Rainbow Electronics MAX1801 User Manual
Page 12
MAX1801
Digital Camera Step-Up Slave
DC-DC Controller
12
______________________________________________________________________________________
1) Choose the compensation resistor R
C
that is equiv-
alent to the inverse of the transconductance of the
error amplifier, 1/ R
C
= G
EA
= 100µS, or R
C
= 10k
Ω.
This sets the high-frequency voltage gain of the
error amplifier to 0dB.
2) Determine the maximum output pole frequency:
where:
R
LOAD(MIN)
= V
OUT
/ I
OUT(MAX)
3) Place the compensation zero at the same frequency
as the maximum output pole frequency (in Hz):
Solving for C
C
:
Use values of C
C
less than 10nF. If the above calcu-
lation determines that the capacitor should be
greater than 10nF, use C
C
= 10nF, skip step 4 , and
proceed to step 5.
4) Determine the crossover frequency (in Hz):
f
C
= V
REF
/ (
π
D C
OUT
)
and to maintain at least a 10dB gain margin, make
sure that the crossover frequency is less than or
equal to 1/3 of the ESR zero frequency, or:
3f
C
≤ Z
O
or:
ESR
≤ D / 6 V
REF
If this is not the case, go to step 5 to reduce the
error amplifier high-frequency gain to decrease the
crossover frequency.
5) The high-frequency gain may be reduced, thus
reducing the crossover frequency, as long as the
zero due to the compensation network remains at or
below the crossover frequency. In this case:
ESR
≤ D / (G
EA
R
C
6 V
REF
)
and:
f
C
= (G
EA
R
C
) 2 V
REF
/ (2
π
D C
OUT
)
≥ 1 / (2
π
R
C
C
C
)
Choose C
OUT
, R
C
, and C
C
to simultaneously satisfy
both equations.
Continuous Inductor Current
For continuous inductor current, there are two condi-
tions that change, requiring different compensation.
The response of the control loop includes a right-half-
plane zero and a complex pole pair due to the inductor
and output capacitor. For stable operation, the con-
troller loop gain must drop below unity (0dB) at a much
lower frequency than the right-half-plane zero frequen-
cy. The zero arising from the ESR of the output capaci-
tor is typically used to compensate the control circuit
by increasing the phase near the crossover frequency,
increasing the phase margin. If a low-value, low-ESR
output capacitor (such as a ceramic capacitor) is used,
the ESR-related zero occurs at too high a frequency
and does not increase the phase margin. In this case,
use a lower value inductor so that it operates with dis-
continuous current (see the Discontinuous Inductor
Current section).
For continuous inductor current, the gain of the voltage
divider is A
VDV
= V
REF
/ V
OUT
, and the DC gain of the
error amplifier is A
VEA
= 2000. The gain through the
PWM controller in continuous current is:
A
VO
= (1 / V
REF
) (V
OUT2
/ V
IN
)
Thus, the total DC loop gain is:
A
VDC
= 2000 V
OUT
/ V
IN
The complex pole pair due to the inductor and output
capacitor occurs at the frequency (in Hz):
P
O
= (V
OUT
/ V
IN
) / (2
π
(L
× C
OUT
)
1/2
)
The pole and zero due to the compensation network at
COMP occur at the frequencies (in Hz):
P
C
= G
EA
/ (4000
π
C
C
) = 1 / (4 x 10
7
π
C
C
)
Z
C
= 1 / (2
π
R
C
C
C
)
The frequency (in Hz) of the zero due to the ESR of the
output capacitor is:
Z
O
= 1 / (2
π
C
OUT
ESR)
And the right-half-plane zero frequency (in Hz) is:
The Bode plot of the loop gain of this control circuit is
shown in Figure 5.
Z
(1 - D) R
2 L
RHP
2
LOAD
=
π
C
C
V
V
- V
I
V
- V
C
OUT OUT
OUT
IN
C OUT(MAX)
OUT
IN
=
R
(
)
2
Z
1
2V
- V
V
- V
C
C
C C
OUT
IN
OUT
IN
LOAD(MIN)
OUT
=
=
2
2
π
π
R C
R
(
)
P
2V
- V
V
- V
C
O(MAX)
IN
IN
LOAD(MIN)
=
OUT
OUT
OUT
R
2
π(
)