MAX8597/MAX8598/MAX8599
When the output capacitor is comprised of paralleling n
number of the same capacitors, then:
C
O
= n x C
EACH
and
R
ESR
= R
ESR_EACH
/ n
Thus, the resulting f
Z_ESR
is the same as that of a sin-
gle capacitor.
The total closed-loop gain must be equal to unity at the
crossover frequency, where the crossover frequency is
less than or equal to 1/5 the switching frequency (f
S
):
f
C
≤ f
S
/ 5
So the loop-gain equation at the crossover frequency is:
G
EA(FC)
x G
MOD(FC)
= 1
where G
EA(FC)
is the error-amplifier gain at f
C
, and
G
MOD(FC)
is the power-modulator gain at f
C
.
The loop compensation is affected by the choice of out-
put filter capacitor due to the position of its ESR-zero
frequency with respect to the desired closed-loop
crossover frequency. Ceramic capacitors are used for
higher switching frequencies and have low capaci-
tance and low ESR; therefore, the ESR-zero frequency
is higher than the closed-loop crossover frequency.
Electrolytic capacitors (e.g., tantalum, solid polymer,
and OS-CON) are needed for lower switching frequen-
cies and have high capacitance (and some have high-
er ESR); therefore, the ESR-zero frequency can be
lower than the closed-loop crossover frequency. Thus,
the compensation design procedures are separated
into two cases:
Case 1: Crossover frequency is less than the out-
put-capacitor ESR-zero (f
C
< f
Z_ESR
).
The modulator gain at f
C
is:
G
MOD(FC)
= G
MOD(DC)
x (f
P_LC
/ f
C
)
2
Since the crossover frequency is lower than the output
capacitor ESR-zero frequency and higher than the LC
double-pole frequency, the error-amplifier gain must
have a +1 slope at f
C
so that, together with the -2 slope
of the LC double pole, the loop crosses over at the
desired -1 slope.
The error amplifier has a dominant pole at a very low
frequency (~0Hz), and two additional zeros and two
additional poles as indicated by the equations below
and illustrated in Figure 7:
f
Z1_EA
= 1 / (2 π x R4 x C2)
f
Z2_EA
= 1 / (2 π x (R1 + R3) x C1)
f
P2_EA
= 1 / (2 π x R3 x C1)
f
P3_EA
= 1 / (2 π x R4 x (C2 x C3 / (C2 + C3)))
Note that f
Z2_EA
and f
P2_EA
are chosen to have the
converter closed-loop crossover frequency, f
C
, occur
when the error-amplifier gain has +1 slope, between
f
Z2_EA
and f
P2_EA
. The error-amplifier gain at f
C
must
meet the requirement below:
G
EA(FC)
= 1 / G
MOD(FC)
The gain of the error amplifier between f
Z1_EA
and
f
Z2_EA
is:
G
EA(fZ1_EA - fZ2_EA)
= G
EA(FC)
x f
Z2_EA
/ f
C
= f
Z2_EA
/ (f
C
x G
MOD(FC)
)
This gain is set by the ratio of R4/R1 (Figure 6), where
R1 is calculated as illustrated in the Setting the Output
Voltage section. Thus:
R4 = R1 x f
Z2_EA
/ (f
C
x G
MOD(FC)
)
where f
Z2_EA
= f
P_LC
.
Due to the underdamped (Q > 1) nature of the output
LC double pole, the first error-amplifier zero frequency
must be set less than the LC double-pole frequency in
order to provide adequate phase boost. Set the error-
amplifier first zero, f
Z1_EA
, at 1/4 of the LC double-pole
frequency. Hence:
C2 = 2 / (π x R4 x f
P_LC
)
Set the error amplifier f
P2_EA
at f
Z_ESR
and
The error-amplifier gain between f
P2_EA
and f
P3_EA
is
set by the ratio of R4/RM and is equal to:
G
EA(fZ1_EA - fZ2_EA)
x (f
P2_EA
/ f
P_LC
)
where RM = R1 x R3 / (R1 + R3). Then:
RM = R4 x f
P_LC
/ (G
EA(fZ1_EA - fZ2_EA)
x f
P2_EA
)
= R4 x f
C
x G
MOD(FC)
/ f
P2_EA
The value of R3 can then be calculated as:
R3 = R1 x RM / (R1 – RM)
Now we can calculate the value of C1 as:
C1 = 1 / (2 π x R3 x f
p2_EA
)
and C3 as:
C3 = C2 / ((2 π x C2 x R4 x f
P3_EA
) - 1)
