Power calculation for rectified AC.
Quentin Grady
his response CC'd by email
On Sun, 16 Nov 1997 10:17:36 -0800, Dave Baldwin <dib...@psyber.com>
wrote:
>If the source is a sine wave, the power for a 'no-loss' half-wave
>output is half of the 'RMS' power.
IMHO
The _average_ power is calculated from rms voltages or rms
currents, so I tell my students that the rms is incorrect when used
with power. They still do it, of course, to distinguish average
power from peak power or worse still peak_to_peak power.
[..]
>100 V peak sine wave = 70.7 V RMS
>
>70.7vrms^2/20ohms = 4998/20 = 249.9 watts
>
>249.9 watts * 0.5 = 125 watts for half-wave output.
Congratulations on joining the select group who get 125 W on the first
attempt.
Just for fun try putting the average power out for the following in
descending order.
A Full wave rectified without any filter.
B Half wave rectified without any filter.
C Full wave rectified with a shunt filter capacitor.
D Half wave rectified with a shunt filter capacitor.
E Full wave rectified with a series filter inductor.
D Half wave rectified with a series filter inductor.
All capacitors and inductor are ideal and sufficient time has been
allowed for steady states to have occurred.
Thanks for your response, Dave. Good to get some confirmation.
Quentin Grady ^ ^ /
New Zealand, >#,#< [
/ \ /\
"... and the blind dog was leading."
Dave Baldwin
Quentin Grady wrote:
> On Sun, 16 Nov 1997 10:17:36 -0800, Dave Baldwin <dib...@psyber.com>
> wrote:
>
> >If the source is a sine wave, the power for a 'no-loss' half-wave
> >output is half of the 'RMS' power.
>
> IMHO
> The _average_ power is calculated from rms voltages or rms
> currents, so I tell my students that the rms is incorrect when used
> with power. They still do it, of course, to distinguish average
> power from peak power or worse still peak_to_peak power.
Power in Watts is time-dependent by definition (1 Watt =
1 Joule per second). I have a program that uses trapezoidal
integration to calculate the power dissipation in the components
of a class B output stage on my web page called PDQ.EXE. I'm working
on another program that will calculate the RMS voltage and power into
a resistive load for Sine, Square, and Triangle waveforms with DC
offset that I will post fairly soon (PWRQ.EXE). I will look at
including the waveforms you mentioned because the next program is a
power supply and regulator analysis program.
Note that for a continuous waveform, there is no such thing as 'RMS'
or 'average' power, it's just power. 'RMS' power is used to indicate
that a sine wave was used to generate it (audio applications usually).
'Average' power should mean power averaged over several different
measurements and/or calculations.
The most complete mathematical definition of RMS voltage I have found
is on page 289 of the 'The Wiley Engineers Desk Reference', formula 5.123.
The formula is independent of waveform. The standard formulas used for
DC, square waves, and sine waves are just special cases of that formula.
--
-=-=-=-=-=-=-=-=- "http://www.psyber.com/~dibsed" -=-=-=-=-=-=-=-=-=-=-
DIBs Electronic Design | Dave Baldwin: dib...@psyber.com
Voice : (916) 722-3877 | Hands-on hardware and software
TCJ/DIBs BBS: (916) 722-5799 | TCJ/DIBs FAX: (916) 722-7480
-=-=-=-=- The Computer Journal "http://www.psyber.com/~tcj" -=-=-=-=-
Quentin Grady
This response CC'd by email
On Mon, 17 Nov 1997 08:03:43 -0800, Dave Baldwin <dib...@psyber.com>
wrote:
>Quentin Grady wrote:
>> On Sun, 16 Nov 1997 10:17:36 -0800, Dave Baldwin <dib...@psyber.com>
>> wrote:
>>
>> >If the source is a sine wave, the power for a 'no-loss' half-wave
>> >output is half of the 'RMS' power.
>>
>> IMHO
>> The _average_ power is calculated from rms voltages or rms
>> currents, so I tell my students that the rms is incorrect when used
>> with power. They still do it, of course, to distinguish average
>> power from peak power or worse still peak_to_peak power.
>
>Power in Watts is time-dependent by definition (1 Watt =
>1 Joule per second). I have a program that uses trapezoidal
>integration to calculate the power dissipation in the components
>of a class B output stage on my web page called PDQ.EXE. I'm working
>on another program that will calculate the RMS voltage and power into
>a resistive load for Sine, Square, and Triangle waveforms with DC
>offset that I will post fairly soon (PWRQ.EXE). I will look at
>including the waveforms you mentioned because the next program is a
>power supply and regulator analysis program.
My students would have enjoyed that. They have been doing the
integration from V_t and I_t graphs of the output of their pulse width
modulated motor speed controllers.
>Note that for a continuous waveform, there is no such thing as 'RMS'
>or 'average' power, it's just power. 'RMS' power is used to indicate
>that a sine wave was used to generate it (audio applications usually).
Go on Dave you're pulling my leg.
When you or I calculate the power dissipated in a household light bulb
we are calculating average power whether we bother to state the fact
or not. I am assuming of course that the power supply is not
idealised steady DC when of course the term average would be
redundant.
>'Average' power should mean power averaged over several different
>measurements and/or calculations.
It could, though I suspect my students would laugh themselves hollow
if I tried to convince them that that was the case. Using the term
average in the context of power more usefully refers to averaging over
time.
>The most complete mathematical definition of RMS voltage I have found
>is on page 289 of the 'The Wiley Engineers Desk Reference', formula 5.123.
>The formula is independent of waveform. The standard formulas used for
>DC, square waves, and sine waves are just special cases of that formula.
Absolutely.
Roy McCammon
Dave Baldwin wrote:
> Note that for a continuous waveform, there is no such thing as 'RMS'
> or 'average' power, it's just power.
There definitely is such a thing as average power, as
distinct from instantaneous power. Even in sine waves,
its customary to compute the average power over one cycle.
> 'Average' power should mean power averaged over several different
> measurements and/or calculations.
Could mean that. Could also mean averaged over time.
Opinions expressed herein are my own and may not represent those of my employer.
D.H. Kelly
Dave Baldwin wrote:
Note that for a continuous waveform, there is no such thing as 'RMS'
that a sine wave was used to generate it (audio applications usually).
measurements and/or calculations.
-----
RMS power is a term once (and still may be) used in the audio industry to
distinguish true average power from "music" power as defined by marketing
people.
Dave is right in that power is energy/unit time but instantaneous power is
not used to a great extent but normally power is stated (as in a 100 w
light ) as time average and this time average power is what is usually used
rather than instantaneous power. When power calculations are done with RMS
voltages and currents, the resulting value is this time average power
(average over a cycle) (P=real(VI*) where V is the rms voltage and I* is
the conjugate of rms current, P is the time average power over a cycle. )
--
Don Kelly
dke...@nanaimo.ark.combull
remove the "bull " to reply
Robert H. Penoyer
Dave Baldwin <dib...@psyber.com> wrote:
<snip>
>Note that for a continuous waveform, there is no such thing as 'RMS'
>or 'average' power, it's just power.
When the term "power" is used it usually means average power. Average
power is the basis for the calculation of RMS voltage and current.
That's where the "M" of RMS comes from: M = mean = average
>'RMS' power is used to indicate
>that a sine wave was used to generate it (audio applications usually).
RMS power is a bogus term no matter where it's used.
>'Average' power should mean power averaged over several different
>measurements and/or calculations.
Whatever you think it should mean, average power is calculated by
standard methods. Once understood, these methods are intuitively
obvious.
>The most complete mathematical definition of RMS voltage I have found
>is on page 289 of the 'The Wiley Engineers Desk Reference', formula 5.123.
>The formula is independent of waveform.
True.
>The standard formulas used for
>DC, square waves, and sine waves are just special cases of that formula.
True.
Don Barr
Perhaps I am mistaken here, but this is how I learned RMS Power;
120VACrms is the voltage (about 167VACp-p) that will dissipate the same
wattage into a given resistive load (power factor=pf=1) that 120VDC will.
In a resistive load current and voltage are "in phase" and full power is
delivered to the load. In reactive loads current may lead or lag the
voltage and this affects the power dissipated in the load. This is true
for mostly symetrical waveforms, in other words it is not specifically
linked to "sine" waveforms. As pf shifts (capacitive or inductive) the rms
power (or "real" wattage) will change. For this purpose often Volt-Amperes
(VA) are used, since pf is not considered, when loads of unknown or mixed
pfs are to be connected. Volt-Amperes-Reactive is a term also used, though
I believe only routinely in the power industry when considering load
sharing among generators.
Thanks for the time.
Later
DBarr
Dave Baldwin <dib...@psyber.com> wrote in article
<34706ADF...@psyber.com>...
> Quentin Grady wrote:
> > On Sun, 16 Nov 1997 10:17:36 -0800, Dave Baldwin <dib...@psyber.com>
> > wrote:
> >
> > >If the source is a sine wave, the power for a 'no-loss' half-wave
> > >output is half of the 'RMS' power.
> >
> > IMHO
> > The _average_ power is calculated from rms voltages or rms
> > currents, so I tell my students that the rms is incorrect when used
> > with power. They still do it, of course, to distinguish average
> > power from peak power or worse still peak_to_peak power.
>
> Power in Watts is time-dependent by definition (1 Watt =
> 1 Joule per second). I have a program that uses trapezoidal
> integration to calculate the power dissipation in the components
> of a class B output stage on my web page called PDQ.EXE. I'm working
> on another program that will calculate the RMS voltage and power into
> a resistive load for Sine, Square, and Triangle waveforms with DC
> offset that I will post fairly soon (PWRQ.EXE). I will look at
> including the waveforms you mentioned because the next program is a
> power supply and regulator analysis program.
>
> Note that for a continuous waveform, there is no such thing as 'RMS'
> or 'average' power, it's just power. 'RMS' power is used to indicate
> that a sine wave was used to generate it (audio applications usually).
> 'Average' power should mean power averaged over several different
> measurements and/or calculations.
>
> The most complete mathematical definition of RMS voltage I have found
> is on page 289 of the 'The Wiley Engineers Desk Reference', formula
5.123.
> The formula is independent of waveform. The standard formulas used for
> DC, square waves, and sine waves are just special cases of that formula.
>
kingsnake
You know: this question is really ambiguous. Just to make those of you
who ask "am I missing something?" feel any better: yes, you are. :-)
You are asked to calculate the power dissipated in the load. A
question like this on an exam would =always= specify average, RMS or
peak power. Since this qualifier is not specified, there is more than
one correct answer --
(Assuming an ideal rectifier,)
125 W. RMS, is of course, _one_ correct answer, since the peak power
of a pure sine wave is twice the RMS power, we have (100^2)/20/2, or
250 watts Rms, modulated by a 50% duty-cycle square wave coming out of
the half-wave rectifier. Since the peak, RMS and average values of a
Square wave are the same, the average, and therefore RMS current
through the load == 250x(50%), or 125 watts.
The peak power would be 500 watts.
The average power would be == 101.33 watts.
Hope this helps.
In article <346bda1c....@news.zippo.com>, que...@inhb.co.nz
wrote:
>G'day G'day Folks,
>
>Students swatting for examinations unearth model answers.
>The problem is some of these answers seem to me to be wrong.
>
>Here is an example.
>
>A transformer has a 100 V pk output voltage which is then half wave
>rectified without significant voltage loss. The load is 20 Ohms.
>
>Calculate the power dissipated in the load.
>By my reasoning, rectification of itself doesn't change the power
>available to a resistor although being half wave does halve it.
>Thus power is a quarter of (100^2 / 20) = 10000/80 = 125 W
>
>The "model" answers suggests the output of the rectifier has an
>average value of 1/(pi) of 100 which is fine.
>Then however it precedes to use the average value to calculate the
>power which I believe is wrong.
>
>Am I missing something?
>
>Thanks,
>
>Quentin Grady ^ ^ /
>New Zealand, >#,#< [
> / \ /\
>"... and the blind dog was leading."
-John <kingsnake> WA6FRN/6
kingsnake photography; a division of Gyro Gearloose Productions
http://www.humboldt1.com/~gyrgrls/
ICQ uin:4604100
Quentin Grady
This response CC'd by email
Thanks John for a truly fascinating response.
>You know: this question is really ambiguous. Just to make those of you
>who ask "am I missing something?" feel any better: yes, you are. :-)
>
>You are asked to calculate the power dissipated in the load. A
>question like this on an exam would =always= specify average, RMS or
>peak power. Since this qualifier is not specified, there is more than
>one correct answer --
OK. I can see two possibilities; average and peak.
To me rms power is a misnomer. IMHO the question would need to
specify average power or peak power to avoid ambiguity.
Good point.
>(Assuming an ideal rectifier,)
>125 W. RMS, is of course, _one_ correct answer, since the peak power
>of a pure sine wave is twice the RMS power, we have (100^2)/20/2,
Many people would disagree with you. They would say that the power
calculated by using rms is the average power. The average power
curve certainly requires the rms values for voltage and current to
enable power to have the formula P = VI rather than P = 0.5 Vp Ip
Naturally rms can also refer to non-sinusoidal voltages but in each
case the rms voltage and current gives the average power ie power
averaged over time. At least that is what I believe. If I am wrong
in any aspect of this I really want to know.
>or 250 watts Rms, modulated by a 50% duty-cycle square wave coming out of
>the half-wave rectifier.
This is an interesting concept. There is a sine wave and the
rectifier modulates the sine wave. Does modulate mean what I think
it does, ie the waves are multiplied together rather than added.
I take it that your modulating square wave has an offset of + Vp so it
is equivalent to a square pulse train with a lower limit of 0 V and
upper limit of Vp.
>Since the peak, RMS and average values of a
>Square wave are the same, the average, and therefore RMS current
>through the load == 250x(50%), or 125 watts.
True the peak, RMS and average values of the square wave you have
described are equal. This appears to be irrelevant to the
calculation but hey, I'm here to learn.
> The peak power would be 500 watts.
Nice point. The peak power for half wave it the same as full wave.
>The average power would be == 101.33 watts.
I see absolutely nothing to justify this statement.
Care to enlighten?
By now I am pretty convinced the average power is 125 W for half wave.
It would take a pretty good argument to dissuade me.
Now there's a challenge.
Quentin Grady
This response CC'd by email
>RMS dictates the effective work done by an AC signal, which can
>be less than, equal to, or greater than the average current through
>the load, depending on the waveform.
Care to give me one example of an RMS value being less than the
average value?
Thanks,
Quentin Grady
>You should read ARRL's "The Radio Amateurs Handbook", available at
>any good library, or for about $30.00 from ARRL.
I could read ARRL's "The Radio Amateurs Handbook" though it is by no
means clear why I should.
Take a look at your suggestion from my point of view for the moment.
We are talking $30 US dollars plus those horrendous US postage rates.
On cheaper books they double the cost. All for a book that appears
to filled with what, based on what you have presented, I believe to
be misleading concepts. Think I'll save my money for the moment.
Perhaps this book contains examples where the average is greater than
the rms? If there really are such examples then I'll welcome your
advice to read this book. For thirty years or so I've been happy
with the idea that the rms value is equal to or greater the average
value depending on the waveform for which the calculation is done.
My students (admittedly only first year technicians) have been happy
with the idea too. If we have been wrong all this time please break
the news to us gently.
Quentin Grady
>Mathematically, it is the square root of the arithmetic mean of the
>squares of the two limits. For 1 volt pk, we get: sqr [ (1^2 + 0^2)/2]
>= sqr.( 0.5) = (sqr.2)/2 , or ~ 0.707 .
If this argument were valid then the form factors = rms/|average| of
sine waves and triangular waves would be equal.
Are they?
These proofs you are offering are seductive in their simplicity.
I can think of a few students who would love them and a few more who
would be less than subtle then I am in pointing out their fallacies.
D.H. Kelly
There seems to be a wide array of answers out there - but few have really
considered the definition of RMS voltage and/or current.
The basis for rms is a DC equivalence - i.e. 1 volt x1 amp gives 1
watt(resistive load)
for this to be true R*(Irms)^2 = R*(Idc)^2 -i.e. To do this
Irms = sqrt((1/2pi)*integral(0 to 2pi) of i(t)^2 .dt
i.e. the root of the mean square. This will depend on the wave shape.
For a square wave Irms = Ipeak , for a sine Irms =0.707 Ipeak, etc.
Go to the definition - surely it is still in textbooks.
Quentin -you were right the first time.
Robert H. Penoyer
king...@dm.net (kingsnake) wrote:
<snip>
Your posts contain so many errors and misconceptions that it's
difficult to correct them, but I will try.
>Oddly enough, a sinusoidal wave has a higher Effective power (heating
>effect when passed through a resistive load) than the computed average
>power.
Effective power and average power are the same thing. When you use RMS
(i.e., effective) voltages and/or currents in standard power
calculations (P = I E, P = E^2 / R, or P = I^2 R) you are calculating
AVERAGE power. After all, the M in RMS is Mean, and Mean = average.
There is no error in referring to this as effective power since it is
the power that causes the same heating whether using a DC voltage or
the effective (i.e., RMS) AC voltage equivalent. But RMS power is a
misnomer since the concept of RMS is applied, as I'll explain below,
to voltage and current, but not power.
Consider what RMS is: a means of comparing AC voltage or current to DC
voltage or current. More specifically, RMS is calculated as follows.
I'll use the case of RMS voltage, so that we are, by implication,
dealing with P = E^2 / R. That is, some representation of AC voltage
is to be squared and divided by R such that the AC voltage has the
same heating effect as a given DC voltage. Here is how the RMS
equivalent is found:
1. At each point along a voltage waveform, square the voltage value.
This is the S of RMS. (When divided by R, the squared value at each
point gives the intantaneous power delivered to R.)
2. Over a complete cycle, take the average of the squared values
calculated in Step 1, above. This is the M of RMS. (This average, when
divided by R, is clearly the AVERAGE power delivered to R.) But we
don't want the square of any voltage; we want the voltage.
3. Take the square root of the average found in Step 2. This is the R
of RMS. (Thus, this result is the voltage that must be squared to find
the AVERAGE power delivered to R.)
Now notice that steps 1, 2, and 3 work for any regular (i.e.,
repeating) waveform, even DC if "complete cycle" in Step 2 is taken as
any desired time since we'll assume the DC we refer to here is
unchanging in amplitude. If, for example, we use these steps to
calculate "effective" DC voltage, we simply get Erms = Edc. (No
surprise here.) If we follow the steps for a sine wave we get Erms =
Epk / sqrt(2). If we follow the steps for a square wave we get Erms =
Epk. And so on.
>RMS dictates the effective work done by an AC signal, which can
>be less than, equal to, or greater than the average current throught
>the load, depending on the waveform.
Don't mix up voltage and current terms, such as RMS, with power.
>You should read ARRL's "The Radio Amatuers Handbook", available at
>any good library, or for about $30.00 from ARRL.
The ARRL handbook is quite a good reference. I think you might want to
carefully review whatever it has to say about RMS voltage and current.
I think you will find that it does _not_ refer to RMS power.
>However,
>Let me try to explain RMS in my own words. Let's consider sine waves,
>since we are talking about power supplies.
>
>A quick look at calculus:
There is no calculus in what you're about to describe.
>A sinewave is nonlinear. On it's leading edge , in time, halfway up to
>the peak (45 degree phase angle) would correspond to the peak value
>(which occurs at 90 degrees) at half the rotation, which is 45
>degrees.
What?
>Since the sine of 45 deg is the sqr.root of 2, or approx
>0.707, this would correspond to the Root-Mean-Square (RMS) value of
>the extremes (limits), which are 0 and 1, respectively.
No. The sine of 45 degrees has nothing to do with the calculation of
RMS except by coincidence. See the description I provided above.
>Mathematically, it is the square root of the arithmetic mean of the
>squares of the two limits. For 1 volt pk, we get: sqr [ (1^2 + 0^2)/2]
>= sqr.( 0.5) = (sqr.2)/2 , or ~ 0.707 .
This is wrong.
<snip>
>Now let's consider _average_ values.
>A sinewave is derived from a circle. Consider a full circle 360 deg,
>and one full AC cycle one full circle.
>OK, since we have ( 2 pi) radians in a complete cycle, the radius (pi)
>would conform to the _average_ of the power dissipated throughout the
>cycle. Therefore, the _average_ voltage (or current) would be
>{V (or I) pk} / pi . The average power would be P/2pi .
None of what you've just described makes any sense. You've mixed up
voltage waveforms and power.
The average value of a sine wave is calculated as follows. Let the
sine wave (voltage or current) have a peak value of 1. Then over a
distance of a half-cycle (i.e., pi) the area under the sine wave is 2
so that the average value is 2 / pi, or approximately 0.637.
>Now let's look at the 100 volt problem again. The output of the
>transformer is 100 Vpk.
>The half-wave rectification effectively chops the duty cycle in half
>(e.g.: 50%), so we divide everything by two.
>Therefore, the peak current is 5 amps, but we use 2.5 amps here,
>because the wave , being chopped, is no longer sinusoidal.
No. In moving from a full-wave rectification to half-wave
rectification, POWER is reduced by half, not current.
<snip>
>>Many people would disagree with you. They would say that the power
>>calculated by using rms is the average power.
>
>They are misled.
No, they're not.
>> The average power curve certainly requires the rms values for
>> voltage and current to enable power to have the formula P = VI
>> rather than P = 0.5 Vp Ip
>
>BOTH EQUATIONS ARE CORRECT!
Bingo! This statement is true.
>Don't forget that V and I are also RMS values. :)
>
>Since Peak voltage = RMS Voltage x sqr.2, and current follows voltage
>in a resistive load, the peak power (in a sine wave) is
> V RMS x sqr.2 x I RMS x sqr.2 = Vrms Irms (sqr.2)^2 = 2VIrms = 2
>Prms.
Peak voltage multiplied by peak current yields peak power. What you've
called RMS power is simply the AVERAGE power. That is Vrms * Irms =
Pave = (Vpk * Ipk) / 2 = Ppk / 2.
<snip>
>No simple formula like P/2 or P/2(pi) exists for complex waves.
Steps 1, 2, and 3 that I outlined above apply to any regular
waveforms, that is, any waveform that repeats.
<snip>
>But if you plug in the formulae for average power, using pi/2 in place
>of the sq.rt.of 2, you will be enlightened.
And wrong.
Tony Williams
In article <3478d536....@news.zippo.com>, Quentin Grady
<URL:mailto:que...@inhb.co.nz> wrote:
This thread is rumbling on , so I did the experiment and actually
measured it. It's a bit crude, but here's what happened....
Variac, diode, 1800-ohm resistor, dvm, surface thermometer.
-----------------------------------------------------------------
Expt1. Diode in series with the R.
At 72vrms from the Variac the R reached a surface temp of 110degC.
-----------------------------------------------------------------
Expt2. Diode shorted out, see what acv is req'd for 110degC.
It took 52vrms to get the diode up to 110degC.
-----------------------------------------------------------------
So the actual power in the R in expt1 was 52^2/1800 = 1.5W.
For 72vrms straight, the R-power would be 72^2/1800 = 2.88W.
Therefore, the power ratio diode/nodiode = 1.5/2.88 = 0.52.
(Posted and cc'd)
--
[Tony Williams, Ledbury, Herefordshire, UK.---Pagewidth=64-----]
Tony Williams
In article <ant22102...@ledelec.demon.co.uk>, Tony Williams
<URL:mailto:to...@ledelec.demon.co.uk> wrote:
> It took 52vrms to get the diode up to 110degC.
~~~~~
Bugger! I saw it 100mS after pressing the transmit button.
For 'diode' read '1800-ohm resistor'.
Quentin Grady
This response CC'd by email
<to...@ledelec.demon.co.uk> wrote:
>In article <3478d536....@news.zippo.com>, Quentin Grady
><URL:mailto:que...@inhb.co.nz> wrote:
>
> This thread is rumbling on , so I did the experiment and actually
> measured it. It's a bit crude, but here's what happened....
Thanks Tony,
Nice experiment.
> Variac, diode, 1800-ohm resistor, dvm, surface thermometer.
>
> -----------------------------------------------------------------
> Expt1. Diode in series with the R.
>
> At 72vrms from the Variac the R reached a surface temp of 110degC.
>
> -----------------------------------------------------------------
> Expt2. Diode shorted out, see what acv is req'd for 110degC.
>
> It took 52vrms to get the resistor up to 110degC.
(edited) No probs.
> -----------------------------------------------------------------
>
> So the actual power in the R in expt1 was 52^2/1800 = 1.5W.
>
> For 72vrms straight, the R-power would be 72^2/1800 = 2.88W.
>
> Therefore, the power ratio diode/nodiode = 1.5/2.88 = 0.52.
G'day G'day Tony,
A wise choice of measurements. Glad you didn't try measuring, for
instance, the rms voltage after rectification. <grin>
Then the thread would really have rumbled on as people argued about
exactly what your meter measured with non-sinusoidal voltages. Also
glad you didn't measure the transformer rms current ... for the same
reasons.
Special thanks for doing the experiment.
I take it the conclusion is too obvious for you to state. <grin>
Quentin Grady
This response CC'd by email
G'day G'day Tony et al,
I liked this experiment so much I think I'll change one of the
introductory exercises for first year AC circuits students.
IMHO it really captures the meaning of rms voltage for sinusoidal and
non-sinusoidal waveforms. It can be performed with easily understood
equipment. The surface temperature device didn't need calibrating etc
etc I many posts I have been exploring options for affordable
wattmeters. This experiment works without a wattmeter.
(The voltages will all have to be less than 32 Vrms to conform with
local regulations. )
On Sat, 22 Nov 1997 10:35:27 +0000 (GMT), Tony Williams
<to...@ledelec.demon.co.uk> wrote:
> This thread is rumbling on , so I did the experiment and actually
> measured it. It's a bit crude, but here's what happened....
>
> Variac, diode, 1800-ohm resistor, dvm, surface thermometer.
>
> -----------------------------------------------------------------
> Expt1. Diode in series with the R.
>
> At 72vrms from the Variac the R reached a surface temp of 110degC.
>
> -----------------------------------------------------------------
> Expt2. Diode shorted out, see what acv is req'd for 110degC.
>
> It took 52vrms to get the resistor up to 110degC. [ed]
>
> -----------------------------------------------------------------
>
> So the actual power in the R in expt1 was 52^2/1800 = 1.5W.
>
> For 72vrms straight, the R-power would be 72^2/1800 = 2.88W.
>
> Therefore, the power ratio diode/nodiode = 1.5/2.88 = 0.52.
Tom Burke
Quentin Grady wrote:
>
> This response CC'd by email
> >squares of the two limits. For 1 volt pk, we get: sqr [ (1^2 + 0^2)/2]
> >= sqr.( 0.5) = (sqr.2)/2 , or ~ 0.707 .
>
> sine waves and triangular waves would be equal.
>
> Are they?
>
> These proofs you are offering are seductive in their simplicity.
> I can think of a few students who would love them and a few more who
> would be less than subtle then I am in pointing out their fallacies.
>
> New Zealand, >#,#< [
> / \ /\
> "... and the blind dog was leading."
jeez... All this from people who should already know... OK, from:
Nilsson, James W., "Electric Circuits Fourth Edition," Addison Wesley,
1993
ISBN: 0-201-54987-5
pp 762
The rms value of a periodic function can be expressed in terms of the
Fourier coefficients; by definition,
/-----------------
Frms = / 1 t0+T 2
/ - [ f(t) dt
\/ T t0
where [ represents an integral sign...
--
to...@siu.edu
t...@engr.siu.edu
tbur...@aol.com
Tom Burke
2520 New Era Rd #19
Murphysboro, IL 62966
(618) 549-3188
http://members.aol.com/tburkeii/index.html
Quentin Grady
This response CC'd by email
>jeez... All this from people who should already know...
Not exactly.
Some of it is from people who do know and yet want to be certain when
conflicting statements from reputable sources while some is from
people who don't know and what's worse don't know they don't know.
If you have difficulty in figuring out which is which then it might be
better for you to find another outlet for your frustrations.
Just a thought.
On the other hand you could share with us the benefits obviously
bestowed on you from you education and demonstrate your method using
Fourier coefficients by applying it to a 20 Ohm load supplied by the
half wave lossless rectification of a 100 Vpk sinusoidal ac supply.
Here is your chance to demonstrate your superiority.
Let's see you do it using Fourier transform coefficients. They sound
so grand they must be better.
Your demonstration of the supposed advantages of this method could be
very enlightning for us poor plebian types armed as we are with simple
arithmetic.
Or perhaps you are having second thoughts about what method would be
most appropriate to the situation.
<grin>
Go on. Give a thrill. Show us how is done.
This thread started when students started quoting some incorrect
"model" answers. It would be great to see another method just to be
sure we have it right.
Come on Tom. We are waiting.
Get over your exasperation at our ignorance and really show us what
you've got.
Tony Williams
In article <3477288f....@news.zippo.com>, Quentin Grady[snip]
> > Variac, diode, 1800-ohm resistor, dvm, surface thermometer.
> >
> > -----------------------------------------------------------------
> > Expt1. Diode in series with the R.
> >
> > At 72vrms from the Variac the R reached a surface temp of 110degC.
> >
> > -----------------------------------------------------------------
> > Expt2. Diode shorted out, see what acv is req'd for 110degC.
> >
> > It took 52vrms to get the resistor up to 110degC.
>
> >
> > So the actual power in the R in expt1 was 52^2/1800 = 1.5W.
> >
> > For 72vrms straight, the R-power would be 72^2/1800 = 2.88W.
> >
> > Therefore, the power ratio diode/nodiode = 1.5/2.88 = 0.52.
>
>
> A wise choice of measurements. Glad you didn't try measuring, for
> instance, the rms voltage after rectification. <grin>
>
> Then the thread would really have rumbled on as people argued about
> exactly what your meter measured with non-sinusoidal voltages. Also
> glad you didn't measure the transformer rms current ... for the same
> reasons.
Bum! A 'keep off the grass' sign. I can never resist it.
Here are the V/I measurements.....
-----------------------------------------------------------------
Expt.4
Same as Expt1, 72vrms from the Variac, into a D and 1800-ohm.
Meters were a Keithley 179 (for I) and a 197.
Both are supposed to be 'true rms-reading'.
Volts across the R = 39V-rms. Amps through = 21.5mA-rms.
-----------------------------------------------------------------
Expt.5
Same as Expt4, 72vrms from the Variac, into a D and 1800-ohm.
Meters were a Keithley 179 (for I) and a 197.
Switched to Average DC-reading.
Volts across the R = 31.8V-dc. Amps through = 17.7mA-dc.
-----------------------------------------------------------------
Note, all measurements are quick and dirty, our 240v mains
seems to be a little variable this morning.
Posted and cc'd.
Tom Burke
First off, sorry this rambles, as it's getting late & I'm getting
tired & cranky...
Quentin Grady wrote:
>
> This response CC'd by email
> On Sat, 22 Nov 1997 17:11:44 -0600, Tom Burke <to...@siu.edu> wrote:
>
> >jeez... All this from people who should already know...
That was supposed to be a friendly jab...
>
> Not exactly.
>
> Some of it is from people who do know and yet want to be certain when
> conflicting statements from reputable sources while some is from
> people who don't know and what's worse don't know they don't know.
Well enough...
>
> If you have difficulty in figuring out which is which then it might be
> better for you to find another outlet for your frustrations.
> Just a thought.
I had no frustrations until I read the rest of this letter... I see it
this way... I was lead to beleive from some earlier posts that some of
the people in this group were actually instructors in technology /
engineering (including yourself - maybe I was mistaken)...
This may not be true at your location, but here, at SIU, you are not
allowed to teach unless you are at least at the PHD (doctorate) level,
or in the last year or so of your PHD work (I, as a Master's student,
am not allowed to teach). So, It appears to me, that either I am
mistaken
in the belief that there are instructors in this group, or that some of
them may be undertrained.
If I (personally), had any doubts to some statement, I would probably
have just reached over to my bookcase, picked out one of my reference
books (many of which are textbooks), and just verified one way or the
other, if I was capable (which I did). If the person was wrong, I
then may have corrected them, or asked them about it, but it's rare
that I try to belittle someone as you are seeming to do to me...
By citing the ISBN, title, and publication date, and page number,
I felt that I had given all the proof I needed, as anyone would
then have an idea (at least) of where to go to verify my statement.
Unfortunately, Someone has to be both rude, and crude (who I get the
impression should know my statement was true), and call a truth into
doubt. This leads me to believe a lot of things, which I will leave
unsaid, but it calls me into doubt, which puts a shadow over any help
I may have given someone (anyone) through my statements. Therefore, I
have wasted an hour this afternoon, developing some answers, etc. to
post in response. Unfortunately, it will take more time to lay
them out in a somewhat readable fashion here, on this piece of
electronic paper, as the font varies in size for spaces, etc...
However, maybe this was unintentional, or mnaybe you just were having
a bad day, so I forgive, and forget. I don't usually hold a grudge
for such a simple thing (especially when it has only occured once).
So, I hope that I will have proven some of my abilities by the end
of this letter, and that then you guys will trust me as being at
least a little bit competent.
>
> On the other hand you could share with us the benefits obviously
> bestowed on you from you education and demonstrate your method using
> Fourier coefficients by applying it to a 20 Ohm load supplied by the
> half wave lossless rectification of a 100 Vpk sinusoidal ac supply.
I would not use Fourier transforms for this, as it would be a waste of
time,
since it is a relatively simple integral... I will give the entire
discussion at the end of the post.
>
> Here is your chance to demonstrate your superiority.
I never claimed superiority, as I had to look up my information in order
to ensure I said something correct. I try not to memorize anything, as
that's a sure way for me to get it wrong...
>
> Let's see you do it using Fourier transform coefficients. They sound
> so grand they must be better.
>
> Your demonstration of the supposed advantages of this method could be
> very enlightning for us poor plebian types armed as we are with simple
> arithmetic.
Again, some advantages will be shown at the end, in the discussion.
>
> Or perhaps you are having second thoughts about what method would be
> most appropriate to the situation.
>
> <grin>
>
> Go on. Give a thrill. Show us how is done.
>
> This thread started when students started quoting some incorrect
> "model" answers. It would be great to see another method just to be
> sure we have it right.
>
> Come on Tom. We are waiting.
>
> Get over your exasperation at our ignorance and really show us what
> you've got.
>
First of all, references:
Staff of Research and Education Association, "Handbook of Mathenatical,
Scientific, and Engineering Formulas, Tables, Functions,
Graphs, and Transforms," New Jersey, 1984.
ISBN: 0-87891-521-4
Samir S. Soliman and Mandyam D. Srinath, "Continuous and Discrete
Signals and Systems," Prentice Hall, NJ, 1990.
ISBN: 0-13-171257-8
Johnson, Hilburn, Johnson, and Scott, "Basic Electric Circuit Analysis,"
fifth edition, Prentic Hall, NJ, 1995.
ISBN:0-13-059759-7
Nilsson, James W., "Electric Circuits," fourth edition, Addison Wesley,
MA,
1993. ISBN: 0-201-54987-5
OK, on with the stuff:
By definition, thr rms value of a periodic function f(t) is as follows:
__________________
/ t0+T 2
Frms = / 1 | f(t) dt (1)
/ - |
\/ T to
If a function is non-periodic, but is a limited time, it may be
represented
as a periodic function.
>From this follows the definition that we use so often for our Vrms
(A*0.707)
as follows;
P=Pi, in all cases...
T=the period of the waveform in all cases...
A=Peak amplitude of the periodic signal in all cases...
if f(t) = A cos (2P/T) (pure sinusoid)
____________________________
/ t0+T 2 2
Frms = / 1 | A cos (2P/T)t dt
/ - |
\/ T to
Which is equal to:
_______________________________
/ 2 { } T
Frms = / A { t sin 2(2P/T)t }
/ - { - + ------------ }
\/ T { 2 4(2P/T) } 0
Which expands to:
________________________________________________
/ 2 { }
Frms = / A { T 0 sin 2(2P/T)T sin (0) }
/ - { - - - + ------------ - ---------- }
\/ T { 2 2 4(2P/T) 4(2P/T) }
This, then, simplifies to (through simple cancellation):
___
/ 2
Frms = / A = A
/ - -------
\/ 2 sqrt(2)
But, then, we already knew that.....
By following the previous steps, you can also see that if a DC
constant is used for f(t), that the rms value is that same DC
value (as expected).
For your half-wave rectifier, (as I said not worth the effort of
a fourier transform), we follow the same procedure as for the
pure sinusoid, but we integrate only from t0 to T/2, as the
signal is equal to 0 for the second half of the period.
Following the same steps, we get that
A
Frms = -
2
or, for a 100 V sinusoid to the circuit, we see that the rms
value is 50 V. Since the power dissipated by a device is
given as P=IV, and V=IR, we do a simple substitution, which
gives that P=(V**2)/R, or that the power dissipated by the
20 ohm load is (50**2)/20, or 120W (an incandescent light bulb).
As I previously stated, this could have been done using a fourier
transform, but not worth the effort, I believe. If a function
is easily represented by a single sinusoid, or a fairly simple
piecewise function, then the above method is the way to go...
Unfortunately, not all functions break down so easily, and then
a fourier transform may well be the way to go about it...
So, on with that....
Any periodic signal may be represented by a fourier transform.
Again, if the signal is not periodic, but limited in time,
multiple copies of the signal may be put together to form a
periodic function.
By representing f(t) by its fourier series, we get the following:
E=sgma notation from this point forward...
TH=angle notation theta from this point forward...
w=radian notation lower-case omega from this point forward...
_______________________________________________
/ t0+T { } 2
Frms = / 1 | { } (2)
/ - | { av + E { An cos (nwt - thn) } }
\/ T to { }
Where:
av represents the DC component of the signal, An is the nth fourier
coefficient, and thn is the phas angle of the nth signal.
The integral of the squared function simplifies, because the only terms
that survive integration over a period are the product of the DC term,
and the harmonic products of the same frequency. All other products
integrate to zero. Therefore, (2) reduces to:
_______________________
/ 2
Frms = / 1 ( 2 inf An T )
/ - ( av T + E ---- )
\/ T ( n=1 2 )
where inf = infinity...
This simplifies to:
__________________________
/ 2
Frms = / 2 ( inf An T )
/ av T + ( E ---- ) (3)
\/ ( n=1 sqrt(2) )
(3) implies that the rms value of a periodic function is the square root
of the sum obtained by adding the square of the value of each harmonic
to the square of the DC valur.
So, if the voltage is a pure DC value, then:
Frms = sqrt(av**2) = av
as expected....
If the applied waveform is a pure sinusoid, then no harmonics exist
except for the 1st (the fundamental frequency), giving
Frms = sqrt( (An/sqrt(2))**2) = An/sqrt(2)
again, as expected....
correspondingly, if the waveform can be represented by:
v=10+30cos(wt-th1) + 20cos(2wt-th2) + 5cos(3wt -th3) + 2cos(5wt-th4)
then
vrms = sqrt( 10**2 +(30/sqrt(2))**2 + (20/sqrt(2))**2
+ (5/sqrt(2))**2 + (2/sqrt(2))**2)
= sqrt(764.5)
= 27.65
The great thing anout this is that a _very_ good representation
can be found by using only the first three coefficients (usually)
of the fourier series, which makes life a lot easier...
So, I hope I haven't made any mistakes, & I hope this satifies you...
Also, I hope it's readable after the spacing gets all screwed up...
Dave Baldwin
Tom Burke wrote:
[snip]
This is the right formula.
> By definition, thr rms value of a periodic function f(t) is as follows:
> __________________
> / t0+T 2
> Frms = / 1 | f(t) dt (1)
> / - |
> \/ T to
>
> If a function is non-periodic, but is a limited time, it may be
> represented as a periodic function.
Calculating power for a periodic waveform is even easier and comes from
that formula. If you divide the waveform into 'n' slices and 'x' is the
current slice, R = the load resistor, and V0=Vn (starting voltage = ending
voltage for a periodic waveform), a good approximation that works for any
waveform is:
n
Power = | (((Vx + Vx-1)/2)^2)/Rn
|
1
or:
For x = 1 to n
Ptotal = Ptotal + (((Vx + Vx-1)/2)^2)/Rn
Next x
Note that if you apply this formula to the original question about a
half-wave rectified sine wave, all the values from n/2 to n are 0.
Obviously, there are a number of special cases (sine waves with no DC
component) that are much easier to calculate.
Quentin Grady
This response CC'd by email
Thanks Tom,
My apologies for being so in your face.
On Sun, 23 Nov 1997 22:26:46 -0600, Tom Burke <to...@siu.edu> wrote:
>First of all, references:
>
>Staff of Research and Education Association, "Handbook of Mathenatical,
> Scientific, and Engineering Formulas, Tables, Functions,
> Graphs, and Transforms," New Jersey, 1984.
> ISBN: 0-87891-521-4
>Samir S. Soliman and Mandyam D. Srinath, "Continuous and Discrete
> Signals and Systems," Prentice Hall, NJ, 1990.
> ISBN: 0-13-171257-8
>Johnson, Hilburn, Johnson, and Scott, "Basic Electric Circuit Analysis,"
> fifth edition, Prentic Hall, NJ, 1995.
> ISBN:0-13-059759-7
>Nilsson, James W., "Electric Circuits," fourth edition, Addison Wesley,
>MA,
> 1993. ISBN: 0-201-54987-5
With a bit of luck I should be able to get some of these via interloan
from the local library. The ISBN numbers will be most appreciated.
>By definition, thr rms value of a periodic function f(t) is as follows:
> __________________
> / t0+T 2
>Frms = / 1 | f(t) dt (1)
> / - |
> \/ T to
[...]
So far I have managed to toggle the screen view and glimpse these
equations as you intended them to be.
Removing long line wrap and using monospace works on screen but will
not print so it will take me a while to re-edit them into MS Word.
I really do appreciated the amount of effort you have put into them
and they do look magnificient on screen. It's just that writing notes
and inbetween steps on the screen doesn't work too well.
>or, for a 100 V sinusoid to the circuit, we see that the rms
>value is 50 V. Since the power dissipated by a device is
>given as P=IV, and V=IR, we do a simple substitution, which
>gives that P=(V**2)/R, or that the power dissipated by the
>20 ohm load is (50**2)/20, or 120W (an incandescent light bulb).
Trade in that Pentium <grin> We still get 125 W.
[...]
>So, I hope I haven't made any mistakes, & I hope this satifies you...
More than satisfied. Thanks,
>Also, I hope it's readable after the spacing gets all screwed up...
It's sortable.
kingsnake
Tony Williams wrote:
>
> In article <3478d536....@news.zippo.com>, Quentin Grady
> <URL:mailto:que...@inhb.co.nz> wrote:
>
> measured it. It's a bit crude, but here's what happened....
>
>
> -----------------------------------------------------------------
> Expt1. Diode in series with the R.
>
> At 72vrms from the Variac the R reached a surface temp of 110degC.
>
> -----------------------------------------------------------------
> Expt2. Diode shorted out, see what acv is req'd for 110degC.
>
>
> -----------------------------------------------------------------
>
> So the actual power in the R in expt1 was 52^2/1800 = 1.5W.
>
> For 72vrms straight, the R-power would be 72^2/1800 = 2.88W.
>
> Therefore, the power ratio diode/nodiode = 1.5/2.88 = 0.52.
>
> --
> [Tony Williams, Ledbury, Herefordshire, UK.---Pagewidth=64-----]
arguments are based upon Ideal circumstances.
Also note that, in practical applications, the efficiency of
rectification for a half-wave circuit is typically low; less than 40%.
--
needs junk mail@mail.com Dale Redford
There can be no doubt about it, a frog with no legs is deaf! The only
question left is "how long does it take a one-legged grasshopper to kick a
hole in a pickle?" Quentin you are a peach, I think the other guy is
working with you.
dale.redford@!airmail.com
Tony Williams wrote in message ...
>In article <3477288f....@news.zippo.com>, Quentin Grady
><URL:mailto:que...@inhb.co.nz> wrote:
>>
>> This response CC'd by email
>> On Sat, 22 Nov 1997 10:35:27 +0000 (GMT), Tony Williams
>> <to...@ledelec.demon.co.uk> wrote:
>>
>> >In article <3478d536....@news.zippo.com>, Quentin Grady
>> ><URL:mailto:que...@inhb.co.nz> wrote:
>[snip]
>> > Variac, diode, 1800-ohm resistor, dvm, surface thermometer.
>> >
>> > -----------------------------------------------------------------
>> > Expt1. Diode in series with the R.
>> >
>> > At 72vrms from the Variac the R reached a surface temp of 110degC.
>> >
>> > -----------------------------------------------------------------
>> > Expt2. Diode shorted out, see what acv is req'd for 110degC.
>> >
>> > It took 52vrms to get the resistor up to 110degC.
>>
>> (edited) No probs.
>> > -----------------------------------------------------------------
>> >
>> > So the actual power in the R in expt1 was 52^2/1800 = 1.5W.
>> >
>> > For 72vrms straight, the R-power would be 72^2/1800 = 2.88W.
>> >
>> > Therefore, the power ratio diode/nodiode = 1.5/2.88 = 0.52.
>>
needs junk mail@mail.com Dale Redford
Hello Quentin,
Glad you're here and I hope your students are following the posts, it will
give them an idea of what to expect as a technician working with
engineers/scientists or people with opinions.
For the record, since no one else will state it, a Vpk of 100 Volts = 70.71
VRMS. The simplest way to calculate the power transmitted to a 20 Ohm load
by a half-wave rectified signal of this voltage is to calculate E**2/R using
the RMS value of the original waveform and then dividing the result by 2 --
as you are well aware, this results in 125 Watts and this is the correct
answer. To other respondees, it is obvious that others understood this but
did not state it.
One small beef Quentin, I have been at this engineering stuff for a while
and since the 70's, I cannot remember encountering a half-wave rectified AC
power circuit. Do you teach your students practical applications like the
following:
A DC motorized conveyor system with AC-DC power supplies (dual voltage,
direct full-wave rectified for motor drives and 2:1 xfrmr
full-wave-rectified for control) in each module has the following input
parameters:
nominal DC motor voltage: 26 Volts DC
surge current: 6 Amps DC (motors)
steady state current: 2.5 Amps
control dropout voltage: 10 Volts DC (the entire system shuts down when this
happens)
This equipment is being installed in a facility with 3 phase 208V, it must
operate from +/- 10% of nominal input voltage, the conveyor sections are 3
meters long and there are 400 pieces.
1) How many power drops/transformers are required?
2) What turns ratio transformer should be used?
3) What wire size should be used from transformer to conveyor section?
4) How many conveyor sections can be connected to a transformer?
4a) What wire size must be used within the conveyor for daisy-chaining
sections?
Now I know this is an unfair set of questions. It is however, a real set of
questions, not a set that a freshly graduated tech would face (hopefully)
but representative of what the guy he goes to work for might face. Do your
students learn enough to help work on questions like these?
Dale E. Redford
dale.redford@!airmail.net
remove the ! to reply by email
Quentin Grady wrote in message <347ad392...@news.zippo.com>...
>This response CC'd by email
>
>Thanks Tom,
>
>My apologies for being so in your face.
>
>
> On Sun, 23 Nov 1997 22:26:46 -0600, Tom Burke <to...@siu.edu> wrote:
>
>
>>First of all, references:
>>
>>Staff of Research and Education Association, "Handbook of Mathenatical,
>> Scientific, and Engineering Formulas, Tables, Functions,
>> Graphs, and Transforms," New Jersey, 1984.
>> ISBN: 0-87891-521-4
>>Samir S. Soliman and Mandyam D. Srinath, "Continuous and Discrete
>> Signals and Systems," Prentice Hall, NJ, 1990.
>> ISBN: 0-13-171257-8
>>Johnson, Hilburn, Johnson, and Scott, "Basic Electric Circuit Analysis,"
>> fifth edition, Prentic Hall, NJ, 1995.
>> ISBN:0-13-059759-7
>>Nilsson, James W., "Electric Circuits," fourth edition, Addison Wesley,
>>MA,
>> 1993. ISBN: 0-201-54987-5
>
>
>With a bit of luck I should be able to get some of these via interloan
>from the local library. The ISBN numbers will be most appreciated.
>
>
>>By definition, thr rms value of a periodic function f(t) is as follows:
>> __________________
>> / t0+T 2
>>Frms = / 1 | f(t) dt (1)
>> / - |
>> \/ T to
>
>[...]
>
>So far I have managed to toggle the screen view and glimpse these
>equations as you intended them to be.
>Removing long line wrap and using monospace works on screen but will
>not print so it will take me a while to re-edit them into MS Word.
>I really do appreciated the amount of effort you have put into them
>and they do look magnificient on screen. It's just that writing notes
>and inbetween steps on the screen doesn't work too well.
>
>>or, for a 100 V sinusoid to the circuit, we see that the rms
>>value is 50 V. Since the power dissipated by a device is
>>given as P=IV, and V=IR, we do a simple substitution, which
>>gives that P=(V**2)/R, or that the power dissipated by the
>>20 ohm load is (50**2)/20, or 120W (an incandescent light bulb).
>
>Trade in that Pentium <grin> We still get 125 W.
>
>[...]
>
>>So, I hope I haven't made any mistakes, & I hope this satifies you...
>
>More than satisfied. Thanks,
>
>>Also, I hope it's readable after the spacing gets all screwed up...
>
Quentin Grady
This response CC'd by email
ma...@mail.com> wrote:
>There can be no doubt about it, a frog with no legs is deaf! The only
>question left is "how long does it take a one-legged grasshopper to kick a
>hole in a pickle?"
Thank goodness you didin't ask how a blind dog leads.
>Quentin you are a peach, I think the other guy is working with you.
G'day G'day to you too Dale,
Tony working for me? I wish.
Commuting could be a problem though.
IMHO it's one of the miracles of cyberspace that people come together
and work together when they share a common interest.
Quentin Grady
This response CC'd by email
ma...@mail.com> wrote:
>Hello Quentin,
>
>Glad you're here and I hope your students are following the posts, it will
>give them an idea of what to expect as a technician working with
>engineers/scientists or people with opinions.
Some technician's were setting up a news reader yesterday.
Up till now they have mostly been browser class. <grin>
>For the record, since no one else will state it, a Vpk of 100 Volts = 70.71
>VRMS. The simplest way to calculate the power transmitted to a 20 Ohm load
>by a half-wave rectified signal of this voltage is to calculate E**2/R using
>the RMS value of the original waveform and then dividing the result by 2 --
>as you are well aware, this results in 125 Watts and this is the correct
>answer. To other respondees, it is obvious that others understood this but
>did not state it.
It wasn't so obvious to the person writing model answers for an exam.
Actually it was rather awkward standing in front of a large class with
them comparing the method we both used to the model answers. Right
before an exam is not a good time for a crisis of confidence.
>One small beef Quentin, I have been at this engineering stuff for a while
>and since the 70's, I cannot remember encountering a half-wave rectified AC
>power circuit.
Totally agreed. My students would typically use a bridge rectifier.
The question arose from an exam question.
Since jobs depend on exam results I can't afford to ignore them.
> Do you teach your students practical applications like the
>following:
[...]
Nothing as complex as quoted.
I teach introductory courses to students.
They have design projects.
Imagine if you will 20 students designing and building their own pulse
width modulated speed controllers for 12 V motors. The scale is
smaller than the practical situation you suggest but they sure get a
kick out of it. They determine the power output from simulataneous
V-t and I-t graphs. The V-t graphs come out rectangular on a good
day. The I-t graphs are exponential growth curves due to motor
inductance.
Thanks Dale for your interest.
Tony Williams
In article <3479F15F...@dm.net>, kingsnake
<URL:mailto:king...@dm.net> wrote:
> But, you're not accounting for diode losses. Remember: these past
> arguments are based upon Ideal circumstances.
Yes, it was a little rough and ready. I felt that 72vrms,
102v pk, was sufficiently above the diode drop to allow me
to ignore it for the sake of a quick experiment.
0.7v in 102v is about 0.7% error and I suspect far smaller
than the error that comes from trying to get a consistent
measurement of the surface temperature of the load.
I was quite pleased with 0.52 on the first swipe.
needs junk mail@mail.com Dale Redford
Hello Quentin,
You have certainly piqued my curiosity, what are V-t and I-t graphs?
Do your students measure values to create these graphs or are these
calculated values?
What kind of school is this, technical vocation, university, high school or
what?
Dale
dale.redford@!airmail.net
remove ! from above to reply
Quentin Grady wrote in message <347e6303....@news.zippo.com>...
>This response CC'd by email