High brightness LEDs are becoming a prominent source of light and often have better efficiency
and reliability than conventional light sources. While LEDs can operate from an energy source
as simple as a battery and resistor, most applications require an efficient energy source not
only for the reduction of losses, but also to maintain the brightness of the LED itself. For
applications that are powered from low voltage AC sources typically used in landscape lighting
or low voltage DC sources that may be used in automotive applications or to meet safety
requirements, high efficiency DC-DC converters configured for constant output current provide
a high efficiency driver that can operate over a relatively wide range of input voltages to drive
series strings of one to several LEDs
This application note describes a DC-DC converter circuit that can easily be configured to dri ve
LEDs at several different output currents and can be configured for either AC or DC input. The
circuit uses the L5973D monolithic step down converter configured to drive a series string of
LEDs in a constant current mode.
AN2259
APPLICATION NOT E
High intensity LED driver using the L5970D/L5973D
L5970D is a step down monolithic power switching regulator capable of delivering 1A while the
L5973D is able to deliver 2A at output voltages from 1.25V to 35V. Both devices use internal PChannel D-MOS transistors (with typical R
minimize the size of external components. An internal oscillator fixes the switching frequency
at 250kHz.
The brightness of the LED (Light Emitting Diode), or light intensity as measured in Lumens, is
proportional to the forward current flowing through the LED. Since the forward voltage drop of
the LED can vary from device to device it is important to drive the LEDs with a constant current
driver to be able to get good matching of the light output, especially when they are located side
by side where variations in light intensity are quickly noticed. A typical way to drive LEDs in the
constant current mode is to use a DC-DC converter configured to give a constant current
output. The circuit shown in Figure 7. uses the L5973D in a constant current configuration to
drive LEDs
Table 6.6 to 12Vdc Inp ut 1W LED Driv e r Bill of Mater ia ls . . . . . . . . . . . . . . . . . . . 25
Table 7.6 to 12Vdc Inp ut 3W LED Driv e r Bill of Mater ia ls . . . . . . . . . . . . . . . . . . . 27
Table 8.6 to 12Vdc Inp ut 5W LED Driv e r Bill of Mater ia ls . . . . . . . . . . . . . . . . . . . 29
Table 9.6 to 24Vdc Inp ut 1W LED Driv e r Bill of Mater ia ls . . . . . . . . . . . . . . . . . . . 31
Table 10. 6 to 24Vdc Inp ut 3W LED Driv e r Bill of Mater ia ls . . . . . . . . . . . . . . . . . . . 33
Table 11. 6 to 24Vdc Inp ut 5W LED Driv e r Bill of Mater ia ls . . . . . . . . . . . . . . . . . . . 35
5/38
1 DESCRIPTION OF BOARDAN2259
1 DESCRIPTION OF BOARD
The evaluation board shown in F igure 1. was designed so that it can be configured to accept
several different input voltages that are common for automotive and lighting applications. The
most common input voltages are 12V ac, 12Vdc (for automotive) and 24Vdc. The board also
allows the user to select the output current using the jumpers J2 and J4 on the board without
having to change any components on the evaluation board. The standard configuration of the
board includes a full wave bridge rectifier that is required for an AC input
6/38
AN22592 INPUT CAPACITOR SELECTION
Vin
o
D
I
n
V
)
2
2
C
2 INPUT CAPACITOR SELECTION
For DC input, the input capacitor, C1, is selected based on its ripple current rating for the
capacitor. The ripple current is calculated based on its duty cycle as outlined below.
V
=
Where D = duty cycle
Vo = output voltage
Vin = input voltage
The RMS current through the capacitor therefore is:
22
DD2
⋅
ripple
µ= efficiency
DIo
−=
+
2
η
η
For an AC input voltage, the input capacitor is selected primarily to have enough capacity to
supply the LED between the peaks of the AC input. The capacitor must be selected so that the
minimum voltage at the input to the L5973D is maintained during each half cycle of the AC
input.
Vi
peak⋅=
2
If the application is driving only one LED, the Vmin is determined by the minimum operating
voltage specification for the L5973D (4.4V). When driving more than one LED in series, the
minimum input voltage is determined by the output voltage and the minimum differential input to
output voltage for the regulator (the drop out voltage). In this case V
+ V
DO
= (x * Vf) + (I
min
out
* R
sense
Where: x = number of LED in series
= forward voltage of one LED
V
f
= LED drive current
I
o
= Drop out voltage
V
DO
The capacitor can then be selected using the equation:
−
3
⋅⋅⋅
=
1
Vpeak
(
VoIo105
1
⋅−⋅⋅η
22
minV
)
The ripple current rating will have two parts where in the low frequency range, the capacitor will
be charged by 120Hz while at the high frequency range the capacitor is discharged by 250kHz.
For the low frequency part, it is approximately the same as the input RMS current and the
power factor is approximately 0.7 for a full wave rectifier.
7/38
2 INPUT CAPACITOR SELECTIONAN2259
7.
I
Io
I
Vav
D
2
n
V
Io
I
I
⋅
in
=
For the high frequency part (ignoring output current ripple), we have:
VoIo
⋅η⋅
0Vin
ripple
Dav
⋅
−=
η
where Dav is the average duty cycle.
We will use the average duty cycle assuming that the voltage on the capacitor changes from
the peak to the minimum voltage linearly.
av =
ripple
Vo
Dav
av
=
⋅
−=
+
η
The equivalent series resistance of an aluminum capacitor has different frequency
characteristics. There is a coefficient associated with different frequencies. Typically, for 120Hz,
Kfl=1; for frequency greater than 10 kHz, Kfh=1.5.
Iin
cap
Therefore, the ripple current rating of the capacitor has to be greater than Icap
=
Kfl
Iripple
+
Kfh
22
DavDav2
+
DavDav2
⋅
2
η
+
mi
VVpeak
22
⋅
2
η
22
8/38
AN22593 CURRENT FEEDBACK LOOP
2R1
R
2
2R1
R
2R1
R
V
2
1
R
V
Rs
e
I
Rs
2
I
Rs
2
I
3 CURRENT FEEDBACK LOOP
To drive LEDs in a constant current mode, the feedback for the regulator is taken by sensing
the voltage drop across the current sense resistor, Rs, as shown in Figure 2. The voltage
divider between the sense resistor and the feedback pin (R
feedback pin so that it equals the internal reference voltage at the desired current level.
Figure 2. Current feedback
and R2) scales the voltage at the
1
In order to get Io = 350 mA, the values of R1, R2 and Rs are selected based on the following
values.
Vref = 3.3V ; Vfb = 1.235V ; Ifb_bias = 2.5 10
Vsense=Rs.Io
Using the superposition method:
2R
fb
Vref
⋅=
Vsense
+
2R
sense⋅−⋅−−=
o
=
Since Vref
Therefore, the equation can be simplified to:
and Vfb come from same band gap, they are directly correlated. K=Vref/Vfb=2.672.
o
=
For 350mA output the selected values are:
)VfbVref(Vfb
)1K(1
⋅−−
-6
A ; Rs=0.68Ω
1R
⋅+
bias_Ifb
+
Rbias_Ifb
2R
)VfbVref(Vfb
o =
Rbias_Ifb
⋅−⋅−−
⋅+
Vsens
R1R
⋅
+
1R
2R
1R
⋅−⋅
Rbias_IfbVfb
R1 = 2.74k
R2 =1.30k
Rs = 0.68
Ω,
Ω and
Ω.
9/38
3 CURRENT FEEDBACK LOOPAN2259
For an output current of 700mA the value of Rs would be 0.34Ω. If R1 and R2 are small
enough, the effect of the bias current can be ignored.
On the evaluation board, the value of Rs is selected by jumpers J2 and J4. When both J2 and
J4 are open, the output current is set to 350mA. Inserting each jumper connects a 0.68
resistor in parallel with the 0.68
Ω Rs. With J2 shorted, the output current will be set to 700mA
Ω
and the output current becomes 1A with both J2 and J4 shorted.
10/38
AN22594 INDUCTOR SELECTION
n
ax
Im
4 INDUCTOR SELECTION
The output inductor is selected to limit the ripple current in the LEDs.
For example, for a given DC input voltage and an output current of 350mA, the peak current
can be fixed to 500mA. This implies a
For an output current of 700mA, the peak current can be fixed to1000mA. This implies a
∆Imax= 600mA
Lmin=
Figure 3.Ripple Current (One 1W LED)
∆Imax = 300mA.
VoVin
−
⋅
To
∆
Figure 3. shows the ripple current measured with one 1W LED (warm white) at the output with
12Vac input. The measured ripple current is 180mA.
11/38
4 INDUCTOR SELECTIONAN2259
Figure 4.Ripple current (One 5W LED)
Figure 4. shows the ripple current driving one 5W LED at 1.05 A from a 12Vac input. The input
current is 269mA.
12/38
Loading...
+ 26 hidden pages
You need points to download manuals.
1 point = 1 manual.
You can buy points or you can get point for every manual you upload.