8 ohm vs 16ohm sensitivity

[kevin@kab]

Hi,
For the same input voltage, how much difference in level would you expect from an 8 ohm diaphragm, vs a 16ohm one? is there any data for this?
for specifics, this is a pair of 291-16A original drivers. but the diaphragms were changed at one time to 8 ohm. One phram is damaged. If I replace it with a 16 ohm one, would the level mismatch be great or small?
I prefer to stay with symbiotik's and I have a chance to get a 16 ohm nos one.

Thanks
Kevin

Jan 09, 2011

[Old Guy]

Output is normally rated by wattage.



All else equal there should be zero difference in sensitivity. Nada. the 16 ohm will need less current and more voltage. The 8 will need more current and less voltage.

But at given watts, it still works out the same.

IOW I don't understand why you are trying to rate the drivers by voltage? Voltage is not power. Voltage times current is.


A given 8 ohm driver is going to draw more current from a solid state amp at a given voltage, therefore be louder, all else equal.

I suggest a basic read of Ohm's Law

Ohm's law - Wikipedia, the free encyclopedia


Now with your spec of a given voltage....lets just say 20 for grins,


20/8 = 2.5 amps = 50 watts


whereas

20/16 = 1.25 amps = 25 watts.


But to say the 8 ohm driver has a higher sensitivity is ridiculous. It is getting twice the power.


OTOH on a tube amp the output tranny steps up the voltage. Current drops when you step up.


But if you rate sensitivity in watts, a 110 dB/watt driver is that regardless. Bear in mind some ratings are at 4 feet, mostly old, and many newer are at one meter, the one meter rated driver has the advantage, the mic is closer to it.

BTW Ohm's Law is also why solid state amps double up from 8 to 4 ohms, the power supply willing. An amp with a week supply might blow at the lower rating if the current exceeds its capacity. I've seen lots of 2 ohm loaded fried Pro Sound amps. Hundreds. Mebbe as many as a thousand....dunno, I was in Pro for 4 decades...
Load em down, clip 'em out, Rawhide! Sorry I couldn't resist...


In your case, are you running tube or SS? Passive or active?

A 16 ohm driver on an 8 ohm passive will mess with your xover. But a 15-16 ohm NON-INDUCTIVE resistor across it will fix that, at a slight loss in gain, which the level control for that side should fix.

If you are running active and a separate amp channel just play with the gains.

I am amazed you want to stay with Symbiotik. I would use any excuse to get 299 frams.

Some may question the need for the non-inductive in the resistor spec, IMHO they do what they want with their stuff, if I am recommending I try to suggest the CORRECT way. Often you can cheat, I cheat on my own stuff but not for sale, or as a recommendation.

In the past my income depended on being right. I got called in to fix dozens of systems where the installer cheated.

[Old Guy]

Blundered across this....may be of interest?

Altec 21531 Diaphragm 8 ohm, used in 291-16A drivers - eBay (item 200563941826 end time Jan-17-11 01:37:37 PST)

[kevin@kab]

Hi,

I think you missed my point.
Here is the direct example:
In a stereo system, one channel/driver has a 8 ohm phragm and the other channel/driver has a 16 ohm phragm
A stereo amplifier with a common volume control drives these 2 speakers.
Will they play at the same level?
No; because less power is being developed in one vs the other.
Why? because the amp is delivering the same voltage to both. This was my point.

The question remains that for a given voltage how much difference in level would one expect to see? Thats all.
Only 1/2 the power will be developed in one vs the other Power = E**2/R.
But would that mean that one is 6db louder?
I was trying to get a feel of that from anyone who may have had this experience.

Kevin
Jan 10, 2011

[Steve Mac]

4315A Impedance Curve?
Maybe this"ll help?.... an impedance plot?
Impedance plots for Altec drivers on particular horns...did I forget where this info is stored? I wish this info for all drivers was collected in an easy to search library here somewhere..

[aditya]

Hi,

I think you missed my point.
Here is the direct example:
In a stereo system, one channel/driver has a 8 ohm phragm and the other channel/driver has a 16 ohm phragm
A stereo amplifier with a common volume control drives these 2 speakers.
Will they play at the same level?
No; because less power is being developed in one vs the other.
Why? because the amp is delivering the same voltage to both. This was my point.

The question remains that for a given voltage how much difference in level would one expect to see? Thats all.
Only 1/2 the power will be developed in one vs the other Power = E**2/R.
But would that mean that one is 6db louder?
I was trying to get a feel of that from anyone who may have had this experience.

Kevin
Jan 10, 2011

Speaking electrically....there will be a power difference of 3db here, as the voltage is constant (assumed to be so). Had the voltage been half but ohms the same, then the difference would have been 6db.

Acoustically speaking.....a difference will be there in the output. But it will rather affect the sound quality rather than the intensity. I mean the perception of bass will be slightly better for the 8 ohms unit, while the 16 ohms will sound a bit thin.

It is because though there is an extra of 8 ohms in one driver, but still it is a part of the motor winding mechanism and not just an external peice of resistance, and hence it will give the motor more sensitivity (more turns per unit gap-area). But due to lack of current (and loudspeakers being current devices), the LF area, which needs a lot of it, will be affected, and thus the thin sound. For HF drivers it should not be so.

Actually this topic in its detailed form is very complicated and vast as it enters into the domain of Loudspeaker engineering, and one can produce a 16 ohms driver having the same no of turns as an 8 ohms unit will have...

Aditya

[Old Guy]

Kevin...you didn't state a common amp in your first post, so I had to cover all bases.

the volume difference is small...half power is 3 Db, NOT 6. 6 is 1/4 power.

OTOH I don't think you read where I said 16 ohms on an 8 ohm crossover will cause a screwed up response. You should do the parallel resistor as stated.

Really, you should get an 8 ohm fram.

What you want to do can be done, as can a lot of things. It just isn't a good way.

It isn't just the volume! The mismatch will mess with your crossover. I don't think you understand that part.

[kevin@kab]

thanks for the responses.
I did see and understand the interaction with a passive xover, my system uses an active crossover with separate amps.

that is why the post was so specific. I really did not expect to invite so much speculation!

it is hard to find nos diaphragms. So a 16 ohm one popped up on ebay and i grabbed it.
I did see the 8 ohm one you mentioned, but it is damaged. And I have found that while a damaged diaphragm can sound ok at low levels, if you sweep it with a sine wave, you can find alot of problems. my one driver has some very minor line like marks in it and below a certain level it sounds fine, but at the right level it begins doubling. So I wanted to replace it.

anyway, If I choose to replace it I will let you know what I find out.

I also now have a pair of nos 291-8L's so I will likely put them in and sell the older 291's with the spare 16 ohm diaphragm.

Kevin

Jan 11 2011

[Old Guy]

Actually your post wasn't very specific.

Had you posted active with separate amps you would have been told very little issues. A slight volume difference of 3 db. Easily fixed with a knob tweak.

It is impossible to have any kind of EXACT formula, you need to understand every amp is different.

Theoretically, an amp should double from 16 to 8, 3 dB. barely audible actually, to most.

OTOH some amps do NOT double when the impedance is halved, so the difference would be less.
Math is for approximations, reality is for test. And of course the final test/reality is are you happy with it.

i still recommend an 8 ohm, if for no other reason should you want to change to a different model driver, say 288-G, the resale will be crippled by the mismatched pair.

[techmecca]

I read the Q&A and want to write out a response to see if I can explain this more clearly to myself, and to anyone else who wants to read me typing to myself...

sensitivity, in this context, is stated as a sound pressure level, in dB_SPL, that is measured at a specific distance from a speaker/transducer, when a specific power level is applied to the device under test... sometimes the rest of the test conditions are described and specified but every manufacturer has differences in the details of how they set up measure and present their specs.

so you have a pair loudspeakers/drivers, in this case 291-16A's

what are those - I don't remember, some sort of compression driver...
let's look it up ... ...

oh yeah - these are sort of like 288's (I used to have a pair of those)

291-16A

Sensitivity is stated as 107dB SPL @ 1 watt input pink noise 500-3KHz bandwidth measured 4' from mouth of 30" horn (ref.: 0.0002 dyne/cm^2 for 1 Watt input)

there is no 8 ohm version of 291...

so we're not going to luck out and get direct data off a spec sheet

so assume you have a power amplifier with a low output impedance so you can call it an AC voltage source... and that this amplifier is connected directly to the driver under test

all else being equal except for the winding impedance of the voice coils, for a given applied voltage you will have twice as much power developed in the 8 ohm driver as you will in the 16 ohm driver

expand on this...

from

ohms law... Volts = Amps x Ohms

and power... Watts = Amps x Volts

Watts = Volts^2 / Ohms


note that none of this takes into account phase angle and other factors that happen when you are talking ac currents flowing through inductors and the various interactive springs and masses and couplings in the magnetic circuits and such ... speakers are not linear devices

so with Watts = Volts^2 / Ohms you have an inverse linear relationship between load/speaker ohms and power when voltage is held constant... double the ohms and half the Watts

okay...

in dB terms - you calculate power dB as 10 x log of a ratio of the difference between one value and another... one of these values could be a reference like 1mW is for dBm or we could be talking about just relative power levels in our different voice coil drivers

so with half the power that is a ratio of ?

10 x log 0.5 = -3.010299956639812 dB

and with twice power

10 x log 2.0 = 3.010299956639812 dB

(gotta love digital calculators - they don't know when to stop)

so... 3dB difference in Watts developed in the two drivers

and it turns out that, when everything is operating within design limits, sound pressure level measurement tracks with the power level such that a 3dB change in power in a driver will result in a corresponding 3dB change in measured dB_SPL

so... answer is yes and the level difference is about 3dB less level from the 16 ohm driver

that was the long way 'round

but... this is a fairly specific situation and assumes two identical compression drivers - other than voice coil impedances - driven directly from a modern power amplifier with practically zero output impedance (voltage source)

if you have a tube amp with a step-DOWN output transformer that has multiple impedance output taps you are in another realm... here we on the path to talking about power transfer, some stuff about tube output coupling through transformer, plate loading, power matching... if you have a stereo amplifier one driver to a channel you can put the 8 ohm driver on the 8 ohm tap and the 16 ohm driver on the 16 ohm tap and you'll be doing okay - that will work - BUT you can't put them both on two taps of one transformer because tap loading is interactive

if you are feeding THROUGH a passive crossover then we are really getting in to reactive interactions that take us away from the simple ohms law figuring... and that makes everything above non-applicable because you are messing with filter response AND level... it's a network and there is really nothing you can do... no you can't put an 8 ohm resistor in series with the 8 ohm driver to make it work with a 16 ohm crossover... it will not match a 16 ohm driver



IF you have a stereo power amplifier and you are directly driving one channel to the 16 ohm driver and the other channel to the 8 ohm driver ... so you are actively filtering and crossing over before the power amplifier ... and there are no voltage gain trims on the power amplifier or level trims on the output of the crossover you COULD put a pad on the input to the power amplifier channel feeding the 8 ohm driver...

this would be a voltage divider so you need to knock the voltage feeding the 8 ohm driver by 3dB

voltage dB's = 20 times the log of the ratio - somebody was thinking when they came up with this stuff because the the "20" multiplier takes into account the V^2 relationship to Watts

so we want to work backwards from 3dB to figure out a voltage divider that will knock down the level BEFORE the input to the power amp

3 / 20 = 0.15 and take the reverse of the log... so 10^0.15 = 1.412537544622754

ooops! forgot this is a loss not a gain so it's really

10^-0.15 = 0.707945784384138 ... yeah that's the number and it's the reciprocal of the gain... makes sense


so if you normalize a voltage divider to 10K ohms

you want to... this will look like crap unless you are using a monospace type

signal->----\/\/\/\/\/--•------input to power amp
|
|
\
/
\
/
\
/
|
|
common->----------------•-------common to power amp

[oh yeah - this doesn't work... - okay it's a series resistor inline with source signal and following that a shunt resistor to signal common - assuming unbalanced bridging input to power amp]

the series R would be close to 3K and shunt R would be close to 7K... or 2K9 and 7K1 - something like that... a trim pot and sound level meter might be better a solution

anyway - with this pad on the input to the power amplifier channel feeding directly to the 8 ohm driver, the power developed in that driver will be knocked down to equal the power developed in 16 ohm driver on the other, unpadded, power amplifier channel, given same signal applied to both the padded and unpadded channels

okay - getting close to the end here...

Back to the 8 ohm resistor in series with 8 ohm driver to replace 16 ohm driver thing that I mentioned in passing above... that does not work because you are making a voltage divider, so the voltage at the driver goes in half and that results in a 6dB change in power in the driver... the crossover would see the 16 ohm load and be a little less unhappy (or not)

BUT!

IF you for some reason need to hang BOTH DRIVERS across the SAME AMPLIFIER OUTPUT... and the amplifier has no problem with this, and there is no passive crossover to deal with, and you really don't want to just spring for a new diaphragm or two so you can actually have matched-impedance drivers... IF all that is true... then I guess you could cheeze-ball it and put a 10 or 20 Watt 3.3 Ohm resistor (non-inductive) in series with the 8 ohm driver, which will knock the power in that driver, and the output SPL, by 3dB... it's not pretty and that driver will not be well-damped but... there you go

how was that?