Audiophile help please - Ohms and power - Page 1
Forum section: electronics / projects

author: DW1961
author_rank: Frequent Contributor
timestamp: July 30, 2020, 07:08:26 am
content: If you have an amp rated at 100 watts RMS total @ 4 Ohms (but can power 4-8 Ohm speakers), if you plug 8ohms speaker into it, do you get 50 watts total in reality?I thought a 4 Ohms speaker would be more sensitive than an 8 Ohm speaker, thus requiring less power for the same volume. I've been reading what sounds like it is the opposite. If a 4 Ohms speaker is less resistance, wouldn't that make it easier to drive with less power?Thanks again.

author: DaJMasta
author_rank: Super Contributor
timestamp: July 30, 2020, 07:39:57 am
content: The answer is not really, if you take the basic ohms law V = IR and the basic calculation for power P = VI, you'll find that to get a given power you need a combination of current and voltage applied. So with a lower rated resistance of your speaker, there needs to be more current applied by the driver, and with a higher rated resistance it needs more voltage to get the equivalent power dissipated by the speaker in either case.Since your driver says it can drive 4 or 8 ohm speakers, it can likely put 100W into either (since it would be sort of fudging its specifications if it can't!), but for any given amplifier, you're going to be limited by your driver supply current when driving lower impedance (resistance) speakers and you're going to be more limited by your drive voltage capability when dealing with higher impedance speakers.It ends up being finding an optimum point on a few curves given your driver circuit and your load (the speaker), and usually the maximum output of the driver comes along with some distortion, so there's also probably a THD concern when driving at the edges of your driver's output ratings.So more generally, a speaker rated for however many watts should sound equivalently loud as another speaker with a different rated impedance properly driven with the same power - though depending on speaker construction and choice, there can be some difference in efficiency of turning the electrical power into vibrations in the air given the physical construction of the given speaker.

author: Psi
author_rank: Super Contributor
timestamp: July 30, 2020, 07:41:24 am
content: The speaker is just a load.If you put a 4 ohm load across a PSU that's outputting 12V you get 36WIf you put a 8 ohm load across a PSU that's outputting 12V you get 18WThe 4 ohms speaker draws more current for the same voltage.So a 4 ohm speaker will pull more power from the amp and be louder than an 8 ohm speaker.Of course there is a limit, the amp must be able to supply the current.You can't put a 1 ohm speaker on the amp and expect even more power. Unless the amp is rated to drive 1 ohm speakers

author: DW1961
author_rank: Frequent Contributor
timestamp: July 30, 2020, 07:48:23 am
content: Quote from: Psi on July 30, 2020, 07:41:24 amOkay, so if you run identical speakers except for ohms, then the lower Ohm speakers would put out the same volume with less power needed?

author: Psi
author_rank: Super Contributor
timestamp: July 30, 2020, 07:54:58 am
content: Quote from: DW1961 on July 30, 2020, 07:48:23 amNope.Volume is power and power is volume.You cant get free power or free volume.If you run two identical speakers except for ohms one will use more power and be louder and one will use less power and not be as loud.You can get more efficient speakers which are louder than other speakers at the same power.But this is not ohms related, this is because they are made out of lighter materials so need less power to move air.Both speakers are getting the same power, but in one of them more of that power is converted into sound.

author: Psi
author_rank: Super Contributor
timestamp: July 30, 2020, 08:05:28 am
content: I think i can see where you might be getting confused.The volume knob on your amp isn't setting the power directly, is not an "amount of power" knob.Its scale changes depending on what speakers you have on it. It's more of a voltage knob.If you have 8 ohm speakers the max position on the vol knob might be 100Wbut if you change to 4 ohm speakers the max position is now 200W and half way will be 100W and be same volume as max was with your 8 ohm speakers.Something like that..

author: magic
author_rank: Super Contributor
timestamp: July 30, 2020, 08:14:49 am
content: I haven't read the whole thread and don't know if this answers the question, butQuote from: Psi on July 30, 2020, 08:05:28 amIt isexactlya "voltage" knob. It simply controls effective end-to-end voltage gain of the whole preamp - power amp chain. At one setting the output volts might be 100x the input volts, at another it might be 1x. The power stage will supply whatever current to the load to maintain the expected voltage.Exception may be some "toob" amps or other audiophile contraptions with crazy high output impedance.

author: Michaelaudio
author_rank: Regular Contributor
timestamp: July 30, 2020, 10:22:30 am
content: Its a combination of efficiency and Impedance (and frequency response) Lets say u have one 8ohm woofer and one 4ohm, 4ohm with 83db @2,83V/m and the other 8ohm with 93db @2,83V/m efficiency. U apply 2,83v to both speakers(and the same frequency), the 4 Ohm speaker will draw more current from the amp than the 8ohm one, but would still be not as loud as the 8Ohm couse the speaker is way less efficient. There are 10db diference of efficiency between the speakers (10db more means double the volume for ur ears)

author: DW1961
author_rank: Frequent Contributor
timestamp: July 30, 2020, 07:43:28 pm
content: I want to get back to the speaker aspect of this, but this might help me understand. Here is the TEXAS INSTRUMENT 3116D amp power specs:https://www.ti.com/document-viewer/TPA3116D2/datasheetSupports Multiple Output Configurations2 × 50 W Into a 4-Ω BTL Load at 21 V (TPA3116D2)2 × 30 W Into a 8-Ω BTL Load at 24 V (TPA3118D2)2 × 15 W Into a 8-Ω BTL Load at 15 V (TPA3130D2)Since they increased the voltage a little for the 8 Ohm spec, it's still about half of what the 4 Ohm spec is given the same voltage.I'm wondering why only 60 watts instead of 72 total, since 8 ohms / 24V / 3 A = 72 watts?For instance, 8 Ohms @ 19V = 45 wattsHow do they get 60 watts at 24V?Also, back to speaker Ohms, if you had that amp and had two identical speakers, but one with 4 Ohms and one with 8 Ohms with the above amp specs, how is that going to effect speaker output?Damn I'm having a hard time with this. It seems like you would want to increase power for the 8 Ohms spec, but that's my confusion dammit. It's like the 4 Ohm will use more power, gets more power, and the 8 Ohm NEEDS more power, but gets less. Shit!

author: Siwastaja
author_rank: Super Contributor
timestamp: July 30, 2020, 08:03:47 pm
content: Efficiency is a completely different parameter.Similarly, you can have a LED light bulb designed for 12V, 110V, or 230V. All can have good, or poor efficiency.The speaker impedance is an arbitrary parameter that can be designed quite freely. A lower impedance allows lower voltages to be used yet still deliver quite some power to the load. If youdohave enough voltage available, the speaker designer can design it for higher impedance. The benefit is less losses in wiring. Unsurprisingly, lower impedances are seen in automotive, where the limitation of 12V battery was relevant (modern high-power amps boost the voltage internally to overcome this limitation, though); in PA, higher impedances are seen, so that higher voltages are used, so that the relative voltage drop over the long wiring isn't a problem - the exact same thing why power delivery networks use higher voltage than what you have at your home.All of this does not relate to the sound level in any way.Sound level relates to the power and efficiency. Whatever power can be delivered at any impedance, it's just about designing the speaker and the amplifier. And, good efficiency can be had at any impedance, as well (ignoring extremes which will make the target difficult; say a 0.1-ohm or a 100000 ohm speaker would likely be inefficient with all the real-world constraints).There is quite some efficiency variation in speakers, and it's not always specified at all. An efficient speaker puts out quite some volume at, say, 10W, while another one might require 50W to sound the same and mostly produce heat.

author: magic
author_rank: Super Contributor
timestamp: July 30, 2020, 08:10:40 pm
content: Quote from: DW1961 on July 30, 2020, 07:43:28 pmFor supply voltage of ±21/24/15V maximum sinewave output is approximately 14/16/10V RMS, respectively. Do the math (P=V²/R) and you get the numbers they give.

author: Benta
author_rank: Super Contributor
timestamp: July 30, 2020, 08:13:11 pm
content: I understand your confusion, and some power specs for amplifiers are extremely misleading. But you're on the right track.First, solid-state audio power amplifiers (quaint 70s expression, no?) are almost perfect voltage sources (as opposed to tube amps with output transformer).This simplifies things, as the electric power absorbed by your loudspeaker (not to be confused with acoustic power!) relates directly to its impedance.If your amp delivers 10 VRMSinto a 4 ohm speaker, this equals 25 WRMS(U2/R). For an 8 ohm speaker, it will be 12.5 WRMS.10 VRMSequals ~14.1 VPEAK, meaning the amp's power supply should be around +/- 20 V (to have headroom)."Marketing" power numbers such as "music", "peak" etc. can safely be ignored.

author: DW1961
author_rank: Frequent Contributor
timestamp: July 30, 2020, 10:00:25 pm
content: Quote from: magic on July 30, 2020, 08:10:40 pmIf I understand, that seems a weird way to explain the output power drop. 24V is RMS. It seems like in order to get the amplifier output, there is another calculation after you get input RMS voltage? It looks like 1/2 peak power is "continuous?"

author: helius
author_rank: Super Contributor
timestamp: July 30, 2020, 10:30:17 pm
content: RMS power refers to a periodic AC waveform, but audio amplifiers are powered from DC rails. So the output voltage waveform will have a peak (not RMS) level limited by the supply voltage minus some diode drops. Converting peak to RMS is the familiar divide by sqrt(2).

author: ejeffrey
author_rank: Super Contributor
timestamp: July 30, 2020, 10:43:30 pm
content: QuoteIt won't. The difference between an 8 ohm and 4 ohm speaker is the voice coil. An 8 ohm speaker will have more turns of thinner wire than a 4 ohm speaker. When driven with the same power (higher voltage but lower current) the coil will produce the same magnetic field, and thus the same sound.Speaker impedance is within reason a matter of choice and what is convenient for an amplifier. Common transistors are happy with 10s of volts and a few amps, so a few ohms is a good target. If you go to the low end (say 1 ohm or less), then the ohmic losses of your speaker wires and PCB traces start to be a problem, and even the inductance of your speaker wires can become a factor. At the high end (say 32 ohms or higher) you start needing uncomfortably high voltage in your amplifier to get reasonable power. This can be an issue for safety as well as limiting your transistor choices.8 ohms is a pretty happy place for general purpose audio use and therefore became the almost universal standard for home stereo. Low resistances became common in car audio to get maximum output out of the low supply voltages available (~12 V total) without a boost converter: a 4 or 2 ohm speaker can get a moderate amount of output power from a single 12 supply, especially in a bridge tied configuration. Even when you have a higher supply voltage, for any given amplifier a lower impedance speaker will draw more power from the amplifier until it gets too low and something breaks. So people looking for MOAR POWR! will generally pick the lowest resistance speakers their amplifier can drive.Quote from: DW1961 on July 30, 2020, 07:43:28 pmThe maximum RMS voltage is ~1/sqrt(2) the DC voltage and from there you get the power in these configurations. These specified in BTL configuration where both wires of the speaker are driven with opposite polarity rather than leaving one terminal at ground. Therefore, a 24 V supply can produce up to + or - 24 V across the speaker. An old school design would use +/- 24 V rails to accomplish the same thing.

author: Dundarave
author_rank: Regular Contributor
timestamp: July 30, 2020, 11:02:51 pm
content: Another, simpler (I think) way to look at this is from an impedance-matching perspective:For maximum power transfer between any amplifier circuit and its load, the output impedance of the amp needs to match the speaker. Any other load impedance will, theoretically, reduce the amount of power delivered to the load. It’s as simple as that.That’s the theory. The reality is that the impedance of anything is dependent on the frequencies involved, but practically speaking, matching an 8 ohm speaker with an 8 ohm amp is always going to deliver the most power to the speaker.

author: DW1961
author_rank: Frequent Contributor
timestamp: July 31, 2020, 12:30:02 am
content: Quote from: Dundarave on July 30, 2020, 11:02:51 pmYes, I do understand that the impedance changes depending on what sound is begin generated, but keeping it static is all my head can handle. Thanks for that simplistic explanation. However, I do need to do some calculations, so I'm missing something in the formula (see post below).

author: retrolefty
author_rank: Super Contributor
timestamp: July 31, 2020, 01:35:29 am
content: Quote from: Dundarave on July 30, 2020, 11:02:51 pmI have the understanding that audio amps normally have a much lower output impedance rating then the speakers they drive by a factor of 50 to 150. In fact there is a quality specification parameter called damping factor covers this topic(https://en.wikipedia.org/wiki/Damping_factor).As most solid state audio amps normally use significant negative feedback (Vs tube amps} to maintain a linear output voltage it should be considered as having a very low output impedance Vs normal speaker impedance. Your point is about when mostefficientpower transfer occurs, but most all audio amps are not designed and used for best efficiency, that doesn't sell amps.

author: Dundarave
author_rank: Regular Contributor
timestamp: July 31, 2020, 02:41:56 am
content: Quote from: retrolefty on July 31, 2020, 01:35:29 amLol, I’m not going to argue the point, as I was only talking about “maximum power transfer”, but an amp driving a 4 ohm speaker to have a reduced impedance factor of 50 to 150 below 4 ohms has got to have a pretty insignificant output impedance!

author: ejeffrey
author_rank: Super Contributor
timestamp: July 31, 2020, 04:24:32 am
content: Quote from: Dundarave on July 30, 2020, 11:02:51 pmThe maximum power transfer theorem only applies to linear (and stable) sources and loads, so while it kind of hints at the right answer (making the load *closer* to the source impedance increases power output), it is wrong if taken literally. If you put a power matched speaker on a typical audio amplifier the power transfer will be zero as you will blow a fuse or power transistor.Quote(Solid state) audio amplifiers are almost always very low impedance -- a small fraction of an ohm, not 8 ohms or similar. An amplifier described as an "8 ohm amp" just means that is the type of speaker it is designed for.

author: DW1961
author_rank: Frequent Contributor
timestamp: July 31, 2020, 04:58:08 am
content: Quote from: magic on July 30, 2020, 08:10:40 pm24V RMS 2*/8 ohms = 72 total Then /2 = 36 watts per channel.Why do you say that 24 volts they use is maximum and not RMS? If it's a 24V power supply, then that's RMS, isn't it? I mean you don't rate power supplies by peak voltage. That's what Where in the link is the voltage being lost? If I can get this, then were done with the Amp discussion. I'll get it.I'm reading this and trying to figure it out:"Now let’s look at the RMS values. In our example, we know the RMS voltage is 10 volts. We saw earlier that 10 squared is 100, and 100 divided by 5  gives us a calculated power of 20 watts."https://geoffthegreygeek.com/understanding-amplifier-power/

author: ejeffrey
author_rank: Super Contributor
timestamp: July 31, 2020, 05:27:14 am
content: 24 volts is the DC supply voltage to the amplifier. It's a constant input voltage, so peak, RMS, average are all the same. When the amplifier has to produce a varying waveform from the 24 V input source, it can produce any voltage up to (slightly less than) 24 V. The voltage is dropped across the amplifier's power transistors. So if you want to produce a sine wave with maximum power output, it would have a 24 V peak. In that case the RMS voltage would be 17 V. (17 V)^2/(8 ohm) is 36 watts into an 8 ohm speaker. In reality it won't be able to go all the way to 24 V, there will always be some voltage drop on the power transistors, hence the 30 W rating.

author: DW1961
author_rank: Frequent Contributor
timestamp: July 31, 2020, 05:38:17 am
content: Quote from: ejeffrey on July 30, 2020, 10:43:30 pm"It won't. The difference between an 8 ohm and 4 ohm speaker is the voice coil. . . . When driven with the same power (higher voltage but lower current) the coil will produce the same magnetic field, and thus the same sound."But they aren't being driven with the same power for that 3116D amp. One gets 50 watts per channel and one is getting 30. They do have diff. voltage and current, but the power output is different.I really apologize if I am just hacking this thing to death. I really am trying. I know you can change voltage and current and get the same power rating. But you did say "same power" so I'm going with that

author: Siwastaja
author_rank: Super Contributor
timestamp: July 31, 2020, 06:03:09 am
content: Quote from: Dundarave on July 30, 2020, 11:02:51 pmNoooooo!!!11This stupid thing, taught in university classrooms, goes around and around misunderstood from decade to decade.Well obviously, from a very limited perspective, it is true. If you have an arbitrary Thevenin equivalent source, say an amplifier which output transistors define output impedance of 10 ohms, then if you do want to get the maximum possible power from it, you match the output load of 10 ohms.But then you make the logical mistake and turn it around - if you have a 10 ohm load, to get maximum power possible, do you use a 10 ohm output impedance amplifier? Obviously, no! Use 1 ohm, and you'll geteven more power. Run the math, it's trivial!The key you didn't think about in school is this:Often:* the power used by the load does actual work (is converted to sound in a speaker, is converted to mechanical energy in a motor, is converted to light in a lightbulb...)* the power used by the amplifier is wasted power (converted to heat in the amplifier)From this perspective, matching the impedances so that power use is shared 50%-50% is a ridiculous idea. Instead, if you choose Zamplifier = 0 and Zload = whatever to bring the voltage & current relationship to the most optimal point for practical design, you have just the amount of power youneed, with perfect load voltage regulation, and best possible amplifier efficiency.The imaginary amplifier with 10 ohm output impedance was probably never meant to drive a 10 ohm load, but maybe a 1000 ohm load. Why?Because amplifiers are often designed to be voltage sources that can drive the output to whatever value within its rails. If it had the equal output and load impedance, then the output voltage would change by factor of 50%. Try to output 10 volts - only 5 volts available to the load. 5 volts wasted as heat in the amplifier (this obviously won't matter in a linear amplifier which wastes power as heat anyway).This is why amplifiers have the feedback, which is designed to bring the output impedance close to zero. The transistors and PCB traces themselves still have some on-resistance, so the feedback needs some voltage leeway to work with. Want to output exact 10V, to a load which hogs 1A? Say, transistors drop 1V because of their internal on-resistance. Is the output dropped to just 9V, and is the output impedance 1V/1A = 1 ohm? No, and no. The feedback measures the output and adjusts the transistors to be "more on" until the output is at 10V, and the output impedance is hence 0V/1A = 0 ohms.So whenever you want to drive a load with a voltage source, the correct impedance is not the load impedance, but zero. In case of linear amplifiers driving intermediate levels, it's about voltage regulation, giving the load the correct voltage. In case of switching amplifiers, it's all about heat as well. If you match the source and load impedance, the efficiency of the amplifier is limited to just 50%, which would be very crappy for such circuit.

author: Siwastaja
author_rank: Super Contributor
timestamp: July 31, 2020, 06:14:06 am
content: Quote from: DW1961 on July 31, 2020, 05:38:17 amTo drive them with the same power, you need to up the voltage. This may mean changing to another amplifier chip designed to handle higher voltage, but now this chip doesn't need to deal with as much current.

author: dzseki
author_rank: Frequent Contributor
timestamp: July 31, 2020, 06:34:33 am
content: Also in practical amplifiers the power supply has a limited power output (this is where they can cut cost), so while in theory the output power ratio for the amplifier at 4 Ohm / 8 Ohm load would be 2 : 1, this rarely happens as the power supply is usualy underspecified and drops the output voltage too soon. Specification for amplifier that is putting 30W in to 4 Ohm and 20W into 8 Ohm is very common.Also some amplifiers (mostly NAD I think?) had switches on the back for 4/8Ohm setting, which increased the supply voltage for the power amplifier, then it could drive 4 and 8 Ohm speakers with roughly the same power

author: DW1961
author_rank: Frequent Contributor
timestamp: July 31, 2020, 06:54:29 am
content: Quote from: ejeffrey on July 31, 2020, 05:27:14 amOK I get that. So is the voltage drop constant or depends on the amp chip? It looks like they are losing 16% from an RMS of 17V.It looks like they lose 16% in the conversion?24V @ 8 Ohms = 72 watts / 2 =36 watts36 -30 = 6 watts lost.16.6% loss? Is that linear for each voltage?

author: DW1961
author_rank: Frequent Contributor
timestamp: July 31, 2020, 07:38:39 am
content: Just one other thing before I pass out tonight.If you need to double power to increase volume 3dBs, then is there any loss using less voltage.e.g. @ 8 ohms24V gets you 60 watts total19 volts gets you 37 watts totalWhat about a difference is bass power needs? Would the extra power help?

author: dzseki
author_rank: Frequent Contributor
timestamp: July 31, 2020, 11:11:08 am
content: At different loudness levels the human ear perceives the bass and treble notes differently, consult with the Fletcher-Munson curves. Some amplifiers until the 90's had a "loudness" button to mimic the human hearing.

author: SiliconWizard
author_rank: Super Contributor
timestamp: July 31, 2020, 03:05:25 pm
content: As said already - as a reasonable estimate for classic audio amps: max power is basically U^2/R, with U the "compliance" voltage (basically the supply voltage of the output stage minus some unavoidable drop-out voltage), and R the nominal impedance of the load. Of course the output stage also has a max power rating, meaning that you can estimate the max power at a given load if you increase the load's impedance, but if you decrease it, you won't be able to exceed the max rated power. Depending on the amplifier's design, it may just limit the power, blow a fuse or even burn if you use a load with too low an impedance.

author: ejeffrey
author_rank: Super Contributor
timestamp: July 31, 2020, 03:57:13 pm
content: Quote from: DW1961 on July 31, 2020, 06:54:29 amNo, it would typically be approximately constant voltage drop for the same topology and is limited by the Vbe of the transistors (or equivalent turn-on voltage of MOSFETs if used). With a Darlington output stage you are limited to about 1.2 volts away from each rail (2x Vbe) for a Sziklai output stage about 0.6 V. MOSFETs have considerably higher Vgs than a BJT Vbe, so unless used in high power (and high voltage) amplifiers where this dropout is negligible they would usually employ a small auxillary supply to allow the gate drive to go over the main supply rail. In this case the output could get to nearly the supply voltage. These factors are basically independent of supply voltage, so they become less significant as the supply voltage increases.

author: DW1961
author_rank: Frequent Contributor
timestamp: July 31, 2020, 05:15:20 pm
content: Quote from: dzseki on July 31, 2020, 11:11:08 amIt increases the top and bottom end when levels are low. I sure do remember those buttons! Always sounded better with it on. It was like a automatic equalizer boost at lower levels that would go away the higher you got it.

author: DW1961
author_rank: Frequent Contributor
timestamp: July 31, 2020, 05:21:56 pm
content: Quote from: ejeffrey on July 31, 2020, 03:57:13 pmSo are my calculations correct given a 16% decrease in RMS voltage due to the DC loss, and I could calculate power using the formula above, mainly getting the new RMS and subtracting 16.6%

author: gman76
author_rank: Contributor
timestamp: August 01, 2020, 07:09:00 pm
content: The loudspeaker typ has a spec (individual drivers in a loudspeaker certainly do) called ‘sensitivity’ that relates a given voltage to an SPL acoustic output. When an amplifier drives a loadspeaker, it is driving a voltage. Current and therefore power is simply a consequence of the voltage applied to an impedance. 8 ohm vs 4 ohm ratings are approximations. The impedance varies a lot with freq and the industry provides these as ‘standard’ values that are roughly the average over the audio freq range. Back to sensitivity - it’s standard practice to quote it as 2.83Vrms into 8 ohms which is 1 watt. Whether the speaker is 4 or 6 or 8 ohms, generally it doesn’t matter. They almost always quote sensitivity as 1W into 8ohms. When you apply 2.83V rms, you get some acoustic output that is measured at 1meter, and that’s the sensitivity in dB.So, you could call a speaker that delivers 90dB more efficient than one at 88dB since both are driven by the same voltage. But is a 90dB 4ohm speaker more or less efficient than the 88dB 8ohm speaker? This is a hypothetical. The 4ohm speaker draws more power but produces more output, so maybe it’s a toss up.The audiophile world generally isn’t concerned about efficiency or power consumed or power required. After all, Class A amplifiers or vacuum tube amps are fairly common in that space. Space heaters.But I have seen virtually the same drivers, 4 vs 8 ohm, that show better sensitivity for the 4 ohm model. More voice coil winding in the 8 ohm. The 8 ohm draws less current, lower power. Seems like a basic trade off. How much output current capability does the amplifier have? The flip side is that your amp needs more voltage headroom to deliver the same SPL (at high volume levels) for the 8 ohm model.

author: DW1961
author_rank: Frequent Contributor
timestamp: August 01, 2020, 08:19:47 pm
content: Quote from: gman76 on August 01, 2020, 07:09:00 pmThat's roughly how my head was processing it. However, My 8 Ohms speakers are 93 dB @ 2.83V/1M, so there's that aspect too. So you can both of world's and less power that way I guess.All fascinating, but back to my question above about calculating power output: for the above amp calculating for power using a 19V driver instead of 24.I'm still not clear on the voltage % drop due to DC to AC and back conversion. Otherwise, it's DC voltage calculated for RMS to AC, then then the % drop from conversion itself in the amp.

author: Siwastaja
author_rank: Super Contributor
timestamp: August 01, 2020, 09:12:05 pm
content: There is no specific %. The more accurately you want to understand the amplifier, the more details you need to fill in.Exact maximum voltage swing you get out of amplifier depends on the engineering details. For example, very low-resistance MOSFETs can go justmillivoltsaway from the rails, and assuming the rails are produced with regulated supply (say switch mode, for example) to a stable value, you get practically +/- 20V swing from +/-20V rails.On the other hand, if the rails are non-regulated and have the rectified mains ripple (100Hz or 120Hz) riding on the top, say, the rails ride between 18V and 20V at full load, then if you try to go any further than +/-18V, youperiodicallyrun out of voltage, butat other timesget higher; meaning, the mains ripple goes through to the speakers. So you need tolimitthe swing to the lowest voltage available to avoid this.Or, a certain power stage might use Darlington transistors with large saturation voltage drop, say over 1V. Then your 20V supply with 2V ripple only gets you up to 17V.Maybe the final stage is a voltage follower (voltage gain = 1), and even if the transistors could get closer to the rails, the previous stage cannot.This is just the tip of the iceberg. It's always case-by-case and requires understanding of the actual circuit used.

author: DW1961
author_rank: Frequent Contributor
timestamp: August 01, 2020, 11:05:46 pm
content: Quote from: Siwastaja on August 01, 2020, 09:12:05 pmI'm talking speicficallya bout teh amp chip I posted above:2 × 50 W Into a 4-Ω BTL Load at 21 V (TPA3116D2)2 × 30 W Into a 8-Ω BTL Load at 24 V (TPA3118D2)2 × 15 W Into a 8-Ω BTL Load at 15 V (TPA3130D2)No other variables. How would I calculate power for 19V?

author: retrolefty
author_rank: Super Contributor
timestamp: August 01, 2020, 11:42:19 pm
content: Just to bring another perspective to the topic of maximum audio output power, I thought I would share my experiences with vintage 1970s audio systems (a golden era) that I started collecting in the 1990s via thrift store finds and some ebay purchases. I would snap up anything near top of the line in Japanese stereo that I could find as people were dumping their gear for 5 channel stuff in the 90s. I would repair anything broken and either keep it if it was 'better' then I had and or sell it off on ebay. I went though maybe a dozen receivers and preamp/amp separates.I worked all the way up to a Kenwood SX-1980 working into some English 12" 3 way speakers. That receiver was rated at 270 RMS watts per channel back when the FTC regulations required much more rigorous and truthful ratings compared to the nonsense of peak/music power ratings many advertise today. I mostly like classic rock and like to feel the music as well as hear it, but I would get plenty of complaints from SWMBO any time the wattage scale on the receiver was peaking above just 2 watts no matter wherever in the house she might be!It is mostly a marketing thing, like never too thin, too light, never to much horsepower, never too much audio output power that sells regardless of any actual practical need. Unless one is building large PA systems or concert setups, one rarely needs more then say 20 wpc when driving normal speakers in normal homes. There are some speaker designs (seal enclosures) that have much lower SPL ratings that could require more power but they are not very popular these days.What are others thoughts and experiences on audio setups they have enjoyed? How much output wattage is too little or too much?

author: DW1961
author_rank: Frequent Contributor
timestamp: August 02, 2020, 05:11:54 am
content: Quote from: retrolefty on August 01, 2020, 11:42:19 pmYou should start a new thread for that. Would be fun.This was my 1980s receiver. Damn it would drive those 70 watts RMS Cerwin Vega HED 12" woofer speakers I had. Had the neighbor come down one Saturday night when we were partying to say he could hear my music 10 houses down, inside his bedroom.https://www.hifiengine.com/manual_library/yamaha/r-900.shtmlEven by today's standards, it was a nice looking piece.But that was nothing compared my friends brother in law and sister who bought a pair of BOSE 901s in 1972. OMG you could hear those things a mile a away. (Yes, literally)

author: Siwastaja
author_rank: Super Contributor
timestamp: August 02, 2020, 06:25:32 am
content: Quote from: DW1961 on August 01, 2020, 11:05:46 pmOpening up the datasheet, Figure 13 on page 10 shows the output power at any input voltage. Yes I know; no calculation involved, you were given the answer directly.Theycalculated, simulated or measured it, because no one else can know all the internal details of the chip that affect the exact value. Do still note this is just a single figure for a single part, there will always be unit-to-unit variation so even when this graphs says 22W at 19V at 1% THD to 8 ohm load, if you measure that on your system, it might be 21W or 23W. Also your external filter network components affect the power, for example, if you have more resistance than TI did in their test system, you are losing part of the power there.

author: magic
author_rank: Super Contributor
timestamp: August 02, 2020, 07:55:51 am
content: This is surely the best answer, but if I were to guess just from the marketing numbers alone, I would assume that saturation voltage (how close the output can get to the rails) depends only on peak load current. Then calculate saturation voltage and peak current for the three examples they give and use them as a very coarse guideline.

author: Siwastaja
author_rank: Super Contributor
timestamp: August 02, 2020, 10:14:03 am
content: Being class D with MOSFET output stage, and claimed 90% efficiency, the voltage over the MOSFETs is not much, and also does not vary much (relatively to the full output swing). If it would, the MOSFETs would dissipate significant amount of the total power, and cooling those tiny integrated parts would be difficult, so it's easier just to minimize the voltage drop by using large enough MOSFETs (with small enough Rds(on)).Some limitation to the analog output voltage swing is likely due to the modulation algorithm, i.e., they do not necessarily let the MOSFETs run at 100% or 0% duty.

author: DW1961
author_rank: Frequent Contributor
timestamp: August 02, 2020, 05:32:25 pm
content: Quote from: Siwastaja on August 02, 2020, 06:25:32 amDude! Thanks. I was close, but the chart looks like abut 21 watts at 19V. I was about 19.6. But, more importantly, there is no way to know without them telling you, and thanks for that.However, one thing: In the power chart, it looks like at 1% THD, the amp puts out actually 34 watts. I'm also assuming the chart is "per channel?"Also, another problem: The specs are not for the same amp chip! I just noticed that. They are all different chips, so what does that do for the table?2 × 50 W Into a 4-Ω BTL Load at 21 V (TPA3116D2)2 × 30 W Into a 8-Ω BTL Load at 24 V (TPA3118D2)2 × 15 W Into a 8-Ω BTL Load at 15 V (TPA3130D2)If the table is for any of the three chips, then the table tells the story.It's more like 34 (x2) watts at THD of 1% at 8 Ohms @ 24V. At 21V it's 25 watts x2. At 19V it's 21 watts x2. Am I reading it correctly?Here is the link to the PDF:https://www.ti.com/lit/ds/symlink/tpa3116d2.pdf?ts=1596389304658&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FTPA3116D2

author: emece67
author_rank: Frequent Contributor
timestamp: August 03, 2020, 11:04:15 am
content: .