參數(shù)資料
型號: MAX1904
廠商: Maxim Integrated Products, Inc.
英文描述: 500kHz Multi-Output, Low-Noise Power-Supply Controllers for Notebook Computers
中文描述: 500kHz、多路輸出、低噪聲、電源控制器,用于筆記本電腦
文件頁數(shù): 24/33頁
文件大?。?/td> 649K
代理商: MAX1904
M
500kHz Multi-Output, Low-Noise Power-Supply
Controllers for Notebook Computers
24
______________________________________________________________________________________
Use a Schottky diode with a DC current rating equal to
one third of the load current. The Schottky diode
s rated
reverse breakdown voltage must be at least equal to
the maximum input voltage, preferably with a 20% der-
ating factor.
Boost-Supply Diode
A signal diode such as a 1N4148 works well in most
applications. If the input voltage can go below +6V, use
a small (20mA) Schottky diode for slightly improved
efficiency and dropout characteristics. Don
t use large
power diodes, such as 1N5817 or 1N4001, since high
junction capacitance can pump up V
L
to excessive
voltages.
Rectifier Diode (Transformer Secondary Diode)
The secondary diode in coupled-inductor applications
must withstand flyback voltages greater than 60V,
which usually rules out most Schottky rectifiers.
Common silicon rectifiers, such as the 1N4001, are also
prohibited because they are too slow. This often makes
fast silicon rectifiers such as the MURS120 the only
choice. The flyback voltage across the rectifier is relat-
ed to the V
IN
- V
OUT
difference, according to the trans-
former turns ratio:
V
FLYBACK
= V
SEC
+ (V
IN
- V
OUT
)
N
where: N = the transformer turns ratio SEC/PRI
V
SEC
= the maximum secondary DC output
voltage
V
OUT
= the primary (main) output voltage
Subtract the main output voltage (V
OUT
) from V
FLY-
BACK
in this equation if the secondary winding is
returned to V
OUT
and not to ground. The diode reverse-
breakdown rating must also accommodate any ringing
due to leakage inductance. The rectifier diode
s current
rating should be at least twice the DC load current on
the secondary output.
Low-Voltage Operation
Low input voltages and low input-output differential volt-
ages each require extra care in their design. Low
absolute input voltages can cause the V
L
linear regulator
to enter dropout and eventually shut itself off. Low input
voltages relative to the output (low V
IN
- V
OUT
differential)
can cause bad load regulation in multi-output flyback
applications (see the design equations in the
Transformer
Design
section). Also, low V
IN
- V
OUT
differentials can
also cause the output voltage to sag when the load cur-
rent changes abruptly. The amplitude of the sag is a
function of inductor value and maximum duty factor (an
Electrical Characteristics
parameter, 97% guaranteed
over temperature at f = 333kHz), as follows:
The cure for low-voltage sag is to increase the output
capacitor
s value. Take a 333kHz/6A application circuit
as an example, at V
IN
= +5.5V, V
OUT
= +5V, L = 6.7μH,
f = 333kHz, I
STEP
= 3A (half-load step), a total capaci-
tance of 470μF keeps the sag less than 200mV. The
capacitance is higher than that shown in the
Typical
Application Circuit
because of the lower input voltage.
Note that only the capacitance requirement increases,
and the ESR requirements don
t change. Therefore, the
added capacitance can be supplied by a low-cost bulk
capacitor in parallel with the normal low-ESR capacitor.
Applications Information
Heavy-Load Efficiency Considerations
The major efficiency-loss mechanisms under loads are,
in the usual order of importance:
P(I
2
R) = I
2
R losses
P(tran) = transition losses
P(gate) = gate-charge losses
P(diode) = diode-conduction losses
P(cap) = input capacitor ESR losses
P(IC) = losses due to the IC
s operating supply current
Inductor core losses are fairly low at heavy loads
because the inductor
s AC current component is small.
Therefore, they aren
t accounted for in this analysis.
Ferrite cores are preferred, especially at 300kHz, but
powdered cores, such as Kool-Mu, can work well:
Efficiency = P
OUT
/P
IN
100%
= P
OUT
/(P
OUT
+ P
TOTAL
)
100%
P
TOTAL
= P(I
2
R) + P(tran) + P(gate) + P(diode) +
P(cap) + P(IC)
P (I
2
R) = I
LOAD
2
x (R
DC
+ R
DS(ON)
+ R
SENSE
)
where RDC is the DC resistance of the coil, R
DS(ON)
is
the MOSFET on-resistance, and R
SENSE
is the current-
sense resistor value. The R
DS(ON)
term assumes identi-
cal MOSFETs for the high-side and low-side switches:
because they time-share the inductor current. If the
MOSFETs aren
t identical, their losses can be estimat-
ed by averaging the losses according to duty factor.
V
I
V
L
×
C
D
V
SAG
STEP
OUT
IN MIN
(
MAX
OUT
=
×
×
×
2
2
(
)
)
-
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