How can I best control various high power LEDs without using PWM

Question

I want to be able to dim various high powered (20mA) LEDs in an array (8X12 LEDs). I do not need individual control, but would like to be able to dim each row (or two rows together) so that I can create gradients.

An important caveat is that I can't use pulse width modulation (PWM) as there should be no flashing whatsoever, even if it is invisible to humans.

How can I best achieve this? (I do not care about energy efficiency)

The fist thing I found which seemed to be the right solution in the beginning is using digital potentiometers as explained in this tutorial. However they only support low currents so I can't use one channel for 8 or 16 LEDs. Maybe I could use various digipots in parallel with one Arduino, each powering 1 LED per channel? But that sounds overcomplicated and a nightmare to wire up. Is there a more elegant solution?

Answer 1

The figure below illustrates a method of current control. It shows four LEDs in a 2x2 arrangement; one would adapt the number of LEDs in series or parallel in accord with how one's array is wired, what voltages are available, LED current requirements, transistor power limitations, etc. [Edit: See transistor-power example calculation at end.]

Each separately-controlled group of LEDs would need one such circuit. Each circuit needs one control voltage (eg, from a DAC or a digital potentiometer voltage divider circuit) and needs one transistor per series chain.

Circuit Operation:

Suppose the following conditions hold.

• Q1 and Q2 are equivalent transistors and V_be is 0.7 V typ. (one diode drop)
• D1, D2, D3, D4 are equivalent LEDs
• Resistors R1, R2 are small and equal (eg 1–30Ω)
• V1 is some fixed DC voltage, eg about 1 V
• V > 2*V_f + V_be + V_R where V_f = typical forward voltage of D1...4

Under these conditions, V1 will drive current into the bases of Q1 and Q2, causing transistor-amplified current to flow through R1 and R2 and maintain the condition V1 = V_be + V_R1 = V_be + V_R2.

Here is a concrete example: Suppose we want 23 mA through each LED and that R1=R2=17Ω. V_R = IR = 0.023 mA · 17 Ω = 0.391 V. If V_be is about 0.7 V, then we set V1 = 0.7 + 0.391 V or about 1.09 V to get 23 mA through each LED. More generally, to get current I through the LEDs we solve V1 = V_be + V_R = 0.7 + IR = 0.7 + I·17.

Base current:

The amount of current required from the V1 voltage source depends on transistor h_FE or β. Eg, if β is about 100 and I_LED is 20 mA then each transistor will need about 20/100 mA or about 200 μA of base drive. This amount of current probably could be delivered by a voltage divider based on an AD5206 (the digital potentiometer chip that's used in the SPIDigitalPot tutorial mentioned in the question).

Power Dissipation:

The AD5206 digital potentiometer datasheet shows an absolute maximum rating for Package Power Dissipation of (T_JMAX-T_A)/θ_JA. If ambient T_A is 25℃ and θ_JA is 70°C/W (as for a SOIC package) this gives (150-25)/70 = 1.8 W max package power. However, it would be reasonable to limit dissipation to under half a watt. In the 10 KΩ version of the AD5206, that means a little less than 2.9 mA maximum per channel. [I = √(0.5 W /(6 channels * 10000Ω)).] If several low-beta transistors are used to drive several strings of LEDs, 2.8 mA might be marginal.

Voltage-divider Linearity:

Only a small fraction of the current through a voltage divider can be drawn off without upsetting the division ratio. A unity gain buffer amplifier [as in the following figure] can be used to increase linearity of voltage-divider output by providing a high-impedance connection to the divider.

(In the figure above, R1 and R2 are not related to R1 and R2 of the previous figure. Here, R2 represents a digitally-variable resistor. Choose R1 and R3 so that V1's range is from slightly below V_be to slightly above V_be + I_maxR, where I_max is the maximum LED current desired. For example: Suppose V=5 V, that R2's total resistance is 10 KΩ, that R is 15Ω, that V_be is 0.7 V, and I_max is 25 mA. Let R_T = R1+R2+R3. Now R3/R_T = V_be/V = 0.7/5 = 0.14, and R1/R_T = (V - V_be - I_maxR)/V = (5 - 0.7 - 0.025·15)/5 = 3.925/5 = 0.785. Thus R1+R3 = 0.925·R_T, whence 0.075·R_T = R2 = 10 KΩ, giving R_T = 133 KΩ, R1 = 105 KΩ, and R3 = 19 KΩ. With these R1 and R2 values, each digital pot step should give a change in LED current of 25/256 mA, or about 98 μA.)

[Edit: When V1 is set to less than V_BE, no current will flow through the LEDs. That is, to turn off the controlled LEDs, set V1 to V_BE or less.]

PWM with RC filter:

It may be worthwhile to experiment with using a PWM-driven low pass RC filter to supply V1, possibly avoiding need for buffer amplifiers and digital potentiometers. See eg “Arduino’s AnalogWrite – Converting PWM to a Voltage” by Scott Daniels at provideyourown.com, and also see the filter-calculation page it refers to. You might need an oscilloscope to verify the design and its quality of results, and might need to use several RC filter stages in a row to adequately filter out ripple, but it can be done, at the cost of slower response to control changes.

Transistor-power example calculations:

Letting V_CE, I_CE, V_BE, and I_BE represent collector-to-emitter and base-to-emitter voltages and currents, transistor power dissipation is V_CE·I_CE+V_BE·I_BE. Because V_BE·I_BE is likely to be sub-milliwatt, we ignore it in following, taking P ≅ V_CE·I_CE.

In the LED-driving circuit, V_CE = V – V_LEDs – V_sense, where V_LEDs is the series voltage across the chain of LEDs the transistor controls, and V_sense is the voltage across the sense resistor (ie, across R1 or R2 in the top figure).

In general, it is a good idea to keep V_sense small (under a volt) to allow a wider compliance range – ie, so that LED current is the same over a wider range of variation of supply voltage V. Thus, the approximation V_CE ≅ V – V_LEDs is reasonable.

Here is a concrete example: Suppose we have eight LEDs in series, and they are white LEDs with about 3.6 V forward drop at 20 mA. [LEDs of different colors may have different drops; eg see LED color chart at oksolar.com.] Then V_LEDs = 8·3.6 V = 28.8 V. We need V_(BR)CEO > V > V_LEDs + V_BE + V_sense, where V_(BR)CEO = collector-emitter breakdown voltage (which is 40 V min. for 2N3904 or 2N2222). Allowing a volt for V_sense and 0.7 V for V_BE we have V > 28.8 + 0.7 + 1 = 30.5 V. Let's suppose V=32 V. Then P < (32–28.8)·0.020 W = 64 mW. Thus a small general purpose NPN transistor will work, for example a 2N3904 (rated at 350 mW to 1 W, depending on package, before derating for ambient and heatsinking) or a 2N2222 (similarly rated).

For a second example, suppose V_LEDs = 8·1.7 V = 13.6 V (for, eg, eight infrared LEDs at 50 mA) and V=32 V. Then P ≅ (32–13.6)·0.050 W = 920 mW, so a larger (and heatsinked) transistor would be needed. Or, V could be decreased, say to 16 V, giving P ≅ (16–13.6)·0.050 W = 120 mW, allowing use of a small transistor.

For a third example, suppose eight LEDs are paralleled (rather than seriesed), that V_LEDs = 2.0 V at 20 mA, that V_sense is about 0.4 V at desired operating point, and that V=5 V. The transistor will need to drop 5-2-0.4 V = 2.6 V at 160 mA, which is 416 mW, probably large enough to require a heatsink. [Note: if LEDs are placed in parallel and driven by a single transistor, some of them probably will be dimmer than others; “current hogging” may occur in some of the LEDs, making them brighter than the others. However, many cheap flashlights connect LEDs in parallel without individual current control.]