Using the NUD4001 to Drive High Current LEDs

AND8198/D
Using the NUD4001 to Drive
High Current LEDs
Introduction
The use of Light Emitting Diodes (LEDs) has increased
dramatically over the past few years. LEDs have become
almost commonplace in applications such as traffic lights,
brake lights, turn indicators and backlighting because of
their longevity compared to incandescent bulbs. To keep the
LEDs light output constant, the current through the device
must be constant. There are many possible solutions to
achieve this, ranging from a series resistor to a linear
regulator to a switching power supply. Each one offers its
own particular advantages. One of the main considerations
with lighting suppliers is the cost. Above all, we must keep
in mind that the competition is an incandescent bulb, and
although the LED solution may be superior in terms of
lifetime, the up front cost must be a close as possible to the
cost of a simple bulb.
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APPLICATION NOTE
NUD4001 Circuit Description
The representative block diagram of the NUD4001
integrated current source is shown in Figure 1.
Vin
Iout
N/C
Iout
Current
Set Point
REXT
Iout
NUD4001
The NUD4001 from ON Semiconductor is a low cost
versatile LED driver. The device allows the user to set the
current with an external resistor and it is capable of
delivering currents of up to 500 mA in a SO8 package. Of
course, the 500 mA output current depends on a number of
factors including input and output voltage, the number of
LEDs driven and the amount of heatsinking that is provided
on the PCB.
As described in the data sheet and application note
AND8156/D, the maximum power dissipation for the
NUD4001 is about 1.13 W, which limits the NUD4001 to
driving 1 W LEDs. There are a number of ways to squeeze
the last bit of heat of the NUD4001, including the use of
a series resistor in situations where the input to output
differential is high, or the use of multiple NUD4001 to share
current. These techniques are described in the documents
mentioned above and in the Design Guide available on the
ON Semiconductor website at:
GND
Iout
Figure 1.
An external resistor connected between Pin 1 and Pin 3
sets the output current. There is a voltage drop of
approximately 0.7 V between these two pins so the output
current becomes:
0.7 V
+ IOUT
REXT
The NUD4001 maintains this current independent of
input voltage as long as the minimum voltage overhead is
available.
A typical NUD4001 circuit is shown in Figure 2.
NUD4001
Vin
1
8
2
7
Iout
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These techniques are very useful in many applications but
do not make up for the limitations of the SO8 package that
houses the NUD4001. We could use a switching power
supply solution at higher power level but that introduces
extra cost, complexity and EMI to the design.
In this article, we use the NUD4001 as a control IC to drive
an external Power transistor that carries most of the LED
current. It offers all the advantages of the NUD4001, an
easily selectable current, low cost, and versatility but with
the capability of delivering higher currents to power 3 W
LEDs.
© Semiconductor Components Industries, LLC, 2013
October, 2013 − Rev. 1
N/C
REXT
GND
3
Current
Set Point
4
6
5
Iout
Iout
Iout
12 V
Figure 2.
1
Publication Order Number:
AND8198/D
AND8198/D
Design Example
If the LEDs are 1 W devices (with a load current of
350 mA), the power dissipation in the NUD4001 is
approximately 0.355 W. The detailed calculations are
available in the NUD4001 data sheet.
If the LEDs are 3 W LEDs as could be expected in an
Automotive tail light application, the power dissipation is
increased to the point where the NUD4001 alone will not be
able to handle the extra heat generated.
To drive three 3.0 W LEDs in series with an input voltage
that ranges from 12 V to 16 V:
VLED per LED = 3.5 V
ILED = 700 mA
VIN = 19 V to 16 V
TA = −40°C to 85°C
1. Calculate REXT
0.7 V
0.7 V
+ REXT +
+ 1.0 W
IOUT
0.7 A
Using the NUD4001 with an External Power Transistor
Figure 3 shows the NUD4001 as a control IC that
regulates the current through an external PNP transistor (in
this case, the MJB45H11). Since the majority of the current
flows through the MJB45H11 power transistor, it dissipates
most of the power and needs additional heatsinking. In this
case, we use additional copper on the PCB, but a TO−220
package can also be bolted to a commercial heatsink or to the
same heatsink material that is used for the LEDs.
2. Define VLED for the LED string.
VLED + 3.5
3. Calculate the Voltage drop across the NUD4001
and associated circuitry.
VDROP + Vin * VLED * VSENSE
VDROP + 12 V * 10.5 V * 0.7 V
R1 (REXT)
VDROP + 0.8 V
Q1
MJB45H11
1.0 W
3 + 10.5 V
At high line:
NUD4001
Vin
D1
Schottky
MBR0540T1
VDROP + 16 V * 10.5 V * 0.7 V
Vin
Iout
N/C
Iout
REXT
Current
Set Point
GND
VDROP + 4.8 V
4. Calculate the power dissipation in the NUD4001
and external transistor at high line.
PD + VDROP
Iout
ILED + 4.8 V
0.7 + 3.36 W
5. Calculate the power dissipation in the NUD4001
control circuitry based on the information in
Figure 4 of the data sheet.
Iout
R2
1.0 k W
There will be some additional current flowing through the
NUD4001 due to the external resistor R2 and Schottky diode
D1. A good estimate is to take 1.1 times the value shown in
Figure 5 as being the new internal power dissipation in the
control section of the NUD4001.
Using the graph, for a Vin of 16 V, the power dissipation
is approximately 0.13 W, so, allowing for the extra current
through the control section of the NUD4001 we calculate the
PD from:
Figure 3.
Looking at the Figures 1 and 2, for the NUD4001, one of
the first things you will notice is that Pin 2 is marked as N/C
(or no connect). Normally this is the case, but Pin 2 is also
available to drive an external transistor.
In the circuit shown in Figure 3 the diode D1 and external
resistor, R2, with the internal circuitry both work to set up
a base current for the external pass transistor, Q1. The
additional voltage drop across the diode D1 ensures that
there is a sufficient VBE to turn on the external transistor Q1.
Although a Schottky diode is used in this schematic, D1 can
easily be replaced with a less expensive signal or switching
diode if required.
Q1 delivers most of the output current to the LEDs without
contributing any additional voltage drops in the circuit. As
long as the circuit has sufficient headroom to operate, the
external transistor will carry most of the current.
We will use the following design example to illustrate the
operation of the circuit.
PD(Cont.NUD4001) + 0.13 W
1.1 + 0.143 W
6. Calculate the Power Dissipation in the Driver
section of the NUD4001. While it is difficult to
calculate the current through the NUD4001, it is
easy to measure with a current probe or meter by
comparing the current flowing through the various
paths to the load. As an example, with
Vin = 13.7 V at room temperature, the current
through the MJB45H11 is measured as 694 mA.
The total output current is measured at 694 mA
also, leaving less than 1 mA to flow through the
NUD4001.
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AND8198/D
The power dissipation is given by:
PD(NUD4001) + 0.8 V
IOUT(NUD4001)
PD(NUD4001) + 0.8 V
0.001 + 0.0008 W
This is well above our calculated PD of 0.144 W, so the
NUD4001 will need only a minimum pad size heatsink area
in this application.
9. Select a heatsink for the external pass transistor.
7. Calculate the total Power Dissipation in the
NUD4001.
As calculated in Step 4, the power dissipation at high line
(Vin = 16 V) is just over 3 W.
Unfortunately, the MJB45H11 data sheet contains little
useful thermal information so we have to depend on other
data. The RqJA is listed at 75°C/W, this is most likely the
figure with a minimum pad size.
Figure 6 shows a graph of Thermal Resistance vs. Pad size
for a typical D2PAK.
The pad size to dissipate 3.3 W will need to be greater than
30 mm square. At higher power dissipation it is advisable to
use a TO−220 package and a heatsink.
PD(TotalNUD4001) + PD(Cont.NUD4001) + PD(NUD4001)
PD(TotalNUD4001) + 0.143 W ) 0.0008 W + 0.144 W
8. Calculate the heatsink requirements for the
NUD4001 based on the information in Figure 11
in the data sheet. This graph is reproduced as
Figure 5 in this paper. From Figure 4, the worst
case RqJA = 170°C/W with minimum board area.
T
* TA
PD + JMAX
+ 150° C * 85° C + 0.382 W
170
RqJA
180
0.500
0.450
0.400
0.350
PD_control (W)
140
120
100
0.300
0.250
0.200
0.150
0.100
80
0.050
60
0.000
2
3
4
5
6
7
8
9
10
0
5
10
BOARD AREA (in2)
3.5
80
PD(max) for TA = +50°C
3.0
Free Air
Mounted
Vertically
60
ÎÎÎÎ
ÎÎÎÎ
ÎÎÎÎ
ÎÎÎÎ
2.0 oz. Copper
L
Minimum
Size Pad
50
40
2.5
2.0
L
1.5
RqJA
30
0
5.0
20
25
30
Figure 5. Power Dissipation vs. Vin for the NUD4001
Figure 4. RqJA versus Board Area
70
15
Vin (V)
10
15
20
L, LENGTH OF COPPER (mm)
25
PD, MAXIMUM POWER DISSIPATION (W)
1
JUNCTION‐TO‐AIR (°C/W)
0
R θ JA, THERMAL RESISTANCE
RqJA (°C/W)
160
1.0
30
Figure 6. Thermal Resistance vs. Pad Size for a Typical D2PAK
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AND8198/D
Current Sharing
Conclusion
As mentioned earlier, the external transistor will carry
most of the current provided the circuit has sufficient
headroom to operate properly. In the circuit described here,
the 3 LEDs have a combined forward voltage of 10.1 V at
700 mA. The NUD4001 needs approximately 1.4 V of
headroom. If the input voltage drops below about 11.5 V
there is insufficient voltage to fully turn on the external
transistor and the NUD begins to carry more of the output
current. This is illustrated in Figure 7.
The NUD4001 works well as a control IC for high current
applications and the user now has to contend with the easier
task of heatsinking an external transistor in a power package
such as a DPAK, D2PAK or TO−220. This circuit is also
applicable where there is a large difference between the
input voltage and the LEDs forward voltage, e.g., where one
or two LEDs are driven from a 12 V source.
References
[1] NUD4001/D − ON Semiconductor Data Sheet.
[2] AND8156/D − ON Semiconductor Application Note.
0.8
0.7
Itotal
CURRENT
0.6
0.5
0.4
MJB45H11
0.3
0.2
I NUD4001
0.1
0
0
5
10
15
20
Vin
Figure 7. Current Sharing
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