Why are low impedance speakers harder to drive than high impedance speakers
I don't understand the electrical reason for this. I look at it from a mechanical point of view. If I have a spring that is of less resistance, and push it with my hand, it takes little effort, and I am not working hard to push it. When I have a stiffer spring (higher resistance) I have to work harder to push it. This is inversely proportional when we are looking at amplifier/speaker values.
So, when I look at a speaker with an 8 ohm rating, it is easier to drive than a speaker with a 4 ohm load. This does not make sense to me, although I know it to be true. I have yet been able to have it explained to me that makes it clear. Can someone explain this to me in a manner that does not require an EE degree?
Ralph, you mentioned 'former' twice in the last part of your post.
Ach!
That should read:
Again, this all comes down to intention. Is your intention to get
the system to sound as good as it can or is it more important to simply
play loudly? If the latter, than some of the lower
impedance speakers and higher power transistor amps will be of interest;
if the former, then you will be very careful to be matching the speaker
to the amplifier (and not the other way ’round) and most likely
avoiding lower impedances in general.
-as Al corrected.
So while I can say with precision that most solid state amps are more accurate, as far as measurements are concerned
Erik, I don't think this statement is correct, and here's why. If you look at the specs, the lower distortion and apparent constant voltage characteristics of most solid state amps looks great! The problem is, that bit of paper ignores how our ears perceive sound.
This takes a bit to grasp! To give you some idea, most of us know that the ear employs a logarithmic approach to sound pressure. This is why we use the VU scale of decibels.
So take this concept, but apply it to harmonics. The ear seems to use something that looks very much like a logarithmic approach when it comes to how sensitive it is to harmonics- being less sensitive to lower orders and far more sensitive to higher orders on what looks much like an inverse logarithmic function.
The fact that the ear is more sensitive to higher orders has been known for decades and should not be a matter of debate! This is very easy to prove with simple test equipment.
Add to that the fact that the ear is tuned to be most sensitive to bird song frequencies (Fletcher-Munson). This fact arises out of evolution and is millions of years old- birds are the first warning of a predator in the area!
So the fact is that if the ear does not care about the lower harmonics so much, then logically we should be designing to eliminate the higher orders, especially since the tools that the amplifier designer has in the tool box all have certain limitations. For example, as I stated earlier (and as been stated by Norman Crowhurst, a universally recognized sage), loop negative feedback is known to add additional harmonics and IM distortions (the harmonics can go as high as the 81st and the intermodulations occurring at the feedback node in the amp). In this way an amp with feedback will usually sound brighter than an amp without, even though on the bench they both measure flat.
So what is more 'accurate'? Low distortion on paper is meaningless unless we also know what it is that makes it 'low'. Its one thing if we can see the lower orders in the harmonic distortion spectrum. But if we are to take how the ear perceives sound into account, the higher orders should really be a lot lot lower than they are currently with all 'low THD' amps. And by that I mean **at least** 2 orders of magnitude!
Just seeing 'low THD' doesn't cut it.
****This is ignoring how the ear works!!!**** (fist bangs tabletop)
The fact is that as far as the ear is concerned, the distortion of most amps with seemingly really low THD is that the distortion is higher. Its easy to hear too- which is why tubes still exist in the marketplace 60 years after being declared obsolete. Its why the tubes/transistor thing has been going on longer than the internet!
(if the tubes weren't doing something right, they would have been gone long ago. How many flathead V8s are still in production? If you got 'none' then you probably also know its because they are obsolete. There is a huge difference between being declared obsolete and actually **being** obsolete!)
In essence, the bench specs are an excellent example of the Emperor's New Clothes. This is because you have to ignore the obvious coloration of brightness/harshness/brittle in order to really say that its more accurate. The bench spec thing still has its roots in the 1960s and has not changed much since then (its mostly based on an idea of low distortion and flat frequency response while totally ignoring what the ear perceives; its actually tuned to the eye rather than the ear).
Put another way- we like to think our amps are low distortion because that is how they look on paper. That appearance is false- we're not measuring the right thing. Try to wrap your head around the fact of the ear's crazy sensitivity to higher ordered harmonics and use **that** as a baseline instead. If you can make that translation, you will see that most amps are fairly high distortion and not accurate at all.
bdp24 1,906 posts 01-07-2017 7:17am In a related matter, Roger Modjeski of Music Reference recommends hooking up your speakers to a tube amp on the lowest impedance tap that provides the power you need. So if an amp puts out 45 watts at 8 ohms and 30 at 4, and 30 watts is enough for your needs (with the combination of speaker sensitivity, room size, listening level, etc.) with an 8 ohm speaker, use the 4 ohm tap for lowest power amp distortion and best sound.
THE last part of the sentence is usually true, however the logic is totally off. No audiophile amplifier will go down in output in that fashion , if its 45 watts at 8 ohms then it should be 90 watts at 4 ohms....If the wattage is dropping then the amplifier should not be trying to drive the lower impedance load in the first place.
bdp24 In a related matter, Roger Modjeski of Music Reference recommends hooking up your speakers to a tube amp on the lowest impedance tap that provides the power you need. So if an amp puts out 45 watts at 8 ohms and 30 at 4, and 30 watts is enough for your needs (with the combination of speaker sensitivity, room size, listening level, etc.) with an 8 ohm speaker, use the 4 ohm tap for lowest power amp distortion and best sound.
THE last part of the sentence is usually true, however the logic is totally off. No audiophile amplifier will go down in output in that fashion , if its 45 watts at 8 ohms then it should be 90 watts at 4 ohms....If the wattage is dropping then the amplifier should not be trying to drive the lower impedance load in the first place.
the impedance of the speaker generally dictates the best sounding amplifier tap. Sometimes speakers have widely varying impedance curves that means experimentation is probably necessary to see which tap sounds best. Depends on the impedance curve of the speaker, though. You can't get something for nothing. There's no free lunch. 😄
I agree with Geoff’s post just above. Also, regarding:
Timber77 1-15-2017 No audiophile amplifier will go down in output in that fashion , if its 45 watts at 8 ohms then it should be 90 watts at 4 ohms....If the wattage is dropping then the amplifier should not be trying to drive the lower impedance load in the first place.
While the maximum power capability of high quality solid state amps will of course often be twice as much into 4 ohms as into 8 ohms, tube amps do not behave in that manner. A tube amp which has an output transformer and provides 4 and 8 ohm taps will generally be designed to have a maximum power capability that is the same or similar when a 4 ohm load is connected to the 4 ohm tap as when an 8 ohm load is connected to the 8 ohm tap. And an output transformerless tube amp will typically have a greater maximum power capability into an 8 ohm load than into a 4 ohm load (and often an even higher capability into 16 ohms).
In the situation bdp24 referred to, where an 8 ohm load is connected to a 4 ohm tap, maximum power capability will usually be reduced in comparison to the amp’s capability when an 8 ohm load is connected to the 8 ohm tap or when a 4 ohm load is connected to the 4 ohm tap. The degree of that reduction will depend on the specific design, as will the desirability of the "light loading" (i.e., 8 ohm load connected to 4 ohm tap) that Mr. Modjeski recommends.
Excellent clarification from Al, with more specifics. No surprise there! Timbre77, you missed the important qualifying adjective of the amplifier scenario I described---"tube". As Al stated, while the power solid state amps create increases with dropping impedance, the opposite is generally true of tube amps, with the notable exception of the unique Music Reference RM-200, which actually behaves more like a ss amp in that regard.
As Al detailed, the different impedance taps on a tube amp allow the amp to provide similar power to all the taps---that's one task of tube amps output transformers. However, if your tube amp has 4, 8, and 16 ohm taps (typical in tube amps, though the RM-200 offers 2, 4, and 8 ohms), and you hook up an, say, 8 ohm speaker to the 4 ohm tap, the power available to the speaker will be less than it would be if connected to the 8 ohm tap. But, says Music References Roger Modjeski, a tube amp so employed will usually be producing not only less power, but also less distortion, and better sound. As Al mentioned, Roger calls this tactic "light loading". In addition to lower distortion, an additional benefit of using a lower impedance tap is that the amps output impedance will be lower---it will have a higher damping factor, and will interact less with the varying impedance characteristics of the speaker load, resulting in a more predictable frequency response.
if your tube amp has 4, 8, and 16 ohm taps (typical in tube amps,
though the RM-200 offers 2, 4, and 8 ohms), and you hook up an, say, 8
ohm speaker to the 4 ohm tap, the power available to the speaker will be
less than it would be if connected to the 8 ohm tap. But, says Music
References Roger Modjeski, a tube amp so employed will usually be
producing not only less power, but also less distortion, and better
sound. As Al mentioned, Roger calls this tactic "light loading". In
addition to lower distortion, an additional benefit of using a lower
impedance tap is that the amps output impedance will be lower---it will
have a higher damping factor, and will interact less with the varying
impedance characteristics of the speaker load, resulting in a more
predictable frequency response.
While generally true, a problem that can turn up when doing something like this is that the transformer can 'ring' if insufficiently loaded. In addition, with such a load, it will not be as flat across its bandwidth, as the transformer will tend to express less of its turns ratio and more of its inter-winding capacitance. The 'lighter' you load the transformer the more of a problem this becomes.
So the result, while possibly reducing distortion in the output tubes, will be to **increase** distortion from the transformer (ringing) and degrade the frequency response. Of course, if the amp employs negative feedback some of this will get sorted by that, but a problem with negative feedback is that while reducing distortion overall (in particular lower ordered harmonics), it actually **introduces** higher ordered harmonics that otherwise may not have been present at all! Its best not to give feedback too many places to screw up.
Excellent additional information from the always illuminating Ralph Karsten! Thanks as always. Modjeski designs his transformers and has them built to his specs (even winding them personally, for those willing to pay him to do so). Perhaps he does so in a way that takes light loading into consideration. I don’t employ it myself, needing all the power I can get for the rather insensitive 8 ohm loudspeakers I use the RM-200 with.
Light loading like many audio techniques has to tried on an individual basis, results will vary depending on circumstances. My amplifier has 8 and 16 ohm taps and my speakers are 14 ohm nominal. The 16 ohm tap sounds better than using the 8 ohm tap.
I was surprised when I learned that some choose their power amp first, then look for a speaker it can drive well. I thought the notion that speakers, being transducers, vary much more in sound that do amps was universally agreed upon, and should therefore be selected first. The notion that power amps vary in character as much or more than do speakers is one I disagree with. I feel the same way about phono cartridges (also transducers) vs. pickup arms and/or turntables, though to a lesser degree.
So, when I look at a speaker with an 8 ohm rating, it is easier to drive than a speaker with a 4 ohm load. This does not make sense to me, although I know it to be true. I have yet been able to have it explained to me that makes it clear. Can someone explain this to me in a manner that does not require an EE degree?
You are in luck. I can explain this in terms that does not require any degrees. In fact due to the inverse law of reciprocal square roots you probably will not understand if you DO have a degree. The first critical thing is The Gnome. Gnomes are small and live inside your squeaker. Gnomes all belong to the Conservative Party - they RESIST change. If you want more resistance you must get more Gnomes. Second critical thing is Impudence. Impudence is the amount of resistance. However unlike resistance in a resister where there are always the same number of Gnomes, in a squeaker the amount of resistance (measured in Gnomes) varies with frequency. A squeaker may be described as having 8 Gnomes. In reality - and this includes my speakers - there is a variation in resistance. So at a lower frequency the impudence dips to as little as 4 Gnomes. The third critical component is Currant. Your amplifier feeds your squeakers with currants. So it goes like this: Your amplifier feeds 8 Gnomes at most levels of frequency. Each Gnome processes one currant at a time. But where the impudence dips to where there are only 4 Gnomes the speaker still needs to process 8 currants. But there are only 4 Gnomes, so what to do? A well designed amplifier, ie one that has a lot of currants, will pass twice as many currants to each Gnome. So the 4 Gnomes will process 2 currants per Gnome. The amplifier therefore has to put out currants at twice the rate. This is all known as Gnome's Lore.
I thought the notion that speakers, being transducers, vary much more in
sound that do amps was universally agreed upon, and should therefore be
selected first. The notion that power amps vary in character as much or
more than do speakers is one I disagree with.
If you have no preference for tube or solid state then this is the way to go. If you have found that you prefer one over the other then you will need to get the amp first and then find a speaker to match. That is why in most cases, you start with the amp. It is tricky- to know that amp you have to hear it in a variety of circumstances to make an informed purchase. But if you do that, you are less likely to flush more $$$$$ down the loo trying to get the system to sound the way you want it.
Interesting thing is you need amp and speakers together to make sound. Either one alone says nothing. So you can judge each on paper or do measurements but absolutely cannot judge how one sounds exactly without the other. Only the two together and each pairing sounds different.
Then as if that’s not bad enough there is this thing called room acoustics that means the same amp and speaker pairing will sound different depending not just on room but location in it.
However natural selection is at work. Only the strong will survive no matter what.
Good point Ralph. It's great when ones taste in speakers and amps happens to allow a synergistic match---a tube amp with the Quad 57 ESL, for instance.
Would it be possible to go back to the original question? This thread contains lots of great information,
but I am still struggling with the basics. I apologize for this “newbie” request, but I'm clearly misunderstanding something and would be grateful for some help!
All other
things being equal (PLEASE SEE full caveats note at bottom), I thought the following were generally true:
If
you double an amp’s wattage, you increase your potential spl by 3db. Thus, an amp which doubles its watts
when the speaker load is halved (not all amps truly do this) driving an 8 ohm
speaker with a sensitivity of 90db would generally have essentially the same
spl potential driving a 4 ohm speaker with a sensitivity of 87db.
In
general, the higher you turn up the volume of your amp to the same set of speakers, the harder it has to
work. All the way up is relatively hard work (and
may cause clipping, etc.), barely on should not be much work.
Theoretically,
if the same speaker system could be made in an 8 ohm version and a 4 ohm
version with the same sensitivity rating (let’s say 90db), then the same amp
driving the 4 ohm version would need less “signal” or a lower volume setting to
attain the same spl in the same room as the 8 ohm version.
If the above are basically right, I don’t understand why an
amp would need to work harder with a 4 ohm load than an 8 ohm load to put out the
same spl in the same room. If the above are not correct, where did I go wrong?
THE HOSE ANALOGY: I’ve
heard the previously referenced analogy of water going through a hose many
times, and each time it sounds backwards to me.
It seems to me that if the flow of water – or amount of water moving
through space over time – is to remain constant, a larger diameter pipe, or
lower impedance, would make it easier for a pump (or amp) to push that water. If the flow remains constant, then as the hose
diameter decreases, the pressure increases and the pump would need to work harder. Why is this not correct?
Clearly, I must misunderstand some fundamental concepts! I’m not an engineer or science “type” so may
need some baby steps.
PLEASE NOTE: in order
to try to understand the basics, all of the above is based on simplistic
and theoretical situations, with all other things such as speaker
configuration and design, other components, room size, etc. being equal AND
with all components properly matched. I
understand that real world implementations may vary.
Hi, If you keep the Ohm’s law equation in mind it’s easy to explain. I = (current), V=voltage and R = resistance (speaker load impedance expressed as ohms). I=V÷R. So the smaller R becomes, the larger I becomes. If for example V=10 and R=10 then I will =1. If R is reduced to 1 then I now =10. Reducing the R (speaker impedance expressed in ohms) will increase the value of I (current demand). So a 2 ohms load (lower R) will demand more current (I) than a 8 ohm load (larger R). Charles
Thanks for the reply. Could you please relate this to my 3 points above and the hose analogy as well? I see the formula, but don't know how that applies to needing less signal for an amp that can produce more watts because there is less resistance. How does that relate to a pump needing to work harder (or less hard in the amp world) to push the same amount of water through a smaller diameter hose?
Sorry, I thought Ohm's Law is self evident in its equation. As R decreases I must increase for a given V I.e. voltage. There's an inverse relationship, more current is needed to maintain the voltage as the resistance diminishes. It all seems straightforward to me. Charles
I’ll add a few comments to Charles’ excellent answers.
Swingfingers, first let’s change the word "sensitivity" in your question to "efficiency." Speaker sensitivity is usually defined on the basis of an input to the speaker of 2.83 volts, rather than 1 watt. 2.83 volts into 8 ohms corresponds to 1 watt, so the resultant SPL (sound pressure level, in db) is the same either way. But 2.83 volts into 4 ohms corresponds to 2 watts, so if the 87 db figure you referred to for the 4 ohm speaker is defined on the basis of a 2.83 volt input that speaker would produce only 84 db in response to 1 watt.
So with the word "sensitivity" (which we’ll define as db SPL at 1 meter in response to a 2.83 volt input) changed to "efficiency" (which we’ll define as db SPL at 1 meter in response to a 1 watt input, although in some other contexts the term "efficiency" may also be used to refer to the ratio of acoustic power out to electrical power in), my answers to your three questions are:
Q1)Yes.
Q2)Yes, with the slight qualification that in the specific case of a class A amplifier the amp will dissipate (consume) less power internally (and therefore have a lower internal operating temperature) when it is supplying large amounts of power to the speaker than when it is supplying small amounts of power (or no power) to the speaker. And in that sense and to that extent (there are other factors that come into play, of course) a class A amp may be working less hard when supplying more power rather than less.
Q3)Yes, a 90 db/1 watt/1 meter/4 ohm speaker will require a lower setting of the volume control to produce the same volume as a 90 db/1 watt/1 meter/8 ohm speaker.
If the above are basically right, I don’t understand why an amp would need to work harder with a 4 ohm load than an 8 ohm load to put out the same spl in the same room. If the above are not correct, where did I go wrong?
Keep in mind that the speakers referred to in Q2 are identical, while in Q3 they are not.
In both situations referred to in Q3, the amp will deliver the same amount of power to produce a given SPL. For a resistive load power = voltage x current. The volume control setting controls the amp’s output voltage, while the impedance of the speaker determines how much current is drawn from the amp at a given output voltage. In the case of the 4 ohm speaker the lowered setting of the volume control that you correctly referred to will result in less voltage being supplied by the amp compared to the 8 ohm case, but the amp will be supplying more current at that lowered volume control setting than at the higher volume control setting of the 8 ohm case. Put simply, it is easy for an amp to supply voltage, as long as it is operated within the range of voltage it is capable of, but less easy for it to supply current.
I’ll leave the hose analogy question to others, as I generally prefer to avoid using non-electrical analogies for electrical things.
Al, THANK YOU very much!
I truly appreciate your full, detailed response that also went to the
heart of the issue. I think I am
beginning to understand. It seems like
the critical part (at least for me) is the following:
Put simply, it is easy for an amp to supply voltage, as long as it is
operated within the range of voltage it is capable of, but less easy for
it to supply current.
Ohm’s law – by itself - doesn’t seem to get at this.
If I truly understand how the relationships work, there are several steps
involved:
Volts (voltage) x Amperes (current) = Watts (power)
The amount of Watts or power required to drive a
speaker of a certain efficiency to a certain SPL in a certain space remains
constant even if you change a speaker’s impedance.
The following example demonstrates
the relationship:
A) 2 Amperes = 16
Volts / 8 Ohms where 2 Amps x 16 Volts = 32 Watts
B) 4 Amperes = 8
Volts / 2 Ohms where Amps x 8 Volts = 32 Watts
Therefore, when you reduce impedance, but keep power constant,
the current increases but the voltage decreases. This is where the crucial piece of
information applies to clarify that the reduction in voltage does not mitigate the
increased energy required by an amp to increase the current.
Am I close????
PS. Still not sure how the pipe analogy works here?
You’re better than close; that’s exactly right :-)
I’ll mention also that the following equations can be derived by substituting some of the terms in equation 2 in your post above into equation 1, and doing some algebraic rearrangements, and these equations may add some further clarity to what has been said:
Power (watts) = (Volts squared) / Ohms
Power (watts) = (Amperes squared) x Ohms
It can be seen from these equations that for power to remain constant, as the number of ohms decreases voltage must decrease, while current must increase.
Finally, to be precise I should mention that we’re simplifying all of this a bit by making the assumption that the load is purely resistive. Volts x Amps = Watts in the case of a resistive load, but things get somewhat more complex when the load has a significant inductive or capacitive component, in addition to its resistive component.
To me it seems like if you need twice the amount of electrons to flow to a 4 ohm speaker than to an 8 ohm speaker, the amplifiers would need to work harder and in return wouldn't this cause more distortion? Also, if it takes twice the amount of current to drive a 4 ohm vs an 8 ohm speaker, wouldn't this mean the 4 ohm speaker is less efficient. However, you would think the 8 ohm speaker would be less efficient because it has twice the amount of resistance to the current.
I was told once a 4 ohm speaker requires twice the amount of current than an 8 ohm speaker. Because a 4 ohm amplifier delivers twice the amount of current, would this in turn supply twice the amount of information to the speaker to create more detail in the music?
Since the OP thinks in mechanical terms, this analogy may best describe how speaker loads affect an amp. Imagine the amp is a somewhat fragile flywheel that will fly apart at a certain RPM. If you apply a resistance (an 8 ohm speaker load in electrical terms) to the flywheel that is sufficient to prevent the flywheel from reaching critical speed it will not break. If you remove part of the resistance (think 4 ohm speaker) the flywheel will speed up. If you remove enough of the load, you eventually reach a point where the flywheel is spinning so fast it fails. Replace the concept of the flywheel speed with power output from the amp. The power an amp will produce is inversely proportional to the resistance to current flow. More resistance keeps the amp in check so to speak.
According to Al's information, having the two extreme examples, I have the exact opposite question: For headphones, why do we need good, high, powered amp to move headphones with very high impedance (300-600 ohms). According to the ohm equasion, for the same power, as long as the amplifier "understand" that the output is not disconnected (as the "100,000 ohms" al's example), than the amp needs only a tiny current to "drive" the headphones. What am I missing here?
For headphones, why do we need good, high, powered amp to move headphones with very high impedance (300-600 ohms).
You don't. Many phones of this impedance are easily driven by a few milliwatts. Some 'phones require a bit of power and others don't. How much power they need is independent of their impedance.
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