Ah, in post #9. That's a long time ago
Yes it was correct apart from the last three words The average of a sine wave is zero. And we know what the RMS value is now.
*FULLY* worked out calculation, including doubts you didn't even ask:
Post #13 .
If the whole post is too complex, the end result is resumed in the last line:
To which it might be added that since you only have half wave rectification, ripple/voltage drop will be much worse than I calculated.
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Which is why this all is mostly just a learning/relearning exercise IMO. Applying it directly to real world systems is of limited value.
Its useful stuff to know, and useful to be able to work forward or back to get some idea of what is possible with actual values.
The Quad amp just mentioned is a specific case. We could look at why the output is limited when theoretically it could deliver more from the available supply.
The theory is really important because it tells you what is and isn't possible.
So, getting clearer to me now. I have been puzzling for months over how to calculate the power output of an amp circuit from the circuit values. I have come across various approaches , ranging in method from simple maths to differential calculus. Are you saying that this can't be easily done? I could imagine that factors like o/p stage biasing arrangements would play a part. But, is there a straightforward way to get a reasonable idea from a circuit diagram of what the amp can do/
BTW - I have to say that threads like this one are hugely informative and helpful for people like me - enthusiastic amateurs. Thanks to all who are contributing!
BTW - I have to say that threads like this one are hugely informative and helpful for people like me - enthusiastic amateurs. Thanks to all who are contributing!
For the bare basics, you're not doing anything wrong. Your value isn't necesarily THE max output power, but the max with some "ideal" amplifier circuit. As already said:
There looks to be a big discrepancy between the two power ratings, but there are many things to be accounted for, one of them being that the manufacturer may want to give conservative specs. Maybe the amp gives worst-case, say, 49 watts with the line voltage 10 percent lower than the nominal 120V - to insure every unit off the assembly line meets the spec (and to give a "nice" number on the spec sheet), they round it down to 45 watts for the claimed value.
Also, power ratings are often given with a "maximum distortion" spec. The amp may be fully capable of 50 or 60 watts at maybe 2 percent distortion, but only goes to 45 watts at 0.01 percent distortion. Many people will look at the distortion figure and think the amp has that much distortion over the whole range of its power output, so the manufacturer might not want to give that 60 watt figure.
I recall seeing a "distortion at 1 watt output" spec, which can be more useful in evaluating an amp than at full output (well, as useful as a THD figure can be, as discussed elsewhere).
So there's lots of reasons, nontechnical as well as technical, for the value of published specs.
If you want a "fair approximation" it can be easily done with what's been discussed in the thread.
If you want a more accurate value, you need to take into account the rectifier diode drops, the transformer secondary voltage drop at full load, the filter capacitor "droop" between charging pulses, at what voltage the driver and output transistors saturate, and maybe a few other things that don't come to mind.
If you know enough about electronics and do it enough, taking all these things into account becomes "straightforward." Kinda sorta.
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assuming no losses in the output trannies, the peak to peak ac possible is your rails, so having a +18, -18 volt rails, peak to peak output possible is 36 volts....sine...
this then becomes 12.7 volts rms, so that power = V^2/R, therefore at 8 ohm load, power is 20 watts....
remember that all this is just approximation, depending on your power supply situation, this could change however slightly....
this then becomes 12.7 volts rms, so that power = V^2/R, therefore at 8 ohm load, power is 20 watts....
remember that all this is just approximation, depending on your power supply situation, this could change however slightly....
Sofaspud and others - Thanks, but I still find a rule of thumb, quick and dirty calculation that gives a difference of 50% a little useless as a practical answer, and not much of an "approximation". Is there no simple way of getting closer to the actual figure for (eg) that Quad circuit? After all, amplifier designers seem willing enough to give their designs "straightforward" power-ratings, and I just wonder how they do this. If the experts can look at a circuit and have a good idea of what power it will give, what is their method? Surely, we are dealing with science here, not intuition...
Not trying to be difficult or confrontational, but this (what is often presented as a useful practical formula) really perplexes me. How does an amplifier designer know what power his circuit will provide? If he doesn't know with any level of accuracy, and has to build it to find out, then we'd be better to just admit that.
Not trying to be difficult or confrontational, but this (what is often presented as a useful practical formula) really perplexes me. How does an amplifier designer know what power his circuit will provide? If he doesn't know with any level of accuracy, and has to build it to find out, then we'd be better to just admit that.
you don't....just looking at an amplifier circuit will never give you the power ratings without considering the physical power supply as well...
the only way to know is with a dummy load, a scope, a sine wave generator and optionally a distortion meter....
you also need to have your ac line fixed as a certain volts, since variation in line voltage while you are doing it affects the result.....
formulas are fine, but what is the correlation to the real world?
the only way to know is with a dummy load, a scope, a sine wave generator and optionally a distortion meter....
you also need to have your ac line fixed as a certain volts, since variation in line voltage while you are doing it affects the result.....
formulas are fine, but what is the correlation to the real world?
Sound advice - and I would perhaps add a Variac to that list. Problem is that manufacturers can afford to build many prototypes, and risk them all catching fire, to get to their final product. DIY'ers don't have those resources behind them, and need to try and get it tolerably right first time, rather than have their o/p xistors fried and their speaker cones splattered on the living-room wall. But, hey, that's the fun of DIY isn't it? be careful, and try again if necessary. It's very good to have clarity that there is no easy answer to my question, other than an informed suck it and see approach.
yes.....actually the quick and dirty approximations presented here will always be lower than actual....this is because the standby rails drops at full power testing.
in my super leach amp build, i have rails of +/-85 volts, then at full power, these rails drop to +/-75 volts....
i have seen commercial amps with rails at +/-90 volts but dropped to +/-71 volts at full power...
so looking at a circuit alone and computing power does not tell you much....
in my super leach amp build, i have rails of +/-85 volts, then at full power, these rails drop to +/-75 volts....
i have seen commercial amps with rails at +/-90 volts but dropped to +/-71 volts at full power...
so looking at a circuit alone and computing power does not tell you much....
Those formula aren't rules of thumb; they are the actual formulae using the simplified ideal numbers of the example.
I've never calculated percentages, but I wouldn't say the presented ratings are that straightforward. benb gave some good examples. We would need to know exactly how Quad determined their ratings.
This is what I meant by my previous comment. Using these formulae with ideal numbers can provide a rough estimate, and understanding of the correlation among the electrical values. For specifics, one must look deeper into the system to find more useful real world numbers to plug into those formulae.
I can't really speak for them, but a designer can create an amplifier circuit, and then he designs a power supply that meets the requirements.
Perhaps less often, the designer may already have a power supply, and must design an amplifier that meets those requirements.
So, yes, it is science. But there is also room for the art and creativity (and DIY) side of it.
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@ Tony - the Quad 303, of course, has a regulated supply rail with tons of headroom across the regulator - 88V regulated to 67V if I remember rightly (tho apparently Peter Walker of Quad admitted to D Self years later that this was unnecessary, and a result of Quad's over-caution in it's first bjt power amplifier design).
Another puzzle I have had is how to do the thermal design/heatsinking, etc. Is it reasonable to think that if there is big difference between the theoretical maximum power given by the simple formulae, and the actual output power, then this might have implications for the heat dissipation of the output transistors? That's, perhaps, where some of that power could be going? Heatsinks are expensive things, and they have big consequences for physical design. Poor old DIY'ers want to minimise trial and error on things like that!
Dang it, before this post I thought I had it. Taking a pair of output transistors, isn't the load positioned at the halfway point of the rails? if one transistor is in cutoff, and the other is saturated, wouldn't that still be only 18v positioned at the load?
I think you are talking tube amps maybe? I see the valve in your avatar, and you mentioned "losses in the output transformer", which to me suggests tubes. In a tube amp with an output trannie, the full rails are indeed presented to the load. I'm pretty sure that isn't true in the simple solid state example I am talking about.
Could someone more knowledgeable chime in? I'm 99% sure I'm correct, but that one percent is making me doubt what I thought I had gained through this thread!
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