I need help, I'm looking to know what's the best load for Audio Technica OC-09 using 1:10 SUT.
I read somewhere on the net is 20 Ohms, is OK or is other value?
TIA
Felipe
I read somewhere on the net is 20 Ohms, is OK or is other value?
TIA
Felipe
How are you proposing getting this 100 ohm load (as seen by the cart) when:
* the default on the MM phono stage is probably 47K, and
* the 1:10 SUT converts this to 470 ohms?
Andy
Andy I guess the impedance seen by the cartridge, so adding 12.702 ohms to the SUT secondary between signal and ground.
Not according to my maths, mem! 🙂
Please let mkane77g elaborate on his recommendation - then we can continue this conversation.
I'm afraid I don't know what is in fact the best R load to use with an AT OC-09 but the simplest way IMO to get the best load for your AT OC-09 is not to use a SUT in front of your MM phono stage but, instead, use a 10x ss 'head amp' (aka phono pre-pre amp). Using a head amp between the cart and your MM phono stage enables you to put loading plugs on its input, to reduce the default 47K to whatever your cart likes. So you can try different load plugs from 20 ohms ... to XK ohms. 😀 (Just as an example, I am using 4,700 ohms as the load on my Benz LP.)
Andy
Last edited:
Andy my maths for 100 ohms load
With a gain of 1:10, you need an impedance of 10000 ohms on yor Moving Magnet input (10^2*100).
The resistors must have a value equal to:
Rload = 1/(1/R1 - 1/R2)
Where
R1 is the impedance you want for your Moving Magnet input and R2 is the actual impedance of your Moving Magnet input
Rload = 1/(1/10000 - 1/47000) = 12658 ohms
If you don't want to post here please send me a PM with your maths.
With a gain of 1:10, you need an impedance of 10000 ohms on yor Moving Magnet input (10^2*100).
The resistors must have a value equal to:
Rload = 1/(1/R1 - 1/R2)
Where
R1 is the impedance you want for your Moving Magnet input and R2 is the actual impedance of your Moving Magnet input
Rload = 1/(1/10000 - 1/47000) = 12658 ohms
If you don't want to post here please send me a PM with your maths.
Felipe we had a conversation on this theme weeks ago, with all ins and outs!
Kevin (and I) provided you with all necessary information.
Now you start it all over again, and again not including all issues.
What's your point??
Kevin (and I) provided you with all necessary information.
Now you start it all over again, and again not including all issues.
What's your point??
Got the information about loading the AT-OC9 with SUT, that's the information I was looking for.
We explained that there is no single load impedance "best"; it depends on other factors too.
Also, your information on the AT-OC9 only specifies a recommended load impedance through headamp or SUT, and does not take into account the specifications of whatever SUT which determines the real load impedance.
We (tried to) explain this over and over.
How stubborn can one be?????
Also, your information on the AT-OC9 only specifies a recommended load impedance through headamp or SUT, and does not take into account the specifications of whatever SUT which determines the real load impedance.
We (tried to) explain this over and over.
How stubborn can one be?????
Could you please link to this discussion?
Would appreciate double checking my math as well - for a standard MM input of 47k, assuming a 1:10 SUT and a desired 100ohm loading, my math suggests adding 12.7kohm parallel to the MM input (or the SUT secondary).
This explains it all:
http://www.diyaudio.com/forums/analogue-source/302821-s-u-t-valve-tube-riaa-s.html, starting with # 47
http://www.diyaudio.com/forums/analogue-source/302821-s-u-t-valve-tube-riaa-s.html, starting with # 47
Last edited by a moderator:
Jensen JT-44K-DX
Primary DC resistance of that SUT is 3 ohm.
Secondary DC resistance in 1:10 mode is 950 ohm.
Now we can calculate the extra load (actually the copper loss), which is the (square of the step up ratio x primary DC resistance) + secondary DC resistance.
In our example the extra load is (10² x 3) + 950 = 1250 ohm.
Now, to calculate a real 100 ohm load resistance, these 1250 ohm must be added to the (simplistic) 10k (10k is 100 (load impedance) x 10² (square of step up ratio).
So we must present the cartridge through the SUT with a load of 10k + 1250 ohm = 11250 ohm.
To get this resistance the standard 47k input resistor of the phono preamp must be parallelled with a 11k300 ohm resistor nearest standard value.
Now Pieter?
Primary DC resistance of that SUT is 3 ohm.
Secondary DC resistance in 1:10 mode is 950 ohm.
Now we can calculate the extra load (actually the copper loss), which is the (square of the step up ratio x primary DC resistance) + secondary DC resistance.
In our example the extra load is (10² x 3) + 950 = 1250 ohm.
Now, to calculate a real 100 ohm load resistance, these 1250 ohm must be added to the (simplistic) 10k (10k is 100 (load impedance) x 10² (square of step up ratio).
So we must present the cartridge through the SUT with a load of 10k + 1250 ohm = 11250 ohm.
To get this resistance the standard 47k input resistor of the phono preamp must be parallelled with a 11k300 ohm resistor nearest standard value.
Now Pieter?
Last edited:
mc step-up transformers explained - http://www.sengpielaudio.com/calculator-paralresist.htm
I may have jumped the gun 'Sir' andyr. You can't use loading plugs with a step up? A 12k resistor possibly
I may have jumped the gun 'Sir' andyr. You can't use loading plugs with a step up? A 12k resistor possibly
Last edited:
- Status
- Not open for further replies.