# RMS and Average Current

Discussion in 'Electronic Design' started by JH, Dec 30, 2005.

1. ### JHGuest

I've not posted to this forum before, hope this is not too dumb of a
question!

Should I use RMS current or Average current in finding the power on a
diode? Say I have a pulse current waveform through the diode. The RMS
value of this waveform is different than the average value. Which do I
use? If I use the RMS value, do I also need to use the RMS value of
the voltage across the diode? If I use the average current, do I then
use the forward diode voltage with it to get the diode average power or
do I need to find the average diode voltage instead? There is some
confusion!

Jeff

Average.

Graham

3. ### John LarkinGuest

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

4. ### Phil HobbsGuest

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

5. ### Fred BloggsGuest

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.

6. ### Phil AllisonGuest

"John Larkin"

** The power produced by such a rectangular pulse can be measured with an
average responding meter.

P = E . I

Where E = diode conduction voltage

and I = average amps.

.......... Phil

7. ### Don LancasterGuest

Not even wrong.

An average responding meter will ALWAYS lie like a rug.

See http://www.tinaja.com/glib/numschip.pdf for a tutorial.

--
Many thanks,

Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552

Please visit my GURU's LAIR web site at http://www.tinaja.com

8. ### Phil AllisonGuest

"Don Lancaster"

** Huh ?

What strange dialect of gibberish is this ?

** Really ?

Not like a * magic carpet * at all then.

** The correlation between " Magic Sinewaves" and " magic carpets " is
blindingly obvious - even to the congenitally tone deaf.

......... Phil

9. ### John LarkinGuest

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

John

10. ### Fred BartoliGuest

It indicates average ...on the average.

11. ### Winfield HillGuest

Fred Bartoli wrote...
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.

12. ### Phil AllisonGuest

"Winfield Hill"

** That is NOT what the pompous ass posted - Win.

** So said the Nazis about the Jews in WW2.

In a GREAT many, VERY useful cases the *average value* is JUST what is
needed.

** You are so full of crap - Win.

Try judging a purely technical matter on the facts alone, instead of

You would not want folk to think you were partial or foolish, now would you
??

......... Phil

13. ### Winfield HillGuest

Phil Allison wrote...
OK, I'm sure you're right, what are some good examples?

14. ### Phil AllisonGuest

"Winfield Hill"

** Snip to pieces, ignore the awkward bits and then post a troll.

What a sickening example of a usenet fake.

......... Phil

15. ### Tim WilliamsGuest

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

16. ### Tim WilliamsGuest

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

17. ### Winfield HillGuest

Phil Allison wrote...

18. ### John LarkinGuest

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

19. ### John LarkinGuest

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

20. ### Don LancasterGuest

The link copier apparently got off track.

The correct cite for the crucial differences between RMS and average
pulse power is http://www.tinaja.com/glib/muse112.pdf

The specific analysis of the trouble this caused with the PE "magic
lamp" is found at http://www.tinaja.com/glib/muse112.pdf

--
Many thanks,

Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552