Edited by Leland Teschler
Linear constantcurrent
drivers, for instance, offer
low EMI, low cost, and use only
a few components. Many designers
favor them for low-current
applications where power loss is
a minor concern. If power dissipation
is a problem, or if the application
needs more current, you can
parallel two or more drivers. Two
drivers in parallel provide twice as
much current and split power loss
into two locations, which makes
for easier heat dissipation.
High-side pass elements are
preferred, especially in automotive
uses, because they let you connect
only a single current-limited
wire (chassis return) to the load.
To configure parallel drivers with
high-side pass elements, however,
the current-sense feedback
circuitry must also be on the high
side, and it must be able to withstand
at least the voltage the LED
load creates. Thus, linear drivers
pose a dilemma if they have either
low-side current-sense feedback
or a requirement for limited common-
mode voltage on the sense
inputs.
Suppose we want to drive an
LED string with 400 mA. Drivers
U1 and U2 (MAX16803) are good
for automotive use because they
have a high input-voltage range,
excellent thermal qualities, and
the automotive temperature range.
Their maximum output current,
however, is 350 mA. Also, with
maximum input voltages of 16 V,
each IC would dissipate 3.6 W,
far too much for one package to
handle.
One solution is to use two drivers,
with each handling half the
current and generating half the
heat. If the drivers are to share current
equally, their current-sense resistors
must be on the high (anode)
side of the LEDs. That arrangement introduces another problem:
Forward voltage for the three LEDs
in series reaches almost 8 V, but the
maximum rating for the currentsense
(CS) pins is only 6 V.
A matched transistor pair (Q1)
solves this problem by translating
the current-sense voltage (developed
across R1) onto R3. Because
the bases connect together, voltages
at the emitters are nearly the
same, and therefore R2 IR2 equals
R1 IR1. High gain in the transistors
means negligible base current,
so IR3 is approximately equal to IR2.
You can set the output current as
follows:
| IOut= |
R2 X V;Sense |
R3 X R1 |
where VSense = the current-sense
voltage between CS+ and CS-
(203 mV). For best accuracy,
choose R4 so the two transistor
currents are about the same:
| R4= |
VLed - 0.7 |
| { |
Vsense |
} |
R3 |
where VLED = the nominal LED
voltage at the desired load. This additional low-cost circuitry keeps
the CS-pin voltages near ground,
and lets the sense resistor (R1) be
at any voltage up to the 45-V limit
of U1.
The second driver must have
an identical circuit. Tie the driver
outputs together to get the full
LED current. If you need more
current or power-handling capability,
you can add more drivers in
parallel.
— Jim Christensen
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