NUD4001 and NUD4011 Current Sources for LEDs Lighting Applications

AND8156/D
NUD4001 and NUD4011
Low Cost Integrated
Current Sources for LEDS
Lighting Applications of
Low and High Voltage
(6.0 V − 120 V)
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APPLICATION NOTE
Prepared by: Alejandro Lara and Mike Girand
ON Semiconductor
ABSTRACT
The increasing usage of Light Emitting Diodes (LEDs) to
address different lighting applications such as traffic
signals, exit signs, backlighting and general illumination
have made them very popular to the point where they are
now an attractive choice for all those lighting applications
that were once the domain of incandescent lamps. LEDs
have several advantages when compared with incandescent
lamps. They offer fast turn−on, lower heat generation, lower
power consumption, higher operating life, and high
resistance to shock/vibration. Nevertheless, LEDs need to be
driven properly to ensure optimal performance and long life.
Designing and implementing an effective driver is key to
obtain all the benefits of LEDs. The driver’s implementation
must be cost effective, which is not usually achieved with
discrete components but with integrated solutions.
high currents would damage the LEDs, so this is why
electronic drivers must be added when LEDs are driven
from any voltage source. The amount of light emitted by an
LED is proportional to the amount of average current
flowing through the device in the forward bias direction. As
the current is varied, the output of the light will vary in a
similar way. Therefore, a Light−Emitting Diode (LED) is
essentially a PN junction semiconductor diode that emits
light when current is applied.
By definition, it is a solid−state device that controls
current without heated filaments and is therefore very
reliable. LEDs have special characteristics that assure high
reliability and compatibility with electronic drive circuits.
LEDs have advantages and disadvantages when
compared with other light sources such as incandescent or
fluorescent lamps. The most significant advantages are fast
turn−on, lower heat generation, lower power consumption,
longer operating life, and high resistance to shock/vibration.
Some of the disadvantages are the narrow viewing angle,
near monochromatic light, limited wavelength selection,
and they require electronic drive circuits for operation.
LEDs, regardless of color, have an extremely long
lifetime, whenever their current and temperature limits are
not exceeded.
Lumileds Lighting LLC [1, 2] has published lifetime
data stating that after 50,000 hours the LEDs will have 70
percent or greater of the original light output. Using an
engineering rule of thumb with data already collected and
plotted on a semi−log graph paper, LEDs are projected to
have 50 percent or greater of the original light output after
100,000 hours. This is approximately 11 years and 5 months
of continuous service with light greater than 50 percent of
the initial output. Remember, in order to obtain maximum
life, the LEDs must be operated within the manufacturers
specified limits of both current and diode junction
temperature. LEDs should be used where extremely long
life is desired and the cost of lamp replacement is very high.
INTRODUCTION
LEDs are created from various doped semiconductor
materials in the form of a P−N diode junction. When
electrical current flows through the junction in the forward
direction, the electrical carriers deliver energy proportional
to the forward voltage drop across the diode junction, which
is emitted in the form of light. The amount of energy is
relatively low for low current (< 30 mA) infrared or red
LEDs. However, for green and blue LEDs which are
manufactured from higher forward voltage materials, the
amount of energy is greater.
The conversion efficiency of electrical energy into light
energy is very important. Today’s LEDs vary between 10
and 20 percent efficiency. The rest of the energy is converted
to heat. This heat must be effectively dissipated, as the
operating junction temperature of the LED die must be
maintained no higher than +125°C.
Since the device is being used in the forward biased mode,
once the voltage applied exceeds the diode forward voltage;
the current through the device can rise exponentially. Very
 Semiconductor Components Industries, LLC, 2004
July, 2004 − Rev. 2
1
Publication Order Number:
AND8156/D
AND8156/D
TRADITIONAL METHODS FOR DRIVING LEDS
USING DISCRETE COMPONENTS
There are several methods to drive LEDs with discrete
components. Figure 2 illustrates one of the simplest using a
resistor in series with the supply voltage to limit the current.
This type of methodology is simple and cheap but has
several disadvantages. The most significant one is that since
there is not any current control device, the variations in the
input voltage will change the average current to the LEDs,
which results in poor illumination quality and sometimes
even in the degradation or total damage of the LEDs for high
line voltage conditions. To better illustrate the problem,
calculations of the current supplied to the LEDs will be made
based on the circuit shown in Figure 2. The normal ac line
can fluctuate by 10 percent and therefore the transformer
output can vary between 10.8 Vac and 13.2 Vac whenever
the normal secondary voltage is 12.0 Vac. Based on this, the
LED’s current calculation for low, normal, and high voltage
ac line is made through the Formula 1:
Figure 1 shows a typical LED V – I curve. This graph in
particular makes reference to the high current LED
technology recently introduced. The maximum forward
current varies with the different type, style, and
manufacturer of LEDs. LED manufacturers have specified
the maximum forward currents at 30 mA, 75 mA, 150 mA,
350 mA, and 700 mA for differently constructed LEDs. The
higher current devices have special thermally designed
packages to transfer the heat to a heat sink. The same rules
can apply to devices having other current ratings by simply
scaling down the current and power designs.
Formula 1:
ILEDs Vin 2) (3 VLED)
R1
Assuming that the characteristics of the LEDs used are:
If = 350 mA and Vf = 3.5 V, then the resulting current
calculation for each of the voltage line conditions is as
follows:
Low line: ILEDs = 238 mA
Normal line: ILEDs = 323 mA
High line: ILEDs = 408 mA
Figure 1. Typical V– I Curve for Different LED’s
Colors (Courtesy: Lumileds)
R1A
20 , 3.0 W
Bridge1
24 V, 1.0 A
V1
12 Vac
60 Hz
LED1
LXHL−MW1D
−
+
−
2
C1
1000 F
+
4
1
LED2
LXHL−MW1D
3
LED3
LXHL−MW1D
0
Figure 2. Discrete LED driver circuit using a resistor in series with the supply voltage.
Application circuit for 12 Vac landscape lighting.
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Another common LED driver method using discrete
components is made through a linear regulator (MC7805,
MC7809 or similar), and a series medium power resistor
(usually 1.0 W or bigger). Figure 3 shows this concept.
As it can be seen, the change of the LED’s current is higher
than 25% for a 10% variation in the ac line. For the low
line case, it causes the LEDs to dim while for the high line
case it may potentially damage them due to the overheating
caused by the high current. This is why, these type of drive
circuits are not recommended nor often used because they
basically eliminate the valuable features of the LEDs.
Linear Regulator
MC7809
1
3
R1A
5.7 , 1.0 W
2
Bridge1
24 V, 1.0 A
V1
12 Vac
60 Hz
2
−
+
+
4
−
1
C1
1000 F
C2
0.1 F
C3
0.1 F
LED1
LXHL−MW1D
3
LED2
LXHL−MW1D
0
Figure 3. Discrete LED driver circuit using a linear regulator and a series resistor.
Application circuit for 12 Vac landscape lighting.
INTEGRATED LED DRIVERS NUD4001 DEVICES’
DESCRIPTION
The LED’s current is set by the regulated output voltage
of the linear regulator and the value of the resistor R1. The
resistor value is easily calculated through Formula 2:
NUD4001 Device Description
This integrated current source is designed to replace
discrete solutions for driving LEDs in low and high ac or dc
voltage applications (6.0 V − 30 V). Its integrated design
technology eliminates individual components by combining
them into a single small surface mount package (SOIC−8),
which results in a significant reduction of both system cost
and board space. Figure 4 illustrates the pin−out of the
NUD4001 LED driver device.
Formula 2:
R1 (Vout VLEDs)
ILEDs
Assuming that the characteristics of the LEDs used are:
If = 350 mA and Vf = 3.5 V, then the resulting resistor
calculation is as follows:
R1 (9.0 V 7.0 V)
5.7 0.350 A
The power dissipation in R1 is given by:
P I2 R (0.350)2 5.7 0.7 W
The current regulation is mostly dependent on the
regulator’s performance and should be expected to be good
since most voltage regulators provide good percentage of
line and load regulation, usually lower than 5%.
Although this type of concept provides good current
regulation to the LEDs, it may not be optimum for
applications where cost is critical, and either for those
requiring enable or electronic dimming functions.
Vin
Iout
N/C
Iout
Rext
Current
Set Point
GND
Iout
Iout
SO−8 Package, Case 751
Figure 4. NUD4001 Integrated LED Driver
Pin−out Description
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4. Define VLED @ ILED per LED supplier’s data
sheet:
a. Per Example in Figure 5,
VLED = 3.5 V + 3.5 V + 3.5 V = 10.5 V
5. Calculate Vdrop across NUD4001:
a. Vdrop = Vin – VLED
b. Vdrop = 12.0 V – 10.5 V
c. Vdrop = 1.5 V (Vdrop must be higher than device’s
overhead, 1.4 V)
6. Calculate Power Dissipation (PD):
a. PD = Vdrop x Iout
b. PD = 1.5 V x 0.330 A
c. PD = 0.495 W
7. If PD > 1.13 W (or derated value based on ambient
temperature. Refer to Figure 3 of NUD4001
device’s data sheet), then select the most
appropriate recourse and repeat steps 1−6:
a. Reduce Vin
b. Reconfigure LED array to reduce Vdrop
c. Reduce Iout by increasing Rext
d. Use parallel configuration of two or more
NUD4001 devices.
The typical current regulation performance of the
NUD4001 device for the circuit of Figure 5 using
Rext = 2.2 is shown in Figure 6:
This device provides a regulated dc current to a LED
array, from an ac or dc input. It can drive arrays of series or
parallel−series LEDs for a wide range of applications. It has
a low voltage overhead (1.4 V) to facilitate its usage in low
voltage applications. Its current regulating principle is made
through the generation of a constant voltage drop (0.7 V)
across an external low power sense resistor (Rext), which
sets the current independently of the input voltage supplied.
This operating principle makes it very simple to design
LED’s circuits around the NUD4001 device. Nevertheless,
there are certain design considerations such as the maximum
device’s power dissipation (1.13 W), operating ambient
temperature range, device’s voltage overhead and LED’s
array configuration that have to be taken into account before
implementing this integrated driver. A general design guide
is explained in the next section and illustrated in Figure 5.
In this case the device is utilized to drive three high intensity
white LEDs (If = 0.350 A, Vf = 3.5 V).
NUD4001
Iout
Vin
Iout
N/C
Rext
Current
Set Point
GND
Iout
0.40
TA = 0°C
Iout
0.35
0.30
TA = 25°C
0.25
Iout, (A)
12 V
TA = 85°C
0.20
0.15
Figure 5. NUD4001 Device Driving Three White
High Current LEDs
0.10
0.05
The steps to calculate the value of the external sense
resistor (Rext), and to validate the operation of the device
within its power dissipation capability are explained in the
following design guide:
0.00
0
2
4
6
8
10
12
14
16
Vin, (V)
Figure 6. NUD4001 Device Regulation Performance
at Three Different Ambient Temperatures
(0C, 25C and 85C)
NUD4001 Device’s Design Guide:
1. Determine Iout – LED’s current:
a. ILED = 330 mA
2. Calculate Resistor Value for Rext:
At 25°C, the change of the LED’s current is only 1% for
an increment of 15% in the input voltage. This regulation
ratings are obtained from the data shown in Figure 6:
For Vin = 12 Vdc, Iout = 328 mA,
For Vin = 13.8 Vdc, Iout = 331 mA
a. Rext = 0.7 V
Iout
b. Rext = 0.7 V 2.2 0.330
3. Define Vin:
a. Per Figure 5, Vin = 12 V
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power dissipation of the device for different ambient
temperatures, which is reduced as the ambient temperature
rises. Figure 7 shows the power derating graph at different
ambient temperatures for the NUD4001 device (mounted
onto FR−4, 2 sq inches copper pad, 1 oz coverage double
sided board):
Similar regulation values are obtained at low and elevated
temperatures. However, it is important to note that at low
temperature, the LED’s current is shifted by a factor of 5%
while at elevated temperature it is lowered by a factor of
11%. These values are also obtained from the data shown in
Figure 6:
For TA = 25°C and Vin = 12.5 Vdc, Iout = 329 mA
For TA = 0°C and Vin = 12.5 Vdc, Iout = 344 mA
For TA = 85°C and Vin = 12.5 Vdc, Iout = 291 mA
Even if it seems strange, this type of behavior is ideal and
usually desired by the LED manufacturers. It is because at
high ambient temperatures the junction temperature in the
LEDs would increase but the reduction in current cancels
this effect. At low temperatures the current may be increased
by a small percentage (usually no higher than 10%) since the
LED’s junction temperature is colder.
PD, POWER DISSIPATION, (W)
1.200
NUD4001 Power Dissipation
Although the basic design considerations of the
NUD4001 device are explained in the design guide, it is
necessary to emphasize and put special attention in the
power dissipation (PD) and thermal parameters (RJA) of the
device as these are the main things to consider for designing
around it.
The power dissipation of the SO−8 package is a function
of the pad size. This can vary from the minimum pad size for
soldering to a pad size given for maximum power
dissipation. Power dissipation for a surface mount device is
determined by the maximum rated junction temperature of
the die (TJ), the thermal resistance from the device junction
to ambient (RJA), and the operating ambient temperature
(TA). Using the values provided in the NUD4001 device’s
data sheet, PD can be calculated through the Formula 3:
0.800
0.600
0.400
0.200
0.000
25
35
55
65
75
85
95
105 115 125
Figure 7. NUD4001 Power Dissipation (PD) vs.
Ambient Temperature (TA)
Based on this information, it is possible to conclude that
in order to optimize the LED’s driver circuit design using the
NUD4001 device, it is necessary to keep the ratio between
Vin and VLEDs as low as possible but also higher than the
device’s overhead value of 1.4 V. A good design window for
this ratio should be between 1.5 V and 2.5 V for a 350 mA
application.
NUD4001 DESIGN CONSIDERATIONS FOR
12 VAC APPLICATION CIRCUITS
Currently, most of the 12 Vac landscape lighting
applications are being designed with the new technology of
high intensity white LEDs. This LED technology needs to
be supplied with currents between 100 mA and 700 mA,
which makes the driver’s power dissipation the main design
consideration. Figure 8 illustrates a typical 12 Vac landscape
lighting application circuit using the NUD4001 device as the
LED’s driver.
(TJ TA)
RJA
The NUD4001 device is rated for 1.13 W at TA = 25°C
whenever it is mounted onto FR−4, 2 sq inches pad, 1 oz
coverage double side board. Its thermal resistance
junction−to−ambient is 110°C/W under the same board
conditions. From Formula 3, it is possible to calculate the
Rext1
D1
45
TA, AMBIENT TEMPERATURE (°C)
Formula 3:
PD 1.000
D2
12 Vac From:
60 Hz Transformer or
Electronic Transformer
+
C1
1
8
2
7
3 4001 6
4
5
Q1
NUD4001
LED1
Luxeon Emitter, 350 mA
LED2
Luxeon Emitter, 350 mA
D3
D4
LED3
Luxeon Emitter, 350 mA
0
Figure 8. 12 Vac landscape lighting application circuit using the NUD4001 device to drive three 350 mA LEDs.
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cases where the input voltage is supplied through a 60 Hz
transformer. Therefore if an electronic transformer is used
then the formula is not applicable and the selection of the
capacitor is most likely to be made through entirely lab
work.
The final design steps for the circuit of Figure 8 are to
calculate the external sense resistor (Rext), and the
NUD4001 device’s power dissipation (PD), which is made
by following steps two and six of the design guide
previously explained. The resulting Rext for a LED’s current
of 320 mA is 2.2 , and the power dissipation for the same
current value is 0.8 W. Based on these calculations, the
resulting Bill of Materials (BOM) is as follows:
The first design consideration for the circuit shown in
Figure 8 is the type of diodes needed for rectification (D1,
D2, D3 and D4), because the 12 Vac input can be supplied
from a 60 Hz transformer or an electronic switching
transformer. Therefore, for the case where an electronic
transformer is used, it is necessary to use rectifier devices
with fast reverse recovery time (trr), otherwise problems will
be experienced. The ON Semiconductor’s MURA105T3
device is the ideal choice for these purposes.
Another design consideration is the voltage drop (Vdrop)
across the NUD4001 device (Q1) as it directly impacts the
power dissipation of the device. As explained previously,
Vdrop is dependent on the average voltage (Vavg) supplied to
the device and the voltage drop of the LED’s array
(Vdrop = Vavg − VLEDS). For an optimum design, this Vdrop
should be kept between 1.5 and 2.5 V for a 350 mA
application. The value of the capacitor (C1) is an important
factor to consider as Vavg strongly depends on it. The value
of the capacitor in farads can be determined by using the
Formula 4 developed by Savant [3].
Formula 4:
C
Vendor
Part Number
D1 − D4
ON Semiconductor
MURA105T3
Q1
ON Semiconductor
NUD4001
C1
VISHAY
470 F, 25 V
R1
VISHAY
2.2 LED1 − LED3
LUMILEDs
LXHL−MW1D
Figures 9 and 10 show two oscilloscope pictures of the
waveforms generated across the circuit shown in Figure 8
once the previous BOM is implemented. Figure 9 refers to
the case where the input voltage is taken from a 60 Hz
transformer (Tamura 3FD−424 or similar), and Figure 10 for
the case of a switching transformer (Cooper LZR−404 or
similar).
The current regulation performance of the circuit shown
in Figure 8 for different average voltages values resulting
from variations in the ac line, is similar than the one shown
in Figure 5. At 25°C, the change of the LED’s current is
expected to be of only 1% for an increment of 15% in the
input voltage.
There are some cases where the application only requires
to light up one LED instead of three, so for those cases it is
necessary to add an external power resistor to reduce the
voltage drop (Vdrop) across the NUD4001 device to have it
operating within its power dissipation (PD) capabilities.
Figure 11 illustrates this circuit concept and Figures 12, 13
and 14 show the circuit regulation performance, the
NUD4001 device’s power dissipation, and the power
dissipation in the power external resistor (Rext2)
respectively.
The BOM for D1–D4, C1 and Rext1 is similar to the one
defined for the circuit of Figure 8. The only addition to the
circuit of the Figure 11 is the power resistor Rext2.
Similar circuits can be made to drive different LED arrays
for low input voltage applications (6.0 V to 24 V). The main
important to consider is to not exceed the power dissipation
of the NUD4001 device to assure reliable operation.
VMAX
VfRRL
C = Value of the capacitor is farad
VMAX = Peak AC line voltage
V = Peak−peak capacitor voltage normal 0.4 Vpeak
ƒR = Twice the ac line frequency (120 for a 60 Hz system)
RL = Effective load resistance
The effective load resistance (elr) is a term used for
converting the LEDs and the driver into an equivalent
resistance. It is the value of this resistance that is used in
selecting the electrolytic capacitor. The elr is the input
average voltage to the driver for a given ac line voltage value
divided by the LED current. Using Formula 4, it is possible
to theoretically calculate the capacitor needed for the wanted
Vavg. In the case of Figure 8, Vavg is selected so that Vdrop
is no higher than 2.5 V. If the Vf of each LED is 3.5 V and
they all are in series, then the resulting Vavg is 13 V
(Vdrop = V(avg) − VLEDS). Based on this and using
Formula 4, the capacitor is calculated as follows:
C
Part Code
13
0.4 12 2 120 37.2
C 428 F
The closest commercial capacitor value is 470 F, which
gives a first approximation of the capacitor (C1) needed.
This value should not be taken as the final one as it is
necessary to do additional lab analysis to validate it. It is
important to notice that Formula 4 is only applicable for
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Channels
Definition
Vac
Vavg
Iavg
Channels
Definition
Vac
Ch1– Vin
Ch1– 10V/div
Ch1– 12Vrms
Ch1 – Vin
Ch1 – 10V/div
Ch1 – 12Vrms
Ch2– Vavg
Ch2– 5V/div
Ch2– 13Vavg
Ch2 – Vavg
Ch2 – 10V/div
Ch2 – 12.8V avg
Ch3– Iavg
Ch3– 200mA/div
Ch3– 320mAavg
Vavg
Ch3 – Iavg
Ch3 – 200mA/div
Ch3 – 315mAavg
Iavg
10V/div
200mA/div
10V/div
200mA/div
5V/div
10V/div
Figure 10. Waveforms generated across the circuit
shown in Figure 7 when an electronic switching
transformer is used for the supply voltage.
Figure 9. Waveforms generated across the circuit
shown in Figure 8 when a 60 Hz transformer is
used for the supply voltage.
Rext2
17 , 3.0 W
D1
D2
Rext1
12 Vac From:
60 Hz Transformer or
Electronic Transformer
+
C1
1
8
2
7
3 4001 6
4
5
Q1
NUD4001
LED1
Luxeon Emitter, 350 mA
D3
D4
0
Figure 11. 12 Vac landscape lighting application circuit using the NUD4001
device to drive one 350 mA LED.
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0.400
2.00
TA = 0°C
TA = 25°C
0.280
TA = 85°C
1.75
POWER, WATTS
0.320
0.240
0.200
0.160
0.120
1.50
1.25
TA = 85°C
1.00
TA = 25°C
0.75
0.50
0.080
0.040
0.25
0.000
0.00
6
8
10
12
14
16
TA = 0°C
6
8
10
Vin, (V)
14
16
Figure 13. NUD4001 device’s power dissipation
when operating in the circuit of Figure 11 for
different Vavg values.
2.80
TA = 0°C
2.40
2.00
TA = 25°C
1.60
TA = 85°C
1.20
0.80
0.40
0.00
12
Vin, (V)
Figure 12. Current regulation performance of the
circuit of Figure 11 for different Vavg values.
POWER, WATTS
Iout, mA
0.360
6
8
10
12
14
16
Vin, (V)
Figure 14. Power dissipation in Rext2 for the circuit
of Figure 11 for different Vavg values.
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NUD4011 DESIGN CONSIDERATIONS FOR
120 VAC APPLICATION CIRCUITS
In addition to the already explained design considerations
for the NUD4001 device, it is necessary to consider power
factor requirements and transient voltage protection when
designing 120 Vac circuits around it. If there is no
requirement for power factor, then the circuit becomes
simpler as it does not need to have a full bridge rectifier nor
a capacitor. Figure 15 shows a schematic diagram to drive a
LED array of 30 white, low current units (30 mA) for a 120
Vac application using the NUD4011 device. This device is
similar than the NUD4001 device but with higher
breakdown voltage.
Q1
NUD4011
1
8
7
Rext1 2
3 4011 6
4
5
D1
V1
120 Vac
60 Hz
LED1
Nichia, 30 mA
+
−
VARISTOR
200 V
LED2
Nichia, 30 mA
LED30
Nichia, 30 mA
0
Figure 15. 120 Vac application circuit using the NUD4011 to drive 30 white low current LEDs (30 mA) in series.
No requirement for power factor.
Formula 5:
For Figure 15, current conduction occurs only after the
peak voltage of the positive ac cycle exceeds the forward
voltage of the LED array. The total forward voltage is given
by the Vf voltage of a specific LED device multiplied by the
number of LEDs connected in series. Assuming that Nichia
white LEDs with characteristics of Vf = 3.6 V and
If = 30 mA are used, then the resulting total forward voltage
of the 30 series LEDs array is 108 V. This indicates that
current conduction is only present during the time that the
positive cycle of the ac input is higher than 108 V. This is
something very important to consider for determining the
value of the external resistor (Rext1) to set the LED’s current
because it is now dependent on the peak current value and
the conduction time. To better illustrate this, the current peak
calculation is made for the circuit shown in Figure 15.
Formula 5 shows the equation to calculate the instantaneous
voltage over time for a sinusoidal waveform:
V Vpeak sin Using this formula and other analogies, it is possible to
determine the time for current conduction during the
positive cycle of the ac input.
As known, Vpeak happens at = 90° of the ac cycle, which
translates to 4.165 msec in time for a 60 Hz frequency.
Formula 5 is then used to find for 108 V:
108 V 120 2 sin 39.52°
and then, since 90° is 4.165 msec, then 39.44° is 1.82 msec.
Based on this, the current conduction time is calculated as
follows:
(4.165 msec 1.82 msec) 2 4.67 msec
Assuming that the 60 Hz current waveform is square
shaped, it is possible to say that since 16.66 msec is 100%
duty cycle, and 4.67 msec is 28%, therefore:
I(avg) Ipeak duty cycle
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If the average LED current wanted is 25 mA:
procedure already explained. The only variation is the
frequency which is increased to 120 Hz because of the usage
of a full bridge rectifier. The frequency affects the duty cycle
and therefore the Ipeak calculation.
I(avg)
Ipeak duty cycle
Ipeak 25 mA
0.28
NUD4001 DEVICE’S PWM AND ENABLE
FUNCTIONS
In addition to the current regulation function, the
NUD4001 device offers the ability to PWM the LEDs for
dimming applications by using an external small signal NPN
transistor connected between pin 4 and ground. This is a
very important feature specially for those applications
where multicolor lighting is required (swimming pools,
bars, etc.). The same small signal transistor can be used for
an enable function for conditioning applications. Figure 17
illustrates the circuit for PWM function in a 12 Vdc
application to drive three high current (350 mA) white
LEDs.
Ipeak 89.2 mA
Upon the calculation of the Ipeak, it is possible to calculate
Rext:
Rext 0.7 V
Ipeak
Rext 7.84 This theoretical procedure provides a way to have a first
approximation for the value of the Rext needed. However it
should not be taken as final until validated through lab
analysis. Figure 11 shows the expected current waveforms
in the LED’s array for the circuit shown in Figure 10, with
Rext = 7.8 .
Q1
NUD4001
Rext1
2 . 1/4 W
0.09
Iout
AMPLITUDE, AMPS
0.08
0.07
1
8
2
7
3 4001 6
4
5
Rext2
110 k, 1/4 W
0.06
LED1
LXHL−MW1D
Vin
12 Vdc
0.05
0.04
Q1
2N2222
I(avg)
0.03
0.02
LED1
LXHL−MW1D
LED1
LXHL−MW1D
PWM
0.01
0.00
0
0.05
0.1
0.15
0.2
0.25
0.3
0
Figure 17. NUD4001 Device Configuration for
PWM Function
TIME, SECS
Figure 16. Ipeak and I(avg) waveforms through the
LED’s array for the circuit shown in Figure 15,
with Rext = 7.8 .
The function of Rext2 is to pull up the pin 4 of the device
when the PWM signal in the base of the NPN transistor is
low. The average current applied to the LED is directly
dependent on the duty cycle (Iavg = Ipeak x duty cycle). And
the LED’s light intensity is directly dependent on the
average current I(avg) applied. In the case of Figure 17, the
current is set to be 350 mA at 100% duty cycle and therefore,
it proportionally decreases for narrower duty cycles. The
PWM circuit is good for frequencies between 100 Hz and
1.0 kHz. Most dimming applications use frequencies within
this range.
The same type of configuration is used for the enable
function. The only difference is the way that the base of the
NPN transistor is driven. The same concepts can be applied
for high voltage applications, the only things to consider is
to select a transistor with high enough breakdown voltage,
and to add an external low power resistor (1/4 W) of 80 k
between pin 4 and the collector of the PWM transistor.
As it can be observed, the Ipeak and I(avg) values do not
exactly match with the calculated ones. So this is why, it is
important to do lab analysis to achieve specific design targets.
The power dissipation of the NUD4011 device for the
circuit shown in Figure 15 is calculated as follows:
PD [(Vpeak VLEDs) I(avg)] 0.28
PD 0.464 W
In order to achieve lower power dissipation in the LED
driver (NUD4011), it is necessary to configure the LED’s
array in a way so that the total Vf is close enough to the
voltage applied to it.
If the application has power factor requirements, then it is
necessary to use a full bridge rectifier across the ac input to
keep in phase the current and voltage waveforms as much as
possible. The Rext calculation is made through the same
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10
AND8156/D
SUMMARY
LEDs are driven from a high voltage supply, then it is
necessary to add an external power resistor to reduce the
voltage drop across the device (see Figure 11).
Selection of the proper copper area according to the
application needs and device’s specifications is key to
achieving optimum device’s operation.
The enable and PWM features as well as the low cost
implementation is what distinguishes the NUD4001 and
NUD4011 devices from linear regulators and discrete
solutions (resistors) to drive LEDs.
Discrete Methods (resistors)
Discrete methods to drive LEDs are not recommended nor
often used because they basically eliminate the valuable
features of LEDs, and sometimes even cause total damage.
Linear Regulators
Although the concept of linear regulators provide good
current regulation for LED’s circuits, it may not still be
optimum for applications where cost is critical, and either
for those applications requiring enable or electronic
dimming functions.
REFERENCES
1. ON Semiconductor site: www.onsemi.com
2. Fitzgerald, A.E. Higginbotham, David E. Arvin
Basic Electrical Engineering. Fifth edition. 1981.
3. Lumiled: www.lumiled.com
4. Nichia: www.nichia.com
5. Boylestad, Robert L. Circuit Networks. Eighth
edition. August, 1988.
6. PSPICE release 9.1. Released notes February
2000.
7. ON Semiconductor’s data sheets NUD4001/D and
NUD4011/D.
NUD4001 Device
The NUD4001 device offers a low cost current regulation
integrated solution for different LED’s circuits of high and
low ac/dc voltage (6.0 V – 24 V).
Design considerations such as the device’s power
dissipation, breakdown voltage and maximum current
capability have to be taken into account before
implementation in application circuits.
Selection of the proper LED’s configuration to drop as
much as possible of the supply voltage in the LED array is
key to achieve design optimization. If a low quantity of
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11
AND8156/D
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