Great video. How does the rpm of the pressing affect the loading? Seems like the faster the record is moving, the faster the cantilever must react when traveling as the groove progresses from the outer edge to the spindle. My guess is the higher load? Perhaps this makes no difference? I’d love to hear your thoughts. Thank you
Thanks for your info. Just to make certain, when the cartridge specs say [>10ohm; 470ohm nominal] which number are I using? This would be the number that I punch in the phono stage that reads load resistance. Thanks
I am now wishing that I never got into this SUT rabbit hole. In the 90s I was using a Denon DL304 into EAR 834 P and everything was fine except for some hard piano notes on some tracks. An impedance or loading mismatch? Now , its a big headache and a heartache (they cost quite a bit😢) with an Audio Technica AT650( 34 years old), an ICON Audio MCTX, a Phasemation T300 and a Ned Clayton( eBay) built to order SUT. Not to mention an ICON Audio PS1 , an Icon Audio PS2 , a NOS EAR834P Deluxe. Plus 2 different models in the Pro Ject Phono stage lineup. All in the hope of getting my Koetsu Rosewood to sound warm. Help!
Cartridge loading has to do with controlling the electrical resonance not the physical resonance (like we do with a loud speaker and the damping factor of an amplifier.) The input impedance would have to be very very low for it to affect the physical movements of the cantilever and at that point there would be no signal output. Lets say the cartridge impedance is 5 ohms, a 5 ohm load would represent a damping factor of 1. A damping factor of 1 does very little to control the movement. To truly control the movements of the cantilever you would need it to be critically damped and that would require a load much lower than 5 ohms. Again, at that point there would be very little signal.
Great explanation. I finally found the answer to my question about my loading.
Best video on this topic
Superb. You are the only person I have ever heard recognise that the loading resistance can have a minor effect on the cartridges tracking ability.
Great video. How does the rpm of the pressing affect the loading? Seems like the faster the record is moving, the faster the cantilever must react when traveling as the groove progresses from the outer edge to the spindle. My guess is the higher load? Perhaps this makes no difference? I’d love to hear your thoughts. Thank you
Thanks for your info. Just to make certain, when the cartridge specs say [>10ohm; 470ohm nominal] which number are I using?
This would be the number that I punch in the phono stage that reads load resistance. Thanks
Can you recommend a source for producing MC Cartridge parts? I will be making my own MC Cartridge.
Yes man, u have understood physics!
Unipivot vs fixed for micro line
What are your thoughts as far as tracking and distortion n the grove?
I am now wishing that I never got into this SUT rabbit hole. In the 90s I was using a Denon DL304 into EAR 834 P and everything was fine except for some hard piano notes on some tracks. An impedance or loading mismatch?
Now , its a big headache and a heartache (they cost quite a bit😢) with an Audio Technica AT650( 34 years old), an ICON Audio MCTX, a Phasemation T300 and a Ned Clayton( eBay) built to order SUT.
Not to mention an ICON Audio PS1 , an Icon Audio PS2 , a NOS EAR834P Deluxe.
Plus 2 different models in the Pro Ject Phono stage lineup.
All in the hope of getting my Koetsu Rosewood to sound warm.
Help!
Fantastic.
Cartridge loading has to do with controlling the electrical resonance not the physical resonance (like we do with a loud speaker and the damping factor of an amplifier.) The input impedance would have to be very very low for it to affect the physical movements of the cantilever and at that point there would be no signal output. Lets say the cartridge impedance is 5 ohms, a 5 ohm load would represent a damping factor of 1. A damping factor of 1 does very little to control the movement. To truly control the movements of the cantilever you would need it to be critically damped and that would require a load much lower than 5 ohms.
Again, at that point there would be very little signal.