THD Hotplate power reduction question...

Discussion in 'Amps and Cabs' started by itgoesto11uc, May 17, 2006.

  1. itgoesto11uc

    itgoesto11uc Member

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    Does anyone know how much power would be getting to the speaker at the various settings of a 16 ohm Hotplate used with a 50 watt Marshall? What would be the least attenuation that would be safe with a single 12" Vox Bulldog rated at 15 watts?
     
  2. markp

    markp Member

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    3 db is like cutting power in half ,so -8 db would be like 7 watts or somethin like that.
     
  3. itgoesto11uc

    itgoesto11uc Member

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  4. hal9000

    hal9000 Member

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    Attenuation and resulting output power is calculated like this:

    Amplifier power = 50 W = 10*LOG(50W) = 17 dBW, and to convert back it's 10^(17 dBW/10) = 50 W, where LOG is base 10, and "^" means to the power of.

    Hot Plate setting and resulting power based on 50 W (17 dBW) of input power:
    -4 dB, 17 - 4 = 13 dBW = 20 W
    -8 dB, 17 - 8 = 9 dBW = 8 W
    -12 dB, 17 - 12 = 5 dBW = 3 W
    -16 dB, 17 - 16 = 1 dBW = 1 W

    So, even if your amp is producing twice the rated power with the Hot Plate set to -8 dB, you're still only going one watt over the rated power of the speaker.
     
  5. GAT

    GAT Gold Supporting Member

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    I was going to do the same calculation for you but I couldn't figure out how to turn on my calculator..:confused:
     
  6. itgoesto11uc

    itgoesto11uc Member

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    Thanks a lot for the info!
     
  7. Orren

    Orren Member

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    Great information!

    But...

    What happens if a setting reduces the dBW to a negative number? How do you calculate the watts output at that point?

    Thanks,
    Orren
     
  8. hal9000

    hal9000 Member

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    Orren, thanks. :)

    Logarithms are a bit strange if you're not used to them, and negative numbers are all part of the deal. In logarithms, addition => multiplication and subtraction => division. So, even if you end up with a negative number for dBW, it just means you're going less than one watt since that's the standard for dBW (Power relative to one watt).

    Example: What's the power output of a 50 W amplifier after 20 dB of attenuation?

    20 dB = 10^(20/10) = 100, or better, 20 dB of attenuation = 100 times less power. So, 50 W/ 100 = 0.5 W

    Let's do the calculation like the first example to show the similarity:
    50 W = 10*LOG(50W) = 17 dBW
    20 dB attenuation => 17 - 20 = -3 dB
    Output power = 10^(-3/10) = 0.5 W

    So there you go. I hope this makes sense.
     
  9. yannis

    yannis Member

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    Yep Log(X) can be negative if X is between (0,1), but X will be always X>0 otherwise Log(X) cannot be defined.
     
  10. hal9000

    hal9000 Member

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    Actually, it can be (imaginary numbers), but that's not going to help this discussion one bit. :)
     
  11. Orren

    Orren Member

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    See, all the above is in English, so I can understand it. It's when we start using numbers I get all out of wack. ;)

    That's where it begins to fall apart for me. I can intellectually understand what you're talking about, but when I plug in my own numbers, I get lost.

    For example, lets say that I have a 100W amp, and I'm applying 32dB of attenuation (basically, a Hot Plate with the Rotary knob at -16dB and the continuous knob half way). Here's the formula, as I see it:
    100W = 10*LOG(100W) = 20 dBW -- so far, so good!
    32dB attenuation => 20 - 32 = -12 dB -- again, so far I'm keeping up.
    Output power = 10^(-12/10) -- that much I understand.

    But....

    Using my Mac's scientific calculator X^Y function, 10^(-12/10) gets me a result of 8.8. So basically, it's figuring out that -12/10 = -1.2, and that 10^-1.2 = 8.8. I have absolutely no idea what to do with that number, or where I went wrong. Can you see where I went astray?

    This is why I became a writer. ;)

    Thanks,
    Orren
     
  12. hal9000

    hal9000 Member

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    Your math is good until the end. 10^(-12/10) = 0.063 W or 63 mW. Basically, with scalar power numbers that end up with negative dBW, you're just going below the reference of 1 W, but it won't go past 0. Also, from algebra, negative exponents mean you can place operation in the denominator to better understand why it's less than one. So, 10^(-1.2) = 1/(10^(1.2)) = 0.063 W = 63 mW.

    Your calculator may also have an ANTILOG(Base 10) function that will do the same. On my HP 48GX, ALOG(-1.2 dBW) = 0.063 W.

    BTW, with that much attenuation, I doubt the speaker is even rounding off the high end per normal operation. So, even with the FM curve, the speaker is also producing a lot more high end than normal. I actually prefer to use my Hot Plate as a load, and reamplify with a clean SS amp for that level of attenuation.
     

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