16 ohm vs 8 ohm

[InclinedPlane]

I asked fullrangespeakers (formerly AER) if they could make a 32 ohm fullrange driver to help balance its output better with lower-efficiency bass drivers in an open baffle. They replied that it is the opposite of what I think. My MD-2B 16 ohm is a full 10db more efficient than their 8-ohm unit. They claim their 8-ohm unit is 89 db.
I don't understand this. Doesn't a 16-ohm coil grab half the power from an amp that an 8ohm does? Wouldn't the 8-ohm be alot louder as a result?

The reason for all this is that I'm looking for a less efficient FR driver without using a resistor to make it so. I already run two Pyle PPA15s (Qt .65) per side parallel with the 16ohm FS driver (Qt .75) also in parallel for 3.2 system impedance and I get good bass but still a little too lean. My goal was that if I could have the bass drivers at 4ohm and the fullrange driver at 16 surely plenty of power would go to the bass drivers; enough to make up for their lower sensitivity and open baffle losses...but it appears the fullrange driver is still too loud. I don't want to use a resistor for signal purity.

If they claim the 8-ohm is 10db less efficient but halving the impedance pulls twice the power what have I accomplished?

[keantoken]

Efficiency is a measure of how much power goes into the device as compared to how much useful power you get out.

The efficiency doesn't have anything to do with the resistance of the speaker coil. The efficiency is the output divided by the input.

Since an 8-ohm speaker has more power going through it than a 16-ohm, it will be louder than a 16-ohm (and it will need twice the power rating!).

So the 16-ohms are probably more efficient simply because of their construction, but it will not make them louder unless you are listening at the same input level as the 8-ohms.

I don't know how much 10db affects the volume, as I haven't gotten into this area of study yet. However, I think the 8-ohms will still be louder.

Also, I'm sure you know that the lower efficiency means that you have to put more power into the cone to make it move - indicating that (theoretically) the material keeping the cone in place will cause more distortion. You would have to ask someone else if this is noticeable or not, but I would go with the higher efficiency speaker.

Hope this helps,

- keantoken

[gmilitano]

[i]I don't know how much 10db affects the volume, as I haven't gotten into this area of study yet. However, I think the 8-ohms will still be louder.
[/B]

10dB is perceived to be about twice as loud.

http://trace.wisc.edu/docs/2004-About-dB/

[despotic931]

Originally posted by gmilitano


10dB is perceived to be about twice as loud.

http://trace.wisc.edu/docs/2004-About-dB/

Excellent article, I wish I had read it back when I was studying these concepts, it would have helped me to grasp them much easier!

-Justin

[InclinedPlane]

Awright so if the 8 ohm one is 89 db and the 16 ohm one 99db then they'll both play at the same level of loudness at the same position on the volume control? Idk why this is confusing for me.

[Cal Weldon]

Let me see if I can make this a little easier. Please understand these are just the basics, there are other factors.

2X the power = 3 dB louder
10X the power = 10dB louder
10dB = double the "volume" to your ear.

This is more clear if you understand logarithms.

To partly answer your question, if the volume control remains the same on an 8 & 16 ohm driver with the same sensitivity (read efficiency) you will hear an increase of 3 dB. 3dB is not a huge amount but you can certainly hear it.

[Cal Weldon]

So as not to confuse, the idea behind leaving the volume control the same spot is that you hypothetically send half the power (-3dB) to the speaker with a 16 ohm load as you would an 8 ohm. There's more to it than that but you get the idea.

[Nanook]

dB(difference) in units of power :=10 × log(P1/P2), so twice the power yields 3dB, and 10X's the power yields 10dB

An example (use a scientific calculator or a widget on your desktop):
If P1=10 watts. P2=20 watts:
dB (power) =10 × log(20/10)
dB= 10 × log (2)
dB = 10 × (0.3010)
dB = 3.01


dB (difference) in SPL:= 20 × log (P2/P1), as sound pressure levels are related to 1/R^2..where R is the distance from the source.
An example :
if P1=30 watts, P2=10 watts

dB (power) =20 × log(10/30)
dB= 20 × log (1/3)
dB = 20 × (-.4771)
dB = -9.5424

so a reduction of nearly 10dB!

Most amplifiers are constant Voltage devices (Nelson Pass' First Watt amps are the only constant current amps I currently know of ), so from Ohm's Law,

V:= I × R, and ;
Power (in watts) := I × V , which is the equivalent to I^2 × R
So if V is constant, then R must increase as I^2 decreases.

An example :
if R=8 ohms, P=10 watts, then I^2=P/R---> so I=√(10/8), ---> I= √5/2
if R=16 ohms P=10 watts then I^2 ----> I=√(10/16), ----> I=√5/(4 × √2) which is clearly smaller than if R=8.

Although power remains the same, the ability to drive current may actually increase, or at least the apparent load to the amp may decrease. Most amps are limited not by voltage, but by current.


stew