Source XLR to Balanced Preamp will increase gain?


I was wondering, if I changed my CDP to a new one that has XLR out and if I can use XLR between the new CDP and the preamp, will it provide me with more gain in the system? I believe this was the case when I changed the connection between my preamp and the amp. I was told that the gain was because each "leg" of the XLR (+ve and -ve) provided 2V each (4V total), as compared to RCA which provided only 2V. In the near future, I do plan to change my CDP.

Thanks!
128x128milpai
This may depend because you are talking about 2 different CD players. On my CD player, which have both RCA and XLR outputs and which is directly connected to the amp, the XLR output is about 6dB louder than the RCA connection.
A properly designed balanced output will be at a higher level than an unbalanced output. I'm not sure it's 6 dB though. I seem to recall once I compared the output of my tuner (it has both balanced and unbalanced outputs) using interstation hiss, and measured the balanced output at 3 dB higher.
Thank You for your perspective on the XLR connections.
The thing that seems tricky is that some CDPs mention the same output level 2.2V (for example) in both RCA and XLR connections while some other CDPs mention exactly twice the output level in XLR compared to RCA. This seem to correspond to the 6db higher output level that you mention here.
From an article on the web, I got this formula:

3dB = doubling the amplitude. If you have a reference signal at 1 volt, 3dB is 2volts, 6dB is 4 volts, and 9dB is 8 volts.

With that in mind, if I choose a CDP with say, 4V output at XLR, will it mean that my amp will ultimately see 8V at the MAX level of the preamp's output?
In many cases, but not always, there will be a 6 db increase going from unbalanced to balanced, everything else being equal. It depends on the designs of both of the interconnected components.

But also, of course, as Jcharvet pointed out different CDPs may have different output levels for the same connector type. So everything else may not be equal.
From an article on the web, I got this formula:

3dB = doubling the amplitude. If you have a reference signal at 1 volt, 3dB is 2volts, 6dB is 4 volts, and 9dB is 8 volts.
That is not correct. A 3 db increase = doubling of power. A 6 db increase = doubling of amplitude/voltage, which corresponds to a 4x increase in power. (As you may recall, power is proportional to voltage squared, for a given impedance).

To be completely precise, this assumes that impedances remain the same as power or voltage is doubled.

Regards,
-- Al