question ?

[mrblob]

someone said that a speaker with higher impedance requires less power to reach the same amplitude. Is this true?

[gridracer]

I would say no but thats just my opinion others may know for sure.

[Jacampb2]

I believe the answer would be, that it can be true. It is by no means universal, but it is possible. It all depends on the drivers in question.

[ttocs]

there are too many things to take into question, that is a really broad statement. Efficiency will determine more about how loud it will be rather then impedence but the impedence does play into it as well. What they probably meant, is that with a lower resistance, with the correct amp, that the speaker would be louder.

[Jacampb2]

Okay, here is a bit of the reasoning behind this myth/statement. The actual statement that I have heard, is in regards to a 8 ohm and 4 ohm driver with the same sensitivity at 2.83V. The 8 ohm driver requires exactly half the power to reach the same output that the 4 ohm driver does. You see, 2.83V corresponds to 1 watt at 8 ohms (2.83*2.83/8=1.0011W), where 2.83V corresponds to 2 watts at 4 ohms (2.83*2.83/4=2.022W), and 2.83V corresponds to 4 watts at 2 ohms, and so on and so forth.

IMO, this is not a very fair way to rate drivers efficiency since it does not take into account their impedance. It used to be that SPL or Sensitivity was measured at 1 Watt at a distance of 1 meter. This way driver impedance did not have any factor in the measurement. The 2.83V thing got standardized because of home audio from what I understand, the fact that it tends to make a lower efficiency low impedance automotive driver look a lot better than it should is just icing on the cake.

So in all actuality, two drivers with identical measured sensitivity using the "new" 2.83V method, one being a 8 ohm driver, one being a 4 ohm driver, the 8 ohm driver is in fact using 1/2 the power of the 4 ohm driver. This will be universal provided that they are both measure with the voltage method, and they both had equal sensitivity. The reason I said earlier that it can be true, but is not always true, is because the OP left off the key points about the sensitivity and measurement method. If the only thing you are comparing is a 8 ohm vs. 4 ohm driver with no other regard to any other specifications, then it's a crap shoot, it might be, but it might not be.

Later,
Jason

[shawn k]

someone said that a speaker with higher impedance requires less power to reach the same amplitude. Is this true?

The question here is far too vague to answer correctly. I'm assuming you are trying to compare two "identical" drivers with the only difference being impedance (4ohm vs 8 ohm) correct? If so, then the short answer is no. Jacampb2 is correct with his explaination (and did a very good job at it). The only concern I have however is that you will not find two identical drivers (other than impedance) with the "same" sensitivity. You will just about always find that the sensitivity of a 4ohm driver is 3db higher than the same driver with a 8ohm impedance. The reason is simple.. the 4ohm driver is extracting twice the power from the amplifier. BUT! Let's not forget that the 4ohm version is still 3db higher... and...3db=twice the output (in an ideal world of coarse) So if we cut the power in half (equal to that of the 8ohm) then we will see that the sensitivity of the 4ohm is "identical" to that of the 8ohm.

I hope that made sense

Here.. take a look at these two identical drivers and check out the sensitivity for each.

4 ohm
https://www.madisound.com/store/product ... cts_id=929

8ohm
https://www.madisound.com/store/product ... cts_id=930