I'm wondering what's the most efficient way to use a cmoy amp + device. Should I turn the knob full on the amp and keep the volume low on the device or vice versa?
I'm using a JDS V2.02 cmoy amp.
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I'm wondering what's the most efficient way to use a cmoy amp + device. Should I turn the knob full on the amp and keep the volume low on the device or vice versa?
I'm using a JDS V2.02 cmoy amp.
You should keep the player volume high. Ideally you'd want to keep the cmoy above 9 or 10 o'clock to minimize any channel balance issues, and it might be a good idea to adjust player volume if you'd be turning the amp lower than that.
Anyone else want to chime in on this (and maybe why this is) because I was wondering the same thing.
My cMoy with a 12V regulated DC power supply is connected to my iPod (running rockbox) through a Fiio L3 LOD cable.
When the cMoy is maxed out (dial at 5 o'clock), I listen to music at about -88dB on the iPod, which is almost as low as it will go.
At about half volume on the cmoy (dial at 12 o'clock) I run the iPod at -77dB.
At a little less than 1/4 volume on the cmoy (dial at 9 o'clock) I run the iPod at about -60dB.
When I used to run it through the iPod's headphone jack, I would listen to it at about -30dB with the cMoy maxed out.
It depends a bit on the equipment.
This is my view:
If the source does not have any distorsion issues when cranked to the max and the amp can handle the input then that will be potentially best for the following reasons:
I have a BSG CMOY with dual inputs. 1 for headphone level and 1 for line level (less gain).
I use the line level for both my USB DAC and Android phone (at max volume). In this case the headphone amp merely works as a buffer. It takes the load (current) of the source and drives the headphones at a level which is usually at a lower or the same volume as when I would plug it straight into the source.
Works best for me but different equipment (and ears) could have a different optimum.
When I used a cheapish android tablet I had better results at 60-70% than 100% as it was probably not as capable at providing the the full output voltage.