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FAN5094
PRODUCT SPECIFICATION
18
REV. 1.0.2 5/13/02
For the high-side MOSFET, the gate charge is as important
as the on-resistance, especially with a 12V input and with
higher switching frequencies. This is because the speed of
the transition greatly affects the power dissipation. It may be
a good trade-off to select a MOSFET with a somewhat
higher RDS,on, if by so doing a much smaller gate charge is
available. For high current applications, it may be necessary
to use two MOSFETs in parallel for the high-side for each
phase.
At the FAN5094’s highest operating frequencies, it may be
necessary to limit the total gate charge of both the high-side
and low-side MOSFETs together, to avert excess power dis-
sipation in the IC.
For details and a spreadsheet on MOSFET selection, refer to
Applications Bulletin AB-8.
Gate Resistors
Use of a gate resistor on every MOSFET is mandatory. The
gate resistor prevents high-frequency oscillations caused by
the trace inductance ringing with the MOSFET gate
capacitance. The gate resistors should be located physically
as close to the MOSFET gate as possible.
The gate resistor also limits the power dissipation inside the
IC, which could otherwise be a limiting factor on the switch-
ing frequency. It may thus carry signicant power, especially
at higher frequencies. As an example, consider the gate
resistors used for the low-side MOSFETs (Q2 and Q4) in
Figure 1. The FDB7045L has a maximum gate charge of
70nC at 5V, and an input capacitance of 5.4nF. The total
energy used in powering the gate during one cycle is the
energy needed to get it up to 5V, plus the energy to get it up
to 12V:
This power is dissipated every cycle, and is divided between
the internal resistance of the FAN5094 gate driver and the
gate resistor. Thus,
and each gate resistor thus requires a 1/4W resistor to ensure
worst case power dissipation.
The same calculation may be performed for the high-side
MOSFETs, bearing in mind that their gate voltage swings
only the charge pump voltage of 5V.
Inductor Selection
Choosing the value of the inductor is a tradeoff between
allowable ripple voltage and required transient response. A
smaller inductor produces greater ripple while producing
better transient response. In any case, the minimum induc-
tance is determined by the allowable ripple. The rst order
equation (close approximation) for minimum inductance for
a two-phase converter is:
where:
Vin = Input Power Supply
Vout = Output Voltage
f = DC/DC converter switching frequency
ESR = Equivalent series resistance of all output capacitors in
parallel
Vripple = Maximum peak to peak output ripple voltage
budget.
One other limitation on the minimum size of the inductor is
caused by the current feedback loop stability criterion. The
inductor must be greater than:
where L is the inductance in Henries, RDS,on is the on-state
resistance of one phase’s low-side MOSFET, RDroop is the
value of the droop resistor in Ohms, Vin is either 5V or 12V,
and Vo is the output voltage. For most applications, this for-
mula will not present any limitation on the selection of the
inductor value.
A typical value for the inductor is 1.3
H at an oscillator
frequency of 1.2MHz (300KHz each phase) and 220nH at an
oscillator frequency of 4MHz (1MHz each phase). For other
frequencies, use the interpolating formula
Schottky Diode Selection
The application circuits of Figures 1-2 show a Schottky
diode, D1 (D2 respectively), one in each phase. They are
used as free-wheeling diodes to ensure that the body-diodes
in the low-side MOSFETs do not conduct when the upper
MOSFET is turning off and the lower MOSFETs are turning
on. It is undesirable for this diode to conduct because its high
forward voltage drop and long reverse recovery time
degrades efciency, and so the Schottky provides a shunt
path for the current. Since this time duration is extremely
short, being minimized by the adaptive gate delay, the selec-
tion criterion for the diode is that the forward voltage of the
Schottky at the output current should be less than the forward
EQV
1
2
---C
+
V2
70nC
5V
1
2
---
+
5.4nF
12V
5V
–
()
2
==
482nJ
=
P
Rgate
Ef
R
gate
R
gate
R
internal
+
()
-------------------------------------------------
482nJ
300KHz
==
4.7
4.7
1.0
+
---------------------------------
19mW
=
L
min
V
in
2V
out
–
f
-----------------------------------
V
out
V
in
-----------
ESR
V
ripple
-----------------
=
L3
10
–
R
DS on
,
R
Droop
V
in
2V
o
–
()
≥
LnH
()
1.86
10
6
×
fKHz
()
---------------------------
240
–
≈