Datasheet

=
p´ ´
ESR
ESR OUT
1
f
2 R C
=
p´ ´
Z1
8 8
1
f
2 R C
( )
87
887
2P
CC
RCC2
1
f +´
´´´p
=
TPS40200
www.ti.com SLUS659F –FEBRUARY 2006–REVISED MARCH 2012
In order to properly compensate this system, it is necessary to know the frequencies of its poles and zeros.
Step 1
The break frequency of the output capacitor is given by:
where
• C = the output capacitor, 221 μF
• R
ESR
= the ESR of the capacitors
Because of the ESR of the output capacitor, the output LC filter has a single-pole response above the 1.8-kHz
break frequency of the output capacitor and its ESR. This simplifies compensation since the system becomes
essentially a single pole system.
Step 2
The first zero is place well below the 1.8-kHz break frequency of the output capacitor and its ESR. The phase
boost from this zero is shown in Figure 40.
where
• R
8
= 300 kΩ
• C
8
= 1500 pF
• F
Z1
= 354 Hz
Step 3
From its minimum gain bandwidth product of 1.5 MHz, and knowing it has a 20 dB/decade roll off, the open-loop
gain of the error amplifier is 33 dB at 35 kHz. This approximate frequency is chosen for a crossover frequency to
keep the amplifier gain contribution to the overall system gain small, as well as following the convention of
placing the crossover frequency between 1/6 to 1/10 the 300 kHz switching frequency.
Step 4
The second pole is placed well above the 35 kHz crossover frequency.
where
• R
8
= 300 kΩ
• C
7
= 10 pF
• C
8
= 1500 pF
• f
P2
= 53 kHz
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