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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

  • 1a.

    Use a Thevenin equivalent looking back at the pot

    Vs/2 x R?/(2.5K+R?) = 0.1Vs

    R?/(2.5K+R?) = 0.2

    R? = 625Ω

    1b.

    Assuming Rsource=100 and Rload+Load=10K

    Use a Norton equivalent looking back at the pot

    Vs/5.1K = 0.1Vs/5.1K + 0.1Vs/5K + 0.1Vs/R? + 0.1Vs/10K

    10K/5.1K = 1K/5.1K + 1K/5K + 1K/R? + 0.1

    9/5.1 – 0.2 – 0.1 = 1K/R?

    R? = 682.7Ω

    2.

    (Vs-Vload)/100 = Vload/5K + Vload/682.7 ;The 5K equals Rpot||(Rload+Load)

    Vs = Vload + 100Vload/5K + 100Vload/682.7

    Vs/Vload = 1 + 0.02 + 100/682.7 = 1.166477

    Vload/Vs = 0.85728 or 85.73% (a decrease of 14.27% from 1a at full volume)

     

    The above results were verified by simulation in LTspice

  • To solve the question, we first need to simplify the circuitry.

    We combine all resistors above the wiper to a single one and do the same for the lower. By doing so, we can formulate the solving equation:

    Transforming this equation leads to:

    As a second generalisation we express the upper and lower effective resistor of the potentiometer:

    Finally we need the parallel of the effective lower resistance and the searched resistor:

     Combining the equations and transforming for the lead to:

    Case 1a:

    With

    we get 

    and the searched resistor is:

    Case 1b:

    With

    we get 

    and the searched resistor is:

    Case 2:

    When the wiper is at the upper end we get:

  • Addendum for case 2:
    If also Rx from case 1b is connected, we get:


  • typo: result for case 2 must be 0.8753 and not 0.3753
    (I accidently took 1kOhm instead of 0.1kOhm for Rsource)

  • Attenuation is defined by Vout/Vin.

    1a: R=625 ohms

    1b: R=2000/3 ohms (approx. 666.67 ohms)

    2 : Vout/Vin is always =1 but the input voltage Vin is reduced to 100/(100+588.24)=0.855.E (with E=efm of the source).To fix, insert A voltage follower between the source and the pot.

  • Wow, impressive  ! All your answers are fully correct, big applause!

  • Big applause to you as well  ! You have also found the right answers!

  • Haha, well catched  , the last question was indeed not easy; but you find it ! Congratulations!

  • Thanks  . Your answers to 1a and 2 are correct. For 1b, you might have a typo somewhere; normally you should come to 683 ohms...I wonder how you came with the formula 2000/3 ?