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KCC's Quizzes AQQ295 about an audio logarithmic volume controller using linear pot - a proposal from Tim O'Brien

1. Quote of the month: "Think like a proton. Always positive" - Unknown

2. New quiz of the month: AQQ295 about a audio logarithmic volume controller using linear pot circuit

A kind proposal from Tim O’Brien (a frequent participant to our quizzes)

Audio volume is usually controlled by a logarithmic (log) potentiometer (pot) to approximate human hearing response.  However true log pots are increasingly hard to source and relatively expensive.  But there is a simple trick to establish an approximation of a log response using a linear pot and a single fixed resistor that’s good for about two decades - provided that the source and load impedances are appropriate to prevent undesirable attenuation at full volume.

One has to build a digital 90 degrees phase splitter circuit giving an in-phase (LO_I) and a quadrature phase (LO_Q) square LO signals phased out 90° one to each other. Among the following 4 circuits, there is one which is not performing that role.

Question: can you localize it and tell why?

1. With a 10K Ohm linear pot what is the value of the parallel fixed resistor R? to establish 10% attenuation at 50% rotation:
     a. assuming zero source and infinite load impedance?
     b. assuming 100 Ohms source and 10K Ohms load impedance?

2. There is no voltage attenuation for 1 a. at full volume, but what is the voltage attenuation at the load at full volume for 1b?

Good luck and try to be among the first ones!

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  • For a potential divider let's say that the two resistors are Rt (top) and Rb (bottom)
    For a linear pot at 50% rotation, Rt would be 5k
    For 10% attenuation bottom resistor, Rb must be 10% of the total value of Rt + Rb, so...
    Rb = 1 / 9 * 5k = 556 ohm

    To work out the resistor needed for the parallel combination to equal 556 ohm
    Rb = Ra x Rp / Ra + Rp
    Ra is also 10k at 50% rotation = 5k

    So parallel resistor, Rp in kohms, would be
    5 x Rp / (5 + Rp) = 0.556
    Multiply both sides by (5 + Rp) then simplify
    5 x Rp = (5 + Rp) * 0.556
    5 Rp = 2.78 + 0.556 Rp
    4.44 Rp = 2.78
    Rp = 625 ohms

    Now assuming 100-ohm source and 10k load
    Rt = 5k + 100 ohm = 5.1k
    So, Rb would need to change to maintain 10% attenuation
    Rb2 = (1 / 9 * 5.1k = 567 ohm

    But Rp is now in parallel with the 5k from the pot and the 10k load
    Their parallel combination is 3.33k
    So the parallel resistor, Rp in kohms, would be calculated from
    3.33 x Rp / (3.33 + Rp) = Rb2
    3.33 x Rp / (3.33 + Rp) = 0.567
    3.33 x Rp = (3.33 + Rp) * 0.567
    3.33 Rp = 1.89 + 0.567 Rp
    2.76 Rp = 1.89
    Rp = 684 ohms

    The attenuation at full volume for 1b would be
    100 ohms / total resistance of Rsource + (Rpot || Rload || Rp)
    total resistance would be 100 ohms + (10k || 10k || 684) = 100 + 601.7 = 701.7 ohms
    So drop would be 100 / 701.7 = 14.25% which would be 85.75% attenuation

Reply
  • For a potential divider let's say that the two resistors are Rt (top) and Rb (bottom)
    For a linear pot at 50% rotation, Rt would be 5k
    For 10% attenuation bottom resistor, Rb must be 10% of the total value of Rt + Rb, so...
    Rb = 1 / 9 * 5k = 556 ohm

    To work out the resistor needed for the parallel combination to equal 556 ohm
    Rb = Ra x Rp / Ra + Rp
    Ra is also 10k at 50% rotation = 5k

    So parallel resistor, Rp in kohms, would be
    5 x Rp / (5 + Rp) = 0.556
    Multiply both sides by (5 + Rp) then simplify
    5 x Rp = (5 + Rp) * 0.556
    5 Rp = 2.78 + 0.556 Rp
    4.44 Rp = 2.78
    Rp = 625 ohms

    Now assuming 100-ohm source and 10k load
    Rt = 5k + 100 ohm = 5.1k
    So, Rb would need to change to maintain 10% attenuation
    Rb2 = (1 / 9 * 5.1k = 567 ohm

    But Rp is now in parallel with the 5k from the pot and the 10k load
    Their parallel combination is 3.33k
    So the parallel resistor, Rp in kohms, would be calculated from
    3.33 x Rp / (3.33 + Rp) = Rb2
    3.33 x Rp / (3.33 + Rp) = 0.567
    3.33 x Rp = (3.33 + Rp) * 0.567
    3.33 Rp = 1.89 + 0.567 Rp
    2.76 Rp = 1.89
    Rp = 684 ohms

    The attenuation at full volume for 1b would be
    100 ohms / total resistance of Rsource + (Rpot || Rload || Rp)
    total resistance would be 100 ohms + (10k || 10k || 684) = 100 + 601.7 = 701.7 ohms
    So drop would be 100 / 701.7 = 14.25% which would be 85.75% attenuation

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