Datasheet

LTC3114-1
27
Rev. C
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APPLICATIONS INFORMATION
Assuming the error amp zero is designed as just described,
at frequencies above P2 (and Z1), the closed-loop gain of
our system simplifies to:
G
CL
=
G
CS
•R
LOAD
• g
m
•R
Z
V
OUT
where:
G
CS
is the inner current loop closed-loop transconduc-
tance = 1.97A/V
R
LOAD
is the minimum load resistance in ohms
g
m
is the transconductance of the error amplifier, 120µS
R
Z
is the compensation zero setting resistor (one of
our design variables)
V
OUT
is the output voltage
Our desired closed-loop frequency (f
CC
) defined earlier is
29kHz. Assuming that we have a single pole response in
our system, we can express the ratio of the closed-loop
crossover frequency to f
P1
in the buck mode of operation
as follows:
f
CC
f
P1
=
G
CS
•R
LOAD
• g
m
•R
z
V
O
We can now calculate R
Z
by rearranging the previous
equation:
R
Z
=
f
CC
• V
OUT
• 2π •C
OUT
G
CS
• g
m
It’s important to note that the value of R
Z
is proportional
to the overall crossover frequency, f
CC
. If we later want to
adjust f
CC
lower, for example, R
Z
can be lowered in value
and C
P1
increased proportionally to keep the compensa-
tion zero at the same frequency.
As mentioned previously
, we will place the zero at fre-
quency P1, yielding:
C
P1
=
1
2π •R
Z
• f
P1
or more simply,
R
LOAD
•C
OUT
R
Z
where R
l
is the minimum load resistance in buck mode,
12Ω in this example.
Quickly substituting our values in the above equations
yields:
R
Z
= 407k, C
P1
= 1.3nF,
but please continue reading as this is not the final answer.
If the inner current loop were an ideal V
CCS
, then the
previously derived compensation would be sufficient to
stabilize the converter. However, the inner current loop
utilizes an operational amplifier with an integral compen
-
sation network, which contributes an additional zero and
pole in the power stage response, the gain peaking, as
described previously
. The effect of the additional zero/pole
pair pushes out f
CC
, our crossover frequency, beyond what
was predicted by the previous calculations. A simplified
approach to calculating our compensation components
then is to re-use the previous equations but scale f
CC
, the
cross over frequency, by a scaling factor (α), which will
account for the gain boost present in the system:
f
CC′
=
f
CC
3
• α
( )
, where α = 0.42
So, in our example, this results in:
f
CC′
=
29kHz
3
• 0.42
( )
= 4.06kHz
Using the new value of f
CC′
in the previous equations for
R
Z
and C
C
yields:
R
Z
=
4.06kHz•12V • 2π • 44µF
1.97A
V
•
120µA
V
R
Z
= 56.9kΩ, use 56.2kΩ
C
P1
=
12Ω • 44µF
56.2k
C
P1
= 9.4nF, use 10nF
C
P2
is usually chosen to be a small value around 10pF as
it is meant to filter out high frequency switching frequency
related components.
Keep in mind that this analysis assumes that the zero pro
-
vided by the output capacitor and its ESR is at a frequency
much higher than f
CC
.
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