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?

Thanks

128x128crazyeddy
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:

  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.

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