Low output MC - Math Question ??


Calculators Ready... I'm presently using a low output- .3mV
MC cartridge with the Manley Steelhead phono stage. I prefer the sound of the cartridge thru the MM section (no step-up transformer) to the MC section w/ step-up. The MM section has a maximum gain of 65db. However, there is not enough gain in the MM section to allow me to run the Steelhead direct into the amps, so I must use a pre-amp (ARC REF 2Mk.1 w/ 18db gain) to achieve the correct gain structure. BTW- This combo sound pretty good! Now the Math Question- What is the relationship between cartridge output and db gain? Specifically how high of a cartridge output would I need to run into the Steelhead MM section set a 65db gain to realize 80+db of total gain, so as to have enough gain to drive the amps directly??? Hopefully I have not oversimplified this equation because of a lack of information.
fbhifi
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...
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.
Herman: Thank you very much for your precise answer to my question, it is exactly what I was looking for.
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