![](http://datasheet.mmic.net.cn/370000/MAX8729_datasheet_16717243/MAX8729_21.png)
M
Constant-Frequency, Half-Bridge CCFL
Inverter Controller
______________________________________________________________________________________
21
Applications Information
MOSFETs
The MAX8729 requires two external n-channel power
MOSFETs to form a half-bridge inverter circuit to drive
the transformer primary. Since the positive half-cycle
and negative half-cycle are symmetrical, the same type
of MOSFET should be used for the high-side and low-
side switches. When selecting the MOSFET, focus on
the voltage rating, current rating, on-resistance
(R
DS(ON)
), total gate charge, and power dissipation.
Select a MOSFET with a voltage rating at least 25%
higher than the maximum input voltage of the inverter.
For example, if the maximum input voltage is 24V, the
voltage rating of the MOSFET should be 30V or higher.
The current rating of the MOSFET should be higher
than the peak primary current at the minimum input
voltage and full brightness. Use the following equation
to estimate the primary peak current I
PEAK
_
PRI
:
where P
OUT
_
MAX
is the maximum output power,
V
IN
_
MIN
is the minimum input voltage, and is the esti-
mated efficiency at the minimum input voltage.
Assuming the half bridge drives four CCFLs and the
maximum output power of each lamp is 4.5W, the total
maximum output power is 18W. If the minimum input
voltage is 8V and the estimated efficiency is 75% at
that input, the peak primary current is approximately
4.3A. Therefore, power MOSFETs with a DC current rat-
ing of 5A or greater are sufficient.
Since the regulator senses the on-state, drain-to-source
voltage of both MOSFETs to detect the transformer prima-
ry current, the lower the MOSFET R
DS(ON)
, the higher the
current limit is. Therefore, the user should select n-chan-
nel MOSFETs with low R
DS(ON)
to minimize conduction
loss, and keep the primary current limit at a reasonable
level. Use the following equation to estimate the maxi-
mum and minimum values of the primary current limit:
Both MOSFETs must be able to dissipate the conduc-
tion losses plus the switching losses at both V
IN_MIN
and V
IN_MAX
. Calculate both terms. Ideally, the losses
at V
IN(MIN)
should be roughly equal to the losses at
V
IN(MAX)
, with lower losses in between. If the losses at
V
IN(MIN)
are significantly higher than the losses at
V
IN(MAX)
, consider increasing the size of the MOSFETs.
Conversely, if the losses at V
IN(MAX)
are significantly
higher than the losses at V
IN(MIN)
, consider choosing
MOSFETs with lower parasitic capacitance. If V
IN
does
not vary over a wide range, the minimum power dissi-
pation occurs where the conduction losses equal the
switching losses.
Calculate the total conduction power dissipation of the
two MOSFETs using the following equation:
where I
PRI
is the primary current calculated using the
following equation:
Both MOSFETs turn on with ZVS condition, so there is
no switching power dissipation associated with the
MOSFET. However, the current is at peak when the
MOSFET is turned off. Calculate the total turn-off
switching power dissipation of the two MOSFETs using
the following equation:
where C
RSS
is the reverse transfer capacitance of the
MOSFETs and I
GATE
is the peak gate-drive sink current
when the MOSFET is being turned off.
PD
x C
xV
I
GATE
x f
xI
SWITCH
RSS
IN
SW
PRI
=
2
2
I
P
xV
PRI
OUT MAX
_
η
IN
=
PD
I
x R
CONDUCT
PRI
DS ON
(
=
2
)
I
mV
R
I
mV
R
LIM MIN
_
DS ON
(
480
MAX
LIM MAX
_
DS ON
(
MIN
)_
)_
=
=
320
I
xP
V
IN MIN
_
x
PEAK PI
OUT MAX
_
_
=
2
η