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Thread: 8ohm, 4ohm, 2ohm - Oh my!

  1. #1
    Indecisive Tuner BowDown's Avatar
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    8ohm, 4ohm, 2ohm - Oh my!

    Ok well after reading a thread on another forum I figured given the level of expertise here I would ask this question here..

    First let me explain what I think I know before starting to ask questions..

    It's my understanding that drivers sensitivity levels are created by using 2.83v. So using some crazy formula it's said that a:
    2ohm driver is actually rated using 4 watts
    4ohm driver is actually rated using 2 watts
    8ohm driver is actually rated using 1 watt

    So in turn an identical driver in a different ohm flavor should be able to produce the same output but with less amplifier overhead the higher the ohm rating? But with the exception of dual voice coil subs wired in different configurations. Wiring a dual in a higher ohm doesn't yield the same results as running it at the lower configuration.


    -So with this being said.. the idea of purchasing the matching ohm version of a driver to match your amps highest output configuration is a moot point provided the amp will output the proper scaled down power to another ohm version?

    -As ohm levels drop the resistance level of the voice coil gets closer to 0ohm or a dead short.. and in turn voltage drops. So with this said running a lower ohm level on an amp really does nothing more than create more current draw? Where does this current go if not turned into sound? Just dissipated as heat?

    -When ohm levels are presented to an amp, do voltage levels stay the same regardless? I know the higher the wattage on an amp, the higher the output voltage capability is. Does this stay constant with different loads even though wattage is doubled with each 1/2 of ohm?

    -Why not rate amps using voltage instead of watts then? Create a minimum load rating, and publish the voltage output?


    The idea that ohm ratings (and their resulted wattage impact on an amp) don't matter in overall output of the driver blows my mind. Am I understanding this correctly? If so, is it just the mass market of 'give me higher wattage numbers' driving the ohm rating of drivers down?
    Last edited by BowDown; 08-17-2011 at 08:47 AM.
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  2. #2
    Indecisive Tuner BowDown's Avatar
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    Let's leave passive xovers out of this discussion. I realize that different ohm loads result in different designs.
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  3. #3
    Founding Member earthtodan's Avatar
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    Not a crazy formula really, just ohm's law. To keep the test voltage the same while the impedance drops, the power has to rise, and vice versa, even though we talk about amp output in terms of power and not voltage.

    The thing is, Ant's argument depends on the sensitivity of an 8 ohm driver being twice that of an equivalent 4 ohm driver (or +3 dB @ 1M) due to some property of voice coil construction. I've combed through Madisound looking for drivers available in 4 ohms and 8 ohms and compared the sensitivity, and the +3 dB trait doesn't appear to exist. Maybe there's something I'm missing, but that's my take.

  4. #4
    Indecisive Tuner BowDown's Avatar
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    Efficiency = ( B^2 * L^2 ) / ( R * Sd^2 * Mms^2 )

    B = magnetic field strength
    L = length of wire
    R = resistance
    Sd = surface area
    Mms = mass
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  5. #5
    Indecisive Tuner BowDown's Avatar
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    You do have a good observation. I am curious how often the 3db trait does exist.
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  6. #6
    Founding Member earthtodan's Avatar
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    Actually, now that I thought about it again, I went back to Madisound and looked at driver specs again. I think the 3db trait does exist.

    When looking at drivers rated at 2.83V, the sensitivity of the 8 ohm driver should be approximately -3dB relative to the equivalent 4 ohm. This is because it is being tested at half the power. So if you doubled the voltage, thus applying the same power, you would get the same sensitivity. This holds true for the driver pairs I found in the last 10 minutes.

    When looking at drivers rated at 1W, the sensitivity of the drivers should be the same for 4 ohm and 8 ohm, but the 4 ohm will be drawing twice the current.

    So, maybe Ant is right. However, there's another twist. Ant's point is that for equivalent power, you get the same output from an 8 ohm driver. However, to get the same power from an 8 ohm driver, your amp has to double the voltage. Amps are voltage limited (say 150V, I'm not really sure), at which point they peak. So maybe your amp will run cooler most of the time, but instead of its rated 50W per channel at 4 ohms, now it only has half that, and the max output of your system is down 3 dB.

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    Founding Member earthtodan's Avatar
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  8. #8
    Indecisive Tuner BowDown's Avatar
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    Well that was not what was implied from my interpretation of reading it. What you're saying now is what I had believed before. That due to the amp producing 1/2 the power at the 8ohm level it would be down 3db from the 4ohm.. but what ANT was saying is that due to the efficiency of the 8ohm driver being superior to that of the 4ohm it would be = in output.
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  9. #9
    Founding Member earthtodan's Avatar
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    Seems like it should be a simple concept, yet it does strain my brain. Anyway, I think the crux of the issue is, driver sensitivity is rated in watts or in volts, but we're concerned about sensitivity to current, because current causes heat. Voltage does not affect temperature. And from driver specs on Madisound, it looks to me like you have to double the voltage on an 8 ohm driver to get the same dB as a 4 ohm driver, thus using the same current. So I'm going to say, ignore Ant's point and don't cut down your headroom for no reason.

  10. #10
    Founding Member capnxtreme's Avatar
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    Higher impedance drivers do tend to be more sensitive, so yeah, it doesn't really matter (unless the impedance is too low for the amp to drive stably).

    There was a classic post by npdang about this back in the good ol' days.

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