The late Bob Pease, a venerated and well-liked engineer at National Semiconductor Corp. (now part of Texas Instruments) once wrote a series of technical musings inspired by questions that mostly younger engineers had to put to him somewhere along the line. We thought back to how Mr. Pease might have reacted when we noticed a recent comment posted in response to one of our teardowns involving 60-W LED bulbs:

To drive LEDs off the mains requires a transformer, bridge rectifier and a resistor! That’s all. The over complication makes for poor reliability when LEDs have an MTBF of around 200 years.

Well, not exactly.

This is a good opportunity to review the basics of why it is inadvisable to power an LED with just a bridge rectifier/transformer/resistor and why makes of LED bulbs are forced to go to some trouble in devising current sources comprised of a bit more than these three components.

It is often said that LEDs must be driven with current sources rather than voltage sources. The primary reason is that LED current varies with forward voltage, and LED light output varies with forward current. For a specific example, consider a Lumileds L150-xxxx500600000 LED, a device optimized for illumination applications. A 200-mV change in its forward voltage causes about a 100-mA change in its forward current. And a 100-mA change in its forward current causes about a 20% change in its light output.

These variations don’t make for high-quality illumination. This is largely why LED drivers are designed to deliver a constant current to the LED rather than a constant voltage.

As the comment writer suggested, the simplest way of setting the current to an LED is to put a properly sized resistor in series with it. However, modern LED supplies don’t do this. A simple example reveals why.

Suppose we have the same Lumileds LED mentioned above. Characteristic curves reveal the LED should carry about 650 mA to produce its rated light output and it should operate with a forward voltage of 6.1 V. For the sake of simplicity, assume the circuit contains only one Lumileds LED. Further suppose we use a 10-V supply. From Ohm’s law, the resistance in series with this 6.1-V LED needed to get 650 mA from a 10-V supply is 3.9/0.65 = 6 Ω because there must be a 3.9-V drop across the resistor. Thus the power dissipated in the resistor is 3.9 × 0.65 = 2.5 W. But the power dissipated in the LED is 6.1 × 0.65 = 4 W (in round numbers). So the combination dissipates 6.5 W, and the resistor dissipates almost 40% of the total power consumed here. Clearly, this isn’t a good state of affairs seeing as the rationale for replacing incandescent lights with LEDs is mainly one of better energy efficiency.

And it gets worse. Another drawback of a series resistance is an inability to closely control the current. LED forward voltage can vary from one unit to the next, so the voltage drop across any resistor put in series with the LED will vary as well. Thus as current can vary from one LED to another, so too will the light output.

Of course, in the real world, most LED lighting products contain multiple LEDs. The drawbacks become more apparent when multiple LEDs are involved. In the example of the 10-V supply, the LEDs would necessarily be powered in parallel with power dissipated in multiple resistors and light output varying from one LED to another. (Note that real LED bulbs configure their LEDs in series, however.)

All these difficulties make it more feasible to drive LEDs with a constant-current power supply rather than a current-limiting resistor with a constant-voltage source. Again for reasons of energy efficiency, switch-mode power supplies are the topology of choice rather than linear supplies. The problem with linear supplies is that they all contain a variable-resistance pass element (i.e. a bipolar transistor or MOSFET) used to meter the load current (or load voltage for constant-voltage supplies). The power dissipated in the pass element can be significant.

For example, suppose we devise a linear supply by first putting ac from a wall socket through a classic full-wave bridge/smoothing-capacitor combo to get a dc level that is near 120 V. If our linear power supply operates in a constant-current mode, we now want to regulate the 120 V down to the 6.1-V/650 mA level needed to run the Lumiled LED without the series resistor. The power loss through the linear regulator is the difference between the input and output voltage multiplied by the load current. In this case, it is (120 – 6.1) ×0.65 = 74 W. Thus, we are losing 74 W in an effort to drive a 4-W LED.

Clearly, then, if your mission is to replace a 60-W incandescent bulb with a 4-W LED, a conventional linear supply won’t give you a bulb that is more energy efficient.

From the standpoint of energy efficiency, the appeal of a switching power supply is that its power loss doesn’t equal the difference between the input and output voltages multiplied by the load current, as in the case of a linear supply. The simplest model of a switching supply is one of a switch with two states, on and off, that chops the input voltage. The switch has a super-low on resistance, unlike the variable-resistance pass element used in the linear supply. To synthesize a steady output from the chopped waveform, an output filter is added. But the filter consists only of temporary energy storage elements, so it consists of nearly lossless components.

The combination of nearly lossless filtering and low-loss switching allow switching supplies to typically display energy efficiencies in the range of 90%. This is good enough for powering LEDs. The real trick is packing the power supply components into a container small enough to screw into an ordinary A19 light socket. As our teardowns of LED bulbs have shown, this is a feat that LED bulb makers achieve with a greater or lesser degree of elegance.