SC4524
POWER MANAGEMENT
Applications Information
wz1 is shown to be less than wp2 in Figure 11. Making
ꢀ
w
= -
w
w
5 &
w
=
=
gives a first-order estimate of C5:
ꢀ
ꢀꢁ
In addition C5 and R5 form a zero with angular frequency:
ꢁꢂ
&
(12)
w 5
ꢀ
w
= -
5 &
Notice that R5 determines the mid-band loop gain of the
converter. Increasing R5 increases the mid-band gain and
the crossover frequency. However it reduces the phase
margin. C6 is a small ceramic capacitor to roll off the
The
output-to-control
transfer
function
Y
Y
Y
=
¼
is also shown in Figure 11. Its mid-
w
Y
Y
Y
loop gain at high frequency. Placing p3 at about
gives:
(13)
ꢃ
Ë
Û
5
ꢀ
Ì
Ì
Ü
Ü
* 5
band gain (between z1 and p3) is
. The
&
5 +5
Í
Ý
pI5
overall loop gain T(s) is the product of the control-to-
output and the output-to-control transfer functions. To
Computed R5, C5 and C6 can indeed result in near optimal
load transient responses in over half of the applications.
However in other cases empirically determined
compensation networks based on optimized load
transient responses may differ from those calculated by
a factor of 3. Therefore checking the transient response
of the converter is imperative. Starting with calculated
R5, C5 and C6 (using n=1 in Equations (11)-(13)), apply
the largest expected load step to the converter at the
maximum operating VIN. Observe the load transient
response of the converter while adjusting R5, C5 and C6.
Choose the largest R5, the smallest C5 and C6 so that
the inductor current waveform does not show excessive
ringing or overshoot.
7ꢁMwꢀ
simplify
Bode plot, the feedback network is
assumed to be resistive. If the overall loop gain is to
cross 0dB at one tenth of the switching frequency
w
ꢁꢂ
pI
ꢀ
w =
=
(
) at 20dB/decade, then its mid-band gain
(between z1 and p2) will be
w
ꢀꢁ
Q
w
w
w & 5
=
=
ꢀꢁQ
& 5
Ë
Û
Feedforward capacitor C11 boosts phase margin over a
limited frequency range and is sometimes used to
improve loop response. C11 will be more effective if
5
Ì
Ì
Ü
Ü
* 5 * 5
This is also equal to
. Therefore
5 +5
Í
Ý
Ë
Û
5
w & 5
.
5 >> 5 ÔÏ5
Ì
Ì
Ü
Ü
* 5 * 5
=
.
5 + 5
ꢃꢁQ
Í
Ý
Example: Determine the compensation components for
the 550kHz 12V to 3.3V converter in Figure 13(a).
Re-arranging,
For the converter, w = ꢊꢄꢃ0UDGV , ,
= ꢅ$ and
Ë
Û
5
5
w &
ꢁꢂQ* *
Ì
Ü
Ü
5 = ꢁ +
(11)
Ì
Í
& = ꢅꢅm) . n is assumed to be 1 in (11) and (12).
Ý
ã 2006 Semtech Corp.
18
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