MAX1531ETJ+ Maxim Integrated Products, MAX1531ETJ+ Datasheet - Page 27

• Manufacturer Part Number: MAX1531ETJ+
• Description: IC PS CTRLR MULTI-OUTPUT 32TQFN
• Manufacturer: Maxim Integrated Products

Specifications of MAX1531ETJ+

• Applications: Five Power Supply Monitor
• Voltage - Supply: 4.5 V ~ 28 V
• Current - Supply: 1.7mA
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 32-TQFN Exposed Pad
• Voltage - Output: 1.25 ~ 16.5 V
• Number Of Outputs: 5
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Voltage - Input: -
• Lead Free Status / Rohs Status: Details

4) If f
f
CROSSOVER
If the frequency of the secondary pole is below the
crossover frequency, the frequency of the secondary
pole must be moved higher, or the crossover fre-
quency must be moved lower. There are two ways to
increase the frequency of the secondary pole:
increase the high-side MOSFET R
the step-down inductance, L. As explained before, for
given input and output voltages, the current ramp sig-
nal is proportional to the high-side MOSFET R
and inversely proportional to the inductance. If the
pole occurs below the crossover frequency, the cur-
rent feedback signal is too small. Increasing R
or reducing the inductance can increase the current
feedback signal. To lower the crossover frequency,
use the methods described in step 3. Repeat steps 1
to 4 after making the changes.
crossover frequency as a function the MOSFET
R
Change one or both of these circuit parameters to
obtain the desired crossover. Recalculate ADC and
repeat steps 1 to 3 after making the changes.
cancel the pole with a feed-forward zero. Determine the
value of C23 (feedback capacitor) using the following:
C23 also forms a secondary pole with R1 and R2
given by the following:
The frequency of this pole should be above the
crossover frequency for loop stability. The position of
this pole is related to the high-frequency current-
mode pole, which is determined by the inductor cur-
rent ramp signal. The inductor current ramp signal
must satisfy the following condition to ensure the
pole occurs above the crossover frequency:
DS(ON)
m
POLE(HIGH)
1
>
Multiple-Output Power-Supply Controllers for
(
f
POLE SEC
R
1
and the output capacitance:
+
C
=
2
R
23
_
π
2
2π
)
× ×
≈
is less than the crossover frequency,
×
D
×
______________________________________________________________________________________
2
f
SW
'
A
π
=
VCS
2
R
×
-
π
2
2
f
×
×
POLE HIGH
×
π
g
V
f
CROSSOVER
m
× ×
OUT SET
(
R
D
1
1
×
(
'
(
||
V
1
R
FB
R
2
)
)
2
×
×
)
×
DS(ON
×
f
C
CROSSOVER
×
R
×
OUT
R
11
1
C
m
23
C
), or reduce
×
R
DS ON
DS(ON)
DS(ON)
(
)
,
5) For most applications using tantalum or polymer
Applications using ceramic capacitors usually have
ESR zeros that occur at least one decade above the
crossover. Since the ESR zero of ceramic capacitors
has little effect on the loop stability, it does not need to
be cancelled.
The following is an example. In the circuit of Figure 1,
the input voltage is 12V, the output voltage is set to
3.3V, the maximum load current is 1.5A, the typical on-
resistance of the high-side MOSFET is 100mΩ, and the
inductor is 10µH. The calculated equivalent load resis-
tance is 1.67Ω. The DC loop gain is:
If the chosen crossover frequency is 20kHz (step 1):
With a 22µF output capacitor, the output pole of the
step-down regulator is (step 2):
Calculate R11 using:
capacitors, the output capacitor’s ESR forms a sec-
ond zero that occurs either below or close to the
crossover frequency. The zero must be cancelled
with a pole. Verify the frequency of the output capac-
itor’s ESR zero, which is:
where R
If the output capacitor’s ESR zero does not occur
well after the crossover, add the parallel compensa-
tion capacitor (C2) to form another pole to cancel the
ESR zero. Calculate the value of C2 using:
f
POLE OUT
A
C
DC
R
2
f
C
ESR
ZERO ESR
11
≈
10
(
≈
≈
≈
is the ESR of the output capacitor C
2
1 238
(
)
π
.
2
=
×
2
π
3 3
π
.
100
)
f
ZERO ESR
×
V
2
V
×
=
4 3
π
LCD Monitors
20
×
μ
.
×
S
×
(
100
kHz
kHz
2π
1 67
22
1
.
×
×
C
m
μ
)
4180
1
×
C
10
Ω
F
×
×
Ω
2000
1 7
OUT
×
.
×
×
R
1 67
1
11
3 5
.
nF
2000
.
×
×
Ω
≈ .
=
R
1 7
C
ESR
22
=
=
10 1
nF
4 3
4180
k
.
Ω
-
kHz
OUT
27
.