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

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:

1. 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.
2. 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.
3. 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.

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?

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?