Cartridge loading explained

[andyr]

The switch positions on the Denon AU-320 don't refer to the load presented to the cartridge. They refer to the impedance of the phono cartridge that should be matched to those switch positions.
So, the 3 ohm position is ideal for cartridges of 3 ohms internal impedance.
The 40 ohm position is ideal for cartridges, like the DL-103, that have an internal impedance of 40 ohms.

If you have the usual 47k ohm phono input resistance, the load presented to the cartridge will be 470 ohms in the 40 ohm position and 36 ohms in the 3 ohm position.

Manual is in the library here.

More info here:
https://www.vinylengine.com/step-ups-an ... dges.shtml

Well, that sure makes a whole lot more sense. Thank you!

Except, see my post above! (So hopefully, it didn't make all that much sense! :D )

When Sunwire posted that "The 40 ohm position is ideal for cartridges, like the DL-103, that have an internal impedance of 40 ohms" what he was implying was that the resulting load of 470 ohms would suit the DL103 ... as 470 is approx 10x the coil impedance (which I understand is 40 ohms).

However, that is not necessarily the load which delivers the optimal sound of the 103. I used to run my Benz LP at 100x the coil impedance - which was 3,300 ohms. Something you can't achieve with a SUT, because of the "square of the turns" impediment. :(

Andy

[jagermonster]

However, that is not necessarily the load which delivers the optimal sound of the 103. I used to run my Benz LP at 100x the coil impedance - which was 3,300 ohms. Something you can't achieve with a SUT, because of the "square of the turns" impediment. :(

Andy

Great - now I'm going to have to fuss with the loading on the MC section of my phono pre and decide the fate of the AU320 once and for all...

[Sunwire]

Sorry, Sunwire, your maths is not quite right.

Previously, jagermonster posted that the '40 ohm' position selects a 1:10 gain ratio, whilst the '3 ohm' position selected the 1:20 gain ratio. The load presented to the cart is the MM phono stage Zin divided by the square of the turns ratio. So:
* yes, the load seen at the 1:10 gain ratio is 470 ohms
* but the load seen at the 1:20 gain ratio is 118 ohms (47,000 / 400).

Andy, you are assuming jagermonster was correct about the turns ratios. He's not, according to the link I posted previously.

[andyr]

Andy, you are assuming jagermonster was correct about the turns ratios. He's not, according to the link I posted previously.

I was - my apologies to you for a wrong assumption. I see from your article that the second position is in fact 1:36 - so are absolutely correct ... this produces a load of 36 ohms for the cart.

Andy

[jagermonster]

Sorry, Sunwire, your maths is not quite right.

Previously, jagermonster posted that the '40 ohm' position selects a 1:10 gain ratio, whilst the '3 ohm' position selected the 1:20 gain ratio. The load presented to the cart is the MM phono stage Zin divided by the square of the turns ratio. So:
* yes, the load seen at the 1:10 gain ratio is 470 ohms
* but the load seen at the 1:20 gain ratio is 118 ohms (47,000 / 400).

Andy, you are assuming jagermonster was correct about the turns ratios. He's not, according to the link I posted previously.

Could have sworn the manual said 1:10 for the 40 Ohm setting. Thanks for proof reading.

[Sunwire]

Yes, the turns ratio for the 40 ohm setting is 1:10.

But the 3 ohm ratio is not 1:20, at least according to that article I posted the link to.

And if you look closely at the manual, you'll see the impedance of the secondary coil is 4 kohm.
If we divide that by the 3 ohm primary, we get 1333.3.
Square root of 1333.3 is 36.5, so we get a ratio of 1:36.5 for the 3 ohm setting.

4 kohm divided by 40 ohms is 100. Square root of 100 is 10. So, turns ratio for the 40 ohm setting is 1:10.

PLEASE CHECK MY MATH, GUYS!

[jagermonster]

Look, pal, if you keep up with the numbers, I'm going to have a stroke - do you want my death on your conscience?!

[Sunwire]

:D