Cartridge Loading and method

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Gerry
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#1 Cartridge Loading and method

Post by Gerry »

This harks back to Andrew's work suggestions at Witham all those years ago and primary vs secondary loading.
The topic has come up elsewhere and I thought I had it all sorted in my mind but I'm beginning to questions myself.

The idea was when loading on the primary, the 47K in the phonostage was replaced and the chosen resistor was placed on the primary side linked between the hot & cold on the input phono plug?
So if you wanted a selection say 20R; 40; 100; 220; etc these were placed on a selector switch (rotary) and wired across the H & C of the input phono?

The reason I ask is someone has done this, then measured the load and got readings from the input phono which don't equate to the resistor values they have used. I've checked mine and I don't either.

The Hagerman site suggests if loading on the primary to:
So let's turn this around. What resistor should you put in parallel with the 47k phonostage to change the loading to a desired value? Note, you can only go lower in value, not higher (negative answer means unrealizable).
Desired Loading:40R , Turns Ratio: 1:18.5, Resistor =19.3 k

You can also put the parallel resistor on the primary side, but then the value is the above divided by the turns ratio squared.
In this instance it would mean a resistor value (for primary loading) of 135.7R to get 40R. Is this because there is 47K in the phonostage rather than 1Meg? Or to achieve 40R on the primary you should in fact use a value of 135R????

Sorry if this is old hat and stupid, I would just like to understand it.

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Gerry
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Mike H
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#2

Post by Mike H »

Yes it's because it's resistors in parallel.


Try this?

http://www.electronics2000.co.uk/calc/s ... ulator.php


There's also the other thing that the impedance ratio is the turns ratio squared (^2).

So if turns ratio = 1:18.5, impedance ratio is 18.5^2, = 1:342.25

His calculation for a 19.3k resistor is exactly right then if it's in parallel with the 47k, the final total resistance being 13.69k.

13.69k = 40 x 342.25


 
 
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Mike H
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#3

Post by Mike H »

Oh yes 47k / 342.25 = 137.32 Ohms

On that basis and according to the resistor calculator the shunt resistor on the primary should then be 56.441R



 
 
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Andrew
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#4

Post by Andrew »

Also, at the time we did this I had removed the 47K from the input and replaced in with 1Meg; mainly to make sure the grid had a reference to ground. 1Meg in parallel with something, well the 'something' usually dominates.

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Mike H
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#5

Post by Mike H »

Oh well, no wonder :D


Can't be completely ignored though. In that case it looks like 2.92k on the primary.

So the 40R becomes 40.555R, not a lot in it is there?


 
 
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Gerry
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#6

Post by Gerry »

Thanks guys, though still a little confused.

The fact we replaced the 47K with 1Meg, to my understanding, meant that the loading resistor on the primary was the value required.

So if you wanted 40R, a 40R resistor was put in parallel across the input?

Why then when measuring do you not get a reading of 40R?
Can't be completely ignored though. In that case it looks like 2.92k on the primary.

So the 40R becomes 40.555R, not a lot in it is there?
Mike not sure what you mean here or where the 2.92K comes from.

Again apologies if I'm asking the obvious!
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Gerry
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Nick
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#7

Post by Nick »

Why then when measuring do you not get a reading of 40R?
By measuring do you mean with a ohmmeter? If so all you will see is the DC resistance of the transformer primary in parallel with the resistor. The values we are talking about with the transformed resistances are actually impedances, so valid only in AC terms, which is what the cartridge sees under signal conditions, but you won't see with a DC meter.
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Gerry
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#8

Post by Gerry »

Nick wrote:
Why then when measuring do you not get a reading of 40R?
By measuring do you mean with a ohmmeter? If so all you will see is the DC resistance of the transformer primary in parallel with the resistor. The values we are talking about with the transformed resistances are actually impedances, so valid only in AC terms, which is what the cartridge sees under signal conditions, but you won't see with a DC meter.
Ah! Now I think I get it. I meant to ask whether the reflected trannie resistance would have an effect in combination with the resistor. Obviously it does.

Yes reading with a standard DMM!

Many thanks Nick.
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#9

Post by Mike H »

Well there you are, I overlooked the "measured the load and got readings from" bit :D



 
 
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