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?

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.

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

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

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.

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.

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

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.