DIY phono loading plugs question


What value resistor (?) would I need to create a 100 ohm phono loading plug for my cartridge? My Ayon Polaris 3 preamp uses RCA plugs for cartridge loading.  I've got some RCA plugs ready and now just need to figure out what value resistors I would need to solder into the plugs to get to 100 ohms.
wcheng2
I purchased a pair of Vishay VAR (I think they're called "naked") resistors and put them into a pair of expensive Furutech 102R RCA plugs.  Only got to compare them for a few songs, but I'm hearing a lower noise floor compared to my old plugs.  Thanks for everyone's help
@wcheng2

The formula for parallel resistors is: R = (R1xR2) / (R1+R2). If R1 is 47K and R2 (your plug in resistor) is 100R, then multiply the two. You will get a large value. Next step, add R1 and R2. You will get a much smaller value. Now divide the larger value with the smaller value to get the final value. Pretty easy.


Thanks for the responses guys.  I think the Ayon Polaris III's loading without any plugs is 47kohms so i will purchase some 100 ohm resistors to make the plugs.
Jjss49, no, the phono section is for low output moving coils.  My post did not imply anything to the contrary.  As I said, I suspect it has a very high input impedance, resulting in the load applied to the LOMC being essentially the same as the value of the resistor on the external loading plug, if that resistor has a low value such as 100 ohms.

Regards,
--Al
 
omg i did not realize he was trying to do that to a 47k ohm mm phono section - LOL

well, never mind..... :)
The input impedance of the Polaris III’s phono input when no loading plug is connected does not appear to be specified. I suspect, however, that it is vastly higher than 100 ohms, perhaps 47,000 ohms or even much more. (I note that an input impedance spec of 1,000,000 ohms is stated, which I suspect corresponds to its line-level inputs).

Assuming that is the case (and you might want to confirm that assumption by contacting Ayon), using a 100 ohm resistor would result in a total load impedance that is within a small fraction of an ohm of 100 ohms.

Regards,
-- Al