I don't understand LM3886 rail calculations - pls help

I am building a powered subwoofer using a Tang Band W6-1136SIF in a vented enclosure. I did my modeling in WinISD, build a prototype box, measured it with REW -- all looks good. Based on my port tuning and anticipated HPF @ 20Hz, I should be able to give this thing about 100W, but I decided to connect it to my big boy Crown PA amp with a tone generator @ 60Hz, to see how much voltage it takes before making unhappy noises. Turns out, that's about 23VAC, so I'm targeting a peak output of 22VAC. If I'm doing my math right, that's ~134W, so I was pretty close.

My plan was to build a plate amp with 2xLM3886 in parallel, but AFAIK, I can think of it as the equivalent of 1x LM3886 with twice the current limit. Still, no matter how I try to wrap my head around the power supply calculations, I just can't make sense of it. I built a Google Docs sheet and plugged in the equations 5 (Vopk), 6 (Iopk), and 7 (+/- vdc), and it seems to want +/-30V (or max +/-39V if I use the 15% regulation and 10% high line figures.)

However, if I don't use the magic equations, and just try to work it out by hand:

22VAC = (22 x 1.414) = 31V PtP. Add 4V dropout on each rail, and that's 39V PtP, or ~ +/-20V.

Why does the data sheet suggest 50% more voltage? Is that just for headroom (WRMS vs. peak)? Or am I missing something else?

I would really like to keep the voltage as low as possible to minimize waste heat, so if my manual calculations are correct based on reaching Xlim @ 22VAC, I would much rather go with +/-20V. But the data sheet, and lots of forum posts about 4R loads all seem to be insisting that I should have more, just for ~40-60W.
 
For 60W rms in 4 ohms you need ~15.5Vrms, right? That's ~22V peak, add some overhead say 26Vpk.
For 60W rms in 4 ohms you need 15.5/4 ~3.9A rms = 5.4A pk.

Does that help?

Jan

Well, that's where I get confused. I get the same numbers you do up to a point. But then, the data sheet puts that number through formula 7.

I'll walk through the process:

Eq. #5: Vpeak = sqrt(2 * RL * PO) -- sqrt(2 * 4R * 60W) = 21.91 Vpk
Eq. #6: Iopeak = sqrt((2 * PO) / RL) -- sqrt((2 * 60W) / 4R) = 5.48 A

I do this differently when I calculate by hand (Vpk = VRMS * 1.414; Ipk = +/- (Vpk / RL)) but it gets to the same end result. So far so good. Now...

Eq. #7: Vsup = +/- (Vpk + Vod) * (1 + xf_reg) * (1 + hi_line) -- (21.91 + 4V) * (if we ignore regulation and high line conditions, this term is '1') = 25.91V

OK, "say, 26Vpk" is pretty much spot on. But... this is PLUS OR MINUS -- +/-26V -- so we're really doubling the voltage.

What makes sense to me is +/- ((VRMS * 1.414) / 2) + Vod = +/- 15V.

So that is my question -- why the discrepancy?

If I'm measuring across the output terminals of my power amp, and I get 22VAC, that's not "+/- 22V" -- that would actually be 44VAC. A bit of a difference there. :)

Are we making room for waveforms that are asymmetric? Is this calculation for music with a 6dB crest factor? (So, if you want 60W, what you really mean is 60W continuous with 240W peaks.) Or is there something else I'm missing?
 
AX tech editor
Joined 2002
Paid Member
Don't overthink it. If need an amp output of 22Vac, that is 22*1.4V = 31V pk, right? That is thus 22*1.4*2 = 62V pk-to-pk.
And if there are no losses etc in the supply and output stage, that would require + and - 31Vdc supplies.
And if you disregard losses in rectifier and droop in reservoir caps, that would require 22Vac transformer windings.
I don't know where you get that 44Vac number at your amp - we started out to assume 22Vac amp output, so how can it now all of a sudden be 44Vac?

Just draw it and pencil in voltages, that will make it clear.

Jan
 
  • Like
Reactions: 1 users
OK - I'm going to boldly display my ignorance here. What is a multimeter actually showing you in AC mode? (Let's, just for a moment, pretend there's no difference between RMS and peak -- let's say we're measuring square waves, and the meter knows that.)

Is 1VAC the difference between the most negative peak and the most positive peak? Or the amplitude of the absolute value? I guess that would be the difference between "peak-to-peak" and "peak."

My assumption is that, when I see 22VAC on my meter, what I'm seeing is the amplitude between the negative and positive peaks, with no consideration given to where 0V is. 22VAC = +11Vp to -11Vp. I might be very wrong about that, and the meter is actually telling me that the voltage is swinging 22V above and below 0V.

It seems the amp calculations all assume that we're measuring the output as what I understand to be the classical "RMS = equivalent to DC" power, where that power will be positive or negative half the time, and so we need to be able to supply that power in either polarity. +22Vp to -22Vp, or 44Vpp.

I guess this is the problem with being self-taught. There is always the potential for a huge gap in fundamental knowledge. I appreciate your patience while I'm an idiot in public. :)
 
Most multimeters/DVMs measure peak, but display rms (under the assumption its a sine wave, even if it isn't). Square waves are just read wrong with such a meter, and a mix of AC and DC will not be handled well!

Some DVMs are "true rms" meters and measure rms, and display rms.


rms makes sense for a generic voltage measurement tool, since AC power is always given as rms (so you can do the average power calculations without confusion)

If you are doing signal processing, sometimes you use amplitude (ie peak) rather than rms, but in most contexts we use rms for ac.

So for instance here in the UK the power we get is 240Vrms, which is 340V peak, 680V pk-pk. Using the rms value makes it easy to see that every 4A taken is about 1kW of power, 240x4 ~= 1000. This is why our plugs are rated at 13A as this allows 3kW devices (the good old fashioned 3-bar electric heater being the canonical example). That's 18A peak, 36A pk-pk...
 
Ah! That's the thing I was having trouble finding a definitive answer to:

So for instance here in the UK the power we get is 240Vrms, which is 340V peak, 680V pk-pk.

I had no trouble finding answers about Peak vs. RMS, and that's something I've had a handle on for a while. I understand that meters assume the input is sinusoidal, and the reading will be incorrect if it's not. I've even read that some meters don't handle signals outside of mains frequency quite as well, so I set my tone generator to a 60Hz sine just to be safe. (That just happened to be in the sweet spot of the subwoofer anyway, with a 30-85Hz target bandwidth.)

But the thing I couldn't find a concrete answer on was: Is the 22VAC reading I'm getting indicative of 11-0-11, or 22-0-22? It's a subtle question, but the difference is significant. If it's indeed the latter, the datasheet equations make A LOT more sense.
 
AX tech editor
Joined 2002
Paid Member
You're on the wrong track. The meter is a black box and if you connect sinusoidal 22Vac to it, it displays 22. Simple as that.
Don't worry about whats in the black box, that will lead you away from the issue at hand.

But the thing I couldn't find a concrete answer on was: Is the 22VAC reading I'm getting indicative of 11-0-11, or 22-0-22?

It is neither. 22Vac rms is a voltage (assuming sinusoidal) that alternates between +22*1.4 and -22*1.4. The peak value is thus ~31, so the pk-pk value is ~62.
But nobody worries about that in practise, but if you insist: 22Vac across 4 ohms need 22/4=5.5A rms. Now, that 5.5A rms alternates between + 5.5*1.4pk and - 5.5*1.4 pk and is 55*1.4*2 pk-pk. As I said - draw it; a picture is worth 1000 words.

Jan
 
A 22VAC RMS sine wave has peaks at 22* sine(45°) = +31.11V and - 31.11V. An amplifier that is capable of 22VAC must have a power supply that is 2 or 3 volts more than the peak output, so you need a power supply that is +/-35VDC. So called "rail-rail" amps saturate very close (~0.5V) to the power supply and others waste as much as 10 Volts on both sides. The LM3886 is a 2EF output with a current source VAS so you lose at least 4x0.65=2.6V on each side. The RMS value of a square wave is essentially the peak value so marketing people use that to rationalize exaggerating their amp power ratings. Note that the RMS power of a sine wave is half the peak power, so you can calculate VP²/R/2 to make the math easy.
The RMS value is the same heating value as the DC voltage, V²/ R. Note that negative voltage squared is positive power. Cheap meters measure peak voltage because that's what a rectifier (diodes) and integrator (capacitor) give you, and scale it *0.71 assuming a sine wave..
To measure an RMS voltage correctly, the voltage must be squared, then averaged, then the square root found, (root-mean-square). This can be done with analog circuits, but today digital calculations are more accurate. Digital measurements are made at many points along the wave.
 
Last edited:
I thought I had made this clear, but I guess not, so here goes: The RMS bit is not my hang-up. I understand RMS =/= Peak, I know that, in a nutshell, Peak = RMS * 1.414, or that RMS = Peak * 0.707. I know that this is only true for a sine wave. For a square wave, RMS == Peak. I know that meters (usually) don't display peak voltage, but either RMS, or some approximation thereof, depending on how sophisticated they are. None of that is an issue. I get it. We're good. :) Yes, I temporarily suspended acknowledging the difference between Pk and RMS to communicate my actual ask a little more clearly -- and I knew that was risky, because so often the explicit part where I said "let's pretend RMS = peak for a moment -- like we're measuring square waves, and the meter knows that" would get lost or ignored or skipped over -- and sure enough it seems to have.

I also fully grok the drop-out voltage thing. Totally fine with that. I've been including that in my "show your work" notes the entire time, in fact. :)

I'm sorry if I'm "overthinking this" -- I don't know how to ignore the fact that there's up to a 2x discrepancy in what I intuit the rails need to be, and what the data sheet says they need to be. That disconnect makes me uncomfortable, because I'm trying to engineer a solution -- and how do you engineer something you don't understand? I could just "do what the data sheet says to do" and it would most likely be fine. But then, if it IS wrong, I have just backed myself into a thermal corner that I need to get out of. Not good. I would rather understand why I'm doing something, particularly something that causes a different problem to solve, than just go with it and hope for the best. Likewise, I'm not sure what a drawing would do to make that more obvious. What is it I'm supposed to draw? I have no problem visualizing a sine wave, with time on one axis and amplitude on the other axis. I just didn't know if the meter was measuring from one hump to the other over the course of a full cycle, or from the mid-line to one hump for one half of the cycle, and mid-line to the other hump for the other half of the cycle.

Anyway, I'm pretty sure I got the answer to my question. I was just mistaken on what amplitude of a waveform the meter measures, vs. what I assumed it measured. Reporting the RMS value of the waveform in absolute value makes a certain kind of sense, I guess. But it didn't seem obviously the right answer, either. Now I know, and it squares with the magic equations in the data sheet, so I think I am good to go.

I appreciate the help.
 
AX tech editor
Joined 2002
Paid Member
OK, then lets look at that datasheet, see here:

For instance: "68W Cont. Avg. Output Power into 4Ω at VCC = ±28V":

Assume about 3V headroom needed, then effective output limits are +/-25V.
That's about 18Vrms, into 4 ohms, that's 79Wrms.
Since TI specified "Cont Avg", that would be ~16V* which gives 63Wrms
Close enough.

OK?

Jan

Average = peak * 0.637.
 

Attachments

  • E0400OSIQR.pdf
    202.5 KB · Views: 59
  • Like
Reactions: 1 user
I understand RMS =/= Peak, I know that, in a nutshell, Peak = RMS * 1.414, or that RMS = Peak * 0.707. I know that this is only true for a sine wave.
That is correct when rectifying a sine wave to obtain "DC" via a transformer.. Look at the signal on the primary. In the uk that is ~240v AC. The signal swings 240v each side of neutral so p to p is 480v. Same as your amplifier, you are missing a factor of 2.

Your meter shows the DC equivalent on it's AC range. A true RMS meter wont care about it being a sine wave it will still show a DC equivalent usually with some max specified frequency range. The crest factors you mention for other waveforms - they can work out with analogue meters as they average. Personally I wouldn't assume they will on a dvm.

On class B this relates to the power dissipation of the output drivers. It is equally shared. Each drive a 1/2 cycle not the entire waveform.
 
I do. To answer the "where does the 2x come from" question, here was my problem. (Maybe a picture WILL help.)

sine.png


In the picture, let's say the blue sine wave is the 22VAC my meter was reading. The dashed lines represent the RMS value. My problem was that I thought the meter was telling me 22VAC is the span indicated by the green bar. Apparently, that is wrong. It is the span represented by the pink bars.

If you think about "RMS is the equivalent of DC power," then the pink bars make sense. It doesn't matter if the voltage is positive or negative with respect to the 0V reference, we're just concerned about the absolute value.

I was thinking about it like, there is a potential between the most negative and most positive (RMS) values -- and that is what the reading is showing you. Well, nope, that was wrong.

I hope that it's clear where that 2x discrepancy comes from now.

If the meter had been reporting the green span, then the pink spans would have been +11 (*1.414) and -11 (*1.414), and my amp would need that, plus maybe ~2.6V drop out voltage on each rail, so rails of a little over +/-18VDC would have done the trick.

BUT, with the pink bars being 22V each, then I need more like +/-35VDC, because I need to be able to swing to 22 (*1.414) + ~2.6V on EACH RAIL.

My ignorance has been rectified (no pun intended...) and the equations now line up with what I would expect the calculated values to be.
 
  • Like
Reactions: 1 users
AX tech editor
Joined 2002
Paid Member
It may be nice to know how an AC rms meter works. Basically, it is a DC meter.
You can make an AC rms meter from any DC meter.
You rectify the AC voltage (let us assume perfect rectifier for now), put the DC meter on it and adjust the DC meter to read 0.707 times the original reading - done!
This is now a calibrated AC rms meter for sinusoidal waveforms.

If the waveform is not a sine, the 0.707 factor is incorrect, and you need other means to convert AC rms to equivalent DC.
Google will show you several interesting ways, for in stance based on thermal techniques.
And of course, the rectifier is not perfect, and even gets worse with increasing frequency and lower voltages.

BTW: A picture IS worth a 1000 words! :cool:

Jan
 
  • Like
Reactions: 1 user
I do. To answer the "where does the 2x come from" question, here was my problem. (Maybe a picture WILL help.)

View attachment 1211570

In the picture, let's say the blue sine wave is the 22VAC my meter was reading. The dashed lines represent the RMS value. My problem was that I thought the meter was telling me 22VAC is the span indicated by the green bar. Apparently, that is wrong. It is the span represented by the pink bars.

If you think about "RMS is the equivalent of DC power," then the pink bars make sense. It doesn't matter if the voltage is positive or negative with respect to the 0V reference, we're just concerned about the absolute value.

I was thinking about it like, there is a potential between the most negative and most positive (RMS) values -- and that is what the reading is showing you. Well, nope, that was wrong.

I hope that it's clear where that 2x discrepancy comes from now.

If the meter had been reporting the green span, then the pink spans would have been +11 (*1.414) and -11 (*1.414), and my amp would need that, plus maybe ~2.6V drop out voltage on each rail, so rails of a little over +/-18VDC would have done the trick.

BUT, with the pink bars being 22V each, then I need more like +/-35VDC, because I need to be able to swing to 22 (*1.414) + ~2.6V on EACH RAIL.

My ignorance has been rectified (no pun intended...) and the equations now line up with what I would expect the calculated values to be.
Seems like you got it. Just to be complete, the lower dashed line "does not exist", an RMS value is always positive. After all it is starts with taking the square of the voltage so everything is positive.
 
  • Like
Reactions: 1 user