CS1600
5.1.3 PFC Boost Inductor
5.1.4 PFC MOSFET
Equation 3 can be rewritten to calculate the PFC boost
The peak voltage stress on the PFC MOSFET is a diode drop
above the output voltage. Accounting for leakage spikes, for
the 460 V output application, a 600 V FET is recommended.
Inductor, LB, as follows:
Vlink
-------------
× 90V × 2
Vlink
–
2
Vlink
400V
The FET should be able to handle the same peak current as
that seen through the inductor. This would amount to 3.96 A.
90V
in(min)
------------- --------------------
---------------------------------------------------------------------
[Eq.6]
α =
×
×
V
400V
V
Vlink
–
×
2
in(min)
The scaling factor to determine the RMS current through the
MOSFET for a 108 V input is about 1.15, and the minimum
RMS current rating, IFET(rms), required for the FET is
calculated as follows:
Vlink
× 90V × 2
-------------
Vlink
–
2
Vlink
400V
90V
in(min)
------------- --------------------
---------------------------------------------------------------------
α =
×
×
=0.937
V
400V
V
Vlink
–
×
2
in(min)
PO
[Eq.9]
-------------------------------------------
IFET(rms)
ILB(rms)
=
× γ
Vin(min)
×
2 × η
V
V
link – (
×
2)
in(min)
2
V
---------------------------------------------------------
[Eq.6]
LB = α × η × ( in(min)) ×
115
2 × fmax × PO × Vlink
-----------------------------------------
=
× 1.15
108 ×
× 0.95
2
2
(460 – 108 × 2)
----------------------------------------------------------------
LB = 0.937 × 0.95× 108 ×
= 431μH
2 × 70 × 103 × 115 × 460
ILB(rms) = 0.91A
where
The RMS current rating for the inductor is estimated using an
scaling factor used to account for variations in the input
current shape across the AC line cycle, over and above the
nominally calculated value. The nominal value before using
the scaling factor is as follows:
γ = FET scaling factor
5.1.5 PFC Diode
The PFC diode peak current is equal to the inductor peak
current:
ID(pk) = ILB(pk)
PO
[Eq.10]
-------------------------------------------
ILB(rms)
ILB(rms)
=
=
× β
[Eq.7]
Vin(min)
×
2 × η
ID(pk) = 3.17 A
115
-----------------------------------------
× 1.35
The PFC diode average current is calculated as follows:
PO
108 ×
× 0.95
2
-----------
ID(avg)
=
ILB(rms) = 1.07A
[Eq.11]
Vlink
where
β = inductor scaling factor
115
460
---------
ID(avg)
=
The peak inductor current, ILB(pk), may be estimated using the
following equation:
ID(avg) = 0.25 A
5.1.6 PFC Output Capacitor
4 × PO
-------------------------------------------
ILB(pk)
ILB(pk)
=
=
[Eq.8]
The output capacitor needs to be designed to meet the voltage
ripple and hold-up time requirements. In the case of a cost-
sensitive ballast application, the hold-up requirement is not a
key requirement.
η × Vin(min)
4 × 115
×
2
-----------------------------------------
0.95 × 108 ×
2
To address the output ripple requirements, the following
equation may be used as a guide:
ILB(pk) = 3.17 A
PO
Inductor tolerances should be considered when estimating the
peak currents present in the application.
--------------------------------------------------------------------------------------
=
[Eq.12]
Cout
2π × fline(min) × Vlink × ΔVlink(rip)
The internal control algorithm of the controller dictates that the
peak inductor current seen in the application could be as high
as a pre-defined threshold of 0.001984 times the inverse of
the inductor, which in this example amounts to 4.72 A. Care
needs to be taken to ensure that the saturation current rating
of the PFC boost inductor factors in this threshold used for the
protection schemes.
where
Cout = Output Capacitance value
Po = Output Power
f
line(min) = Minimum Line Frequency
link = PFC Output Voltage
ΔVlink = Peak-Peak Voltage Ripple on the PFC Output
V
For a 40 V ripple and minimum line frequency of 45 Hz, the
12
DS904A5
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