MAX5941BCSE(2003) データシートの表示（PDF） - Maxim Integrated

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MAX5941BCSE
(Rev.:2003)

IEEE 802.3af-Compliant Power-Over-Ethernet Interface/PWM Controller for Power Devices

Maxim Integrated 

IEEE 802.3af-Compliant Power-Over-Ethernet

Interface/PWM Controller for Power Devices

4) The reset winding turns ratio (NR/NP) needs to be

low enough to guarantee that the entire energy in

the transformer is returned to V+ within the off cycle

at the maximum duty cycle. Use the following equa-

tion to determine the reset winding turns ratio:

NR

≤

NP

×

1-DMAX ′

DMAX ′

where:

NR/NP = Reset winding turns ratio.

DMAX’ = Maximum value of maximum duty cycle:

NR

≤ 14

×

1- 0.5

0.5

=

14

Round NR to the nearest smallest integer.

The turns ratio of the reset winding (NR/NP) deter-

mines the peak voltage across the N-channel MOS-

FET.

Use the following equation to determine the maxi-

mum drain-source voltage across the N-channel

MOSFET:



VDSMAX ≥ VIN_MAX × 1 +

NP

NR





VDSMAX = Maximum MOSFET drain-source voltage.

VIN_MAX = Maximum input voltage:

VDSMAX ≥ 67V × 1 +

14 

14 

= 134V

Choose MOSFETs with appropriate avalanche

power ratings to absorb any leakage energy.

5) Choose the tertiary winding turns ratio (NT/NP) so that

the minimum input voltage provides the minimum

operating voltage at VDD (13V). Use the following

equation to calculate the tertiary winding turns ratio:

VDDMIN + 0.7

VIN_MIN

× NP

≤ NT

≤

VDDMAX + 0.7

VIN_MAX

×

NP

where:

VDDMIN is the minimum VDD supply voltage (13V).

VDDMAX is the maximum VDD supply voltage (30V).

VIN_MIN is the minimum input voltage (30V).

VIN_MAX is the maximum input voltage (67V in this

design example).

NP is the number of turns of the primary winding.

NT is the number of turns of the tertiary winding:

13.7

30

× 14

≤ NT

≤

36.7

67

× 14

6.39 ≤ NT ≤ 7.67

Choose NT = 7.

6) Choose RSENSE according to the following equation:

where:

RSENSE

≤

NS

NP

VILIM

×1.2 × IOUTMAX

VILIM is the current-sense comparator trip threshold

voltage (0.465V).

NS/NP is the secondary side turns ratio (5/14 in this

example).

IOUTMAX is the maximum DC output current (10A in

this example):

RSENSE ≤

6

0.465V

×1.2 ×10

=

90.4mΩ

14

7) Choose the inductor value so that the peak ripple

current (LIR) in the inductor is between 10% and

20% of the maximum output current:

L ≥ (VOUT + VD ) × (1-DMIN)

2 × LIR × 275kHz × IOUTMAX

where VD is the output Schottky diode forward volt-

age drop (0.5V) and LIR is the ratio of inductor rip-

ple current to DC output current:

(5.5) × (1- 0.198)

L≥

= 4.01µH

0.4 × 275kHz ×10A

8) The size and ESR of the output filter capacitor deter-

mine the output ripple. Choose a capacitor with a

low ESR to yield the required ripple voltage.

Use the following equations to calculate the peak-to-

peak output ripple:

VRIPPLE = VR2IPPLE,ESR + VR2IPPLE,C

______________________________________________________________________________________ 19

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