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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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  • 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

Reply
  • 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

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