# Power conducted in Triac

Discussion in 'Homework Help' started by Dan_DG, Jun 12, 2015.

1. ### Dan_DG Thread Starter New Member

May 18, 2014
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0
Hi everyone,

I have run into an issue during my exam study. I do believe I understand some of the concept but not all of it. For Part (a), I can see that the peak current through the triac is going to be:
$(V_p - V_{triac})/R_2$
as the current will no longer flow through R1.
However I am unsure as to how to calculate the power. The solution states that the average current is given by:
$I_{peak} /2{Pi}$
but i am unsure how this is arrived at. It goes on to say that the power is given by:
$2 * I_{average} * V_{traic}$
Why is the power multiplied by 2? Is it due to the sinusoidal input?

I would really appreciate some clarification as to the how the average current and power is calculated. I do apologize if this is a simple question, this topic was barley covered during lectures so my knowledge is a little shaky. Thanks in advance for you help!

Part (b) is not an issue, so no need to look at that.

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2. ### MrAl Distinguished Member

Jun 17, 2014
2,573
523
Hi,

The current is:
I(t)=Vp*sin(w*t)/R

and the average current is then:
Iavg=(1/T)*integrate(I(t)) for t from t1 to t2

Now doing it the easy way we could take t1 to be zero and t2 to be 1/f/4 (90 degrees) even though the actual wave would go from 90 to 180 approximately, and T=1/f/2 because we have to integrate over the whole half cycle. The result is:
Iavg=Vp/(pi*R)

and note there is no '2' in there.

Another short cut would be to multiply this times the triac voltage of 1.5, and we'd get a value close to the stated result, but slightly higher.

Instead of using Vp=169.7 we could use Vp=169.7-1.5=154.7
because the triac voltage subtracts from the peak voltage.
Using this would give a closer result.

The actual waveform however does not go from 0 to pi/2 nor from pi/2 to pi, it actually will start out slightly after 0 or slightly after pi/2, and to solve for that we would use the inverse tangent function, but this would lead to a more messy solution which i dont think they want you to do. It would be more exact that way however because the actual waveform is slightly less than a quarter cycle because of the triac voltage drop.
Also note that the power is Iavg*Vdrop not 2*Iavg*Vdrop. If you divide Iavg by 2 (in error) then yes you need the 2, but if done right you dont need the 2.
This assumes that Vp is the peak voltage, because if Vp is the peak to peak voltage then you do need the 2.

Try it and see what you get...you should at least be able to get close to their value.

Last edited: Jun 12, 2015
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3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It is confusing to speak of average triac current. Presumably under steady state conditions with a purely resistive load and symmetrical conduction over each half cycle, the average current would be zero.

4. ### Dan_DG Thread Starter New Member

May 18, 2014
2
0
Thanks for the reply, it all makes sense now! From what I can see there may have been an error in the solutions that were provided by my lecturer which threw me off. Thanks!

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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This is the problem of depending upon rote learned formulae. In this instance garbage in = garbage out.
Does no one teach derivation from fundamentals anymore?

6. ### MrAl Distinguished Member

Jun 17, 2014
2,573
523
Hi there tnk,

Well for AC circuits the idea is almost always to deal with it as if it was a full wave rectified wave. So we dont integrate over an entire full cycle, we integrate over one half cycle. This is almost always the case because a result of zero does us no good. That's how we get RMS values too really, except there we happen to square the wave first so it all ends up above the horizontal axis anyway.
I am sure you could recall that the average voltage of a sine wave is 2*Vp/pi (0.6366*Vp) and so if we really integrated over the full sine we'd get zero, which does us no good. Ditto for the power in a resistor driven by an AC sine signal, where it is obvious that the resistor MUST dissipate energy as heat. If we did not do this right with a voltage of 120vac and resistor 1k ohm we would end up with zero watts dissipated but one heck of a very hot resistor

This is a good example of why we need geometry to go along with analytical techniques. Analytical calculations are sometimes not enough on their own without a geometrical foundation to base them on. I could guess that this is because geometry is closer in abstraction to reality than pure math is.

Just one more little clarification, and that is even if the load was capacitive or inductive the triac would still dissipate real power because it conducts with a real current and real voltage so it is a resistive element at that point, which of course still dissipates energy.

Also, to be a little more accurate with my previous post, the waveform we would integrate would be:
(Vp*sin(w*t)-1.5)/R

as that is the more accurate voltage across the resistor for the time it dissipates energy, and we'd have to solve for the zero crossing near t2=1/(2*f) and integrate from t1=1/(4*f) to that solution for t2. The result comes out a little less than the given text example, so i am guessing that they took a few short cuts to get to 1.605 watts.

Last edited: Jun 13, 2015
7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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@MrAl
You are telling me things I already understand perfectly well. I spent many years working on power electronics design.
But thanks for the effort anyway.....

8. ### MrAl Distinguished Member

Jun 17, 2014
2,573
523
Hi again,

Ok maybe i misunderstood you in post #3 then. You said something about average power being zero and it had to be resistive and we are talking about the triac not the load so i thought you meant only the load.

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I actually said something quite different. I was making the point that the average triac current would be zero provided certain conditions were met. This was at odds with the TS's mention of average (triac) current in the original post. The terminology "average" is confusing as used in the original post when one considers the usual interpretation of what is meant by "average". If one placed an average responding ammeter in the circuit what would be indicated? Certainly not what the formula suggested....
I did not state that the average power (either in the load or the triac) would be zero.
I was hoping for (or at least hinting at) greater rigor in the use of established terminology.
Perhaps I need to be less cryptic and simply state the obvious - but where's the fun in that?

Last edited: Jun 14, 2015
10. ### MrAl Distinguished Member

Jun 17, 2014
2,573
523
Hi,

Well, the average triac current will never be zero, unless it never turns on.
You have to think about what you are suggesting here it seems. You are suggesting that we use a DC meter in an AC circuit. A DC meter will give the purely analytical average which should come out to zero, but we dont do that because we know better. That's why i stated that analytics arent enough, we need geometry too, and that gives us a better view of the reality we are dealing with. When we look at the waveform, we see right away that it averages to zero, yet that cant be right, so we start to apply AC principles. An AC meter would register the AC current, regardless of load, unless the triac never turned on. A DC meter would register zero current, but we dont use that in an AC circuit. So we cant integrate over the full cycle, we have to integrate over a chosen portion of it, and in this case the waveform is a little more complicated so we have to solve for the upper and lower limits of integration (the lower is simpler but the upper requires a calculation).

I hope i made this clear, and i hope i understood you correctly but if not please elaborate.
I realize you more than likely have a *LOT* of experience here, so it is not that i am not trying to step on any toes just trying to be clear and complete about what i am stating myself.

I can offer a guess however at what you were intending to convey, and that might be that you just wanted to make the difference between the true average and the AC average more transparent. That's entirely cool

Mar 6, 2009
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