# speaker impedence **URL:** \ **Category:** Factual Questions\ **Created:** [May 9, 2005, 11:30pm UTC](https://boards.straightdope.com/t/speaker-impedence/303133 "2005-05-09T23:30:32Z")\ **Posts on this page:** 4\ **Page:** 2
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [May 12, 2005, 11:11pm UTC](https://boards.straightdope.com/t/speaker-impedence/303133/21 "2005-05-12T23:11:02Z")
David: Trust me on this, I _know_ you can’t stick a 0 Ω load on a perfect voltage source, as the current would be undefined. Note where I said, > [@](#): > > The current spike will cause the power supply to operate outside of its operating range, and the source impedance will no longer be 0 Ω. What I was _trying_ to say was that, if you stuck a superconductor onto a real, regulated power supply, the current would very rapidly (almost instantaneously) reach the power supply’s current limit. A number of things could happen once this occurs, depending on the design of the power supply. At any rate, once thing is for sure: the source impedance will no longer be zero. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [May 13, 2005, 1:41am UTC](https://boards.straightdope.com/t/speaker-impedence/303133/22 "2005-05-13T01:41:30Z")
> [@Crafter\_Man](#): > > David: Trust me on this, I _know_ you can’t stick a 0 Ω load on a perfect voltage source, as the current would be undefined. I beliefve you. The thing is that you can’t show that _any_ resistance greater than zero gets the maximum power available out of an ideal voltage source. That is, one having a constant voltage and 0 internal impedance. For any power that you compute for an assumed resistance greater than zero, someone can always find a greater power by assuming a smaller resistance and the process of assuming smaller resistors is endless. The thing is that for sources having very low output impedance you can’t get maximum theoretical power out of them without running the source out of its linear operating range. And then all bets are off. ---
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [May 13, 2005, 12:24pm UTC](https://boards.straightdope.com/t/speaker-impedence/303133/23 "2005-05-13T12:24:19Z")
David: I think part of the confusion is that I was referring to _ **real** _ regulated voltage sources in all my previous posts. An _ **ideal** _ voltage source is a different beast. And you’re correct… if you stick a variable load to an ideal voltage source, the power consumed by the load will continue to increase as the load resistance is decreased. It’s a hyperbolic relationship, and both the power and the current are undefined when the load resistance equals zero. > [@](#): > > The thing is that for sources having very low output impedance you can’t get maximum theoretical power out of them without running the source out of its linear operating range. And then all bets are off. Agree. I also made this point earlier when I stated, > [@](#): > > For many [real and unregulated] sources, as the current goes up the source impedance also goes up due to self-heating and a PTC. If this occurs, maximum power to the load is achieved when the load impedance is somewhat higher than the source impedance. ---
**Author:** ![casdave](https://avatars.discourse-cdn.com/v4/letter/c/c6cbf5/32.png) [@casdave](https://boards.straightdope.com/u/casdave)\ **Post date:** [May 13, 2005, 1:00pm UTC](https://boards.straightdope.com/t/speaker-impedence/303133/24 "2005-05-13T13:00:54Z")
The idea that a lower impedance somehow translates into more sound, as it appears to be hinted at in earlier posts is wrong. More power does not itself necessarily translate to more sound, connect up a pair of 8ohm speakers, and you may get just the same sound pressure as connecting up 2ohm speakers. Whats missing here is the sensitivity of the speakers, often specified in terms of dBm/watt at a specific frequency, at a specified distance from the sound source ie the speaker. [Previous page](https://boards.straightdope.com/t/speaker-impedence/303133.md?page=1)