Datasheet

( )
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
=
´ ´
TPS54160
www.ti.com SLVS795E –OCTOBER 2008–REVISED SEPTEMBER 2013
For the example problem, the gain of the modulator at the crossover 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 crossover frequency. For
ceramic or low ESR tantalum output capacitors, the zero will usually be located above the crossover frequency.
For aluminum electrolytic and tantalum capacitors, the modulator zero is usually located lower in frequency than
the crossover frequency. For cases where the modulator zero is higher than the crossover frequency (ceramic
capacitors).
(48)
(49)
(50)
For cases where the modulator zero is less than the crossover 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 crossover frequency so
Equation 50 through Equation 53 are used to calculate the compensation components. In this example, the
calculated components values are:
• R
C
= 76.2 kΩ
• C
C
= 2710 pF
• Cƒ =6.17 pF
The calculated value of the Cf capacitor is not a standard value so a value of 2700 pF is used. 6.8 pF is used for
C
C
. The R
C
resistor sets the gain of the error amplifier which determines the crossover frequency. The calculated
R
C
resistor is not a standard value, so 76.8 kΩ is used.
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