# Controlled rectification question

• Connorm1
In summary, an asymmetric waveform will have a DC component that will show up as the first term of the Fourier series.f

## Homework Statement

a) If the firing angle is set for alpha=(pi/3) estimate the power dissipated in the bulb if it is rated at 100 W and the voltage source is 230 V @ 50 Hz.

b) An anomaly that can occur in controlled rectification is drift of the firing
angle on one half cycle, so causing an asymmetrical waveform, as illustrated opposite. State the effect, if any, this would have on the harmonic content of the waveform.

## Homework Equations

P=Vrms^2/R Vpeak=Vrms/0.7071

## The Attempt at a Solution

1a) So first of all i found the Vpeak of a normal sine wave in this case 230/0.7071 = 325V and found R=230^2/100=529ohms. I then found the Vrms for any waveform and set my boundaries and got 325*0.63423.
so Vrms = 206.12475V for this wave form including the firing angles (where those sections would = 0). I then used PT (or total dissipated) =Vrms2/R=(206.12475)2/529=80.31w which is my answer.
1b) I've found through my notes that if the phase angle of a harmonic changes it has no reference on determing the average power so that would be the same. However it would change the instantaneous power as it'd would be delayed by whatever the shift is. Is the correct or do i need to show more?

Show the "asymmetrical waveform, as illustrated opposite. " so we can see what it is the author has in mind ?.

I can tell you from experience that it takes surprisingly little asymmetric firing to produce a DC component in the controlled waveform that'll burn up relay coils.

Try evaluating your integrals from (pi/3 ± a few degrees) to (the next zero crossing) and find a relation between degrees of asymmetry in firing and DC content of resulting controlled waveform..

An incandescent lamp will be indifferent but an inductive load won't.

old jim

Apologies i didn't add the image. So are you saying to integrate the sine wave between T/4 & T? Sorry only thing i found in my notes was very little other than the statement above.

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Apologies i didn't add the image. So are you saying to integrate the sine wave between T/4 & T? Sorry only thing i found in my notes was very little other than the statement above.
@jim hardy I'm still unsure on this question... sorry.

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Apologies i didn't add the image. So are you saying to integrate the sine wave between T/4 & T? Sorry only thing i found in my notes was very little other than the statement above.
Harmonic content: how about Fourier series.?

Apologies i didn't add the image. So are you saying to integrate the sine wave between T/4 & T?

The image you posted shows asymmetric firing?
positive for one-quarter cycle duration T/4, from T/4 to T/2
negative i can't make out looks like from 1.5T to T which is backward.

Here's what i meant to suggest
Clearly the black quarter circles have equal area and if you integrate them individually you'll get two equal magnitude but opposite polarity numbers that'll sum to zero. That's what average value is, area uder the curve.

If you offset your firing points by n degrees to show asymetric firing you'll get unequal areas that'll no longer sum to zero.
So that wave will have a DC component or ofset. That'll show up as the first term of your Fourier series should you choose that approach...

In my day i'd have written a Qbasic program to generate a table of DC offset vs firing angle asymmetry. You young guys have Excel and Matlab to do the same thing.

I'd add another column to show the integral of voltage because for an inductor current is the integral of voltage.
An asymmetrically firing thyristor will burn up an AC relay coil connected to it ,
I've experienced it myself.
We had a batch of temperature sensitive thyristors that went asymmetric as they heated up.
That resulted in several trips of a nuke plant, and at a million bucks per trip it was well worth tracking down the cause and learning how to prevent it..
We found by observing the relay coil currents with a clamp around current probe and oscilloscope that asymmetric firing showed up as distorted sinewave current with high peaks on only one side of the wave.
We looked at every such relay driver in the plant and fixed those showing asymmetry . That stopped our epidemic of relay coil failures.
Needless to say management was very happy.

I hope that little tip helps you someday.
Make yourself painfully aware of the inductor's integral-derivative relation between volts , current & flux. It has so many practical applications that you need it to be intuitive.

old jim

ps sorry to be so slow getting back to you. sickness in family, i was away for a few days..

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Klystron
Hi All,

Can anyone suggest any reading material that I can study to better improve my knowledge in answerin question 1 b)?