Hello again,

I think i can prove crutschow's result mathematically now that i understand his viewpoint better. What threw me was when he said "sine wave" i was assuming he was doing an AC analysis but really he wasnt. Sine waves are different than sine waves with offset because of the very different mathematical forms, and for this application the difference is very important.
In other words, in a mathematical expression sin(wt) is very different than 1-sin(wt). I believe now that he knew the difference just did not convey that soon enough. This led me to believe he was using a sinusoidal voltage drop across the transistors when really it is of the form 1-sin(wt) in it's most basic form.

With that in mind, we would still want to prove that the average current Ia times the supply voltage Vs is equal to the power in the amplifier. For example, how do we know it's not the RMS current times the supply voltage.
The proof can go two ways, but the most direct way is as follows...

We know that the output is Vp*sin(w*t). Since the drop across the transistor is Vs-that, we have:
Vdiff=Vs-Vp*sin(w*t)
The current if we assume is inductuve will be:
Iout=Ip*-cos(w*t)
so the instantaneous power is:
Pt=Vdiff*Iout=(Vs-Vp*sin(w*t))*(-Ip*cos(w*t))

Solving for one cycle time we can find one cycle to be 1/2 cycle of the fundamental. The positive portion goes from 1/(f*4) to 1/(f*4/3) so we integrate over that interval and divide by the time period and we get:
Pavg=2*Ip*Vs/pi

and that is 2*Vs*Ia which is similar to crutschow's formula.
We'd have to look into why we need 2*Vs, but it could be because this analysis is with a dual supply.

 

Dear Friends, I have calculated what happens in time-domain. The result I got was extremely simple.

P (dissipated) = 2 * Vs * Ip

Where;
Vs : rail voltage (i.e. rails are +Vs and -Vs).
Ip: Magnitude of output current (i.e. i(t)=Ip.Sin (wt))
Load assumed fully inductive, R3 (current feedback resistor) neglected.

The power is not related with Vo.

In the graphichs I took;
Vs=20V (Rail to Rail 40V)
Vp=10.sin(wt) (At the end, it will have no effect on the power unless Ip is constant and amplifier operates in its limits)
Ip=5A

The load voltage and current is as below;


There are two BJT transistors (push-pull) within the amplifier. I named them as TRA (positive rail NPN) and TRB (negative rail PNP).

Each transistor conducts the output current in its turn (regardless of the voltage at the output of amplifier), i.e. no reverse current flows through the BJT (the circuit would probably be damaged if it would so).

Take only TRA. The



The instantenous power over TRA is as follows;



The Integral form of the calculation as;



(The integral model solution is here checked from Wolfram ).

The result of the definite integration is simply 2 x Ip x Vs in half period, the second period is the same for the TRB and negative voltage rail, since the second half period completely repeats same power while TRA is off, this power can be said dissipating from the package (as TRA+TRB dissipation).

I believe that this problem is solved; Especially MrAl and Crutschow, thank you very much for your contributions. Please inform me, if I made a calculational mistake.

 

MrAl, the same result we reached, I was working on my post when you were editing yours.

 

In my initial question, I was in doubt, was not unclear about how complementary pairs can drive an inductive load.

Actually it was simple; A complementary output does not need to be at high voltage (with respect to the ground) to source a current from output and vice versa. Each transistor is providing its own directional current from relevant Vs rail whilst Vo voltage generated by inductor. So it is very normal for a current feedback circuit, the output is current, not the voltage.

I was not familiar indeed with working current mode circuits.

 

we would still want to prove that the average current Ia times the supply voltage Vs is equal to the power in the amplifier. For example, how do we know it's not the RMS current times the supply voltage.

RMS only applies when both voltage and current are sinusoidal.
When the supply voltage is a constant, then the supply power is simply the supply voltage times the supply current (V*I).
It doesn't take a bunch of math to prove that.

 

RMS only applies when both voltage and current are sinusoidal.
When the supply voltage is a constant, then the supply power is simply the supply voltage times the supply current (V*I).
It doesn't take a bunch of math to prove that.

The question in my first post was actually about the value of the power dissipation, and I had not realized the circuit can be a simple current node regardless which rail drives in which quadrant. The solution was based on the amplifier as a node, regardless of the load characteristic, applicable to any load which the amplifier can work with.

Anyway, studying more about the analogy inside the amplifier was worth doing.

 

The solution was based on the amplifier as a node, regardless of the load characteristic, applicable to any load which the amplifier can work with.

And my solution will work for that, as long as you know the power supply voltages, the average load current, and the power dissipated in the load.

 

And my solution will work for that, as long as you know the power supply voltages, the average load current, and the power dissipated in the load.

I agree, it will work. Basically, there is no principal difference from a Linear LDO regulator dissipation calculation and it will work as long as (lets sum up assumptions for future reference to this thread);

- The circuit operates correnctly (without any oscillation, without saturation, or principal problems etc.)
- There is no energy storage element within the amplifier stage, i.e. it is a linear solution
- The amplifier does not deliver energy back to the source (such as Class-D / PWM-MOSFET amplifiers those switching returning power to the supply rails)

Thank you very much Crutschow.

 

MrAl, the same result we reached, I was working on my post when you were editing yours.

Hi,

Just one more little thing though. How come you got a formula with Ip in it and i got a formula with Ip/pi in it? That means mine is using average current while yours is using peak current.
Did you make a mistake or did it make a mistake?
I'll go over my derivation and post the entire argument, and if you could go over yours and post each step too we can compare notes.

 

RMS only applies when both voltage and current are sinusoidal.
When the supply voltage is a constant, then the supply power is simply the supply voltage times the supply current (V*I).
It doesn't take a bunch of math to prove that.

Really? How did you prove your result?
Sometimes i do have to wonder if you have the temperament required for a civil discussion of technical matters.

 

Really? How did you prove your result?

I really don't have to prove that for you, but you're welcome to do that yourself.
If you don't know that RMS for sinewaves only applies to to situations where both the current and voltage are sinewaves, then I suggest you do some basic study on that.
Power equals I*V.
You want me to prove that?

Sometimes i do have to wonder if you have the temperament required for a civil discussion of technical matters.

So now we are getting personal?
Okay, you're on.

So no, I suppose I don't really have the temperament for inane technical discussions about things that are obvious, especially with one who seems to be a little slow to understand what I am saying, and posts to me in a condescending manner (as in this post).
So if you don't like my temperament (note the title after my name), please don't respond to any more of my posts.
I certainly won't miss them.

 

I really don't have to prove that for you, but you're welcome to do that yourself.
If you don't know that RMS for sinewaves only applies to to situations where both the current and voltage are sinewaves, then I suggest you do some basic study on that.
Power equals I*V.
You want me to prove that?
So now we are getting personal?
Okay, you're on.

So no, I suppose I don't really have the temperament for inane technical discussions about things that are obvious, especially with one who seems to be a little slow to understand what I am saying, and posts to me in a condescending manner (as in this post).
So if you don't like my temperament (note the title after my name), please don't respond to any more of my posts.
I certainly won't miss them.

Hello again,

A quote from one of your posts:
"It doesn't take a bunch of math to prove that. "

There's nothing wrong with a little math. You sound like there is no place for math here and seek to put that view down. Taking a formula from a book is not the same thing either. "I*V" is math also, so you used math too you naughty boy
It's always good to have two different ways to calculate something anyway. That ensures you get the right result. The OP was looking for a clear and exemplary proof.
Look up the word "exemplary":
1. Serving as a desirable model; representing the best of its kind.

So simply stating a formula isnt as good anyway, i am sure you will agree.
I think we agree on most things, but sometimes when we dont you seem to take offence and start putting down the alternate approach.

BTW, waveforms of all types have RMS values. They can have peak values, average values, and RMS values.

 

Hello again,

Ok i went over my results and proved that the correct theoretical solution is in fact:
Pavg=2*Ip*Vs/pi

This comes from:
p(t)=-Ip*cos(wt)*(Vs-Vp*sin(wt))

and solving for the zeros we get:
t1=1/(f*4)
t2=3/(f*4)

and this is the portion above zero and since it is symmetrical above and below zero and is an integer multiple of the cycle period we can then integrate p(t) for t from t1 to t2 and then divide by t2-t1 and we get the final result for the average power Pavg shown above. This happens to match the result shown in some literature also.

It's nice to have solutions like this because they match exactly what we are looking for without pulling any shortcuts. When i worked in the industry we had to prove very nearly everything. I worked on a design for a 100 amp three phase static switch once and part of the work took up three pages.

 

Ok i went over my results and proved that the correct theoretical solution is in fact:
Pavg=2*Ip*Vs/pi

MrAl, you are right, I made a mistake by omitting the coefficient of t inside sin() function. I appologize. Thank you for pointing it out.

 

For general discussion here;

Some people like math as a powerful tool while some people like shortcut solutions based on basic principles of theory more.

I like math, but I have less time to focus on its details. That was why I started the thread in order to get if an easy idea exists or not.

As seen, "there is" an easy way and now, moreover, it is proven.

Please think positive, we all need this ; -)

 

Some people like math as a powerful tool while some people like shortcut solutions based on basic principles of theory more.

Math is certainly a powerful tool and not much technically can be done without it.
But I find using math very tedious and only do it as much as I have to.
In this case it was apparent from basic principles (power dissipated equals power in minus power out) the problem could be solved with a minimum of math.
That's not a shortcut solution, it's the simplest solution, and I don't see that it needs any further proof.
If you want to use a bunch of math to prove that basic principle, you are welcome to do that, but it's a lot of unnecessary work.

 

MrAl, you are right, I made a mistake by omitting the coefficient of t inside sin() function. I appologize. Thank you for pointing it out.

Hi,

Oh no problem. I also forgot something, i forgot to put the minus sign in front of the "Ip" because we should be using -cos(wt) instead of just cos(wt). I corrected that now though in my previous post.

Yes some people like math and some dont. But i do feel it is necessary when someone gets a formula thrown at them and they cant be sure if it is right or not and we all know that not everything on the web is 100 percent accurate, so when we do a proof we have more confidence that it is the right formula.

I dont mind doing math, and this problem did not require much math even for the proof. Anybody can do an integration like this these days because of automated software so there's no excuse anymore.
But i have my limits these days too because my time is divided up so much. Back some years ago i came up with a solution for ANY rectifier circuit under the sun (such as for a power supply that is) that required a partial differential equation as the general solution. So we now have the general EXACT solution for any recitifer circuit, but to get from the general solution to the particular solution takes a lot of work, much more than we've seen here, much much more, so even i dont get into that as much as i used too anymore but resort to other ways to solve it in most cases. So i have my limits too.

 

Math is certainly a powerful tool and not much technically can be done without it.
But I find using math very tedious and only do it as much as I have to.
In this case it was apparent from basic principles (power dissipated equals power in minus power out) the problem could be solved with a minimum of math.
That's not a shortcut solution, it's the simplest solution, and I don't see that it needs any further proof.
If you want to use a bunch of math to prove that basic principle, you are welcome to do that, but it's a lot of unnecessary work.

I think the real problem these days is that so many things we find on the web are not correct, like the formulas or they may be missing some detail. This brings the confidence level down for formulas that are just thrown at people as they tend to question the validity. So when you say, "unnecessary work" you speak only for yourself and anyone else that might be already familiar with the formula and the shortcut method.

I run into these kinds of doubts quite a bit with different people and different applications. Another common one is with one of the transformer formulas for the maximum flux density using a sine or square wave. One of the constants comes into question now and then by people who are not familiar with the formula already. They dont feel comfortable trusting a formula that is just thrown at them. They dont know you, they dont know me, so they like to see some proof. If you yourself dont need the proof, that's fine, but other people might. Still yet others, just like to see the proof because it comes from basic principles and that adds to the confidence level also. A secondary method is also a good idea.

As i was saying i think in another thread, when i worked in the industry we could never get away without doing a proof on almost everything. I worked with professors and domestic and foreign government military and i can tell you we could not get away with anything. One such client was Israeli Aircraft, and the guy i worked with insisted on 5 digit measurements for example. He was an interesting person though with some great ideas. Ditto with a professor from Perdue University. He had some fresh ideas which made that interesting too.

Once we get too "set in our ways" we forget how it is for the people just getting into the field.