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