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AN-1077

PFC Converter Design with IR1150 One Cycle Control IC

International Rectifier 

Application Note AN-1077

Maximum Input Power and

Currents

Majority of the converter design is based on low

line current. That’s the worst case condition for effi-

ciency and input currents.

(Assume PF to be 0.99 or greater at low line)

Maximum input power can be calculated assuming

a nominal efficiency at low line:

PIN( MAX )

=

PO( MAX )

η MIN

= 300W

0.92

= 326W

The maximum rms AC line current is calculated at

the minimum input AC voltage:

η I = IN (RMS )MAX

PO(MAX )

MIN (VIN (RMS )MIN )PF

I IN (RMS )MAX

=

300W

0.92(85V )0.998

= 3.8A

Assuming sinusoidal AC current, the peak value of

the AC current can be calculated:

I = IN ( PK )MAX

2( PIN( MAX ) )

VIN ( RMS )MIN

I IN ( PK )MAX

= 1.414( 326W

85V

) = 5.4A

The AC Line input average current can be calcu-

lated assuming sinusoidal waveform:

π I IN( AVG )MAX = 2 × I IN( PK )MAX

I IN ( AVG )MAX

=

2 × 5.4A

π

= 3.4A

High Frequency Input Capacitor

Requirements

C IN

= k∆IL

2π ×

I IN (RMS )MAX

f SW × r × VIN (RMS )MIN

C IN

= 0.3

3.8 A

2π ×100kHz × 0.06 × 85V

= 0.335µF

Where:

k∆IL = Inductor current ripple factor (30% in this

design)

r = maximum high frequency voltage ripple factor

(∆VIN/VIN), typically between 3% – 9%, 6% used for

this design.

CIN = 0.330µF, 630V

High frequency capacitor is typically a high quality

film capacitor rated at beyond the worst case peak of

the line voltage. Care must be taken to avoid too

large a value as this will introduce current distortion.

This capacitor can be considered part of the EMI

input filter: its main purpose is to bypass the high fre-

quency component of the input current with the short-

est possible loop.

Boost Inductor Design

Power switch duty cycle must be determined at

VIN(PK)MIN. This will represent the peak current for the

inductor, at the peak of the rectified line voltage at

minimum line voltage.

V = IN( PK )MIN 2 ×VIN( RMS )MIN = 120V

D = VO VIN( PK )MIN = 385V −120V = 0.69

VO

385V

∆I L = 0.2 × I IN( PK )MAX = 0.2 × 5.4 A ≅ 1.1A

I L( PK )MAX

=

I IN ( PK )MAX

+

∆I L

2

= 5.4A + 1.1A

2

≅ 6A

LBST

V × IN( PEAK )MIN

fSW × ∆I L

D

=

120V × 0.69

100kHz ×1.1A

=

752.7µH

∆IL is based on the assumption of 20% ripple cur-

rent. This is another area where design tradeoffs

must be considered.

A smaller value of ripple current would be benefi-

cial in terms of reduced distortion, output capacitor

International Rectifier Technical Assistance Center:

USA ++1 310 252 7105 Europe ++44 (0)208 645 8015 6 of 18

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