Balanced stereo step attenuator ladder type question!
How do I figure out which values of resistors to use in a balanced stereo step attenuator. One friend gave me two very nice ITT 4-pole switches, and I'm planning to use it for my Aleph P 1.7 project.
I found few scripts on the internet that are calculating ladder type configuration for unbalanced mode. But I would like to use both, balanced and unbalaced. So I need 8 decks. Can I use same value of resistors for both + and - balanced, as I use for unbalanced?
So, lets say I want 39dB attenuation. For unbalanced ladder I have to use 99878 ohms for the input (Rx) and 708 ohms for the ground (Ry). Can I use same value of resistors for input of + and - balanced and just use the same ground resistor?
How do I figure out which values of resistors to use in a balanced stereo step attenuator. One friend gave me two very nice ITT 4-pole switches, and I'm planning to use it for my Aleph P 1.7 project.
I found few scripts on the internet that are calculating ladder type configuration for unbalanced mode. But I would like to use both, balanced and unbalaced. So I need 8 decks. Can I use same value of resistors for both + and - balanced, as I use for unbalanced?
So, lets say I want 39dB attenuation. For unbalanced ladder I have to use 99878 ohms for the input (Rx) and 708 ohms for the ground (Ry). Can I use same value of resistors for input of + and - balanced and just use the same ground resistor?
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If you want to do it balanced, you can use just 3 resistors: 2 resistors from the two source phases, and at the end one resistor connecting them. The ends of this resistor goes to the two inputs. This last resistor is the one you switch. You really need only a single switch deck for it.
Jan Didden
Jan Didden
Check out figure 2 of this page:
Balanced attenuator
Such an arrangement with a fixed input resistor may give problems in a passive preamp because of the variable output resistance, so best to use short cable to power amp.
Balanced attenuator
Such an arrangement with a fixed input resistor may give problems in a passive preamp because of the variable output resistance, so best to use short cable to power amp.
Note that doing this kind of balanced volume control changes the calculation for the resistances you need to get a given attenuation.
Here's a standard dB step sequence as used in the Goldpoint controls.
http://www.goldpt.com/taper.gif
Here's my calculation, with disclaimer that I haven't doublechecked it:
Let each series resistor be S ohms and shunt resistors be H ohms.
Output voltage is input voltage times H/(H+2*S).
In dB, attenuation A=20*log_10(H/(H+2*S)).
Solving for H, for a given attenuation the shunt resistor H=(-2*S*10^(A/20))/(10^(A/20)-1), where '^' indicates exponentiation.
If there are any other resistances in series or parallel with this circuit they need to be considered in the calculation.
To get 0 dB H is infinite, i.e. no resistor; a short of 0 ohms for H will not completely eliminate sound because it's unlikely the non-inverted and inverted lines are exactly identical.
Here's a standard dB step sequence as used in the Goldpoint controls.
http://www.goldpt.com/taper.gif
Here's my calculation, with disclaimer that I haven't doublechecked it:
Let each series resistor be S ohms and shunt resistors be H ohms.
Output voltage is input voltage times H/(H+2*S).
In dB, attenuation A=20*log_10(H/(H+2*S)).
Solving for H, for a given attenuation the shunt resistor H=(-2*S*10^(A/20))/(10^(A/20)-1), where '^' indicates exponentiation.
If there are any other resistances in series or parallel with this circuit they need to be considered in the calculation.
To get 0 dB H is infinite, i.e. no resistor; a short of 0 ohms for H will not completely eliminate sound because it's unlikely the non-inverted and inverted lines are exactly identical.
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