MAX5003 Datasheet, PDF (12/16 Pages) Maxim Integrated Products – High-Voltage PWM Power-Supply Controller

High-Voltage PWM
Power-Supply Controller
+48V
(36V TO 72V)
V_IN
1M
0.1µF
39k
0V
62k
0.1µF
470nF
XFACOILTRCTX03
1 _V+
2
INDIV
3
ES
V_DD ¹⁶
15
V_CC
NDRV
14
4 _FREQMAX5003
10µF
13
PGND
5
12
SS
CS
6
11
REF
AGND
7
10
CON MAXTON
8
9
COMP
FB
2
L_P
65µH
8
CMSD4448
4.7µF
33µF
5
IRFD620S
0.1µF
7
11, 12
MBRS130L
100Ω
9, 10
22µF
22µF
51k
0.1µF _CF
R_F
390pF ^200k
R_CS
0.1Ω
Figure 2. Application Example 1: Nonisolated +48V to +5V Converter
+5V
1A
R_A
41.2k
R_B
17.4k
3) The main factors influencing the choice of the turns
ratio are the switch breakdown voltage and the duty
cycle. With a smaller turns ratio, the secondary
reflected voltage and the maximum voltage seen by
the switch during flyback are reduced, which is
favorable. On the other hand, a smaller turns ratio
will shorten the duty cycle and increase the primary
RMS current, which can impact efficiency. A good
starting figure is the ratio of the input voltage to the
output voltage, rounding to the nearest integer. To
keep the flyback voltage under control, choose an 8-
to-1 ratio for the 48V to 5V system. The maximum
duty cycle allowed without putting the device in con-
tinuous-conduction mode can be found using the fol-
lowing formula:
where:
DC_MAX
=


1
V_MIN
V_SEC⋅ N


+1
N = N_P/N_S = Turns ratio
V_SEC = Secondary voltage
DC_MAX = Maximum duty cycle
V_MIN = Minimum power-line voltage
For a 48V to 5V system with an 8-to-1 turns ratio, the
maximum duty cycle before putting the device in
discontinuous mode is 55%. Assume that V_IN min is
36V (minimum input voltage, neglecting drops in the
power switch and in the resistance of the primary
coil) and V_SEC is 5.4V (5V plus a Schottky diode
drop). The MAX5003 maximum duty cycle is internal-
ly limited to 75%. Generally this parameter must fall
between 45% to 65% to obtain a balance between
efficiency and flyback voltage while staying out of
continuous conduction. If the value exceeds these
bounds, adjust the turns ratio.
4) Assuming 80% efficiency, a 6.25W input is needed
to produce a 5W output. Set an operating duty cycle
around 12% below the maximum duty cycle to allow
for component variation: 55% - 12% = 43%. Use the
following formula to calculate the primary inductance:
L_PRI
=
(^DC⋅V_MIN)²
2⋅ PWR_IN ⋅ ƒ_SW
=
(^{0.43⋅36V})²
2⋅6.25W⋅300kHz
≅ 65µH
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