How Much Current Capacity do we Need from our Amplifiers?

[MaxwellsEq]

I've always been math tarded , how do I know which of the 3 pies in a color I should pick?

Solving for Power is top left, Voltage top right, Current bottom left and Resistance bottom right.

So let's focus on one quadrant as you requested, I'll pick the Voltage top right quadrant. The first segment says "I x R" so if I know the current and I know the resistance I can multiply them to get Voltage. The next segment says that if I know the Current and Power, I can divide Power by Current to get Voltage. The final segment is if I know Power and Resistance, I can multiply them and then take the square root of the result to get the Voltage.

So each quadrant has 3 different ways to get an answer (e.g. Voltage for top right quadrant) depending on what you have information about.

 

[mhardy6647]

PV = nRT

Can't argue with that!

 

[mhardy6647]

Where can I get one of those manly amps?

Manley Laboratories Inc | Professional Tube Audio Equipment

Manley Laboratories is an American manufacturer of professional tube microphones, preamps, equalizers and compressors. Trusted by studios for VOXBOX and Variable Mu.

www.manley.com

oops, wrong Manley.

 

[mhardy6647]

Yeah well I don't think you're going to compete with the peak SPL of a tank cannon... even a pistol is louder than any speakers

Challenge. Accepted.

 

[solderdude]

I want to identify "the point" just before the speakers start to go non-linear, long before I would trust my ears to identify distortion.

That is difficult to do as this is frequency dependent and is a form of 'compression' a.k.a. soft clipping.
And even if you have that it takes a dynamic compressor that has to be tuned to properties of the drivers.

Can that be done with

> normal equipment + say REW ?

There is a substantial difference between continuous power handling and music power handling.
To gauge distortion at a point where it isn't clearly audible yet and to 'map' the distortion profile at various frequencies is not easy nor quickly and reliably done using REW and a UMIK for instance.
What would you consider a distortion level that you would find problematic yet not yet audible ?
How to determine what that point is for various frequency bands.
Short peaks for instance can have a lot more power than continued power at a certain level.
It would be silly to 'limit' the power of short duration peaks that cannot destroy the driver based on how much continuous power the driver can handle.
How do you plan to determine those values for short and longer period power levels in various bands ?
The 'exact point' differs under different circumstances... this is the problem.

comparing the GFA-555 to small DC amps.

The amp does not matter as distortion of them will be dwarfed by that of the speaker.

comparing with HPF and without, with DSP and without, etc

Should that point be according to acoustic SPL? Or measuring Watts (and Amperes?) at the binding posts?

When you measure at the binding posts you can see the input power of the speaker.
When you measure the sound of the speaker (at a certain distance and angle in certain conditions) you can see how the speaker reacts to the signal.

Obviously a line level protection mechanism needs to be separately dialed in, once the amp is selected, but that is later on down the road.

Yes, that would be the difficult part and it would

> Don't throw parties with these speakers. For outdoor/parties use outdoor/party speakers.

Again, no, these ARE the speakers I will be using, they are the one GIVEN in my system (& various iterations, there is no space for bigger ones) and I don't want less performant ones wrt SQ - critical listening will be 90+% of their usage.

The point of this topic fo me is, how to protect them, and I accept that that protection will limit how loud they go when used outdoors and for occasional dance / parties.

You can protect them by determining the max continuous power rating of the drivers in the speakers and make an adaptive compressor limiter that takes the continuous and peak levels into consideration.
This circuit should always be active.

 

[rjp]

Pro tip: Ohm's Law also works for impedance (Z)! Just bear in mind that Z is generally frequency-dependent.

Yep, not only frequency dependent, but also complex.
Here you can see how inductance (L) and capacitance (C) contribute to the imaginary component (which changes the phase angle).
The real part is always just the resistance R.

Z(f) = R + j(wL - 1/wC)

where w = 2*pi*f

 

[levimax]

Basically, it is not capacity to handle current, but capacity to deliver enough voltage to drive high current.

There coul be a few reasons for design, that double power:
Amp is using stabilized power supply. IMHO the only reasonable solution is to use SMPS supply, which provide stabilisation without losing efficiency.

Amp is overdesigned for testing. While it can be well engineered and powerful, it will be expensive.

Manufacturer lowers specification for 8 Ohm.

The OP's focus on "amps" (electrical amps not power amplifiers) is misleading as it is about maintaining voltage into a load at the end of the day. It is an interesting question though about the audibility of an amp that has more "headroom" compared to another amp with the same continuous output but less peak output. Amir has shown over and over that most modern amps with regulated power supplies have very little if any headroom for peaks as would be expected. In exchange you get lower distortion and crosstalk and other advantages.

Old fashioned unregulated linear power supplies on the other hand, with their transformers and capacitor banks do provide higher voltage for a brief amount of time before the voltage "sags" down to what can be provided continuously. This brief peak voltage is what is measured by Amir's "headroom" measurement. When making DIY class AB amps with linear supplies I would use a transformer with a higher voltage but a VA rating that would not take out the output transistors and then add extra capacitors to the filter bank. My theory was I could get more "peak headroom" without having to use more or bigger output transistors or larger heat sinks. It "worked" to an extent on the test bench but I was never able to confirm any audible advantages but it made me feel good.

To me the OP is really talking about the behavior of regulated power supplies vs unregulated power supplies for amps with the same continuous power rating. While intuition would say having "more peak power" (more short term amps in the view of the OP) is better, that is a gross oversimplification. There are trade offs and complications and any audibility effects are speculative at best, but it is an interesting question.

 

[Sokel]

For complex distortion patterns with REW, one can test with FSAF for fast and dirty in a room and with complex signals or music.
The bare minimum is UmiK 2 though, the better the gear (specially mic+interface) the most reliable the results.

I wouldn't try it on a non-solid rock system though, there's too much at stake and the charts won't look pretty as it treats the whole system (room+gear+etc) as a black box, so complex gain-staging for example make things vary a lot.

 

[Imakeller]

Challenge. Accepted.
View attachment 541121

I think those come stock in Mister Softee trucks!

 

[MRC01]

Usually in 2 ways. You can test for compression. For example if I increase the output by 3 dB, does the SPL increase by 3 dB? The second method is harmonic distortions.

Those are usually the criteria to tell when you speaker has gone "non-linear", and what its usable max SPL is.

These are related (I know you know this, and many readers do, but perhaps not everyone does). For example, a compressed sin save is flattened, and adding odd order harmonics is one way to flatten it.

For example:

The big dotted wave is a pure sin wave. The small dotted wave is the 3rd harmonic of the big wave (10% amplitude). The solid blue line is the sum of those two waves. Note that it is a compressed/flattened version of the original. Put differently, the solid blue wave is what you get when the big dotted wave has 10% of 3rd harmonic distortion.

 

[EJ3]

must be nice

Considering that I did not start my journey of attempting to aquire the ability to pay for everything until I married at age 48 and realized that my $76K in credit card debt was an issue (and that yes, I needed to pay my parents for the homes because that was how they could afford their elderly care, when it came [for my father] and comes for my mother).
So, 22 years later, (with my wife's help), I have paid (and am paying my mother) to the point where I have ZERO debt (but only a pittance in the bank). But 2 cars and everything else is paid for already. Only, utilities, insurances, vehicle operating/maintenance and taxes are expenses.
Yep, as you said. it IS nice.
But it was NOT easy to make happen. It took more than 20 years of very diligent working on it (and after the first 6 years, my wife jumping in with both feet, to help).
Now we are working on what's IN the bank for an emergency fund for ourselves.
I'm not sure: but it seems, that even though one is momentarily comfortable, one must keep at it, just perhaps at a somewhat lesser than "FULL TILT BOOGIE" level.

 

[EJ3]

Petition to krautfund him a fat outdoor PA with wall of bass and line arrays. Plus power infrastructure like nice solar system and all. Perhaps 48V so there need not be AC inverter...

"Wall of Bass" If that might affect my brain like the "Wall of Sound" seemingly affected Phil Spector's brain, I'll politely decline.

 

[somebodyelse]

Challenge. Accepted.
View attachment 541121

NASA's Reverberant Acoustic Test Facility at Plum Brook, good for 163dB

 

[NTK]

For rocket payloads, they usually need to go through a direct-field acoustic test.

Direct-field acoustic testing - Wikipedia

en.wikipedia.org

https://flic.kr/p/EFiYY8

 

[M.C. Logan]

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.

All good for dynamic speakers, not quite so for electrostatics which very much do swing down to, sometimes 1, commonly 2 ohms and present a capacitive load. Current is definitely required in spades if you want low distortion and realistic volume.

 

[somebodyelse]

All good for dynamic speakers, not quite so for electrostatics which very much do swing down to, sometimes 1, commonly 2 ohms and present a capacitive load. Current is definitely required in spades if you want low distortion and realistic volume.

They only tend to swing that low at high frequencies don't they? For the ESL57 it only drops below 4R above ~10kHz, by which time there's less demand for power in music, so the actual current demand still won't be outrageous. The bigger question is whether the amp can deliver that current into a capacitive load.

 

[EJ3]

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.

Each of my pair of custom, homebuilt subs uses a 12" Pioneer woofer with dual 4 Ohm voice coils. Since I am running stereo bass, that means that I tied the dual 4 Ohm voice coils together so that I have a 2 Ohm load for each woofer (or I can wire them so that I have a 4 Ohm load [which is the way that I use them]) The woofer's free air FR is 20Hz-80Hz). So, I either a bridged mono NAD 2200 or a bridged mono PROTON D1200 on each one.
As to the current being used, I never tried to measure it. All I know is that either amp seems to work just fine in this setup.
And I need to eventually build 2 more subs and get two more bridgeable PROTON D1200's.

 

[markk]

Whether or not a DVC coil in a sub, wired in parallel, will need spectacularly high current really, really depends on the woofer, enclosure volume and resultant in box impedance curve.

Though old, Sigfreid Linkwitz has a really nice page including an excel spreadsheet that will calculate the voltage and current for any different SPL for any woofer in a closed box.. But, the gist of it is that, unless you have a 2 ohm load into a very, very low compliance enclosure, you will likely hit Xmax well before you run out of amplifier voltage or current. It does not predict the full range output of a ported system but you can see that even in the case of a straightforward closed box single driver, the issue of how much current (and voltage) you need is not easily summarized.

Anyway, if you know the TS parameters of your woofer and your enclosure volume, you can run the numbers for yourself. See the following link.


Linkwitz Lab excel spreadsheet page

 

[EJ3]

Whether or not a DVC coil in a sub, wired in parallel, will need spectacularly high current really, really depends on the woofer, enclosure volume and resultant in box impedance curve.

Though old, Sigfreid Linkwitz has a really nice page including an excel spreadsheet that will calculate the voltage and current for any different SPL for any woofer in a closed box.. But, the gist of it is that, unless you have a 2 ohm load into a very, very low compliance enclosure, you will likely hit Xmax well before you run out of amplifier voltage or current. It does not predict the full range output of a ported system but you can see that even in the case of a straightforward closed box single driver, the issue of how much current (and voltage) you need is not easily summarized.

Anyway, if you know the TS parameters of your woofer and your enclosure volume, you can run the numbers for yourself. See the following link.


Linkwitz Lab excel spreadsheet page

I am glad to see folks still using Siegfried Linkwitz's information.
My system is ported (and the current cab ports are tuned to 29Hz for each of the pair [maybe: one day I will build cabs that are tuned to 20 Hz, maybe]). And the FR of the subs high pass is 55 HZ and Low Pass is 70 Hz.
Many times have the volume up around 112 Db, as I listen with the windows open while working (outside) around the house (obviously it is lower when I am working inside of the house).
My amps (in bridged mono mode) can deliver between 1600-1800 watts RMS at 2 ohms, depending on which amp I am using
And, for those concerned about my neighbors, people cannot hear my stereo unless they are actually within my property. My wife tells me that she doesn't hear it until she is about 6 meters inside of the property line.
NAD 2200 AMP Dyno Test:

PROTON D1200 AMP Dyno Test:

 

[M.C. Logan]

They only tend to swing that low at high frequencies don't they? For the ESL57 it only drops below 4R above ~10kHz, by which time there's less demand for power in music, so the actual current demand still won't be outrageous. The bigger question is whether the amp can deliver that current into a capacitive load.

For sure, Martin Logans tend to swing low quite a bit earlier than Quads and go lower. But yeah, the big thing is current delivery. I use a pair of biamped Quad 606s into my MLs, which are pretty unburstable, although about to go off to Huntingdon for a service and new trafos.