Question about passive attenuation

[abbey road d enfer]

I have seen circuits using a dual pots but it seems hard to find one with somewhat matching values... I am not sure how that would affect the performance on the following grids making the matching uneven?

  Well-designed circuitry takes care of that (generally via some king of cross-feed) .

Please correct me if I am wrong when saying that the total resistance of a dual pot (lets say 2 x 50k) is still seen as 100k even though you alter the pot?

  If there is no load (wiper direct to grid) and you look at the "input" of the pot (across top-to-bottom), the resistance doesn't change with rotation.
Now if you add a load between wiper and bottom, the "input" resistance will change with rotation.

 

[johnheath]

Not the total resistance, the resistance seen when probing its output. It involves analysing the network including the source impedance, the series resistance (which is actually two resistors in the case of a balanced attenuator)  and the shunt resistor; you must work your Kirchoff theorem.

Aah - thank you sir

I think I am beginning to see the light

Do you calculate the source impedance in parallel with the pad network (I expressed it wrongly about being the total resistance) for example an U-pad?

Best regards

/John

 

[johnheath]

  If there is no load (wiper direct to grid) and you look at the "input" of the pot (across top-to-bottom), the resistance doesn't change with rotation.
Now if you add a load between wiper and bottom, the "input" resistance will change with rotation.

I mean a dual pot connected in a "push-pull configuration... positive lead from input transformer to top of one of the pots and the negative lead from input transformer to the top of the other pot and both bottoms of the dual pot to ground... both wipers to one grid each.


"Well-designed circuitry takes care of that (generally via some king of cross-feed) ."

I do not know what that is... could you perhaps give an example of it?

Best regards

/John

 

[abbey road d enfer]

I mean a dual pot connected in a "push-pull configuration... positive lead from input transformer to top of one of the pots and the negative lead from input transformer to the top of the other pot and both bottoms of the dual pot to ground... both wipers to one grid each.

Yes, in that specific case, the impedance presented by the pot to the xfmr's secondary does not vary. Some would say it does because of the Miller effect (capacitive impedance of a tube), but that should be negligible.

"Well-designed circuitry takes care of that (generally via some king of cross-feed) ."

I do not know what that is... could you perhaps give an example of it?

I don't want to give you too much food for thought, things must come in their own sweet time, but think of the phase-inverter in a push-pull amp, that converts an unbalanced signal in two equal and opposite signals. By duplicating such a circuit, it is possible to take two signals of different amplitude, such as those coming from two pots that don't track well, and convert them in two signals of exact same amplitude, but opposite in polarity.

 

[johnheath]

I don't want to give you too much food for thought, things must come in their own sweet time, but think of the phase-inverter in a push-pull amp, that converts an unbalanced signal in two equal and opposite signals. By duplicating such a circuit, it is possible to take two signals of different amplitude, such as those coming from two pots that don't track well, and convert them in two signals of exact same amplitude, but opposite in polarity.

Thanks

I think I get the idea about it all

Best regards

/John