>Subject line says it all. I'm curious what the term damping factor means >in relation to amplifiers. Is more better? What is an ideal >specification etc... In its simplest, most honest form, damping factor is NOTHING more than the ratio of the rate load impedance (usually 8 ohms) to the amplifiers output impedance. That means if the amplifier has an output impedance under the measuring conditions (which are almost NEVER specified, by the way) of 0.1 ohms, the amplifier has a damping factor of 8/0.1 or 80. Beyond that, damping factor is one of those numbers that has taken on undeserved and mythical proportions as an indicator of the quality of an amplifier. People claim (as do manufacturers of amplifiers and speaker cables) that low damping factors "destroy the tightness of loudspeakers, turning well controlled bass into mush." The reality is far, far from that. In fact, it can be shown that huge variations in damping factors themselves have absolutely miniscule effects on the control of the loudspeaker, effects which are FAR smaller than manufacturing variations in the components in the loudspeaker, variations in response due to atmospheric changes and so forth. I've done this analysis SO many times, I can do it in my sleep, but here goes again. The whole concept of driver control rests on "losses." These are elements withing a speaker that dissipate energy. Controlling the motion of the cone requires that we get rid of the energy stored either in the motion of the cone or in the stretched suspension The place where the energy is stored alternates between the two, just like in any other resonant system). There are three mechanisms for dissipating this energy: acoustical (that's the radiated sound: without an acoustical loss, the speaker would produce NO sound whatsoever), mechanical (in the suspension, energy is turned to heat, or in the absorptive linings of the cabinet), and electrical (everything else). The relative sizes of these loss mechanisms are crucial to understanding the whole damping factor debate. By far the smallest of the loss mechanisms is the acoustical one. In direct radiator loudspeakers (sealed box, acoustic suspension, vented box, bass reflex, passive radiator, etc.) the acoustical losses are by far the smallest and the least significant. This is why such speakers are so inefficient. The total acoustical losses are on the order of 0.1% to 1% of the total losses. Higher in magnitude are the mechanical losses, they typically comprise about 10% of the total losses in the system. But exceeding the combined acoustical and mechanical losses are the electrical losses. The electrical losses completely dominate the losses that contribute to the control of the drivers. The electrical losses typically account for 75%-90% of the loss mechanisms in a speaker system. Thus, we can concentrate our focus on the electrical losses. The damping factor proponents start throwing things like cables and plugs and even the inductors into the equation. Unfortunately, they ALL ignore the largest single resistance in the whole chain: the one the utterly swamps ALL other resistances combined: the DC resistance of the voice coil. Assume (which is reasonable) an amplifier with a rated 8 ohm damping factor of, say, 1000. Add .25 ohms for speaker cables, .25 ohms for inductors, .25 ohms for plugs (pretty awful, but what the hey), for a grand total series resistance of 0.008 (amp) + 0.25 (cables) + 0.25 (coil) = 0.758. Now, it's argued, the damping factor has been spoiled to a whopping figure of 10.6. That's awful, right? Well, it's a factor 0f 100 poorer than the spec of 1000, for sure, but neither the damping factor NUMBER of 1000 nor of 10 HAS ANY SIGNIFICANCE IN DETERMINING THE BEHAVIOR OF THE LOUDSPEAKER ITSELF. The amount of ACTUAL damping of the driver is determined by a DRIVER figure called the total system Q, or Qtc. This is the ratio of the amount of energy stored in the resonance to the amount of energy dissipated, and is a DIRECT determination of how well the speaker is damped. It is made up of two major parts, the system Q due to mechanical losses Qmc and the system Q due to all electrical losses Qec'. The total Q is related to these by the following relation: Qmc * Qec' Qtc = ----------- Qmc + Qec' Now, the part the "damping factor" or ALL series resistance play is in changing the electrical Q, Qec'. There is a figure, Qec, for the electrical Q with 0 source resistance. The two are related: Re + Rg Qec' = Qec --------- Rg where Re is the voice coil DC resistance and Rg is the TOTAL source impedance. Let's look at the change in Qec' and thus Qtc win the difference between a damping factor of 1000 and 10, due to a change in source impedance of .75 ohms. Let's assume a PERFECT 2nd order maximally flat Butterworth sealed box system, with a Qtc of 0.7071. Assume (which is reasonable) the mechanical losses in the system (including absorption and frictional losses) result in a Qmc of 4.0. We can derive the electrical losses as Qmc * Qtc Qec = ----------- Qmc - Qtc which means a Qec of 0.859. Now, the typical resistance of a voice coil on an 8 ohm driver is around 7 ohms (go measure it). Now, we can recalculate the new Qec' with the added 0.75 ohms of series resistance: 7 + 0.75 Qec' = 0.859 ---------- 0.75 or a new Qec' of 0.951. Put that back into the total Q: 4.0 * 0.951 Qtc = -------------- 4.0 + 0.951 With a new system Q of 0.768. That's an increase of about 9% in the system Q. It's VERY important to note at this point that this is less variation then you find in manufacturing tolerance of high quality woofers, or variation in performance due to atmospheric changes, so from this viewpoint alone, the change is swamped by other effects. Now, how much is the response of the speaker changed at resonance due to this? In otherwords, how has the new series resistance, which lowered the damping by a factor of 100 (oh gosh) decreased control of the speaker at resonance? How much is the speaker UNDERDAMPED? Well, the original system is maximally flat, it's response show no rise to cutoff and then rolls of smoothly. A higher Qtc results in a bump in the response at resonance. The magnitude of that bump is both measurable and calculatable: 4 Qtc Gmax = sqrt [-------------] 2 Qtc - 0.25 Thus, this new series resistance generates a peak of 0.1 dB, FAR less than perturbations caused in manufacturing tolerances, environmental tolerances, room response and so on. Now, this is not to deny that there may be an audible difference between amplifiers with different damping factors, but until you rigorously eliminate all other possible causes, then you cannot point at damping factor as a cause of such differences because damping factor, over the ranges we have explored has an influence that is FAR lower than any one of a vast number of other influences. -- | Dick Pierce | | Loudspeaker and Software Consulting | | 17 Sartelle Street Pepperell, MA 01463 | | (508) 433-9183 (Voice and FAX) | ------------------------------------------------------------------------------ Richard D Pierce (DPierce@world.std.com) wrote: : In its simplest, most honest form, damping factor is NOTHING more than : the ratio of the rate load impedance (usually 8 ohms) to the amplifiers : output impedance. Don't listen to Dick and his techno mumbo-jumbo. Damping factor is a purely subjective term. It describes the degree of "wetness" to the sound, i.e., the "liquidity" (subjectively, a combination of smoothness and sweetness) of the sound as rendered by the amplifier in question. A high damping factor is good; single-ended amplifiers have a high damping factor, and hence give very sweet, liquid sound. High negative feedback results in a very, very low damping factor, and a corresponding dry sound. -Henry -- Henry A. Pasternack hap@bnr.ca Member Scientific Staff (514) 761-8734 (phone) Bell Northern Research, Montreal (514) 761-8509 (fax) ------------------------------------------------------------------------------- As a designer of hi-fi amps, I wish to point out that Dick is right (as always?). Damping factor merely indicates by how much the output of an amp would fall under "loaded" conditions, in relation to its unloaded value. Damping factor used to be of significance during the times of tube amplifier technology, wherein the output was usually taken from the anodes with relatively high impedance, through a matching transformer. Now to get maximum power, the transformer had to be chosen so that the overall output impedance of the amplifier is equal to the load (speaker) impedance. This means a damping factor of 1, corresponding to maximum power. However, this also means poor damping of the speaker and consequentially worse transient response than what would be if the speaker were voltage driven (0 ohm source). The better quality amps invariably had a lower output impedance at the expense of lower power output for the same tube design. Hennce to get more power output, bigger (and more expensive) amplifiers were needed. With modern solid-state amps having emitter or source follower outputs, the above criterion is not true. With feedback amps, the output impedance is typically 0.1 ohm or less. Even at 0.1 ohm, using a 4 ohm load, the power output varies by only 0.2 dB. Shankar Ramakrishnan shankar@vlibs.com ------------------------------------------------------------------------------- Gerard Malecki (vlsi_lib@netcom.com) wrote: : Damping factor used to be of significance during the times of tube : amplifier technology, wherein the output was usually taken from the : anodes with relatively high impedance, through a matching transformer. : Now to get maximum power, the transformer had to be chosen so that : the overall output impedance of the amplifier is equal to the load : (speaker) impedance. This means a damping factor of 1, corresponding : to maximum power. Untrue. For maximum power using triode output tubes, it is necessary to select a transformer primary impedance equal to the plate resistance of the output tubes (or 1/4 this for push-pull stages). In practice, a triode amp will typically run a higher than optimal primary impedance in order to reduce distortion, this imposing a relatively minor loss of power. A second benefit will be improved dampling factor. In the case of pentodes, the plate resistance of the tubes is much higher than the output transformer primary impedance. It is necessary to use feedback to lower the output impedance seen at the secondary to an acceptable level. The secondary impedance is fixed and equal to the load resistance, and the resulting output impedance is determined by the reflected impedance of the output tube plates and the amount of feedback employed. It's quite common to have tube amplifiers with significantly better damping factors than 1; for instance, the amps I just built have modest feedback and a damping factor of about 6. -Henry -- Henry A. Pasternack hap@bnr.ca Member Scientific Staff (514) 761-8734 (phone) Bell Northern Research, Montreal (514) 761-8509 (fax) ------------------------------------------------------------------------------- Gerard Malecki (vlsi_lib@netcom.com) wrote: : This is exactly what I meant originally. Btw, a transformer has no : intrinsic 'primary resistance' as such. The primary resistance is : equal to the secondary load times the square of the turns ratio. : For maximum power output, the plate resistance should equal the load : impedance reflected at the primary. Another equivalnt way of saying it : is that the plate resistance reflected at the secondary should equal : the load. This leads to a damping factor of 1. Two points of disagreement: 1) A transformer is a pure ratio device in theory, but in practice a real-world transformer is designed for lowest losses, widest bandwidth, and highest power handling when operating between specified impedances. If you try to "match up" or "match down" your transformer, you will run into performance problems. This has been discussed in detail in print and on the net by "Famous Mike" LaFevre, professional audio transformer winder. 2) This impedance matching business creates a great deal of confusion. In my opinion, your argument can be extremely deceptive if it's not qualified carefully. If an amplifier acts as an ideal voltage source with fixed source resistance, the maximum power dissipation in the load will occur when the load resistance equals the source resistance. At that point, the efficiency will be 50% and the amplifier will dissipate power equal to the load power. A modern solid-state amp with fractional-ohm output impedance can double its output power for each halving of the output impedance down to some reasonable limit. The theoretical point of maximum power might be a level of hundreds of thousands of watts -- clearly an impossibility. What you forget to mention is that efficiency of power transfer is proportional to the load resistance (although not linearly so). The higher the load resistance relative to the source resistance, the greater the fraction of power that ends up in the load, rather than as heat in the amplifier. It makes a lot more sense to design an amplifier for a low output impedance, and the heck with the fact that the power output is 100W, and not a theoretical maximum of 160,000W (assuming 0.05 Ohm output impedance). The advantage is not only related to power, but also frequency response into complex loads, and damping factor of real-world loudspeakers. To understand why efficiency is an issue when interfacing a triode to its load, but is not an issue for a pentode, it's necessary to think in terms of Thevenin equivalents. Only then will it be obvious why the pentode amplifier is typically more efficient than a comparable triode amplifier, even though the output impedance of the pentode is an order of magnitude higher (or more) than that of the triode. The answer is that the incremental output impedance of the pentode is quite high, but at the clipping point, the effective impedance is very low. Yes, for a given triode and B+ voltage, the most powerful amplifier will be obtained when the output impedance at the 8 Ohm tap is 8 Ohms. But it's not the case that the best strategy, in general, is to design amplifiers with 8 Ohm output impedance. -Henry -- Henry A. Pasternack hap@bnr.ca Member Scientific Staff (514) 761-8734 (phone) Bell Northern Research, Montreal (514) 761-8509 (fax)