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TPS54040Q1
TPS54040Q1  

Manufacturer Part Number  TPS54040Q1 
Description  The TPS54040 device is a 42V 0 
Manufacturer  Texas Instruments 
TPS54040Q1 datasheet 

Specifications of TPS54040Q1  

Iout(max)(a)  0.5  Vin(min)(v)  3.5 
Vin(max)(v)  42  Vout(min)(v)  0.8 
Vout(max)(v)  39  Iq(typ)(ma)  0.116 
Switching Frequency(max)(khz)  2500  Switch Current Limit(typ)(a)  0.6 
Topology  Buck,Inverting BuckBoost  Operating Temperature Range(c)  40 to 125 
Pin/package  10MSOPPowerPAD, 10SON  Duty Cycle(max)(%)  95 
Regulated Outputs(#)  1 
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TPS54040Q1
SLVSA26C – JANUARY 2010 – REVISED AUGUST 2011
Output Inductor Selection (L
O
To calculate the minimum value of the output inductor, use
K
is a coefficient that represents the amount of inductor ripple current relative to the maximum output current.
IND
The inductor ripple current will be filtered by the output capacitor. Therefore, choosing high inductor ripple
currents will impact the selection of the output capacitor since the output capacitor must have a ripple current
rating equal to or greater than the inductor ripple current. In general, the inductor ripple value is at the discretion
of the designer; however, the following guidelines may be used.
For designs using low ESR output capacitors such as ceramics, a value as high as K
When using higher ESR output capacitors, K
part of the PWM control system, the inductor ripple current should always be greater than 30 mA for dependable
operation. In a wide input voltage regulator, it is best to choose an inductor ripple current on the larger side. This
allows the inductor to still have a measurable ripple current with the input voltage at its minimum.
For this design example, use K
IND
design, a nearest standard value was chosen: 47μH. For the output filter inductor, it is important that the RMS
current and saturation current ratings not be exceeded. The RMS and peak inductor current can be found from
Equation 30
and
Equation
31.
For this design, the RMS inductor current is 0.501 A and the peak inductor current is 0.567 A. The chosen
inductor is a MSS1048473ML. It has a saturation current rating of 1.44 A and an RMS current rating of 1.83A.
As the equation set demonstrates, lower ripple currents will reduce the output voltage ripple of the regulator but
will require a larger value of inductance. Selecting higher ripple currents will increase the output voltage ripple of
the regulator but allow for a lower inductance value.
The current flowing through the inductor is the inductor ripple current plus the output current. During power up,
faults or transient load conditions, the inductor current can increase above the calculated peak inductor current
level calculated above. In transient conditions, the inductor current can increase up to the switch current limit of
the device. For this reason, the most conservative approach is to specify an inductor with a saturation current
rating equal to or greater than the switch current limit rather than the peak inductor current.
Vinmax
Vout

Lo min =
´
Io
K
´
IND
(
V
´
Vin max
OUT
I
=
RIPPLE
Vin max
L
´ f
´
O
æ
V
1
2
OUT
( )
I
=
I
+
´ ç
L(rms)
O
ç
12
Vinmax
è
Iripple
ILpeak
=
Iout
+
2
Output Capacitor
There are three primary considerations for selecting the value of the output capacitor. The output capacitor will
determine the modulator pole, the output voltage ripple, and how the regulators responds to a large change in
load current. The output capacitance needs to be selected based on the more stringent of these three criteria.
The desired response to a large change in the load current is the first criteria. The output capacitor needs to
supply the load with current when the regulator can not. This situation would occur if there are desired holdup
times for the regulator where the output capacitor must hold the output voltage above a certain level for a
specified amount of time after the input power is removed. The regulator also will temporarily not be able to
supply sufficient output current if there is a large, fast increase in the current needs of the load such as
transitioning from no load to a full load. The regulator usually needs two or more clock cycles for the control loop
30
)
Equation
28.
= 0.2 yields better results. Since the inductor ripple current is
IND
= 0.3 and the minimum inductor value is calculated to be 42 μH. For this
Vout
Vinmax
ƒsw
´
)
 V
OUT
SW
2
(
)
Vinmax  V
ö
´
OUT
÷
÷
L
f
´
´
O
SW
ø
www.ti.com
= 0.3 may be used.
IND
Copyright © 2010–2011, Texas Instruments Incorporated
(28)
(29)
(30)
(31)
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