Datasheet

( )
OL
f6
I
OL
IN
O
Av1
)S(Z
10
R
R
)S(Z
1
Av
)S(
V
V
+´++
-
=
6
10
fb
R
R
K =
( )
D1RDR)S(Z)S(Z
)S(Z
)S(X
SRSWOUT
OUT
LC
-´+´++
=
( )
( )
)S(XKSKKT
LCPWMEAFBSV
´´´=
TPS40200-EP
SGLS371A –JANUARY 2007–REVISED JANUARY 2013 www.ti.com
Step 5
Calculate the three other gain elements of the system to determine the gain required by the error amplifier at 35-
kHz to make the overall gain 0 dB:
(23)
Where:
K
FB
= Output voltage setting divider
K
EA
= Error amplifier feedback
K
PWM
= Modulator gain
X
LC
= Filter transfer function
The output filter transfer characteristic is given by the following:
(24)
Where:
Z
OUT
= Parallel combination of output capacitor(s) and the load
Z
OUT
and Z
l
should include parasitic R and L.
Evaluating the response at 35-kHz gives the following:
• The full current output load at 3.3 V is 1.32 Ω, and is in parallel with the 0.4-Ω ESR of the output capacitor.
• Including the 400 mΩ of ESR, the capacitive impedance is 14 mΩ, and Z
OUT
= 414 mΩ.
• The impedance of the inductor is Z
l
= 1.659 Ω.
• X
LC(S)
= 0.033, or –29.6 dB
The feedback network has a gain to the error amplifier given by:
(25)
Where:
R6 = 26.7 kΩ
Using the values in this application, K
fb
= 11.4 dB.
The modulator has a gain of 10 that is flat to well beyond 35 kHz, so K
PWM
= 20 dB.
The amplifier gain, including the feedback gain, K
fb
, can be approximated by this expression:
(26)
Where:
Z
I
= R
10
Z
f
= Parallel combination of C
7
in parallel with the sum of R
8
and the impedance of C
8
The gain required to achieve 0-dB system gain is simply the sum of the other three gains: –(–29.6 + 11.4 + 20) =
1.8 dB. With an open-loop gain of 33 dB, the closed-loop gain of the amplifier is 0.8, or –1.66 dB, which gives a
0.13-dB gain at 35 kHz.
Figure 40 shows the result of the compensation. The crossover frequency is 35 kHz, and the phase margin is
45°. The response of the system is dominated by the ESR of the output capacitor and is exploited to produce an
essentially single-pole system with simple compensation.
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