Cirrus Logic AN376 - Manual (Page 3)

AN376
AN376REV2 3
For a given capacitance C
OUT
and resistance R
LED
, the amount of attenuation can be calculated by recognizing
that the output stage can be modeled as an ideal AC current source with an RC filter for the output stage.
Equation 4 defines the transfer function of the RC network:
Since the AC line voltage is rectified, creating a ripple current with a frequency that is double the line frequency,
the RC filter in Equation 4 is evaluated at two times the line frequency f
line
:
To determine the required output capacitance that provides a specified output ripple current, the output charac-
teristics of the LED string must be understood at the full current operating point. To calculate the attenuation of
the RC circuit with a chosen LED resistive load, the magnitude of Equation 5 needs to be determined:
Solving for capacitance C
OUT
using Equation 7:
The important parameters for capacitor C
OUT
are the RMS current rating and voltage rating. For a single stage
LED driver, it is probable for high RMS currents to flow into the RC filter. As a result, the output capacitor C
OUT
may need to be selected for its RMS current rating, as over-current stress can degrade the output capacitor’s
reliability. Although the voltage rating is directly related to the RMS current rating, verify that the output capacitor
C
OUT
is rated for an output voltage plus some guard band.
2.2 Design Example
The LED driver’s input AC line voltage is 120VAC with a line frequency f
line
equal to 60Hz. The RC filter transfer
function defined in Equation 5 on page 3 is evaluated at two times the line frequency f
line
:
For a design specification of 250mA RMS output current I
RMS
, there is a DC component and a peak sinusoidal
component. Since the amplitude of the DC current I
DC
is equal to the amplitude of the sinusoidal peak
current I
PK
, Equation 3 on page 2 is used to calculate the peak output current I
PK
:
H   
1
1 j RC +
------------------------ = [Eq. 4]
H 4 f
line
 
1
1 j 4 f
line
 RC +
------------------------------------------- = [Eq. 5]
H 4 f
line
 
1
1 j 4 f
line
 RC +
-------------------------------------------
1
1 4 f
line
R
LED
 C
OUT
  
2
+
--------------------------------------------------------------------------------- = =
[Eq. 6]
C
OUT
1
H 4 f
line
 
2
-------------------------------- 1 –
4 f
line
R
LED

---------------------------------------------------- =
[Eq. 7]
H 4 f
line
 
1
1 j 4 f
line
 RC +
-------------------------------------------
1
1 j 754  RC +
----------------------------------- = =
[Eq. 8]
I
PK
0.82 = I
RMS
 0.82 250mA  205mA = =
[Eq. 9]