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

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.

Hope that helps. Regards,
-- Al

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:

1. Volts (voltage) x Amperes (current) = Watts (power)
2. Ohms Law: Amperes (current) = Volts (voltage) / Ohms (resistance or impedance)
3. 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?

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