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
Hi crazyeddy,
I can't explain the math to you, but the inverse proportionality that you spoke of has to do with the fact that impedance is the AC version of resistance. While it is also measured in ohms, it is not the same as DC resistance.
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

How about this simple example. A regulator on an alternator of a car stops the alternator from putting out more power than it is designed to do. If not regulated the alternator will push as much power as you demand of it until it melts.

A speaker load that has less resistance or push against it or back pressure, will allow an amp to keep putting out more than it can handle. (over heat or melt outputs.) If an amp can do 100 watts pushing against 8 ohms it will want to do 200 into 4 and 400 into 2. If the heat sinks or power supply wasn’t built to handle that it melts down.

The reason someone else in this discussion said it is not necessarily true is because other factors include efficiency of the speaker, but this is only a factor if your taking into account that your trying to fill a room with a certain SPL level, and woofer size because you need more amperage to move a larger motor.
For the same reason you lose water pressure in the shower when the toilet is flushed. More water is drawn, lowering the pressure in the tank. When the speaker has a high impedance, it’s like a faucet that is slightly open. The pressure (voltage) in the tank (amplifier) is maintained as there is little flow (amperage) through the piping (speaker cables). When the impedance drops, the faucet opens which draws more water (amps) causing the pressure (voltage) to drop. If the amplifier does not have enough ’pressure’ it cannot supply the current for the speaker driver to move.

That is why amplifiers that maintain voltage throughout the impedance swings of speakers are heavy and expensive. They have to act as a voltage source regardless of the current demands of the speaker. This requires a bigger transformer, bigger power supply capacitors and high current output devices to put it all togther.
gs5556,
That is a very good analogy!
I’ve often wondered why given the reality of Ohm’s Law, why do many speaker designers revert to 4 ohms rather than 8 ohms (or higher) speaker impedance. As Al noted in his reply, Ralph (Atmasphere) has pointed out that the higher the impedance load ,the lower the amplifier  distortion produced (which is a desirable result you'd think).

It would seem that a concerted effort to design easier to drive high impedance speakers would yield obvious benefits. The most logical explanation for the plethora of lower impedance speakers is that the higher impedance speaker is probably more difficult to implement to get things right and presents more design challenges.
Charles