error in electronic tutorial analysis of power in half wave rectifier circuit 1

[electricpete]

https://www.electronics-tutorials.ws/power/single-phase-rectification.html

Rectification Example No1
A single phase half-wave rectifier is connected to a 50V RMS 50Hz AC supply. If the rectifier is used to supply a resistive load of 150 Ohms. Calculate the equivalent DC voltage developed across the load, the load current and power dissipated by the load. Assume ideal diode characteristics.

First we need to convert the 50 volts RMS to its peak or maximum voltage equivalent.
a) Maximum Voltage Amplitude, VM
VM = 1.414*VRMS = 1.414*50 = 70.7 volts
b) Equivalent DC Voltage, VDC
VDC = 0.318*VM = 0.318*70.7 = 22.5 volts
c) Load Current, IL
IL = VDC ÷ RL = 22.5/150 = 0.15A or 150mA
d) Power Dissipated by the Load, PL
PL = V*I or I2*RL = 22.5*0.15 = 3.375W ≅ 3.4W

… [neglected diode voltage drop]

Here's my analysis of items a) through d):
[ul]
[li]a) Vmax = sqrt2 times the rms of the sinusoid [ I AGREE][/li]
[li]b) "equivalent dc voltage".
Vdc = int(VM*sin(w*t) from wt=0..Pi) / (2*pi) = [-cos(Pi) – -cos(0)] / (2*pi) = 2 / (2*pi) = 0.318*VM – [ I AGREE, although I would stay away from the terminology "equivalent" without clarifying for which purposes it is equivalent][/li]
[li]c) load current – they call it IL but what they’re calculating is the average (dc) value IDC = <i(t)>. IDC = VDC / RL [I agree, with the caveat that their IL is really IDC][/li]
[li]d) Power dissipated by the load PL = VDC*IDC or IDC^2*RL = 22.5*0.15 = 3.375W ≅ 3.4W [I disagree]. That is the power dissipated by the DC component of the voltage and current. There are other components in the fourier decomposition of the voltage and current that also cause power dissipation in the resistor.[/li]

[/ul]
Compare a half-wave rectified voltage feeding a resistive load to a sinusoidal voltage feeding the same resistive load, where the peak voltage of both are the same (i.e. the sin would represent the input to the hwr). The hwr circuit provides half the power of the sin circuit (with diode voltage neglected, same assumption as stated on the webpage). That conclusion is evident because the instantaneous power of the hwr circuit is the identical to that for the sinusoid circuit during the half-cycle that the hwr is conducting, while it is zero for the other half. In a given period of time, the energy delivered by the hwr circuit is half that of the sin circuit. Since average power is energy per time, the average power in the hwr circuit is half that of the sin circuit.

For the particular example listed:
Power of the sin circuit would be Psin = Vrms^2 / R = 50^2 / 150 = 16.667 watts.

Power of the hwr circuit would be Phwr = 0.5*Psin = 8.333 watts.

My answer for part d is 8.3 watts which does not agree with their answer 3.4 watts.

I say the link is wrong (It has become a source of controversy in another thread). What do you say?


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(2B)+(2B)' ?

 

[waross]

Honestly Pete.
The tough love answer is that it doesn't look as if anyone will change their mind.
It's time to let this dead horse rest in peace.
I understand your position, but I suggest that you accept the support that you have and move on.
Yes, text books have mistakes.
I suggest deleting this thread and moving on.
You have too much knowledge and ability to waste time like this.
Your friend
Bill

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

Thanks Bill

Here's the way I look at it:

There are people who’s opinion I respect and value on many things that hold a contrary opinion on this particular topic.

Part of the problem is this referenced link which contradicts what I (and others) are saying. Apparently there is a credibility gap… who do you believe if you read two different things on the internet.

The other thread has been going on for 3 weeks with multiple rounds of saying the same thing. I do not think that repeating it more times in that thread will help.

But I do think that most people would want to learn when they are relying on incorrect links, especially if it can be done in a non-confrontational manner. How to do that? Starting a new thread focusing on this incorrect link seemed to me a good way to do it (certainly preferable to continuing the original thread). I was hoping multiple people would weigh in to say the link was incorrect in it’s approach to part d). That would help eliminate the credibility gap.


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(2B)+(2B)' ?

 

[waross]

I think that it is more a point of view than a disagreement.
That tutorial doesn't help.
Some people looked at the problem and looked at the effect of the diode on the power. (1/2)
Some people looked at the problem and looked at the effect of the diode on the voltage. (1/root 2)
If either group follow their calculations through they will see that the heater will probably burn out.
It's more a case of a problem that is so simple that people are not looking closely at the other's units.

It's all good, Pete.



Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

I would disagree with your characterization of the old thread. There was plenty of disagreement in the final answer watts predicted by various methods.

Whatever your personal opinion about the old thread and its relationship or lack of relationship to the tutorial link, the subject of this thread is the tutorial link.

I am interested to hear people weigh in on whether or not the tutorial link method outlined for solving d) is correct.


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(2B)+(2B)' ?

 

[IRstuff]

The tutorial is incorrect, but more for ignoring the whole purpose of knowing the rms voltage, since the square of the rms voltage divided by R is the average power dissipation per cycle, so 50^2/150 = 16.67W per half cycle. Since it's a half-wave circuit, the average power is 1/2 of that, so 8.33W

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[LionelHutz]

The fundamental failure is believing that a half-wave rectified voltage without any filtering is DC.

But, it is not DC. It is still a periodic waveform. You always use the RMS voltage of a periodic waveform to calculate the power in a circuit. This goes right back to the first thing taught in AC theory classes and something I initially missed when I posted my first answer.

The second fundamental failure is believing that supplying voltage half of the time is the same as supplying half of the voltage.
V^2/R/2 does not equal (V/2)^2/R

Overall, it's been beaten to death. If you still don't believe the above then there is nothing the believers can do to help you any further. Calculating the relevant values using a period of a sinewave in a spreadsheet proved it without a doubt for me.

 

[crshears]

For any contribution I made to acrimony in the old thread, I apologize; that was not my intent.

As to IRStuff's and LionelHutz' takes on the tutorial, I agree with both of you.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[LionelHutz]

I suggest contacting the site and pointing out the error.