Measuring voltage controlled by an SCR

[mull982]

When a voltage waveform is chopped by an SCR how is the quanity of the measured voltage determined when measured?

Lets say the SCR is only conducting for 50% of the top half of the waveform, then how would this voltage appear when measuring. Because you are measuing half the positive half, it will still reach peak magnitude, but it will only be half this top portion of the waveform. So would the RMS voltage measurment still remain the same? If you were looking at it with a meter would you still see nominal voltage?

Again just looking at the top half of the waveform, what if the SCR was conduting for the 1st 1/4 of the top half, then went off, and didn't start conducting again until the last 1/4 of the top half. How would this voltage value be measured?

 

[Smart $]

When a voltage waveform is chopped by an SCR how is the quanity of the measured voltage determined when measured?

Regarding an RMS voltage measurement? I believe the most basic of methods is to rectify and filter to DC voltage... after all, that is what rms voltage is about, the equivalent DC voltage, in short.

Lets say the SCR is only conducting for 50% of the top half of the waveform, then how would this voltage appear when measuring. Because you are measuing half the positive half, it will still reach peak magnitude, but it will only be half this top portion of the waveform. So would the RMS voltage measurment still remain the same? If you were looking at it with a meter would you still see nominal voltage?

No, the rms voltage would not be the same. A "true" rms voltage reading would indicate 1/4 the full-wave rms value... 50% of the top half is 25% of the full wave, and a purely resistive load would dissipate only 1/16th the power through one full cycle.

Again just looking at the top half of the waveform, what if the SCR was conduting for the 1st 1/4 of the top half, then went off, and didn't start conducting again until the last 1/4 of the top half. How would this voltage value be measured?

Very carefully

Seriously, though. In my ancient and liited knowledge of SCR's, once they start conducting, they do not stop until a zero crossing... thus it wouldn't be able to start, stop, start again, before stopping at the zero crossing.

But let's say it could... or we've added circuitry which does make that output waveform. This is where even passive circuitry has an advantage over the math involved... it "calculates" instantaneously, whereas we do not Anyway, this is where some of my math skills are quite rusty, and would take a bit of time to do a refresher... so I'll give way to someone better suited for that discussion

 

[dbuckley]

What you see depends on what you measure with.

Assume a SCR dimmer set at 50% (thats 50% time, so the triac triggers at the peak of the half-cycle), with a 120V supply.

The bulb sees 60V. So a 60V lamp would be fully illuminated. A 120V lamp would look a lot less than 50% illuminated.

An RMS meter sees 60V. (Edited to note that the RMS value of an AC waveform is the equivalent DC value.)

A scope sees the voltage stay at zero for a while then jump to about 170V and decline to zero over the half cycle.

SCRs conduct from when triggered until both the trigger current is less than the threshold trigger value, and the current through the device is less than the holding current. Then the SCR turns off.

 

[winnie]

Regarding an RMS voltage measurement? I believe the most basic of methods is to rectify and filter to DC voltage... after all, that is what rms voltage is about, the equivalent DC voltage, in short.

The RMS voltage is the equivalent DC voltage into a linear resistive load. Rectifying and filtering are tremendously non-linear, and the value obtained from rectifying and filtering is very much _not_ the RMS value. If you _assume_ a particular waveform (say a sine wave), then there is a known relationship between the rectified/filtered value and the RMS value. This technique is used in some meters, and such meters give incorrect readings if the waveform is not sinusoidal.

No, the rms voltage would not be the same. A "true" rms voltage reading would indicate 1/4 the full-wave rms value... 50% of the top half is 25% of the full wave, and a purely resistive load would dissipate only 1/16th the power through one full cycle.

If you apply 1/4 of a full wave to a resistive load, then the resistor will be dissipating 'full' power for 1/4 of the time. (Of course, with a sine wave, the dissipation is continuously changing, but 1/4 cycle from peak to zero has nice symmetry with the full waveform. If you prefer, image a system that switches _entire_ AC cycles, and only lets through 1 out of every 4 cycles.) Since the resistor is dissipating 1/4 of full power, the RMS voltage is 1/2 of the full wave reading. Similarly, putting a diode in series with an AC supply, thus cutting out half the cycle, gives an RMS output of 0.707 * full.

-Jon

 

[mull982]

No, the rms voltage would not be the same. A "true" rms voltage reading would indicate 1/4 the full-wave rms value... 50% of the top half is 25% of the full wave, and a purely resistive load would dissipate only 1/16th the power through one full cycle.

So if we measured a 480V circuit with an SCR operating as above and measured with a digital voltmeter we would only see 120V?

The bulb sees 60V. So a 60V lamp would be fully illuminated. A 120V lamp would look a lot less than 50% illuminated.

Wouldn't the lamp be 25% illuminated based upon V^2/R? Since voltage is halved wouldn't power to lamp only be 1/4?

Is the conduction angle another term for stating where in the waveform the SCR turns on?

 

[rcwilson]

Depends on the meter.

A "True RMS" meter reads the actual RMS value. An SCR triggering for 1/4 cycle would read 25 % voltage. Most digital meters are True RMS.

A meter that measures the peak value and scales it to the RMS sine wave would read incorrectly. It might show full voltage A meter with a full wave rectifier and a DC movement would read the average value which is not not the RMS value. (Some old vacuum tube meters used this method.)

In an EE labs we were given a black box with four terminals, various meters and a DC and AC supply. We had to figure out what circuit was between the terminals. (No o'scopes allowed. ) We had to know which meters were true RMS, peak pickers, average reading ,and which used a different method that I don't recall. (Thermal?).

Knowing the method and the ratios of the various meters scales to the true RMS value and the math behind the ratios, we could come up with the circuit in the box as an RC with a 1/2 wave rectifier with a resistor in series. (Too many yearrs ago. I don't know which method is which anymore.)

Most modern digital meters calculate and display true RMS values. A cheap $3 meter with an analog scale and needle probably reads average volts which will have some error on a dimmer circuit. A Wiggy doesn't read; it just hums.

 

[mull982]

If you apply 1/4 of a full wave to a resistive load, then the resistor will be dissipating 'full' power for 1/4 of the time.

-Jon

How would it be dissipating full power 1/4 of the time if the RMS voltage was only 1/4 of full voltage? I would think as stated above, that at 1/4 voltage the power would be 1/16.

What would the effects of the voltage be on a transformer primary? The same as a resistor as stated above?

 

[Smart $]

The RMS voltage is the equivalent DC voltage into a linear resistive load. Rectifying and filtering are tremendously non-linear, and the value obtained from rectifying and filtering is very much _not_ the RMS value. If you _assume_ a particular waveform (say a sine wave), then there is a known relationship between the rectified/filtered value and the RMS value. This technique is used in some meters, and such meters give incorrect readings if the waveform is not sinusoidal.



If you apply 1/4 of a full wave to a resistive load, then the resistor will be dissipating 'full' power for 1/4 of the time. (Of course, with a sine wave, the dissipation is continuously changing, but 1/4 cycle from peak to zero has nice symmetry with the full waveform. If you prefer, image a system that switches _entire_ AC cycles, and only lets through 1 out of every 4 cycles.) Since the resistor is dissipating 1/4 of full power, the RMS voltage is 1/2 of the full wave reading. Similarly, putting a diode in series with an AC supply, thus cutting out half the cycle, gives an RMS output of 0.707 * full.

I did say the most basic of methods, did I not? Perhaps I should have said one of the crudest methods???

On that last bit, that'd be .707 * peak for one-half cycle, but .5 * fullwave(rms) over one cycle.

I have several thoughts on your statements, but I can sum it up in one. A true rms meter has to base its reading on some sort of time or periodic domain, otherwise it would just be providing instantaneous values, right?

One cyle is a good time domain, but what if the waveform is non-repititous, non-periodic, such as an "mainstream-audio" waveform. Even readings provided visually at 60 per second would change so fast you would only be able to glean digits (digital readout, of course) that stay relatively the same and that'd be unlikely (I know I don't need I even get into analog and needle deflection..???). So these so-called true rms meters must "average" over a time period of some length... or is this what you are saying, that they simply aren't as true as one is led to believe

 

[winnie]

How would it be dissipating full power 1/4 of the time if the RMS voltage was only 1/4 of full voltage? I would think as stated above, that at 1/4 voltage the power would be 1/16.

1/4 the voltage would yield 1/16 the power into a resistive load.

My point is that the situation described, where you chop things to get the 1/4 of a complete cycle from peak to zero, then the resultant RMS voltage is 1/2 of the original value.

This is easier to see if you look at duty cycle modulated DC.

1V continuous DC has an RMS voltage of 1V

2V DC pulsed with a 50% duty cycle has an RMS voltage of 1.414V

2V DC pulsed with a 25% duty cycle has an RMS voltage of 1V.

What would the effects of the voltage be on a transformer primary? The same as a resistor as stated above?

With a transformer you need to consider both the magnetic circuit of the primary and the load circuit. It is easier to consider a simple inductor.

For a simple inductor, what matters is voltage * time; in other words the _average_ inductor current is related to the _average_ applied voltage, not the root mean square voltage.

-Jon

 

[rattus]

Maybe someone has said it, but Vrms in this case is 1/2 its full wave value, not 1/4.

Then P = 1/4 of the full wave value.

PS: Oh yes, Winnie just said that.

 

[Smart $]

How would it be dissipating full power 1/4 of the time if the RMS voltage was only 1/4 of full voltage? I would think as stated above, that at 1/4 voltage the power would be 1/16.

Jon is correct, and my earlier statement was/is not. I messed up in that my first calculation was ? the power and not voltage.

Taking math shortcuts, during the ? cycle the scr is triggered, last half of the positive half-cycle. The power is the same for that portion of time as if the full cycle was "delivered". So the power would be P??t (where P is power, t is time it takes for one cycle to occur), which is the same as ?Pt. Therefore Vrms over the ? cycle would be the same as the full wave, but over a full cycle it would be Vrms?√??t or ?Vrms-t

 

[Smart $]

Maybe someone has said it, but Vrms in this case is 1/2 its full wave value, not 1/4.

Then P = 1/4 of the full wave value.

PS: Oh yes, Winnie just said that.

I've been trying to post it for what seems like a couple hours now. For some reason this site and quite a few others are extremely slow in responding on my computer or connection. Don't know what the problem is...

 

[rattus]

Jon is correct, and my earlier statement was/is not. I messed up in that my first calculation was ? the power and not voltage.

Taking math shortcuts, during the ? cycle the scr is triggered, last half of the positive half-cycle. The power is the same for that portion of time as if the full cycle was "delivered". So the power would be P??t (where P is power, t is time it takes for one cycle to occur), which is the same as ?Pt. Therefore Vrms over the ? cycle would be the same as the full wave, but over a full cycle it would be Vrms?√??t or ?Vrms-t

Smart, you are confusing us. Vrms must be computed for a full period, T. You should say that the instantaneous value of the voltage, v(t), when the SCR is ON, is the same as it is for full power.

We are looking for average power, P. Then if we plot the instantaneous power, p(t), over the period, T, we see instantly that P is reduced to 1/4 its maximum value.

Why don't you give us one of your nice plots showing that?

 

[steve66]

Again just looking at the top half of the waveform, what if the SCR was conduting for the 1st 1/4 of the top half, then went off, and didn't start conducting again until the last 1/4 of the top half. How would this voltage value be measured?

It looks like we have your answer to the first part of your question.

For the second part listed above, as Smart money mentioned, if we had a circuit that only gave us the first 1/4 and last 1/4 of the top half of a cycle, its not as easy to calculate.

First, we have to decide if we are calcuating the "Average voltage" or RMS voltage. To find the Average, we integrate the voltage over one period, and divide by that period.

Assume a peak voltage of 1 volt, and assume a cycle lasts 2pi radians. We want the integral of sin (x) from 0 to pi/4, and from 3pi/4 to pi. Then we divide all that by 2pi, and we should have the average voltage.

We do a similar process to get the RMS voltage, but we have to square the voltage first (RMS stands for root-mean-squared). So we have the integral of sin^2 (x) from 0 to pi/4 and from 3pi/4 to pi. We then get the average of that (divide by 2 pi again), and finally we take the square root to get our answer.

Steve

 

[steve66]

Here is what I get for average and RMS voltages. Feel to correct the errors: I'm sure there have to be some in here:

http://i49.photobucket.com/albums/f252/Sragan/AverageandRMSvoltage.jpg

Edit: Like for example, I forgot to take the square root of the last number. So the RMS voltage should be sqrt (.159) = .398 volts.

 

[Besoeker]

Here is what I get for average and RMS voltages. Feel to correct the errors: I'm sure there have to be some in here:

http://i49.photobucket.com/albums/f252/Sragan/AverageandRMSvoltage.jpg

Edit: Like for example, I forgot to take the square root of the last number. So the RMS voltage should be sqrt (.159) = .398 volts.

Yes, I had noted that but I think there is also be a problem with the integration. In calculating the average you integrate sin x. The correct integral is actually -cos x. When you evaluate that the result is:
(1/pi)*(1-1/sqrt2) = 0.0932.

You may have done the same for the RMS. I took the relationship:
sin^2x = 1/2(1-cos2x) and used that in the integration
V^2 = (1/2pi)*((pi-2)/4) = 0.213

Expressing maths here is awkward - I should have just scanned in my workings and I might still.

Anyway, here is a picture of the waveforms, self explanatory I hope.

 

[steve66]

I thought the answers looked too large. I completely forgot to do the integration part.

Steve

 

[Besoeker]

I thought the answers looked too large. I completely forgot to do the integration part.

Steve

Easily done.
This is what I did for the RMS calculation:

http://i36.photobucket.com/albums/e39/Besoeker/RMS02.jpg
I took it as one conduction period over zero to pi rather than two over zero to 2pi - makes it a bit simpler.

 

[Smart $]

Smart, you are confusing us. Vrms must be computed for a full period, T.

While I agree our final goal is to determine Vrms over all time (i.e. continuous cycles), Vrms can be applied to any portion of a cycle... and it is quite easy to visualize that a quartered peak to crossing portion of a sinusoid waveform has both the same instantaneous voltage and the same Vrms as the full cycle, only the time domain has changed. Likewise, for power over that same quarter cycle. Knowing this, one can apply DC principles to those values (i.e. pulsed DC at 25% duty cycle) to arrive at the final goal. It's a shortcut.

Why don't you give us one of your nice plots showing that?

Started drawing one up, and it doesn't appear as revealing as you may be thinking.

 

[mull982]

Yes, I had noted that but I think there is also be a problem with the integration. In calculating the average you integrate sin x. The correct integral is actually -cos x. When you evaluate that the result is:
(1/pi)*(1-1/sqrt2) = 0.0932.

You may have done the same for the RMS. I took the relationship:
sin^2x = 1/2(1-cos2x) and used that in the integration
V^2 = (1/2pi)*((pi-2)/4) = 0.213

Expressing maths here is awkward - I should have just scanned in my workings and I might still.

Anyway, here is a picture of the waveforms, self explanatory I hope.

So with this can we assume that when chopping first and last quarter of pos half cycle then RMS voltage will be aprox 1/4 of full waveform? And have we then confirmed that this 1/4 voltage will produce 1/16th power lin load? I'm still a bit confused.