LED Driver Efficiency

One limit to LED driver efficiency is that current flows through rectifiers (diodes used as one-way gates), and diodes have a forward voltage drop and thus waste power. For example, if a 1A driver goes through a diode with 0.6V drop, then the wasted power is P = I2 R = 0.6W. If the driver is running in a mode where the diode is ``on'' 1/3 of the time, then the average power loss is (0.6W)(1/3) = 0.2W. If the driver is driving 10W, then the loss is 0.2W/10W = 2%.

My bad-but-almost-good inspiration was to replace each diode with an LED. Then, the power is no longer wasted. Ta-da!

Well, not quite so fast. A standard diode and an LED are different in many ways, leading to loss of the supposed good properties.

Briefly, a boost regulator works by charging up an inductor by shorting it to ground, then disconnecting the ground end. As the inductor field collapses, it tries to maintain the same current, and in doing so will express as high a voltage as it needs to in order to get current to flow.

[boost-converter.jpg]
Boost (step-up) converter. It switches between ``charging'' and ``driving'' modes. The left shows the ``charging'' mode, where a current is established in the inductor. The right shows the ``driving'' mode, where the inductor current is dumped into the load portion of the circuit. Note the voltage entering diode D is VIN plus the voltage expressed by inductor L in order to keep current flowing through L. From [SR05].

See the diode? Bad diode. Bad, bad diode. That's 2% efficiency. Wasted. Wasted! We just replace it with an LED, and -- bling! -- there was more efficient light!

Except...

Okay, that's not such a good idea after all. Maybe it works better on a buck converter? Here's a buck converter. Note that it has an evil power-wasting diode!

[buck-converter.jpg]
Buck (step-up) converter. It switches between ``building'' and ``decaying'' modes. The left shows the ``building'' mode, where a current is established in the inductor. The right shows the ``decaying'' mode, where the inductor current is dumped into the load portion of the circuit. Note the voltage exiting diode D and entering inductor L is below zero. The voltage drop is created by inductor L in order to get a voltage across diode D in order to keep current flowing through L. From [SR05].

Unfortunately, it looks like no win here, either.

So, um, points for thinking ``outside the box'', but it's back to germanium diodes or using MOSFETs as the diode.

References

[SR05] ``Switching Regulators''. No date, authorship, or other identification given. From www.national.com/appinfo/power/files/f5.pdf as of 2005/07, linked from www.national.com/kbase/category/Power.html.