 Datasheet, PDF (24/37 Pages) Texas Instruments – 2.95 V to 6 V Input, 2 A Output, 2MHz, Synchronous Step Down

TPS54218
SLVS974A – SEPTEMBER 2009 – REVISED AUGUST 2012
www.ti.com
R3 = ^{2p × ¦c ´ VOUT ´ COUT}
gm ´ V_ref ´ V_Igm
(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.
C3 = ^{ROUT ´ COUT}
R3
(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
100
95
90
85
80
75
70
65
60
55
50
0
EFFICIENCY
vs
LOAD CURRENT
V_I = 3.3 V
V_I = 5 V
0.25 0.5 0.75 1
1.25 1.5 1.75 2
I_O - Output Current - A
Figure 34.
EFFICIENCY
vs
LOAD CURRENT
100
90
V_I = 3.3 V
80
70
60
Vin = 5 V
50
40
30
20
10
0
0.001
0.01
0.1
1
10
I_O - Output Current - A
Figure 35.
TRANSIENT RESPONSE, 1 A STEP
V_OUT = 50 mV/div (ac coupled)
TRANSIENT RESPONSE, 2 A STEP
V_OUT = 50 mV/div (ac coupled)
I_OUT = 1 A/div,
0.5 to 1.5 A step
I_OUT = 1 A/div,
0 to 2 A step
Time = 2 ms/div
Figure 36.
Time = 2 ms/div
Figure 37.
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