1A, 4.5V to 40V Input Buck Converters with
50mA Auxiliary LDO Regulators
f C occurs between f Z2 and f P2 . In this region, the com-
pensator gain (G EA ) at f C is due primarily to C6 and R5.
C8
Therefore, G EA (f C ) = 2 π x f C x C6 x R5 and the modula-
tor gain at f C is:
C6
R6
R5
C7
G MOD ( f C ) =
G MOD _ DC
( 2 π × f C ) 2 × L × C OUT
V OUT
R3
R 5 = C OUT
Since G EA (f C ) x G MOD (f C ) = 1, R5 is calculated by:
f × L × C × 2 π
C 6 × G MOD _ DC
The frequency of f Z1 is set to 0.5 x f LC and now we can
calculate C7:
GAIN
(dB)
R4
EA
REF
CLOSED-LOOP
GAIN
COMP
EA
GAIN
C 7 =
1
0 . 5 × 2 π × R 5 × f LC
f P2 is set at 1/2 the switching frequency (f SW ). R6 is
then calculated by:
f Z1 f Z2
f P2
f C
f P3
FREQUENCY
R 6 =
1
2 π × C 6 × ( 0 . 5 × f SW )
Figure 4. Error Amplifier Compensation Circuit (Closed-Loop
and Error-Amplifier Gain Plot) for Higher ESR Output
Capacitors
Note that if the crossover frequency has been chosen
as 1/10 of the switching frequency, then f P2 = 5xf C .
The purpose of f P3 is to further attenuate the residual
switching ripple at the COMP pin.
If the ESR zero (f ZESR ) occurs in a region between
f C and f SW / 2, then f P3 can be used to cancel it. This
way, the Bode plot of the loop gain plot will not flatten
out soon after the 0dB crossover, and will maintain
its -20dB/decade slope up to 1/2 of the switching
frequency.
frequency is higher than f LC but lower than the closed-
loop crossover frequency. The equations that define
the error amplifier’s poles and zeros (f Z1 , f Z2 , f P2 , and
f P3 ) are the same as before. However, f P2 is now lower
than the closed-loop crossover frequency. Figure 4
shows the error amplifier feedback as well as its gain
response for circuits that use higher-ESR output capac-
itors (tantalum or aluminum electrolytic).
Again, starting from R3, calculate C6 for zero f Z2 :
If the ESR zero well exceeds f SW /2 (or even f SW ), f P3
should in any case be set high enough not to erode the
phase margin at the crossover frequency. For example,
C 6 =
1
2 π × f LC × R 3
it can be set between 5 x f C and 10 x f C .
The value for C8 is calculated from:
and then place f P2 to cancel the ESR zero. R6 is calcu-
lated as:
C 8 =
C 7
( 2 π × C 7 × R 5 × f P 3 ? 1 )
R 6 =
C OUT × ESR
C 6
Compensation when f C > f ZESR
For larger ESR capacitors such as tantalum and alu-
minum electrolytic, f ZESR can occur before f C . If f ZESR
< f C , then f C occurs between f P2 and f P3 . f Z1 and f Z2
remain the same as before however, f P2 is now
set equal to f ZESR . The output capacitor’s ESR zero
If the value obtained here for R6 is not considerably
smaller than R3, then recalculate C6 using (R3 + R6) in
place of R3. Then use the new value of C6 to obtain a
better approximation for R6. The process can be further
iterated, and convergence is ensured as long as f LC <
f ZESR .
20
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