Datasheet

VO
IO
VIN
VOUT
EN
IL
( )
C
Z mod
1
C
2 R
f
f
=
p ´ ´
( )
C
C
P mod
1
C
2 R f
=
p ´ ´
( ) ( ) ( )
OUT
C
REF
MOD c Z mod EA
V
R
G gm V
f
f
=
´ ´ ´
OUT ESR
C
C R
C
R
f
´
=
( )
C
C
P mod
1
C
2 R f
=
p ´ ´
( ) ( )
OUT
C
REF
MOD c EA
V
R
G gm V
f
=
´ ´
TPS54140A
SLVSB55B –MAY 2012–REVISED JANUARY 2014 www.ti.com
For the example problem, the gain of the modulator at the cross over frequency is 0.542. Next, the compensation
components are calculated. A resistor in series with a capacitor is used to create a compensating zero. A
capacitor in parallel to these two components forms the compensating pole. However, calculating the values of
these components varies depending on if the ESR zero is located above or below the cross over frequency. For
ceramic or low ESR tantalum output capacitors, the zero will usually be located above the cross over frequency.
For aluminum electrolytic and tantalum capacitors, the modulator zero is usually located lower in frequency than
the cross over frequency. For cases where the modulator zero is higher than the cross over frequency (ceramic
capacitors).
(48)
(49)
(50)
For cases where the modulator zero is less than the cross over frequency (Aluminum or Tantalum capacitors),
the equations are:
(51)
(52)
(53)
For the example problem, the ESR zero is located at a higher frequency compared to the cross over frequency
so Equation 50 through Equation 53 are used to calculate the compensation components. For the example
problem, the components are calculated to be: R
C
= 76.2 kΩ, C
C
= 2710 pF, and Cƒ = 6.17 pF.
The calculated value of the Cƒ capacitor is not a standard value so a value of 2700 pF will be used. 6.8 pF is
used for C
C
. Rc resistor sets the gain of the error amplifier which determines the cross over frequency. The
calculated R
C
resistor is not a standard value, so 76.8kΩ will be used.
APPLICATION CURVES
Figure 52. Load Transmit Figure 53. Startup With EN
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