Low output MC - Math Question ??

voltage gain = inverse log (dB gain / 20)
inverse log X is 10 raised to the power of X

You are looking to make up the difference of 15dB from the 80dB MC stage to the 65dB MM stage.

15dB divided by 20 = .75
take the inverse log of .75 = 5.6
.3 mV times 5.6 = 1.7 mV

So .3 mV amplified 80dB will have the same output as 1.7 mV amplified 65dB.

80dB is a gain of 10,000. So 10,000 times .3 mV = 3V
65dB is a gain of 1,778.. So 1,778 times 1.7 mV = 3V

Herman: Thank you very much for your precise answer to my question, it is exactly what I was looking for.

Another (perhaps simpler for quick mental calculation) way of looking at the math is to take off from the fact that each 6dB increase corresponds to a doubling of the voltage (and volume). So a 6dB gain increase represents a 2X voltage increase, a 12dB gain increase = a 4X voltage increase, a 18dB gain increase = an 8X voltage increase and so on. Therefore the 15dB+ gain increase you are looking for corresponds to about a 6X voltage increase, giving .3mv X 6 = around 1.8mv or greater needed from a cartridge to drive your phonostage to a 80dB+ output (yielding basically the same answer as Herman's formula above).

It's a pretty safe bet to assume that any phonostage which doesn't feature built-in attenuation control is designed with the assumption that it will be partnered with an active preamp (has gain), and not fed directly into the amp or passively attenuated, meaning it's conceivable that doing so (even with a HO cart) may not necessarily result in improved sound.

However, The Steelhead does have an attenuator, so I am surprised that with 65 dB, you can't drive your amps satisfactorily. Your amps must have low sensitivity(2V?) a. I run a Groove (65dB) direct into my Thor amps via a passive attenuator. Unless you are using a MC into MM inputs...