I have a Little Dot MK VI+ and love it. I first connected my HD 650's (300Ω), followed by my LCD2's (50Ω), and both had lots of power at relatively low volume (10o'clock position). Today I connected my LCD4's (200Ω) and the MK VI+ sounds great, but the volume position is now at the 2o'clock position. I understand the LCD4's are very power hungry, requiring 14 watts. This brings me to my question.
Is there any way to predict the power output in watts of the Little Dot MK VI+, given a particular impedance? For example, the reference manual states the power output as "5W + 5W (120 + 120 ohms)", and I've seen another power specification as 1W at 50 ohms. Given my albeit rusty memory of Ohm's Law, this doesn't seem to represent a perfect voltage or a perfect current source, so there's some other math involved in coming up with these specs.
What is the power output of a Little Dot MK VI+ into a 200Ω load, and how do I calculate it?
Thanks!
 Dave
Calculating MK VI+ output power for a given load?
Moderator: sword_yang
 ViperGeek
 Posts: 15
 Joined: Oct 28th, '15, 08:47
 Location: Merrimack, NH, USA
 Contact:

 Posts: 1718
 Joined: Mar 14th, '06, 01:25
Re: Calculating MK VI+ output power for a given load?
You're right there's no perfect current source since we are dealing with tubes and an OCL/OTL circuit. You need to check if there's enough voltage swing (which is greater with high impedance headphones) and Ohm's Law can't be used directly to calculate. The MK VI+'s highest voltage swing is 30V, so PV/R would be 30*30/200 = 4.5W per channel into 200 ohms.
 ViperGeek
 Posts: 15
 Joined: Oct 28th, '15, 08:47
 Location: Merrimack, NH, USA
 Contact:
Re: Calculating MK VI+ output power for a given load?
Perfect! P = V²/R
Thanks David.
 Dave
Thanks David.
 Dave