Phono cartridge left / right output voltage differ

Just received a new Denon DL-S1 cartridge and noticed that the output voltage for the left and right channels are different.The left is .16 mv and the right is .18 mv.Though this is only a .02 mv difference it turns out to be about a 12% difference between channels.

Will I hear this channel imbalance ? Should I send it back ?

Are you sure set-up is correct?
Sorry I didn't mention that this reading is straight off the spec sheet that came with the cartridge.My last DL-S1 spec sheet read .15 mv for both channels.So I expected this new one to have the same output in both channels too.

Not sure if there will be a big enough difference in volume to hear between the right and left channel.So seeking advise.
I doubt that they would have let it pass QC if they thought it would be a problem. Did you think to try it and listen for yourself? My guess is that you will not, but now that you know they are different you will convince yourself that there is a difference and will forever be in doubt.
That translates into (almost) exactly a 1.0 decibel imbalance. It'll be quite audible, and some (myself included) will find it extremely annoying. Others are oblivious to channel imbalance until it's more severe.

Unfortunately - until you get into the big money cartridges - many models are spec'd by the manufacturer at no better than a 1.5 dB channel differential. Some models are spec'd at 1.0dB, and once you get into the high end, 0.5dB is more common. So your cart is almost certainly considered normal - especially since it was stated right on the spec sheet!

You could try to mitigate this by playing off other imbalances that may exist downstream. E.g. if you have an imbalance due to asymmetric room acoustics (and you don't mind swapping L/R). Or, you could have a custom 1.0dB attenuator built and place it between the weaker channel's phono -> preamp link. Or you could return/sell that cart and keep looking.
It's irritating when this happens. We expect "new" to be perfect and free of problems. I bought a DL-103 from AudioCubes years ago and it happened to me, but I don't remember hearing a difference between channels.

Should you complain to the vendor? I don't have a strong opinion one way or the other, but I wish you luck.
The difference is ~.1 dB. No, I don't think you can hear that.
This difference is about a .5 db difference. That is well within most cartridge specifications. One can hear a .5 db difference as a subtle shift in the position of the center image in a set up where the center image is narrowly drawn, but, the shift will be hardly an issue (if it is an issue to you, then you MUST have a balance control, preferably one that can be operated by remote control.
I had a 103r that had about the same percentage channel imbalance. I bought it from a discout store so I figured it was how denon filtered out the better cartridges from the poorer ones and still sell them hence making a profit versus reworking them or scraping them; my guess (opinion). Can't say I could hear the imbalance during play back but I was always concerned about it.

I sold the cartridge after having snapped off the cantilever and replaced it with a Benz Wood SL.

That is a difference of almost exactly 1 db, which is quite small. But I would be hesitant to say that it is completely insignificant. If your preamp has a balance control that is either continuously adjustable or has small steps near the center position, you should be able to determine if there is any audible significance, and compensate if necessary.

-- Al
The left is .16 mv and the right is .18 mv.

That is great for a Denon.
Just to reiterate - Almarg (plus my own posting) is correct - the difference in your cart is *definitely* 1.0 dB, because:
20 * Log_base_10(16/18) = -1.023

^ You get the same magnitude of result, but with a positive sign, if you flip the numerator and denominator. The multiplying factor here is 20, not 10, because we're dealing with voltage proportions, and power (as eventually driven through your speakers) is proportional to the *square* of the voltage (hence the extra factor of 2).

And that's absolutely quite audible, though I've been surprised by how tolerant some folks (even hi-end folks) are about imbalances up to ~ 1.5dB. It's FAR easier to discern a 1dB difference in L/R channel level than it is to notice a 1dB change in level for both channels. In my system, center image would seem shifted by about a foot. I've found myself annoyed by imbalances down to 0.5dB or even less (an almost certainty with vinyl).
Well, we have three different statements of what the difference is in db. I would re-assert that it is almost exactly 1.0 db, as I and Mulveling indicated. 20log(0.18/0.16) or 20log(0.16/0.18).

-- Al
Al is correct.

1.02305044895... dB difference

First I'd like to thank all of you for help in making a decision on what to do.All your comments have been appreciated.

I finally got something I could sink my teeth in after being told that the difference between the channels would be somewhere around 1 db or less.Thanks for that.That doesn't sound like a lot and if it turns out to be a problem I do have a balance control on my preamp.

So I've mounted and aligned the cartridge and had time to casually listen to one side of a Lp.Of coarse in the back of my head I think the right channel is louder but I'm really not sure.I will have some time over the weekend to listen more critically.Plus I can compare it to another cartridge.

Thanks again for your help
Yes, Al and Dre win the cupie doll. I glanced past the milli. :-)
I must have remember the formula incorrectly--I thought it was 10 log (P1/P2).

In any case, one db is not that big a difference if you have a balance control.
I must have remember the formula incorrectly--I thought it was 10 log (P1/P2).
It is. However we are dealing with voltage, not power, and it is 20log(V1/V2) as Mulveling explained. Since for a given load power is proportional to the square of voltage, the logarithm gets multiplied by an additional factor of 2 when calculating from voltages.

log(X squared) = 2(log(X))

-- Al
It is also mathematically correct to compute 20 log for each voltage and subtract the results. However, you have to get the decimal points correct for millivolt values. That was my mistake. Using the quotient method as Al posted avoids the pit I fell into. :-)