Datasheet

Vin = 10 V / div
Vout = 2 V / div
EN = 2 V / div
SS/TR = 2 V / div
Time = 5 msec / div
Vout = 50 mv / div (ac coupled)
Output Current = 1 A / div (Load Step 1.5 A to 2.5 A)
Time = 200 usec / div
1
sw
f p
=
´ ´
C8
R4
Re´
=
o
C sr
C8
R4
p
1
C5
2 R4 mod
=
´ ´ ´ fp
2 f
gmps gmea
p
æ ö
æ ö
´ ´ ´
= ´
ç ÷
ç ÷
´
è ø
è ø
co out out
ref
C V
R4
V
2
sw
p
f
f f= ´
co
mod
p z
f f f= ´
co
mod mod
1
z mod =
2 Resr × Cout
¦
´ p ´
Ioutmax
p mod =
2 × × Vout × Cout
¦
p
TPS54240
www.ti.com SLVSAA6A –APRIL 2010–REVISED DECEMBER 2012
(41)
(42)
(43)
(44)
To determine the compensation resistor, R4, use Equation 45. Assume the power stage transconductance,
gmps, is 10.5A/V. The output voltage, Vo, reference voltage, VREF, and amplifier transconductance, gmea, are
3.3V, 0.8V and 310 μA/V, respectively. R4 is calculated to be 20.2 kΩ, use the nearest standard value of 20.0
kΩ. Use Equation 46 to set the compensation zero to the modulator pole frequency. Equation 46 yields 4740 pF
for compensating capacitor C5, a 4700 pF is used for this design.
(45)
(46)
A compensation pole can be implemented if desired using an additional capacitor C8 in parallel with the series
combination of R4 and C5. Use the larger value of Equation 47 and Equation 48 to calculate the C8, to set the
compensation pole. C8 is not used for this design example.
(47)
(48)
Discontinuous Mode and Eco Mode Boundary
With an input voltage of 12 V, the power supply enters discontinuous mode when the output current is less than
337 mA. The power supply enters EcoMode when the output current is lower than 5 mA.
The input current draw at no load is 392 μA.
APPLICATION CURVES
Figure 50. Load Transient Figure 51. Startup With VIN
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