Electronic Design
Graphical, Numerical Techniques Size LED Array’s Drive

Graphical, Numerical Techniques Size LED Array’s Drive

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LED-based lighting is replacing incandescent and gas-discharge lamps in many situations. As practical lighting sources, and considering their directional nature, LED fixtures usually use multiple LEDs arranged in an array, with “m” LEDs in a serial string and “n” such strings in parallel (Fig. 1).

1. Establishing the correct resistor size for each string of an array of n strings, where each strings has m LEDs, is challenging due to the nonlinear nature of the diodes.
1. Establishing the correct resistor size for each string of an array of n strings, where each strings has m LEDs, is challenging due to the nonlinear nature of the diodes.

For the array configuration and the exponential I-V characteristics model of a single LED,1 it’s also possible to have a similar model representing the loading of such an array.

 

Since the single LED is a diode with a nonlinear forward-voltage/current, driving it with a fixed voltage source is generally not recommended. Rather, a serial (current-limiting) resistor is needed to equalize the power source and the load,(Fig. 2).

2. Even for a single LED driven by a voltage source, the exponential current-voltage relationship complicates selection of the series-resistor value.
2. Even for a single LED driven by a voltage source, the exponential current-voltage relationship complicates selection of the series-resistor value.

However, sizing this resistor requires a significant analytical manipulation. Using the single-LED exponential model, a Kirchoff’s voltage law (KVL) equation around the circuit loop gives:

 

 

which can be rearranged to:

 

where a, b, and c are model parameters for a selected LED, as defined in the reference.

This is a challenge to solve symbolically, but it can be solved numerically with mathematical software tools. Alternatively, it can be solved graphically by separating the two key I-V relationship equations, v+ RxIF = Vs and IF = aebv + c.

For example, using a Lumileds Luxeon Rebel ES LED with a 3.2-V dc source and a desired drive current of 0.5 A, the horizontal current line intercepts the LED curve at 2.92 V (Fig. 3). Therefore, the series resistor will be Rx = (3.2 – 2.92)/0.5 = 0.56 Ω.

However, the graphical approach would not handle an LED array effectively, and the numerical method works better. To represent the array current iA, the exponential model for the (m × n) array is modified, since voltage drops across current-limiting resistor Rn and sense-resistor RSense are usually small compared with driving-source Vs:

With the array model, the switch-mode buck-regulating current driver in a closed-loop configuration (Fig. 4) can be described by a set of equations starting at the feedback point, circling around the control loop, and ending at the driver output:

4. Analysis of the switch-mode buck-regulating current driver in a closed-loop configuration for driving an array yields a set of equations that are better solved using numerical computation rather than an analytical, closed-form approach.
4. Analysis of the switch-mode buck-regulating current driver in a closed-loop configuration for driving an array yields a set of equations that are better solved using numerical computation rather than an analytical, closed-form approach.

• Current sensing:

Vf = iA • RSense (4)

• Error amplifier:

Ver = A(VRef – Vf) (5)

• Effective error:• PWM ramp:• Duty cycle, D:

VRamp(D • Ts) = Vef (8)

• Power stage:With effort, all the equations can be consolidated into a single, closed-loop equation:This is a transcendental equation with an exponential term that prevents us from solving it analytically, but it can be solved using computational techniques and software tools.

Reference        

1.“Generate Realistic Models for LED Current Versus Voltage” Keng C. Wu; Electronic Design, Vol. 61, No. 1, p. 52, Jan. 10, 2013, http://electronicdesign.com/power/generate-realistic-models-led-current-versus-voltage

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