Calculating MK VI+ output power for a given load?

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ViperGeek
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Calculating MK VI+ output power for a given load?

Post by ViperGeek »

I have a Little Dot MK VI+ and love it. I first connected my HD 650's (300Ω), followed by my LCD-2's (50Ω), and both had lots of power at relatively low volume (10-o'clock position). Today I connected my LCD-4's (200Ω) and the MK VI+ sounds great, but the volume position is now at the 2-o'clock position. I understand the LCD-4's are very power hungry, requiring 1-4 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

DavidZheZhe
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Re: Calculating MK VI+ output power for a given load?

Post by DavidZheZhe »

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 P-V/R would be 30*30/200 = 4.5W per channel into 200 ohms.

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ViperGeek
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Re: Calculating MK VI+ output power for a given load?

Post by ViperGeek »

Perfect! P = V²/R

Thanks David.

- Dave

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