Why are low impedance speakers harder to drive than high impedance speakers

First, I wouldn’t say that 4 ohm speakers are **necessarily** harder to drive than 8 ohm speakers, as there are many other variables involved. Including the efficiencies of the speakers; how the magnitudes of their impedances (the number of ohms) vary over the frequency range; the phase angles of their impedances at various frequencies (which describe the degree to which the impedance becomes partially inductive or partially capacitive at various frequencies, rather than purely resistive); etc.

But yes, typically a speaker having a low nominal impedance such as 4 ohms will be more difficult to drive than one having a higher nominal impedance such as 8 ohms. Adding to what has already been said, perhaps a good way to envision that is to consider a pair of extreme examples.

On the one hand let’s say that all the amp is driving is the input impedance of another amplifier, as would be the case if the amp were only connected to the speaker-level input of a powered sub. It might then be seeing a load of perhaps 100,000 ohms, which would result in the sub responding to the voltage being put out by the amp in question at any instant of time, but per Ohm’s Law (thanks for bringing that into the discussion, Ghosthouse) drawing essentially negligible current from that amp. In that situation the amp in question would hardly know that it is connected to anything at all, and the power it would be putting out would be essentially zero. (Power into a resistive load equals voltage x current).  (In saying this, btw, I'm putting aside the fact that tube amps having output transformers should not be operated unloaded while processing a signal, that being a separate issue).

At the other extreme let’s apply a load of essentially zero ohms to the amp, by shorting its + and - output terminals directly together with a heavy gauge jumper. I think most will recognize that the amp would be incapable of putting any kind of reasonable signal into that near zero ohm load, because per Ohm’s Law creating a non-zero voltage across a zero ohm resistance requires infinite current. And as the amp attempts to do that the result is likely to be either that it goes into a self-protective shutdown, or a blown fuse, or damage.

Obviously a 4 ohm load comes closer to being a direct short than an 8 ohm load, and an 8 ohm load comes closer to being a negligible load than a 4 ohm load, so there you go!

One additional point: As Ralph/Atmasphere has stated here many times, for various reasons both solid state and tube amplifiers will exhibit measurably better distortion characteristics when driving 8 ohms than when driving 4 ohms.

Regards,
-- Al

bdp24 1-8-2017
For instance, the original Quad ESL's nominal impedance was 16 ohms, but it’s impedance rose to 60 ohms at low frequencies, and fell to 1.8 ohms at high frequencies---anything but an easy load! That impedance characteristic is one reason the sound of the Quad ESL is so affected by the amp driving it, and why almost no solid state amp is a good match---it makes for overblown bass and missing highs.

BDP, a minor correction to your characteristically excellent inputs.  The last phrase should be "it makes for missing bass and overblown highs."  As you no doubt realize, and aside from some rare exceptions, in comparison with a tube amp a solid state amp will deliver more power into low impedances and less power into high impedances, for a given input signal level.

Best regards,
-- Al
 

Most typical speakers will decrease/increase their sensitivity in direct proportion to the increased/decreased impedance changes.

Electrostatics being a notable exception, though, including the ESL-57 which bdp24 was referring to.  Some other exceptions are referred to in the Paradigms In Amplifier Design paper Ralph has often referred to.

In the case of speakers that have been designed to sound their best when driven by tube amplification, such as the ESL-57 (which was designed before solid state amplification existed), frequency response at the output of the speaker may very well be most flat when the frequency response of the signal provided to the input of the speaker is not flat.  In voltage terms, that is.

Best regards,
-- Al
 

I agree with Unsound’s post just above. And while I consider myself to be a "speakers first" kind of person, as he is, I don’t see that as being inconsistent with Ralph’s statements, including:

Often people have a preference about tubes and transistors- the speaker **must** be chosen to take that preference into account!!

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 [correction by Al], 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 I see it this is saying essentially that what kind of amplification one anticipates using, now or in the future, is one of the major factors to consider in choosing a speaker. That is not quite the same as saying "amplifiers first," rather than "speakers first." Or so it seems to me.

It should also be noted, btw, that there are some speakers that will be equally suitable, or at least comparably suitable, for use with nearly all types of amplification. In those cases the resulting sonics will depend mainly on the intrinsic sonic characters of the speakers and the amp (as well as on speaker-room interactions, of course), with amplifier-speaker interactions contributing minimally if at all. Such speakers will typically have impedance curves that are relatively flat and do not have severely capacitive phase angles at any frequency, and combine relatively high efficiency with the ability to cleanly handle copious amounts of power.

Best regards,
-- Al

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.

Regards,
-- Al

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.

Hope that helps. Regards,
-- Al

Am I close????

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.

Regards,
-- Al