Simple Power Question

"HardySpicer"

** Neglecting losses in the diodes - the exact same power as if the rectifier was not there. ** No way.

** Correct.

** Nope.

Vav and Vrms are only the same for smooth DC.

** Using rms value of a sine wave gives the right answer - ie average power.

..... Phil

y

Approximately zero, assuming a silicon bridge rectifier (1.4V forward drop -- two diodes in series at any given time).

If you meant an ideal rectifier, then it is equal to the AC value. RMS doesn't care about sign.

Tim

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.)

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.

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.

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

ey

/f) ]

] 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

The actual power behavior is related to summing up instantaneous energy and then averaging that, not averaging some value of voltage and then squaring it outside the integral. Not to mention the fact that you were using theta and d-theta in your integral, when you'd need a change of variable (if you weren't already off the mark for other reasons.)

The fourier of a sine continuous from -infinity to +infinity is a dirac spike located at '+/-f'.

h(t) = A*sin(2*pi*f0*t) H(f) = j*(A/2)*(impulse(f+f0)-impulse(f-f0))

At f=0 the dirac impulses are 0. Aren't they? Even a discrete fourier transform wouldn't put much energy into DC. Oh, well. Maybe I'm missing something.

I hope it helped.

Jon

What's the forward drop of your diode?

Has the thought of reading the textbook or doing the coursework entered your little pea-brain?

Good Luck!Rich

It's a full wave rectified sine wave!! abs(sin(theta))

This has a Fourier series that includes a dc term + fundamental + harmonics. The fundamental is at twice the original frequency of the un-rectified wave. You can consider the total power to be the dc power

• Power in all the other terms.

Hardy

It's not course work and unlike yourself I have the brain the size of a planet.

What I was looking for was the power calculation when the waveform is PWM in fact. The rectified since wave was just a jog to the memory. So what's your take on that, a PWM signal of period T with pulse width tau amplitude A going to zero. What is the power to the load?

Hardy

Ah. Okay. As I said, I might have missed a connection.

Jon

This will all end in tears.

The power will be ( tau / T ) * ( A^2 / R )

--
Paul Hovnanian     mailto:Paul@Hovnanian.com
------------------------------------------------------------------
You can discover what your enemy fears most by observing the
means he uses to frighten you. -- Eric Hoffer

tor

ed

Great thanks

H.

tor

ed

oops - sorry - 1 final question. Would it be possible to measure that with a DVM or must I calculate it from the mark/space ratio on the scope? ie will a DVM measure the true power if the signal is not sinusoidal?

H.

It depends upon the DMM. Some can calculate "true RMS".

s/calculate/measure/