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

Showing 1 response by audiobunker

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.