Re: Simple Power Question

On Mar 30, 9:01 pm, Jon Kirwan <j...@xxxxxxxxxxxxxxxxxxx> wrote:
On Sun, 29 Mar 2009 23:00:46 -0700 (PDT), HardySpicer

<gyansor...@xxxxxxxxx> wrote:
Ok, I have a full wave rectifier (with no smoothing) fed to a resistor
of value 1 ohm. The sine wave has amplitude 1 volt. What is the
average power delivered to the load?

Now is the average power defined as Vav^2/R = (2/pi)^2=0.407W

No, for AC sine wave voltages that use the standard designations where
the voltage specified is Vrms, which is Vp-p divided by 2*sqrt(2), the
average power is stated just as in usual DC laws -- Vrms^2/R.  Not
Vavg.  And Vavg for a sine wave is zero, not whatever you were writing
there (which I can't say I entirely fathom.)

or is it defined from the rms value ie  Vrms^2/R  - and shouldn't they
be the same?

No, Vrms is nothing like Vavg.  Vavg is the mean value and since the
sine wave spends equal time at equally opposing voltages throughout
its cycle, they all add up to zero as an average.  (Assuming no DC
offset, of course.)  Vavg is used where you need to know if there is
some DC offset and would like to put a value to it.  Vrms is used for
power calculations, which is what you are talking about here.

Now the rms value of a full wave rectified sine wave is 1/sqrt(2) so
the power is 1/2 =0.5W

The rms value of a sine wave _before_ rectification is just the value
usually given for AC voltages.  I think your confusion (or mine, were
it just you and me talking and no one else in the world existed) is
that AC sine wave voltages can specify their value in one of three
ways:  Vp-p (voltage, peak to peak), Vrms (we'll get to that), and Vpk
(which is 1/2 of Vp-p.)  When you say "1V AC" it is usually taken to
mean 1Vrms, not 1Vp-p or 1Vpk.

   Vrms = Vp-p/(2*sqrt(2))
   Vrms = Vpk/sqrt(2)

Your 1/sqrt(2) looks to me as though you are trying to compute Vrms
from Vpk.  So maybe you meant 1Vpk when you spoke, earlier.  But then
you need to convert that to Vrms before applying the standard DC power
formulas.  If you were talking about Vpk, then the power would be
(1Vpk/sqrt(2))^2/1ohm, which is indeed 1/2 watt.

All that assumes no bridge rectifier, though.  The bridge rectifier
will not pass anything much for a significant part of the cycle and
that calculation isn't entirely trivial.  Certainly, it takes more
effort than you've supplied above.

So which is the average power?

It depends on how you specify the AC voltage.  If you followed
convention, you'd have said 1V meaning 1Vrms.  However, it seems you
may have meant Vpk (V's peak positive value above the midpoint
reference line.)  Before getting mired into a discussion of the
meaning of words, it's best to look at the mathematics.  That's
precise and clear, always.

Let's take the position of specifying the voltage as the peak positive
side value... in other words, Vp-p/2.  So writing 1Vpk means that the
peak positive value reaches 1V at the very top of the positive side of
the AC sine wave cycle.  It will also reach -1V at the negative side,
which is why this is also called 2Vp-p.  But let's stick with the 1V
as meaning 1Vpk.

Using calculus, and an appropriate period of time (1/f):

  P = (1/t) * integral([Vpk*sin(2*pi*f*t)]^2/R, dt, 0, 1/f)

Here Vpk is just 1/2 of the Vp-p voltage.  This will be, and wait and
see below, the same as sqrt(2)*Vrms.  But that's for later.  Some
explanation of the fancy integral is in order, though.

The instantaneous voltage is Vpk*sine(2*pi*f*t).  You need to make
sure you understand why.  Usually, 2*pi*f is given the designation 'w'
or actually small-omega if I could type Greek in here.  It's taken to
be a constant in most cases, since the AC sine wave usually operates
at a fixed frequency.  If you look closely, each time 't' is some
integer multiple of (1/f), the sine value goes to zero.  As it should.
Also, since sine() only goes between +1 and -1, the value of Vpk times
that function should never go above +Vpk or below -Vpk.

Inside the integral, this instantaneous voltage is squared and then
divided by R.  This gives the finite power value at that moment in
time.  That power value, multipled by dt (the instantaneous moment of
time for which that voltage value is exactly correct), is the
instantaneous bit of energy involved.  (In calculus, this value is
always between the smallest possible finite number and zero, and must
be summed up an infinite number of times in order to amount to a
finite value.)  The integral adds up each of these tiny values of
energy into a total energy value.  That energy must then be divided by
the time to get back to "averaged power" during that interval.  That's
why the integral is then multiplied by (1/t) -- to convert total
energy in the period into average power for that period.

Time 't' outside the integral is just 1/f (the time over which the
integral is being performed) and so (1/t) can be replaced by (f):

  P = f * integral([Vpk*sin(2*pi*f*t)]^2/R, dt, 0, 1/f)

R can also be extracted out as it is also a constant value:

    = (f/R) * integral([Vpk*sin(2*pi*f*t)]^2, dt, 0, 1/f)

The Vpk squared is also constant and can be taken outside.  So:

    = (Vpk^2*f/R) * integral(sin(2*pi*f*t)^2, dt, 0, 1/f)

Solving this can be done by recognizing that sin(x)^2 can be replaced
by (1 - cos(2x))/2.  A calculator may solve it using two steps of
integration by parts, but it is easier to just use the half-angle
substitution I just mentioned:

    = (Vpk^2*f/R) * integral([1-cos(2*2*pi*f*t)]/2, dt, 0, 1/f)

The division by 2 inside the integral can be brought out:

    = (f*Vpk^2/R)/2 * integral([1-cos(4*pi*f*t)], dt, 0, 1/f)

Distributing the integral:

    = (f*Vpk^2/R)/2 * [ integral(1, dt, 0, 1/f) -
                    integral(cos(4*pi*f*t)], dt, 0, 1/f) ]

The first integral is easily solved:

    = (f*Vpk^2/R)/2 * [ 1/f - integral(cos(4*pi*f*t)], dt, 0, 1/f) ]

The second integral is then solved:

    = (f*Vpk^2/R)/2 * [ 1/f - 0 ]
    = (f*Vpk^2/R)/2 * [1/f]
    = (Vpk^2/2)/R

But notice that nasty factor of (1/2) in there?  The rule would be the
same as for DC, if only that (1/2) factor wasn't present.  By defining
Vrms as Vpk/sqrt(2) for sine wave voltages (which is all we analyzed
above and so it only applies to sine wave voltages), we can change the
equation to:

  P = Vrms^2/R = (Vpk^2/2)/R

Which is why AC voltages are given that way, as Vp-p/(2*sqrt(2)) or
Vpk/sqrt(2).  It allows you to forget about keeping calibration
factors in mind when working with AC and instead just use the same old
DC laws you are used to using.

Also note that if the varying voltage doesn't follow a sine curve, but
is a sawtooth, square wave, or some other continuous but
non-sinusoidal form, the factor sqrt(2) may no longer apply.  However,
the concept of finding a single voltage value to talk about that can
be used in the usual Ohm's laws without special factors remains.  So
Vrms remains an important concept, even if it is sometimes hard to
calculate exactly.

And then you introduce the bridge rectifier, which only conducts
significantly during a portion of the cycle.  You'd need to know the
current being drawn (because the rectifiers' voltage drops will depend
on that value) in order to perform a proper calculation.  The integral
equations get much nastier looking.

For AC voltages of some magnitude, you could just use the usual
equation.  Flipping the negative side of the sine wave over onto the
positive side by use of a bridge rectifer doesn't much affect the
sin^2 term in the integral, which is positive anyway because it is
squared.  The only difference is the small loss of conduction angle
and that can be ignored for large AC voltages.  But for 1V, whether
Vrms or Vpk, a lot of things matter than didn't matter before.  So a
simple calculation is probably less useful here.  In fact, for 1Vpk
the conduction angle would likely be pretty short.

I'm not sure where your (2/pi)^2 comes from.  I've seen pi used for
certain cases of determining peak diode current in half-wave and
full-wave rectification.  But not for what you used it for.

Jon

2/pi is the average value of abs(sin(theta)) ie a full wave rectified
waveform.
In fact that power I calculated is only the power in the dc term of
the Fourier series.

Thanks for your work.


H.
I
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