Cirrus Logic AN374 - Manual (Page 14)

AN374
14 AN374REV2
In Mode 2 the phase synchronizer MOSFET is turned ‘ON’ and the output power P
MODE2
is:
2. In the parallel configuration, the voltages of the two strings should at least have a 20% difference between
the two strings in order for the CS1630/31 to distinguish between the two strings. The most common
application of a parallel string is to achieve low CCT at low dim (<2000K) and maintain ~2700K at full
brightness using a combination of mint/blue/white on one channel and amber/red on the other channel.
Depending on the configuration, the current I
Red
in the red channel may be higher than the current I
White
in the white channel at low dim, whereas the current I
White
may be higher than current I
Red
at full
brightness. When the phase synchronizer MOSFET is ‘OFF’ the output power P
MODE1
delivered to the
load in a parallel design is as follows:
and when the phase synchronizer MOSFET is turned ‘ON’ the output power P
MODE2
is as follows:
3. If the power delivered to the load is similar in both configurations, then the magnetic utilization will be the
maximum and the inductor size will be the smallest. For an LED configuration, the power can be computed
in the four possible scenarios and it can be determined which design would be the optimum if the size of
the inductor were the dominant factor in deciding between series and parallel configurations.
4. To quantify magnetic size, consider analyzing the peak current I
PK(FB)
, turns ratio N, and primary
inductance L
P
of a system designed to drive a single LED string. For example, examine a design with the
single string CS1610/11/12/13 driver IC. For a comparable CS1630/31-based system converting the same
level of output power, let peak current I
PK(FB)
for Mode 1 be I
PK1(FB)
and peak current I
PK(FB)
for Mode 2
be I
PK2(FB)
. It is assumed that I
PK1(FB)
> I
PK2(FB)
. Let turns ratio N be such that the effective reflected
voltage be the same so that their peak currents would be comparable between the two systems under
comparison.
P
MODE2
I
CH1
I
CH2
–   V
CH1
 = [Eq. 8]
P
MODE1
V
CH1
I
CH1
 = [Eq. 9]
P
MODE2
V
CH1
I
CH2
 = [Eq. 10]