Infineon ICL8001G, Design Manuallines

Page: 9 / 26

ICL8001G / ICLS8082G
Design Guidelines
Design Parameters
Application Note 9 Version 2.0, 2011-04-14
It can be seen in the non-dimming case that a sinusoidal waveform of the primary peak current is dominant, which
defines the almost sinusoidal shape of the input current
Iin(t) . Beside the PFC function Equation (2) above shows
that during phase cut dimming the input current will follow the phase cut-modified input voltage, which assures the
dependence of the driver output power on the RMS input voltage
V
rms
according to the law described by
Equation (24) on page 15 .
4 Design Parameters
This chapter presents the essential design process and illustrates it in parallel with concrete values which are
considered for a sample design with the main target magnitudes VLED = 27 V, ILED = 360 mA, PF = 98 %,
η = 90 %, f < fmax = 100 kHz for the input parameters Vin = 230 V, fin = 50 Hz and Pin = VLED ILED / (η PF). The
values chosen for the 10 W/230 V LED driver demo board are close to the following calculation results.
4.1 Transformer Parameters
For QR operation, a sufficiently large reflected voltage from the secondary to the primary side present during the
flyback time
tf has to be provided. After discharge of the transformer by means of the secondary winding current
the energy coupled to the reflected voltage is needed to reduce the input voltage at switch-on instantly in the VDS
valley to Vdsmin = Vin - Vr. With the selection of Vdsmin = 0 a value Vr = Vinp, where
(3)
is received according to
(4)
The ratio q = Vr / Vinp with q ≤ 1 determines the maximum input voltage Vinp to be processed with zero-voltage
switching. The winding ratio r
n
= np / ns is then calculated as r
n
= 12.
(5)
The driver operation frequency f varies over the line voltage phase and reaches its maximum values in the line
voltage phase close to the voltage peak Vinp . Consideration of a QR mode condition defines the primary
inductance L = 6.1 mH (6.3 mH chosen) based on
(6)
With the selection f = fmax, the equation for the minimum on-time
(7)
yields ton = 5.0 µs and so the minimum duty cycle becomes Dmin = 0.5 as
(8)
Vin Vinp 2

min Vds Vinp Vr
 
Vdo VLED
V
r
r
n


2
2
1
2




 

Vr
Vinp
f Pin
Vin
L
Vin f
Pin L
ton
1 2

Vin
f Pin L
D
2
min
