PX-25 tube amp with Merlin VSM speakers?

Neglecting room effects, it can be calculated that at a 10 foot listening distance 6 watts into two 89 db 8 ohm speakers will result in an SPL (volume) of 90 db. That will be loud enough for most listeners with most recordings, but it will certainly not be loud enough for many listeners with recordings having particularly wide dynamic range (such as many classical symphonic recordings).

Although many and probably most pop and rock recordings are compressed to dynamic ranges (the difference in volume between the loudest and the softest notes) of less than 10 db, lots of music in other genres can have dynamic ranges of 30, 40, or even 50 db. A 30 db dynamic range means that 1,000 times as much power is required for the loudest notes as for the softest notes. A 40 db dynamic range means that 10,000 times as much power is required for the loudest notes as for the softest notes.

Also, I would by no means count on the 6 watt SET as being able to sound its best when putting out close to that number of watts.

Finally, I note that the **minimum** power recommended at the Merlin site for your speakers, for a tube amplifier, is 17 watts. Unless essentially all of your listening is to dynamically compressed recordings at moderate or low volumes, why "upgrade" your amp to something that is likely to disappoint on at least a few (and perhaps many) of the recordings you listen to?

IMO. Regards,
-- Al

Thank you Saki.

I wouldn't be surprised if there were online calculators for figuring this kind of thing, but I'm not specifically aware of any. I just do the calculations myself, using a scientific calculator that can do logarithms. If you don't have one, the calculator that is built into Windows (under "Accessories") can be set to scientific mode via its "View" menu.

These are the steps:

1)Compute how many db greater than 1 watt the amp's specified power rating is, based on the relation db = 10log(P1/P2), where "log" is the base 10 logarithm.

In this case 10log(6 watts/1 watt) = just under 8 db.

2)The OP's speaker is rated at 89 db/1 watt/1 meter. Therefore 6 watts will produce 89 + 8 = 97 db at 1 meter.

3)Add 3 db for the second speaker. (If identical signals are received by both speakers, i.e., if the image is centered between them, and if the listener is perfectly centered, I believe that a 6 db increase would result. But it is safer to count on a 3 db increase, reflecting twice as much power being delivered in total compared with the power being delivered to one speaker). 97 + 3 = 100 db at 1 meter.

4)Calculate the reduction in SPL at distances greater than 1 meter as 20log(1 meter/distance in meters). The example I described of a 10 foot listening distance corresponds to 3.05 meters. 20log(1/3.05) = approximately -10 db. 100 - 10 = 90 db, as I indicated.

Some caveats:

1)For planar speakers, such as Maggies and electrostatics, SPL will decrease at a significantly slower rate than this as distance increases. The 20log(1/distance) figure applies to box-type speakers.

2)Speaker sensitivity/efficiency is often spec'd based on an input of 2.83 volts rather than 1 watt. 2.83 volts into 8 ohms corresponds to 1 watt, so the calculation comes out the same either way in the case of an 8 ohm speaker (assuming the 8 ohm impedance spec is reasonably accurate). 2.83 volts into 4 ohms corresponds to 2 watts, though, which is 3 db greater than 1 watt. So if the sensitivity of a 4 ohm speaker is spec'd on the basis of an input of 2.83 volts, subtract 3 db from its specified sensitivity to determine the SPL the speaker will produce in response to 1 watt at 1 meter.

3)As I indicated in my previous post, these calculations neglect room effects.

This all may sound fairly complicated, but it's pretty easy once you get the hang of it.

Happy holidays to you as well! Best regards,
-- Al