Musings on Headphone Amplifier Output Impedance
Ed Note: Dr. Jan Meier, of Meier Audio, recently submitted this article for me to consider publishing. He and I have had many a discussion, and quite a few disagreements, about the nature and effects of amplifier output impedance over the years. While I still have some concerns with his view brought to light here (I'll intersperse some of my comments in italics through the course of the article), I have moved away from the simple view that lower output impedance is always better. The following article does a lot to show a bit more sophisticated view of factors involved, and I thought it would be useful for InnerFidelity readers.
I also need to note, for ethical purposes, that no compensations have exchanged hands, the article was freely submitted with no promise of compensation or publication. The publication of this article should in no way be taken as an endorsement for Meier Audio Products. (Though, I do have his Corda Rock here, that Skylab reviewed and placed on InnerFidelity's "Wall of Fame", and find it a dandy amp for the price.) I am publishing it because I found it interesting and stimulating to my desire for more knowledge on the topic.
And now, Dr. Meier....
Greetings InnerFidelity readers!
Occasionally I meet people who tell me that a lot of what I do is totally wrong. In my amplifiers I use the wrong capacitors, crossfeed is a hoax, and it is impossible to built a decent amplifier with transistors and without tubes. When I ask them whether they have ever heard one of my amplifiers or ever designed/made an amplifier of their own, the answer is "No". But they have read in this or that article that capacitors of brand X are much better than the ones that I use; that crossfeed only hampers the purity of the signal; and that transistors only create tons of TIM-distortion (whatever that might be). It is written down, it must be true, right!?
But the written word is not all-knowing and, in my experience, sometimes can be quite wrong. Authors are just human after all.
One of these statements repeatedly made in articles on amplifiers and speaker-control is that a low output impedance (high damping factor) should result in the best control over the driver action since energy stored in the driver system should be removed as quickly as possible. However, in those articles a real explanation is never given why a lower output impedance indeed results in faster energy decay. To be honest, thinking about it this statement never made real sense to me. Dumping current into a zero Ohm resistance doesn't dissipate any extra energy at all. So what's really going on?
The network in the figure below presents a (very) simplified model of a driver system that contains three elements only:
- A conductor L that is the equivalent of the inductivity of the driver's coil.
- A resistor R that is of the sum of the coil's ohmic resistance (Rl) and the output impedance of the amplifier (Ro).
- A capacitor C that represents the mechanical elements of the driver that store energy (like the suspension and compressed air).
Ed Note: I'd like to not that this simplified circuit is very far away from what a modled headphone might look like. However, the concept being explored here is about the operating principles of reactive loads, and for purposes of conveying the concept being discussed, I find it perfectly adequate.
The output voltage Vo should follow the applied amplifier voltage Vi as close as possible. The mechanical energy stored in the system should stabilize as quickly as possible after the input signal reaches a fixed level.
The lumped network in the figure is a so-called linear system. Any input-signal can be considered as the summation of a long series of successive impulses. Any output-signal can be considered as a summation of the correlated impulse responses. The response to a single impulse is a good indication how closely the system can follow the input signal.
I will not bother you with a detailed analysis of the network and how we can calculate its response to an input impulse. I will merely present you the final results. Basically there are four different situations to be distinguished:
- The first situation is when the resistor has zero resistance. The network then contains no elements that dissipate energy. The output signal continuously oscillates at a fixed frequency. This is a virtual system as it is not possible to produce a driver/coil that has zero impedance and (normally) amplifiers do have an output impedance that is higher than zero Ohm.
- For small values of R during each cycle of the oscillation a little bit of energy is dissipated by the current through the resistor (producing heat). The signal-amplitude of the cycles thereby continously decreases. This is called underdamped oscillation. Amplitudes decay with a time-constant 2L/R. The red line in the figure below shows the impulse-response for R = 33 Ohm, L = 4 mH, C = 40 nF.
- When the value of R is increased to a value where in the formula above w=0 the signal no longer oscillates. After an initial bump the signal continously decays and goes to zero relatively quickly. This situation is called critical damping and is shown by the green line. R = 200 Ohm.
- If R is increased even further the signal decay becomes more slowly. The large resistor value doesn't allow for a fast (dis)charging of the capacitor. We now have overdamping. The blue line shows the situation for R = 600 Ohm.
It is easily recognized that critical damping provides the fastest and most stable response to an impulse signal. R should be closely matched to the values of L and C. However, be aware that its value is the sum of the output impedance of the amplifier and the ohmic resistance of the coil. If the latter is (too) small then the system may profit very well from a higher output impedance.
(Note for the mathematicians: the three impulse-responses shown in the figure are scaled such that they will result in the same over-all sound pressure. That is, the surface-integrals of all three curves have the same value!)
Have you ever studied the various headphone measurement diagrams as measured by Tyll? Many headphones show very distinct oscillations when fed with short impulses. However, these oscillations do not necessarily have to be caused by inadequate electrical impedance matching. They could well be caused by factors like driver break-up or cavity-resonances that are not necessarily shown in the electric parameters/characteristics of the headphone. The presence or ringing doesn't necessarily imply inadequate electrical damping. For this we have to investigate any ringing in the electrical current as delivered by the amp.
Ed Note: This is a very important paragraph. What we're talking bout here is the electrical signal going into the headphones. Many headphones do show ringing in various ways, but I do think most of that is due to acoustic issues and not electrical issues. The electrical impedance matching issues talked about here, in my opinion, will have relatively small effects on headphone listening fidelity relative to the acoustic design of the headphone. None the less, based on my past experience with headphones, I would think these electrical impedance matching issues may be audible enough to warrant investigation.
The next picture shows the current responses of the LCR-circuitry discussed above for underdamping, critical damping and overdamping.
To measure the current response of real headphones an amplifier was build that allows to monitor the current flow while applying a signal to the headphone. The effective output impedance of this amp can be varied in discrete steps, negative output impedances included, and ranges from -81 ohm to 208 Ohm.
Ed Note: As odd as it might seem, there is such a thing as a negative output impedance. See half way down this page.
Axel Grell, the chief-designer at Sennheiser, once told me that their top-of-the-range headphones are always designed to sound best at a low output impedance. The figure below shows the current flow of a HD-800 with the amplifier having an output impedance of 0 Ohm. There is no oscillation. Just a nice bump after the initial pulse-current, as could be expected from a critically damped system. Increasing the output impedance to 120 Ohm decreases the bump, indicating a slower impulse response, as expected from the theoretical example. This is as good as it can get. True textbook delivery. People at Sennheiser know their stuff!
The HD-800 has a relatively high coil-resistance of 300 Ohm. Thus there is enough resistance available for proper damping. However, over the years there has been a clear tendency towards headphones with much lower impedance values. The reason is simple. They go louder with lower signal voltagesthe output voltages of modern portable players is very limited...and for many the louder the better.
The next figure shows the current response of the AKG K420. This headphone has an impedance of only 32 Ohms and a very high efficiency. At zero output impedance clear oscillations can be distinguished at a frequency of around 5 kHz. This doesn't look nice, does it?
And the situation gets worse at a negative output impedance of -18 Ohm. Brrr. Not pretty at all, right?
On the other hand, when increasing the output impedance the oscillations get smaller and smaller and are completely gone at +88 Ohm.
It is clear that an increased output impedance does improve the electrical response of the K420 considerably. Whether it also shows in the acoustic impulse response unfortunately I can't tell. Maybe Tyll is able to do some measurements.
Ed Note: Unfortunately I don't have an AKG K240 to try. I did, however, take a quick stab at measuring output current to headphones to see if I could replicate Jan's findings and was unable to do so. I'm not sure my gear has the appropriate differential inputs to look at the current signal through a resistor. Additionally, I did a quick and dirty look at a couple of mediocre performing headphones to see if I could observe a change in square wave response as I varied output impedance. Just a gross test looking visually at square waves, and didn't observe any changes. That's not to say they weren't there, just that my method wasn't very sensitive. The only thing I can say for sure after doing my experiment is that if electrical ringing is occurring, it's not of a magnitude similar to that of acoustic resonant problems seen in my measurements. I'll take another stab at it one of these days though, because this a pretty interesting subject, in my mind, and Jan's article has gotten me quite curious.
(Note: the three figures above where measured at similar sound levels. The amplitude of the input pulses Vi is increased with increasing output impedance Ro of the amplifier to compensate for changes in sound level created by this impedance.)
The membranes of magneplanar drivers have a large surface and very low mass. Therefore they are already highly dampened by air. Proper electrical damping is of less importance. This figure shows the current for the Hifiman HE-500 driven at 0 Ohm output impedance.
Changing the effective output impedance only had very little effect on the current. It is the reason that current-amplifiers (instead of voltage amplifiers), like the Bakoon, which by nature have an extremely high effective output impedance, do work well with magneplanar headphones.
From the examples above it becomes clear that the best output impedance is not necessarily a low one. Increasing the output impedance may well reduce any oscillatory behaviour of the driver. Sure, it can slow down the response of the driver but sometimes that's a good thing. If you feel your headphone is a little bit 'hot' then increasing the output impedance using an adapter between headphone and amp (or by soldering resistors into the headphone or the headphone-plug) may well be a solution. I've had various customers who reported very good results with a large variety of headphones. Trust your own ears and experiment with various resistor values. Simply use the one that you personally like most.
With portable players the use of resistors/adapters can have an additional positive effect. Internally these players often have coupling capacitors at the headphone output that prevent offset voltages reaching the headphone. However, in combination with the impedance of the headphone these capacitors make a high-pass filter. Low frequency signals do not reach your headphones at all.
With a capacitor value of 47 uF and a headphone resistance of 30 Ohm the cut-off frequency is a whopping 113 Hz! Adding 47 Ohm to the headphone will reduce the cut-off frequency to 44 Hz. This is as good as a bass-boost. The only drawback is, that you now have to turn-up volume control a little bit further (which actually can be benificial of its own if it is a digital control as this will increase the effective resolution of the digital signal).
Many headphone-addicts prefer tube amplifiers over solid-state equipment. These preferences may well be (partly) explained by the fact that by nature tube amplifiers do have higher output impedances than transistor amps. With many headphones this slightly slows down responses but also suppresses oscillatory behaviour. These people may well try adapters when using transistor-gear.
Ed Note: While I agree that the suggestions in Jan's summary have significant merit, I think there are mitigating circumstances that may argue against that approach. The most important one to my mind is the case of headphones that have widely varying impedance with changes in frequency. Let me explain...
First, let's have a look at the impedance response of a Sennheiser HD800.
In the above plot, you can see that the Sennheiser HD800 has a peak impedance of about 650 Ohms at 100Hz, and a low impedance of about 350 Ohms at 3kHz. I'm going to try to keep the math easy here for illustrative purposes, so let's say it's 600 Ohms at 100Hz and 300 Ohms at 3kHz.
Now, you're going to need to know about how a voltage divider works. here's an illustration.
Here's a series of very simple circuits. The left side is the headphone amp and the resistor models the output impedance of the amp, and on the right is the headphone with the resistor modeling the headphone impedance. To the far left of each circuit is a signal generator putting out a 10Vrms signal. The 10Vrms signal has to go through both resistors to get to ground on the other side.
In case "A", there is a total of 10 Ohms resistance between the two resistors. With a 10Vrms signal across 10 Ohms it will loose 1Vrms for every Ohm. So it will loose 1 Vrms across the 1 Ohm resistor and 9Vrms across the 9 Ohm resistor. so there will be 9Vrms between the two resistors. This is called a voltage divider.
In case "B", the value of both resistors is the same at 9 Ohms. Because they're the same each will loose half the total voltage, so there will be 5Vrms measured between them. Now if each resistor was 18 Ohms, or 100 Ohms, you would still get 5 Vrms between the two resistors because it's the ratio or resistances that count, and not the absolute value.
Case "C" is to model a typical low output impedance, solid-state amp with 1 Ohm output impedance driving an HD800 at 3kHz and 10Vrms. (10Vrms is way too high, but it's a convenient number so let's just use it for now.) Because the output impedance is so low, almost all the voltage is dropped across the headphones. This is good because it means all the drive from the amp is going to the headphones and little energy is being lost as heat across the output impedance of the amp. (The power lost as current goes through a resistor is done so in the form of heat.)
Case "D" would be typical of an output transformerless (OTL) tube amp that have output impedances in the range of 100 Ohms. Here, about 1/4 of the voltage is dropped within the amp across the output impedance, and only 7.5Vrms is left to drive the headphones.
Case "E" is the same amp, but this time is putting out 10Vrms at 100Hz. At 100Hz the impedance of the HD800 is 600 Ohms, and now there is 8.66Vrms across the headphone coil. In other words, because of the high output impedance of the amplifier and the changing impedance of the headphones, they get more drive voltage when their impedance is higher...without any change to the volume control.
Case "F" and "G" show the same headphone impedances, but this time with a 200 Ohm output impedance. The important thing to note here is that the higher the output impedance of the amp, the more effect changes in headphone impedance with frequency will have.
Again, look at the impedance plot of the HD800. With a high output impedance amp, the headphones will get a little more gain as the impedance goes up. Therefore, you can think of the impedance curve as an EQ curve with high impedance amplifiers. The higher the amp output impedance, the more the headphones will be EQ'd toward the shape of the impedance curve. In the case of the HD800, the higher the output impedance of the amp, the more you'll get a mid-bass boost centered at 100Hz.
To prove that I measured an HD800 with 0, 100, and 200 Ohm output impedances. You can see in the plot below that as the headphone amp output impedance changes, you get an ever increasingly large bump in the bass that's related to its impedance curve.
Fortunately, the HD800 has a very high impedance to start with, but, as Jan mentioned, many of todays portable headphones have very low impedances, and that means that they'll be even more sensitive to amplifier output impedance. For example, here's the Sennheiser PX100 impedance curve.
It's roughly 10% of the impedance of the HD800, and therefore you're likely to see changes in the frequency response curve like the HD800 with the differences in output impedance of only 0,10, and 20 Ohms... and that's near the normal variation of portable devices. Fortunately the PX100 could probably use a little more bass.
But it does get worse, here's a somewhat typical multi-balanced armature IEM. You can see that because of the crossovers and multiple drivers of varying impedance, the overall impedance curve (violet) swings wildly any where from 19 Ohms to almost 90 Ohms. Put a 30 Ohm resistor in series with that bad boy and your going to have a pretty severe peak in response at 1.8kHz.
What I'm trying to get across here is that simply increasing the output impedance of your amplifier by putting series resistance in line with the headphones (which is effectively the same thing) has many effects and those effects will vary based on a variety of issues. It's a fairly safe thing to do when the impedance response of your headphones is very flat, but if your headphones have large swings in impedance (usually multi-balanced armature and open dynamic headphones) you need to be careful.
Jan's idea is quite stimulating to me. (Thanks so much for writing this, Jan.) I can see someone building a headphone amp that has a built in analyzer that is able to optimise electrical impulse response to the headphonesnot only by varying output impedance but possibly also adjusting the reactance (inductance and capacitance) of the output. In fact, you might be able to vary the output impedance for EQ and then adjust reactance to critically damp.
My current guess, however, is that the built-in acoustic damping of the headphones is dramatically more responsible for headphone's acoustic impulse response than is the damping of the electrical drive signal. I'd love to investigate this further, but until then I'm betting the acoustics are far more important.
None the less, you are now armed with a little more knowledge. I'd suggest a little box with an in and out jack, and a variable resistor mounted in it that you can adjust between zero and about 200 Ohms in series with the signal. And then play around a bit. It is a hobby after all, nothing wrong with a little playing around.