CS1600

12

DS904A7

5.1.3

PFC Boost Inductor

Equation 3 can be rewritten to calculate the PFC boost

The RMS current rating for the inductor can be estimated as

follows:

following equation:

Inductor tolerances should be considered when estimating the

peak currents present in the application.

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.

5.1.4

PFC MOSFET

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.

The FET should be able to handle the same peak current as

that seen through the inductor. This would amount to 3.17 A.

FET is calculated as follows:

5.1.5

PFC Diode

The PFC diode peak current is equal to the inductor peak

current:

The PFC diode average current is calculated as follows:

5.1.6

PFC Output Capacitor

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.

The CS1600 has been designed to operate with a low output

capacitance of approximately 0.2

μF per watt of output power.

For this specific application:

The 120 Hz ripple on the output capacitor may be estimated

using the following equation:

where

α

V

link

400V

--------------

90V

V

in min

()

--------------------

×





2

V

link

V

link

400V

--------------

90V

×

2

×





–

V

link

V

in min

()

2

×

–

---------------------------------------------------------------------

×

=

[Eq.6]

α

V

link

400V

--------------

90V

V

in min

()

--------------------

×





V

link

400V

--------------

90V

×

2

×





–

V

link

V

in min

()

2

×

–

---------------------------------------------------------------------

×

0.937

==

L

B

αη

V

in min

()

()

2

×

×

V

link

V

in min

()

2

×

()

–

2f

max

P

O

V

link

×

×

×

---------------------------------------------------------

×

=

[Eq.6]

L

B

0.937

0.95 108

×

2

×

460

108

2

×

–

()

27010

3

×

115

460

×

×

×

----------------------------------------------------------------

×

431

μH

==

I

LB rms

()

P

O

V

in min

()

η

×

------------------------------

=

I

LB rms

()

1.12A

=

I

LB rms

()

115

108

0.95

×

----------------------------

=

[Eq.7]

I

LB pk

()

4

P

O

×

η V

×

in min

()

2

×

--------------------------------------------

=

I

LB pk

()

3.17 A

=

I

LB pk

()

4

115

×

0.95

108

×

2

×

-----------------------------------------

=

[Eq.8]

I

FET rms

()

P

O

V

in min

()

η

×

------------------------------

=

I

LB rms

()

1.12A

=

I

LB rms

()

115

108

0.95

×

----------------------------

=

[Eq.9]

I

Dpk

()

I

LB pk

()

=

I

Dpk

()

3.17 A

=

[Eq.10]

I

Davg

()

P

O

V

link

------------

=

I

Davg

()

0.25 A

=

I

Davg

()

115

460

----------

=

[Eq.11]

C

out

0.2

μF

W

----------------

115W

×

23

μF

==

[Eq.12]

ΔV

link rip

()

P

O

2

π f

line min

()

×

V

link

×

C

out

×

------------------------------------------------------------------------

=

115

2

π 45

×

460

×

23

×

-------------------------------------------------

=

40.2V

=