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MAX8745 Datasheet(PDF) 33 Page - Maxim Integrated Products

Part # MAX8745
Description  High-Efficiency, Quad Output, Main Power-Supply Controllers for Notebook Computers
PDF  36 Pages
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Manufacturer  MAXIM [Maxim Integrated Products]
Direct Link  https://www.maximintegrated.com/en.html
Logo MAXIM - Maxim Integrated Products

MAX8745 Datasheet(HTML) 33 Page - Maxim Integrated Products

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High-Efficiency, Quad Output, Main Power-
Supply Controllers for Notebook Computers
______________________________________________________________________________________
33
The output capacitor and the load resistance create the
dominant pole in the system. However, the internal ampli-
fier delay, the pass transistor’s input capacitance, and the
stray capacitance at the feedback node create additional
poles in the system, and the output capacitor’s ESR gen-
erates a zero. For proper operation, use the following
steps to ensure the linear-regulator stability:
1) First, calculate the dominant pole set by the linear
regulator’s output capacitor and the load resistor:
where COUTA is the output capacitance of the aux-
iliary LDO and RLOAD is the load resistance corre-
sponding to the maximum load current. The unity-
gain crossover of the linear regulator is:
fCROSSOVER = AV(LDO)fPOLE(LDO)
2) The pole caused by the internal amplifier delay is at
approximately 1MHz:
fPOLE(AMP)
≈ 1MHz
3) Next, calculate the pole set by the transistor’s input
capacitance, the transistor’s input resistance, and
the base-to-emitter pullup resistor. Since the tran-
sistor’s input resistance (hFE/gm) is typically much
greater than the base-to-emitter pullup resistance,
the pole can be determined from the simplified
equation:
where gm is the transconductance of the pass tran-
sistor, and fT is the transition frequency. Both para-
meters can be found in the transistor’s data sheet.
Therefore, the equation can be further reduced to:
4) Next, calculate the pole set by the linear regulator’s
feedback resistance and the capacitance between
FBA and ground (approximately 5pF including
stray capacitance):
5) Next, calculate the zero caused by the output
capacitor’s ESR:
where RESR is the equivalent series resistance of
COUTA.
6) To ensure stability, choose COUTA large enough so
that the crossover occurs well before the poles and
zero calculated in steps 2 through 5. The poles in
steps 3 and 4 generally occur at several MHz, and
using ceramic output capacitors ensures the ESR
zero occurs at several MHz as well. Placing the
crossover frequency below 500kHz is typically suf-
ficient to avoid the amplifier delay pole and gener-
ally works well, unless unusual component
selection or extra capacitance moves the other
poles or zero below 1MHz.
A capacitor connected between the linear regula-
tor’s output and the feedback node can improve
the transient response and reduce the noise cou-
pled into the feedback loop.
If a low-dropout solution is required, an external p-
channel MOSFET pass transistor could be used.
However, a pMOS-based linear regulator requires
higher output capacitance to stabilize the loop. The
high gate capacitance of the p-channel MOSFET
lowers the fPOLE(CIN) and can cause instability. A
large output capacitance must be used to reduce
the unity-gain bandwidth and ensure that the pole
is well above the unity-gain crossover frequency.
Applications Information
Duty-Cycle Limits
Minimum Input Voltage
The minimum input operating voltage (dropout voltage)
is restricted by the maximum duty-cycle specification
(see the Electrical Characteristics table). For the best
dropout performance, use the slowest switching fre-
quency setting (200kHz, FSEL = GND). However, keep
in mind that the transient performance gets worse as
the step-down regulators approach the dropout volt-
age, so bulk output capacitance must be added (see
the voltage sag and soar equations in the SMPS Design
Procedure and Transient Response sections). The
absolute point of dropout occurs when the inductor cur-
rent ramps down during the off-time (
ΔIDOWN) as much
as it ramps up during the on-time (
ΔIUP). This results in
a minimum operating voltage defined by the following
equation:
f
CR
ZERO ESR
OUTA ESR
()
=
1
2
π
f
CR
R
POLE FBA
FBA
()
(||
)
=
1
25
6
π
f
f
h
POLE CIN
T
FE
()
≈
f
CR
C
g
f
POLE CIN
IN IN
IN
m
T
()
≈
=
1
2
2
π
π
f
CR
POLE LDO
OUTA LOAD
()
=
1
2
π



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