RMS and Average Current

There are no dumb questions... only dumb people. (Smilie)

Neither. A diode is nonlinear, so neither works. There's no "R" to plug into P = I^2 * R

The power is the average of the instantaneous series of E*I samples, which can get messy to compute.

If it's a clean rectangular pulse, measure pulse current and simultaneous diode voltage drop, and do

P = E * I * n

where n is the duty cycle.

John

That isn't a dumb question at all--probably in the 75th percentile, if we include politics.

Electrical power dissipation in any device is the time average of V*I. For a resistor, which is very linear, V = IR, so the power dissipation is the time average of I^2*R. RMS is short for "root mean square", i.e. you time-average the square of the current, and take the square root of the average value. For a linear resistive load, the RMS value is convenient because when you plug it into the I^2*R formula, it gives you precisely the time average of V*I, which you can easily verify from the definition. (Reactive loads are a bit more complicated because V and I get out of phase with each other.)

For a nonlinear device such as a diode, the power dissipation is still the time average of V*I, but V is no longer I*R, so the RMS current will give the wrong answer. Diodes tend to behave like an resistor in series with an ideal diode (Vf = I*R + gamma*ln(I/Is), where gamma = 25 to 60 mV at room temperature). To predict the dissipation accurately, you'd need to compute the Vf waveform from your I waveform, multiply the two together, and time-average. Of course, that would require you to have a decent idea of what R and gamma are for your diode.

For most engineering purposes, you can assume that (to one significant figure) a signal diode will drop 0.7 volts whenever it has forward current, so your power dissipation will be the time average of (0.7 I), which is of course 0.7 * I_avg. Rectifiers and high-speed switching power supply diodes will drop more than this--sometimes more than 1 V--so you'll probably have to measure Vf if you need any sort of accuracy.

Cheers,

Phil Hobbs

Average.

Graham

It depends. To a first approximation the diode characteristic is a piecewise linear V=VF+I*R for I>0, and R is reciprocal slope of the I/V characteristic at the breakpoint. The power dissipation is then VF*I + I^2*R and this averages to VF*Iavg + Irms^2*R over one cycle of the rectified current. At large currents, the RMS current will be a very significant factor in estimating the power dissipation.

You mean, it won't indicate average? OK, what *does* an average-indicating ammeter indicate?

John

Not even wrong.

An average responding meter will ALWAYS lie like a rug.

See

:-) I think Don Lancaster meant, average indication isn't very useful, in that it's rarely what you really want, so relying on its value is likely to give you the wrong answer. For example, an average-indicating meter may be calibrated to show rms, but of course doesn't. But we all knew that.

Hey, Phil, cool your jets!

OK, I'm sure you're right, what are some good examples?

Phil's example a few posts ago is one such:

Which corresponds to my version,

P = E * I * n (E and I being the 'on' values)

since

Iavg = Ion * n

Sort of clever of him, in fact. An RMS ammeter would be the wrong one in this case.

John

One might also note that it's rare to have a ammeter that actually indicates RMS current on a DC range, unless you have an ancient Weston electrodynamic thing with a mirrored scale and brass lugs. As far as DC current measurements, average is usually more meaningful.

John

"John Larkin" a écrit dans le message de news: snipped-for-privacy@4ax.com...

It indicates average ...on the average.

Well, DC consumption, with non-constant current draw across the meter. As long as there's a good capacitor (constant voltage) in the circuit, V * I = P.

You can calibrate the scale in square root volts and read off the average of a numerically squared AC voltage to get RMS. ;-)

Tim

-- Deep Fryer: a very philosophical monk. Website:

Well, DC consumption, with non-constant current draw across the meter. As long as there's a good capacitor (constant voltage) in the circuit, V * I = P.

You can calibrate the scale in square root volts and read off the average of a numerically squared AC voltage to get RMS. ;-)

Tim

-- Deep Fryer: a very philosophical monk. Website:

Sorry about that. The link copier apparently got off track.

The correct cite for the crucial differences between RMS and average pulse power is

The specific analysis of the trouble this caused with the PE "magic lamp" is found at

It indicates an incompetent individual deluding themselves while making a measurement.

Sheeesh! The has-been is the delusional one.

...Jim Thompson

I'll eventually get it right. The PE cite is muse113.pdf

For those of you not up on the story: An EE-challenged epsilon minus created a very late duty cycle half wave dimmer circuit (straight out of a 1938 industrial electronics book) and connected a 32 volt light bulb to it. When compared to a 110 volt light bulb of the same brightness, their average responding meter showed a 3:1 voltage difference and a 3:1 current difference.

Multiplying the two together gave an obvious 10:1 power difference. Which they patented (!) and published (probably in 1997) in a (coincidentally) April issue of Popular Electronics as a way to save 90 percent on your power bills.

Naturally, they never TOUCHED the 32 volt bulb to see if it was any COOLER.

Even years later, that individual refused to believe they had committed EE Lab student blunder 0001-A. Or that their patent was utterly worthless on countless different levels. (even ignoring the illegal waveform, brightness sensitivity, and the ease with which the bulbs could be inadvertently burned out.)

Sure enough, at a 138 degree or so halfwave phase delay angle, the RMS to average difference is a surprisingly high 3:1.

None. But I was talking about DC meters, not about DC current.

Suppose you have an average current of 1 amp, with 0.2 amps RMS of ripple superimposed. An average-reading ammeter will show 1.0, but a true RMS meter would indicate 1.02 amps. There are a few DVMs that do real dc-coupled RMS measurement, but they're kinda rare.

All the DVMs I've been able to see the guts of, seem to have a slow integrating (dual slope, delta-sigma, or some variant) DC section, with a switchable, usually AC-coupled averaging or RMS ac-to-dc converter ahead of it. So why not use a single fast (successive approximation) ADC and then DSP the data? That way you could extract anything and feature-itus it to your heart's content. And get superb wideband AC accuracy.

John