Audiophile help please - Ohms and power

[dzseki]

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

[DW1961]

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.

OK 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 watts

36 -30 = 6 watts lost.
16.6% loss? Is that linear for each voltage?

[DW1961]

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 ohms
24V gets you 60 watts total
19 volts gets you 37 watts total

What about a difference is bass power needs? Would the extra power help?

[dzseki]

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.

[SiliconWizard]

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.

[ejeffrey]

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.

OK 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 watts

36 -30 = 6 watts lost.
16.6% loss? Is that linear for each voltage?

No, 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.

[DW1961]

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.

It 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.

[DW1961]

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.

OK 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 watts

36 -30 = 6 watts lost.
16.6% loss? Is that linear for each voltage?

No, 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.

So 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%

[gman76]

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.

[DW1961]

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.

That'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.

[Siwastaja]

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 just millivolts away 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, you periodically run out of voltage, but at other times get higher; meaning, the mains ripple goes through to the speakers. So you need to limit the 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.

[DW1961]

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 just millivolts away 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, you periodically run out of voltage, but at other times get higher; meaning, the mains ripple goes through to the speakers. So you need to limit the 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.

I'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?

[retrolefty]

 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?


 

[DW1961]

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?

You 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.shtml

Even 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)

[Siwastaja]

I'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?

Opening 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.

They calculated, 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.

[magic]

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.

[Siwastaja]

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.

[DW1961]

I'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) Diff chip
    2 × 15 W Into a 8-Ω BTL Load at 15 V (TPA3130D2) Diff chip

No other variables. How would I calculate power for 19V?

Opening 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.

They calculated, 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.

Dude! 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

[emece67]

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