How Much Current Capacity do we Need from our Amplifiers?

[Transistor]

The importance of amplifier power is often overstated on this forum—frequently by Amir himself. The claim is that massive headroom is needed to handle dynamic peaks. In reality, however, for a typical room of 20–25 square meters, 50 watts into 8 ohms is usually more than enough, and you won't hear any noticeable improvement in sound quality with a more powerful amplifier.

 

[EJ3]

No mystery here, it is unstabilized power supply. SMPS can behave differently.

I was letting them find out on their own, seeing that it is their problem and they are asserting that we do not know what we are talking about when they get our answer.

 

[Ropeburn]

The importance of amplifier power is often overstated on this forum—frequently by Amir himself. The claim is that massive headroom is needed to handle dynamic peaks. In reality, however, for a typical room of 20–25 square meters, 50 watts into 8 ohms is usually more than enough, and you won't hear any noticeable improvement in sound quality with a more powerful amplifier.

Likewise, it's all too easy to underestimate how quickly power demand escalates, due to its exponential nature. Both is true at the same time: the over- and the underestimation. Change just a few factors a bit and suddenly 2x100W into 4Ohm isn't overkill at all anymore but barely enough.

You also don't want to use most of your amp's max power all the time. It's bad for longevity generally. So why not leave amp-le headroom when it's very affordable these days?

 

[NTK]

The importance of amplifier power is often overstated on this forum—frequently by Amir himself. The claim is that massive headroom is needed to handle dynamic peaks. In reality, however, for a typical room of 20–25 square meters, 50 watts into 8 ohms is usually more than enough, and you won't hear any noticeable improvement in sound quality with a more powerful amplifier.

It is easy to underestimated the dynamic nature of naturally (and unnaturally) occuring sound. I borrowed the screenshot below from the "FREE SPLMeter app" thread (post #23).

The measured dBA SPL (the orange line and the number usually used for reporting SPL) is a time weighted average. For standard noise exposure levels, the time constant is 1 second (slow). The highest level the dBA SPL (orange) curve went up to was in mid to high 70's. The 30 minutes average was 58 dBA.

However, the peak C-weighted of the measurement was 106 dBC, which is ~30 dB over highest level the orange curve had reached. If you want your reproduced sound to be undistorted, the peak SPL is the SPL level your equipment needs to be capable of reproducing.

 

[Ropeburn]

It is easy to underestimated the dynamic nature of naturally (and unnaturally) occuring sound. I borrowed the screenshot below from the "FREE SPLMeter app" thread (post #23).

View attachment 540476

The measured dBA SPL (the orange line and the number usually used for reporting SPL) is a time weighted average. For standard noise exposure levels, the time constant is 1 second (slow). The highest level the dBA SPL (orange) curve went up to was in mid to high 70's. The 30 minutes average was 58 dBA.

However, the peak C-weighted of the measurement was 106 dBC, which is ~30 dB over highest level the orange curve had reached. If you want your reproduced sound to be undistorted, the peak SPL is the SPL level your equipment needs to be capable of reproducing.

And of course that difference of 30dB means power factor 2¹⁰=1024 compared to the dBA average. A good example of escalation.

 

[NTK]

And of course that difference of 30dB means power factor 2¹⁰=1024 compared to the dBA average. A good example of escalation.

A small correction -- it is 10³

 

[Ropeburn]

A small correction -- it is 10³

Wait a minute. 10x double is 1024.

 

[NTK]

Wait a minute. 10x double is 1024.

For power, from dB to ratio is: 10^(dB/10) = 10^(30/10) = 10^3.

 

[eliash]

You can find numerous posts on this and other audio fora touting the importance of high current output capacities of amplifiers. For example, in this post the poster wrote about the alleged ability of the NAD 2200 to output 60 A peaks. If you have a speaker with 2 Ω impedance, 60 A current means I²×R = 60²×2 = 7200 W! And the required voltage to drive 60 A across a 2 Ω load would be 120 V! Few speakers have 2 Ω or lower impedance, and per Ohm's law, the "peak power" and voltage that go along with the 60 A current increase proportionally with impedance. Therefore clearly these 60 A peak output current claims (if these claims were actually made) cannot have any real world relevance and are enitrely bogus. At best these claims of 60 A peak current are for short circuit output currents, or into loads that are almost shorts, conditions that are far from (I hope) anything close to real life. Competently designed loudspeakers aren't shorts. (OK, car subwoofers can get close but I am not talking about them. They are designed to work with 12 VDC electronics.)

But the question of how much current a normal speaker will draw remains and is a valid one, especially when the signal is not a simple sine wave. Let's see if we can answer that.

People have developed reasonably faithful loudspeaker electrical models, at least when they are operating in their (mostly) linear operating ranges which ideally we don't want to exceed. It should be straight forward to simulate the current draw for any arbitrary signal with these electrical models.

There are almost infinite variations of speakers and audio signals, and nobody will have the time to analyze them all. In this exercise I will use the combination of the Stereophile simulated speaker load and the AES75 music noise (aka M-noise). M-noise is supposed to be a good surrogate of the high dynamic range audio typically encountered in live sound situations.

The electrical model of the Stereophile simulated speaker load is shown here.
View attachment 539927

The simulated impedance of this circuit matches the Stereophile measurements of their actual implementation quite well.
View attachment 539929

As a second verification test of this circuit I simulated the current draw in response to a sine sweep (chirp). The amplitude envelop of the current draw closely followed the inverse of the impedance, as it was expected. The voltage amplitude to current amplitude ratio should be the same as the impedance magnitude, as Z = V/I. There was a small difference between the maximum voltage to current ratio of 3.69 to the minimum impedance of 3.72 Ω. The transfer function was converted into a discrete time version to make the simulations more efficient, but discrete time model is a non-exact approximation of the continuous time model. (See this MATLAB documentation page for more details.) Also, the "sampling rate" of slightly higher than 15x the chirp bandwidth was used for the discrete time model. This sampling rate should be able to capture the system dynamics with adequate accuracy. Increasing the sampling rate while keeping the total simulation duration the same can improve accuracy (higher sampling rate, same simulation duration = more time steps to compute). Note that this sampling rate is not the same as and unrelated to the Nyquist sampling frequency of the Shannon-Nyquist sampling theorem.
View attachment 539933

Here are the results from the M-noise simulation. The peak voltage to peak current ratio was 7.79, significantly higher than the 3.72 Ω minimum impedance. This means if you estimate the peak current draw as the peak signal voltage divided by the minimum impedance, you would have overestimated the peak current draw by nearly 2x.
View attachment 539977

Below are the histograms of the M-Noise signal voltage and current samples. The vast majority of samples have voltages <0.35 and currents <0.04, giving a ratio of ~8.8, which is close to the peak voltage to peak current ratio of 7.8.
View attachment 539976

In the case of the Stereophile simulated load, the minimum impedance occurs at ~4 kHz, a frequency range which the M-Noise (and most music/audio signals) does not contain a lot of energy. The peak voltage to peak current ratio depends heavily on the spectral energy distribution of the signal, and the speaker impedance to frequency characteristics. If a speaker has low impedance for a substantial portion of the low frequencies and the signal has a lot of energy in the low frequencies, this ratio could potentially be significantly lower and be closer to the minimum speaker impedance, meaning a higher demand for current from the amplifier.

What if we don't have the electrical models of our speakers? Theoretically we should be able to estimate a model when we have the frequency response (magnitude and phase). This is the topic of system identification. I haven't tried and have no experience with these tools yet. I leave you with a link to MATLAB's documentation page for "Estimating Models Using Frequency-Domain Data". May be some day I'll dip my toe into this topic.

So back to the question of how much current capacity do we need. The answer still depends on the signals and the speakers. But it is something that we can find out relatively easily and is no mystery. We just need access to some technical details and/or measurements of the gear. But it is not going to be some huge ungodly number, and it is not going to exceed the maximum anticipated amplifier output voltage (readily calculable from speaker sensitivity and listening SPL requirements) divide by the minimum speaker impedance, regardless of the phase angle.

ZIP file contains the Julia source code for the simulation in jmd (Julia markdown) format. The html file is the viewable output.

[Edit] Found a small error in my code that had a negligible effect on the results. I left out the resistor R5 (0.6 Ω) which meant it was 0 Ω in the previous calculations. Uploaded the updated ZIP file.
[Edit 2] In the off chance that someone may run my Julia code , I forgot to include the M-Noise WAV file. So here it is as a separate attachment.

...When the amp is connected in balanced mode, i.e. one stereo amp driving one speaker, the regular 50V peak output is doubled, as well as the current: 100V/2R=50A
When considering complex loads, as can be seen in several speaker measurements, the impedance may also go below 2R causing a higher reactive current...

Dimensioning a stereo amp for balanced mode is reflected e. g. in my T+A R-series power amp. The bipolar voltage follower output stage is doubled for more current, in contrast to their integrated amp, which has just a single output stage (otherwise similar amp design)...both are rated for the same stereo output power though...

 

[Ropeburn]

For power, from dB to ratio is: 10^(dB/10) = 10^(30/10) = 10^3.

So the doubling thing is a (fairly accurate) rule of thumb? Didn't know that lol

 

[Buckchester]

Seems like a great post but far too technical for me! It would be nice for this industry to have an easier way of figuring this out. Makes me wonder how many average people cheap out and buy something sub sufficient, and on the flip side how many audiophiles who think they have it all figured out spend far too much for what they actually need.

 

[CDMC]

Likewise, it's all too easy to underestimate how quickly power demand escalates, due to its exponential nature. Both is true at the same time: the over- and the underestimation. Change just a few factors a bit and suddenly 2x100W into 4Ohm isn't overkill at all anymore but barely enough.

You also don't want to use most of your amp's max power all the time. It's bad for longevity generally. So why not leave amp-le headroom when it's very affordable these days?

I think this is the real question. A Buckeye MCM502 stereo amp is $750 and gets you 350 watts/channel into 8 ohms. It will power 99% of speaker up to their limits. Why worry about 50 or 100 watts being sufficient when good clean power is inexpensive.

 

[Ropeburn]

I think this is the real question. A Buckeye MCM502 stereo amp is $750 and gets you 350 watts/channel into 8 ohms. It will power 99% of speaker up to their limits. Why worry about 50 or 100 watts being sufficient when good clean power is inexpensive.

Yeah seriously. You can easily spend over 1000 on those 50W and enough people do. Nevermind the 5W tube amp faction lol

 

[wwenze]

The importance of amplifier power is often overstated on this forum—frequently by Amir himself. The claim is that massive headroom is needed to handle dynamic peaks. In reality, however, for a typical room of 20–25 square meters, 50 watts into 8 ohms is usually more than enough, and you won't hear any noticeable improvement in sound quality with a more powerful amplifier.

The claim is that Amir listens at 110dB for 0dBFS. Or was it 120dB

 

[Keith_W]

Wait a minute. 10x double is 1024.

I haven't forgotten.

 

[andyc56]

Back in 2006, Bob Cordell did an interesting demo at RMAF using a custom instrument he designed that measures peak and average power output of a power amplifier when music is playing. A similar demo that he did in 2007 is described in an article on his site, under the heading "The Peak Power Demands of Well-recorded Music". He also did an article about the meter itself, showing a block diagram, but not a schematic. The instrument was calibrated in such a way that with a sine wave input, both peak and average power displayed the same value.

I was at the 2006 demo, which was a lot of fun. One track by Rickie Lee Jones showed average power on the playback system of 1-2 Watts, but reached 250 Watts on a snare drum hit.

I don't know what the conclusion is or should be, but it was interesting and fun to see the data on a real playback system.

 

[marin.weigel]

I've seen this mentioned all too often, but everyone needs to realize, that there's a price to pay when generously over-provisioning amp power, which is much lowered S/N values when usually listening at a comparatively tiny fraction of full scale output from pre/power amps.

Crippled by this shortcoming, there's these audiophiles that then sweat about their DACs specs, upsampling algorithms, modulators and what not - but it's all swamped by comparatively high noise, and I'm not even going to discuss actual audibility of such measures under optimum conditions.

The ticket to best sound is to get the system chain's gain structure optimized to wring the maximum transparency from it during one's most usual listening scenario.

 

[DamianW]

The 240A figure is based solely on the short-circuit current of the power supply capacitors; the power supply and capacitors themselves wouldn't even manage 10A.

The NAD 2200 uses two pairs of transistors per channel, each capable of handling up to 15A—that is presumably where the illusory 60A figure comes from (4 x 15A).

Nothing in this amplifier is designed for those stated maximum currents—not even the capacitors; that’s the ridiculous part.

NAD 2200 service manual specifies 45A limit and 50A nominal for 1msec, (I think they had this backwards ie a typo) manual attached see page 2 specification 16.

The output devices are 2 x 2SC3281 datasheet attached, they are rated at per device 15A continuous or 20-25A pulsed for for 1msec at the supply rail voltage.
Thus the pair of output devices could provided 50A for 1msec if they shared the current equally.

 

[john61ct]

A Buckeye MCM502 stereo amp is $750 and gets you 350 watts/channel into 8 ohms. It will power 99% of speaker up to their limits. Why worry about 50 or 100 watts being sufficient when good clean power is inexpensive.

For some people $750 is a huge investment for just two channels. Better to be able to afford better speakers if a 100W power amp is enough

 

[john61ct]

Back in 2006, Bob Cordell did an interesting demo at RMAF using a custom instrument he designed that measures peak and average power output of a power amplifier when music is playing.

Isn't it now possible to do this with OTS gear?