Re: That Lowe's flashlight, again

Clive George wrote:
"Peter Cole" <peter_cole@xxxxxxxxxxx> wrote in message news:pMGfk.359$gH4.56@xxxxxxxxxxx
SMS wrote:

A bridge rectifier is only 81.2% efficient (maximum with zero-voltage drop diodes)
You keep repeating this, but it's not true. You misunderstand (and misquote) a reference on rectification.

I will admit to being flummoxed by this one too. It's unfortunate that there are quite a few places out there describing it as power efficiency, when that simply isn't the case. To help me understand, I took two approaches.

1) Simple thermodynamics. Heat is work, work is heat,

Careful, this isn't exactly true in electronics. Power is precisely the product of voltage and current, but the vector, not scalar, product. Inductive loads, for instance, change the picture.

All that aside, when comparing incandescent loads to diode loads, both can be modeled as non-linear resistors, so a thermodynamic approach isn't unreasonable. From that perspective, considering efficiency as power out/power in, one has to answer the question of where the power difference goes. In this case (resistive loads), as you say, if not into heat, then where?

The heat dissipated by the rectifier diode(s) is (exactly) the voltage times the current (averaged). The waveform (voltage) you are averaging looks like a constant level (diode forward drop) with little "notches" -- the notches are the parts of the cycle where the forward conductive voltage hasn't been reached (by the generator). Since the voltage waveforms of the LED and rectifier have the same form (exactly), the power ratios will identical to the forward drop ratios. Although the rectifier diode reaches forward conduction voltage earlier in the cycle, no current will flow (hence no power) until the LED begins to conduct (slightly later in the cycle). When they are both conducting, they are passing *the same current* (even though the current waveform is complex), so the ratio of power is the ratio of voltage (relative diode drops). Waveform shape doesn't enter into efficiency calculations.

The reference SMS cited explains that the "regulation efficiency" of a full wave bridge is 81.2%, and a half wave bridge is half of that. This makes perfect sense in the actual context of the definition of "regulation efficiency", but no sense in power efficiency (the usual context for the word).

2) Ask somebody who actually understands these things. Worked for me. You'll have to find your own physics expert - the guy I picked isn't available to the general public, but I suggest to take care they can do all the relevant maths. The basic end result is what SMS is seeing described as power efficiency isn't, and its to do with RMS calculations.

I have to take a bit of offense here. The person to ask is an electrical engineer (OK, we're *applied* physicists). This is the kind of stuff they spent years cramming into our heads. The principle complexity here is non-linearity. Power is still (instantaneously) volts times amps, but when the waveforms are complex (not perfect sinusoids) computing the product and integrating it over the cycle becomes non-trivial. This is not to say the SMS explanation is reasonable, accurate or pertinent. With diodes for both loss and load elements, the ratio of the voltage drops is enough to define efficiency.

What SMS is talking about is the theoretical maximum constant voltage that can be obtained with various rectification circuits relative to the input (RMS) voltage. This is a useful thing to know when designing DC power supplies (something I have done many times), but is not relevant to an analysis of generator/LED system efficiencies, or power efficiency in general. That is "mis-applied" physics.
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