Dan, don't take this thread a post at a time, now, at page 4. Tony just reiterated what has been posted previously.
Mike, there is a similarity with choosing a heatsink.
Power dissipated and not delivered to the load must be determined and dealt with.
You will note that in virtually any stage of design, after all the calculations and measurements, a safety factor of maybe 10 or 20% is added in. Looking for the 1 right answer or the 1 right way will only find frustration.
I have read post #11, and I think I am understanding the confusion, which is probably my fault for poorly stating my thoughts.
Tell me if this is right.
The highest instantaneous voltage an output transistor can have across it is the full rail to rail (in this case 36) voltage
The load does indeed swing from +18v through 0 to - 18v but...
the highest instantaneous voltage it can ever see is 18v, either + or -. correct?
I think that is where the confusion cropped up. And since this is AC, the power output reflects the fact that it does indeed swing rail to rail, even though there is never more than plus or minus 18v across it at any given time. Like what was mentioned earlier, power is the same if the voltage is positive or negative.
Am I doing good here, and more importantly, has the confusion over where I am at here sorted out? I am trying to take the thread as a whole, not just posts in isolation. Sorry if my misunderstanding seems to make it seem like I'm not listening.
Tell me if this is right.
The highest instantaneous voltage an output transistor can have across it is the full rail to rail (in this case 36) voltage
The load does indeed swing from +18v through 0 to - 18v but...
the highest instantaneous voltage it can ever see is 18v, either + or -. correct?
I think that is where the confusion cropped up. And since this is AC, the power output reflects the fact that it does indeed swing rail to rail, even though there is never more than plus or minus 18v across it at any given time. Like what was mentioned earlier, power is the same if the voltage is positive or negative.
Am I doing good here, and more importantly, has the confusion over where I am at here sorted out? I am trying to take the thread as a whole, not just posts in isolation. Sorry if my misunderstanding seems to make it seem like I'm not listening.
The phrase you didn't write which explains the "apparent difference" and which seems to scratch you the bad way (you don't *sound* convinced yet) is that:
a) The load has one end connected to ground/0V.
No point in this circuit example is further than 18V from ground.
Is this clear?
So there's no way that the load receives more than 18V (instantaneous/peak/whatever).
b) the transistors are NOT connected to ground in any way, but in series, from -18V to + 18V, so 36V end to end.
Is this clear?
So when one transistor is shorted .... how much does the other one receive? = 36V
And what if the other one is now shorted?
The unshorted one now receives the full 36V.
I used the word "shorted" instead of the proper word "saturated" to make things clearer.
Hope this is clear.
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Surprisingly, I don't think anyone "looked at it" from this point of view:
Your question was, "if I have a square wave", and presumed [-18 / 0 / +18] supply, what is the power?
During the positive side, the load (8 ohms) sees +18 / 0 and were that it was entirely resistive, 18 x 18 / 8 = +40.5 watts.
During the negative side, the load (8 ohms) sees -18 / 0 and again that is (-18 x -18) / 8 = +40.5 watts.
Since the wave is square, 50% of the duty cycle is positive, and 50% is negative, with a negligible gap between the phases. The total power therefore is:
P = 50% (40.5 watt) positive + 50% (40.5 watt) negative side
P = 40.5 watts.
Not divided by 2 or the square root of 2 (1.414) or anything else. Just 40.5.
Its harder with a sine-wave, but using a spreadsheet (what a wonderful tool!) you can approximate the same thing. At each instant of a half-sine wave, compute the power through the resistor. Simply "take the sum" of powers and there you are. Here's a little table ... showing one sin() of 18 volt amplitude, power to 8 ohms ... and using JUST the "average()" function at the bottom to see what the power is:
deg volts power
0 0.0 0.0
10 3.1 1.2
20 6.2 4.7
30 9.0 10.1
40 11.6 16.7
50 13.8 23.8
60 15.6 30.4
70 16.9 35.8
80 17.7 39.3
90 18.0 40.5
100 17.7 39.3
110 16.9 35.8
120 15.6 30.4
130 13.8 23.8
140 11.6 16.7
150 9.0 10.1
160 6.2 4.7
170 3.1 1.2
180 0.0 0.0
190 -3.1 1.2
200 -6.2 4.7
210 -9.0 10.1
220 -11.6 16.7
230 -13.8 23.8
240 -15.6 30.4
250 -16.9 35.8
260 -17.7 39.3
270 -18.0 40.5
280 -17.7 39.3
290 -16.9 35.8
300 -15.6 30.4
310 -13.8 23.8
320 -11.6 16.7
330 -9.0 10.1
340 -6.2 4.7
350 -3.1 1.2
mean: 20.25
See that 20.25? Now ... if you go thru the calculus theory of this (or just look up equations in the past postings), the recommendation is
P = ((18 / 1.414)^2 ) / 8
P = 20.25 watts
Note that using EXCEL to calculate every 10 degrees agrees exactly with the recommended formula. This is because "the formula" was noted, and designed for pure sine waves. But my analysis of the square wave above also shows that ... things are not as they might conveniently seem on the surface.
One can even do the same thing (albeit lots, and lots of times) with tiny sound clips from WAV files. Of course, you don't drop them in a spreadsheet (though I guess you could, now with excel allowing 250,000 lines!) but rather, you build a tiny program in Basic or C, or Perl, or your favorite scripting language ... that pulls out each (voltage) value, squares it, divides by your resistance, keeps a running sum, then at the end divides by the total number of samples. in other words, an average.
And "what is this average?" Why... it is the average (mean) of the squares of voltage (divided by resistance, to give 'power'). No square-root involved, or needed.
So there you are.
GoatGuy
EDIT: if instead, you take the squares of the volts (each sample in the WAV file), don't divide by resistance (i.e. not try to figure out power), but compute the average, you get Mean of Squares. Now, take the square ROOT of that, and you get Root, Mean, Squares. This "voltage" number, which applied back into the power equation (V²/R) gives the same output value. This is why the RMS A/C voltage was computed this way to begin with!!!
Your question was, "if I have a square wave", and presumed [-18 / 0 / +18] supply, what is the power?
During the positive side, the load (8 ohms) sees +18 / 0 and were that it was entirely resistive, 18 x 18 / 8 = +40.5 watts.
During the negative side, the load (8 ohms) sees -18 / 0 and again that is (-18 x -18) / 8 = +40.5 watts.
Since the wave is square, 50% of the duty cycle is positive, and 50% is negative, with a negligible gap between the phases. The total power therefore is:
P = 50% (40.5 watt) positive + 50% (40.5 watt) negative side
P = 40.5 watts.
Not divided by 2 or the square root of 2 (1.414) or anything else. Just 40.5.
Its harder with a sine-wave, but using a spreadsheet (what a wonderful tool!) you can approximate the same thing. At each instant of a half-sine wave, compute the power through the resistor. Simply "take the sum" of powers and there you are. Here's a little table ... showing one sin() of 18 volt amplitude, power to 8 ohms ... and using JUST the "average()" function at the bottom to see what the power is:
deg volts power
0 0.0 0.0
10 3.1 1.2
20 6.2 4.7
30 9.0 10.1
40 11.6 16.7
50 13.8 23.8
60 15.6 30.4
70 16.9 35.8
80 17.7 39.3
90 18.0 40.5
100 17.7 39.3
110 16.9 35.8
120 15.6 30.4
130 13.8 23.8
140 11.6 16.7
150 9.0 10.1
160 6.2 4.7
170 3.1 1.2
180 0.0 0.0
190 -3.1 1.2
200 -6.2 4.7
210 -9.0 10.1
220 -11.6 16.7
230 -13.8 23.8
240 -15.6 30.4
250 -16.9 35.8
260 -17.7 39.3
270 -18.0 40.5
280 -17.7 39.3
290 -16.9 35.8
300 -15.6 30.4
310 -13.8 23.8
320 -11.6 16.7
330 -9.0 10.1
340 -6.2 4.7
350 -3.1 1.2
mean: 20.25
See that 20.25? Now ... if you go thru the calculus theory of this (or just look up equations in the past postings), the recommendation is
P = ((18 / 1.414)^2 ) / 8
P = 20.25 watts
Note that using EXCEL to calculate every 10 degrees agrees exactly with the recommended formula. This is because "the formula" was noted, and designed for pure sine waves. But my analysis of the square wave above also shows that ... things are not as they might conveniently seem on the surface.
One can even do the same thing (albeit lots, and lots of times) with tiny sound clips from WAV files. Of course, you don't drop them in a spreadsheet (though I guess you could, now with excel allowing 250,000 lines!) but rather, you build a tiny program in Basic or C, or Perl, or your favorite scripting language ... that pulls out each (voltage) value, squares it, divides by your resistance, keeps a running sum, then at the end divides by the total number of samples. in other words, an average.
And "what is this average?" Why... it is the average (mean) of the squares of voltage (divided by resistance, to give 'power'). No square-root involved, or needed.
So there you are.
GoatGuy
EDIT: if instead, you take the squares of the volts (each sample in the WAV file), don't divide by resistance (i.e. not try to figure out power), but compute the average, you get Mean of Squares. Now, take the square ROOT of that, and you get Root, Mean, Squares. This "voltage" number, which applied back into the power equation (V²/R) gives the same output value. This is why the RMS A/C voltage was computed this way to begin with!!!
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Slightly related side note:
If I restructured my power supply to have full wave rectification, I cut my DC roughly in half, right? Can the much reduced ripple performance of halfwave rectification be somewhat mitigated with larger reservoir caps? I'm not talking about creating a high performance amp, but I just want to get to the point where I can plug my iPod into the thing, and get sound that is good enough that I can hear the difference when I make some of the really basic improvements shown in the early chapters of Bob's book, using the cheap trannies I have on hand for now.
Maybe I would be better served to have the lesser power but the better quality rectification scheme. I will certainly try both for the learning experience.
If I restructured my power supply to have full wave rectification, I cut my DC roughly in half, right? Can the much reduced ripple performance of halfwave rectification be somewhat mitigated with larger reservoir caps? I'm not talking about creating a high performance amp, but I just want to get to the point where I can plug my iPod into the thing, and get sound that is good enough that I can hear the difference when I make some of the really basic improvements shown in the early chapters of Bob's book, using the cheap trannies I have on hand for now.
Maybe I would be better served to have the lesser power but the better quality rectification scheme. I will certainly try both for the learning experience.
@goat guy:
actually a similar scheme is show in the Radiotron Designer's Handbook to figure average power dissipation in a power tube. I think it doesn't use every ten degrees though, but rather uses a smaller selection that isn't evenly incremented in order to take into account the fact that the slope isn't constant throughout.
actually a similar scheme is show in the Radiotron Designer's Handbook to figure average power dissipation in a power tube. I think it doesn't use every ten degrees though, but rather uses a smaller selection that isn't evenly incremented in order to take into account the fact that the slope isn't constant throughout.
@JMFahey:
all that you say is clear, and actually has been for awhile. I think I just asked the follow up questions poorly is all. To be honest, I think the think the thing that screwed me up in the beginning was my misunderstanding that while each signal half is only a 50% duty cycle, 2 duty cycles that are 50% and do the same thing powerwise equate to a %100 duty cycle. My divide by 2 error came from this.
all that you say is clear, and actually has been for awhile. I think I just asked the follow up questions poorly is all. To be honest, I think the think the thing that screwed me up in the beginning was my misunderstanding that while each signal half is only a 50% duty cycle, 2 duty cycles that are 50% and do the same thing powerwise equate to a %100 duty cycle. My divide by 2 error came from this.
IMO it's a bit pessimistic to suggest that calculation is not useful when determining the output of an amplifier.
Let's look at the situation backwards. Suppose we want to design an amplifier to deliver 100W into 8 ohms.
We're talking 100 watts continuous sinewave power here, that's the only measure that makes sense. Not squarewave power (or peak power), but continuous sinewave power, because that gives the lowest figure that can still be regarded as a measure of power for any given amplifier, so you can't be accused of exaggerating.
Let's think about DC first.
V=I*R. I=V/R
P(ower)[watts]=V*I
so, substituting, P=(V/R)*V, or (V*V)/R (vee squared upon ahr, or eye squared ahr, which I didn't derive).
We've got R = 8 ohms and P = 100 watts, so
100=(V*V)/8 or V*V= 8*100=800 ...so
V=sqrt(800) = 28.28V.
But that was for DC. To get a sinewave with an average value of 1V we need a peak value of 1.41V [sqrt(2)]. So the peak value we need is 28.28*1.41, or 40V, and the peak-to-peak value is 80V, so the amplifier needs to swing +/-40V into 8 ohms.
So this tells us that the minimum value for the supply rails is +/-40V. Now we can look at the amplifier and PSU topology and decide how close to the rails a given design will swing, and add a margin to accommodate that; how much ripple the PSU will have and how much it is likely to sag under load, and add further allowances to accommodate. Then we can add a couple of diode drops for the bridge rectifier, all of which takes us closer to knowing the output voltage of the mains transformer, which itself will have an AC (rms) rating. Some of these things can be calculated, and some are known from experience, so an experienced designer can sit down and draw up a design that will meet a specification, at least in terms of the output power, without the necessity for extensive prototyping. The more innovative the design, the greater the possibility that it may depart from the specification in prototype form, but it is untrue to suggest no amplifier will meet a specification without extensive prototyping.
Let's look at the situation backwards. Suppose we want to design an amplifier to deliver 100W into 8 ohms.
We're talking 100 watts continuous sinewave power here, that's the only measure that makes sense. Not squarewave power (or peak power), but continuous sinewave power, because that gives the lowest figure that can still be regarded as a measure of power for any given amplifier, so you can't be accused of exaggerating.
Let's think about DC first.
V=I*R. I=V/R
P(ower)[watts]=V*I
so, substituting, P=(V/R)*V, or (V*V)/R (vee squared upon ahr, or eye squared ahr, which I didn't derive).
We've got R = 8 ohms and P = 100 watts, so
100=(V*V)/8 or V*V= 8*100=800 ...so
V=sqrt(800) = 28.28V.
But that was for DC. To get a sinewave with an average value of 1V we need a peak value of 1.41V [sqrt(2)]. So the peak value we need is 28.28*1.41, or 40V, and the peak-to-peak value is 80V, so the amplifier needs to swing +/-40V into 8 ohms.
So this tells us that the minimum value for the supply rails is +/-40V. Now we can look at the amplifier and PSU topology and decide how close to the rails a given design will swing, and add a margin to accommodate that; how much ripple the PSU will have and how much it is likely to sag under load, and add further allowances to accommodate. Then we can add a couple of diode drops for the bridge rectifier, all of which takes us closer to knowing the output voltage of the mains transformer, which itself will have an AC (rms) rating. Some of these things can be calculated, and some are known from experience, so an experienced designer can sit down and draw up a design that will meet a specification, at least in terms of the output power, without the necessity for extensive prototyping. The more innovative the design, the greater the possibility that it may depart from the specification in prototype form, but it is untrue to suggest no amplifier will meet a specification without extensive prototyping.
To make matters worse, by the time I send my next post, often someone has already replied to a previous one, and my new post often appears to be a reply to something I hadn't yet read.
Trust me, I have learned a great deal from this thread, and understand more than I thi9nk you guys realize. Its often my poor framing of my questions that makes it sound like I haven't been listening.
I am completely self taught on this stuff. I stared out with tube guitar amps, and successfully designed a guitar amp from scratch that sounds awesome, and I have been gigging with it for years now. That's why I have been so puzzled at times. I'll misunderstand something to the point it cause me to doubt what I learned years ago.
Tackling solid state has required me to learn some fundamentals that I must have missed when learning tubes.
Trust me, I have learned a great deal from this thread, and understand more than I thi9nk you guys realize. Its often my poor framing of my questions that makes it sound like I haven't been listening.
I am completely self taught on this stuff. I stared out with tube guitar amps, and successfully designed a guitar amp from scratch that sounds awesome, and I have been gigging with it for years now. That's why I have been so puzzled at times. I'll misunderstand something to the point it cause me to doubt what I learned years ago.
Tackling solid state has required me to learn some fundamentals that I must have missed when learning tubes.
I don't think anyone did that.
Maybe I came closest, unintentionally, and followed up to try and make the point more clear.
So I definitely agree with you, cc. It could be said now that the better the calculations, the better the ver1.0 prototype.
In an earlier post it was stated that what was being considered was "a perfect amp, just to understand the theory" , and in that case, the 70W result is correct.
But then you compare that "unreal" result with what's stated for a very real amp and worry about the difference
Ok, fine, but please don't change game rules unannounced and in the middle of a game
Let's do a little Math applied to what the schematic actually shows.
The Quad 303 schematic is:
An externally hosted image should be here but it was not working when we last tested it.
for a simple analysis of just one half of the output circuit (the other half will be roughly the same):
upper half output transistor is Tr1L, old trusty 2N3055, driven by Tr105, BC143 (145?)
To begin with, we must roughly estimate what current are we passing through that 8 ohms load.
We can late re-adjust that value, but at least we'll already be in the ballpark from the beginning.
We have "perfect" peak voltage of 67V/2= 33.5V into 8.3 ohms.
Why 8.3?
Look at R124, 0.3 ohms, in series with the load .
So 33.5V/8.3 ohms=4A peak.
Now we must calculate what voltage is lost across the output transistor , which must be substracted from theoretical peak voltage.
The 2N3055 datasheet 2N3055 pdf, 2N3055 description, 2N3055 datasheets, 2N3055 view ::: ALLDATASHEET ::: gives the designer a lot of useful info; for this analysis we'll concentrate on :
As we see, the datasheet mentions a 4V drop (loss) across the transistor when passing 4A, so we'll use that value.
The situation is more complex, we have a driver to consider, there's also a predriver Tr103, plus base current passes through R122, but trust me, those 4V I accepted as loss across 2N3055 covers most of these "sins".
So peak (available) voltage is not 33.5V but 33.5-4=29.5V .
Further, the actual load does not receive them in full, because a 0.3 ohm resistor R124 is in series with it, so actual load peak voltage now is 28.4V.
RMS value is: 28.4V * 0.707=20V RMS.
Which yield: 20^2/8=50W RMS.
But wait a second, this is just on the edge of clipping, easily reaching 1% to 5% distortion, which is unfair, because the amplifier is capable of doing *much* better than that.
What would you think about "downgrading" , say, 10% of rated power, to be certain we are working within the best area of operation?
Which would give us 50W -10%= 45W.
Would you believe 45W RMS at very low distortion, as QUAD303 "official" spec?
They should state that in ads and brochures.
Oh, they *do*?
In a nutshell: trust Math and calculations, just don't forget to correct for real world stuff
Can I suggest Dan reads an article or a book chapter that does take you through design and calculations as requested - in fully worked sequences and examples? Those also considered here are the supplementary issues like cap size, ripple, losses, building techniques and many other things you haven't yet raised here. (but probably will do as you come across the need)
Considering how long it's taken to get this far, you have nothing to lose. Assuming you haven't already read it in depth, try this one, it is pitched at beginners by forum member Rode. If you want only to look at your present requirement, surely you can substitute your power supply data in the example data fields if need be. Good reading
Elliott Sound Products - Linear Power Supply Design
Considering how long it's taken to get this far, you have nothing to lose. Assuming you haven't already read it in depth, try this one, it is pitched at beginners by forum member Rode. If you want only to look at your present requirement, surely you can substitute your power supply data in the example data fields if need be. Good reading
Elliott Sound Products - Linear Power Supply Design
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have you thought of the possibility that as one trannie is fully off and the other fully on, the voltage on the load is the rail? this is peak value, now while the load is at rail, the cutoff trannie can be twice the rail or very near it, about 36 volts in this case, this is one reason that your output trannie must have at least twice the Vceo rating of the rail....confused you yet again?
you can click the link on my signature and you may find more....
i am a tube guy, but before that many years ago, i built the Tigersaurus amps of Daniel Meyer and the Super Leach amp of the late professor WM Leach of Georgia Tech...i have more than enough experience to be talking about this topic.....
this is what i use for testing....those tapes are able to stand 200*C and is used for heat tracing on pipes....there are those that post here and yet has nothing to show in terms of completed amplifiers that work....
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