MAX1585ETJ+T Maxim Integrated Products, MAX1585ETJ+T Datasheet - Page 23

• Manufacturer Part Number: MAX1585ETJ+T
• Description: IC DGTL CAM PWR-SUP 5CH 32TQFN
• Manufacturer: Maxim Integrated Products

Specifications of MAX1585ETJ+T

• Applications: Controller, Digital Camera
• Voltage - Input: 0.7 ~ 5.5 V
• Number Of Outputs: 5
• Voltage - Output: 1.25 ~ 5.5 V
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 32-TQFN Exposed Pad
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

If Z
ceramic output capacitors) and continuous conduction
is required, then cross the loop over before Z
In that case:
Place:
Or, reduce the inductor value for discontinuous operation.
It is expected that most AUX3 step-down applications
employ continuous inductor current to optimize induc-
tor size and efficiency. To ensure stability, the control-
loop gain should cross over (drop below unity gain) at
a frequency (f
frequency.
The relevant characteristics for voltage-mode step-
down compensation are as follows:
• Transconductance (from FB3 to CC3), g
• Oscillator ramp voltage, V
• Feedback regulation voltage, V
• Output voltage, V
• Output load equivalent resistance, R
• Characteristic impedance of the LC output filter, R
The key steps for AUX3 step-down compensation are
as follows:
1) Place f
2) Calculate C
3) Calculate the complex pole pair due to the output
4) Add two zeros to cancel the complex pole pair.
5) Add two high-frequency poles to optimize gain and
If we assume V
300mA, then R
and L = 10µH, select the crossover frequency to be
1/10 the OSC frequency:
For 3.3V output, select R14 = 30.1kΩ and R15 =
18.2kΩ. See the Setting Output Voltages section.
1 / (2π x R
V
= (L / C)
(f
LC filter.
phase margin.
COUT
OUT3
OSC
C
C
= (V
/ 10).
is not less than Z
/ I
C
f
1/2
C
C
LOAD
IN
sufficiently below the switching frequency
x C
< f
C
OUT
LOAD
/ V
) much less than that of the switching
R
f
0SC
C
C
IN
RAMP
C
) = 1 / (2π x R
______________________________________________________________________________________
.
= f
= R
= 5V, V
OUT3
/ 10, and f
= 11Ω. If we select f
AUX3 Step-Down Compensation
OSC
)(V
LOAD
, in V
FB
/ 10 = 50kHz
RHP
OUT
/ V
RAMP
5-Channel Slim DSC Power Supplies
x C
C
OUT
LOAD
OUT
/ 10 (as is typical with
< Z
= 3.3V, and I
FB
(1.25V)
)(g
RHP
/ C
(1.25V)
x C
M
C
/ (2π x f
/ 10
OUT
OSC
MEA
LOAD
RHP
), so that
= 500kHz
(135µS)
, in Ω =
C
and f
OUT
))
0
O
=
:
Calculate the equivalent impedance, R
where R
(this is the output impedance of the step-up converter
when the AUX3 step-down is powered from the step-
up), R
capacitor equivalent resistance, and R
on-resistance of the external MOSFET.
The output impedance of the step-up converter
(R
R
R
Choose C
Choose C4 = 470pF.
Cancel one pole of the complex pole pair by placing
the R4 C4 zero at 0.75 f
the following:
Choose R4 = 1 / (2π x C4 x 0.75 x f
Choose R4 = 61.9kΩ (standard 1% value). Ensure that
R4 > 2 / g
R14 and R15.
Cancel the second pole of the complex pole pair by
placing the R14 C20 zero at 1.25 x f
Choose C20 = 560pF.
Roll off the gain below the switching frequency by plac-
ing a pole at f
Choose R22 = 1.2kΩ.
If the output filter capacitor has significant ESR, a zero
occurs at the following:
Use the R4 C22 pole to cancel the ESR zero:
If C22 is calculated to be <10pF, it can be omitted.
L
EQ
SOURCE
C20 = 1 / (2π x R14 x 1.25 x f
+ ESR + R
R22 = 1 / (2π x C20 [f
= 1Ω. Choose C
f
0
C
C4 = (V
L
OUT
= 1 / [2π(L x C
= 1 / (2π x 30.1k x 1.25 x 7.345kHz) = 576pF
= 1 / [2π(10µH x 47µF)
R
is the inductor DC resistance, ESR is the filter-
SOURCE
= 1 / (2π x 560pF x 250kHz) = 1.137kΩ
EQ
= (5 / 1.25)(1/ [2π x 30.1k x 50kHz) = 423pF
OUT
) is approximately 1Ω at f
MEA
> L / [(R
= 1 / (2π x 470pF x 0.75 x 7.345kHz)
z
Z
= R
IN
OSC
ESR
DS(ON)
= 47µF:
C22 = C
/ V
SOURCE
= 14.8kΩ. If it is not greater, reselect
is the output impedance of the source
/ 2:
= 1 / (2π x C
RAMP
EQ
OUT
OUT
is small compared to 1Ω, assume
/ 2)
OUT
)(1 / [2π x R14 x f
+ R
0
OSC
)
so R
. The complex pole pair is at
1/2
2
] = 10µH / 0.25 = 40µF
L
x R
]
1/2
/ 2])
O
+ ESR + R
OUT
0
ESR
is less than R
] = 7.345kHz
)
0
x R
0
0
)
/ R4
.
. Since the sum of
ESR
EQ
DS(ON)
C
:
DS(ON)
)
])
EQ
/ 2:
is the
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