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Thread: Impedance matching for DYI RCA attenutators

  1. #11
    Join Date: Mar 2017

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    I'm Dennis.

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    I haven't read every post but a case in which it is required to attenuate the O/P of a device because it overloads the I/P of the recipient device is the clearest and easiest to deal with.

    In general terms an O/P impedance should as said above be about 1/10th of the I/P impedance it feeds into.

    By merely adding a resistor in series with the I/P this effectively raises the I/P impedance 'seen' by the feeding device, and this lowers the current demand on it, which is good.

    There is no need to consider impedance matching, this is applicable to transmission theory in which maximum power transfer is required, and power transfer is not an issue in voltage signal stages of amplification.

    You could use a pot in series with the I/P to gauge the required attenuation to taste, and then substitute a resistor of that value.

  2. #12
    Join Date: Mar 2017

    Location: Seaford UK

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    I'm Dennis.

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    I haven't read every post but a case in which it is required to attenuate the O/P of a device because it overloads the I/P of the recipient device is the clearest and easiest to deal with.

    In general terms an O/P impedance should as said above be about 1/10th of the I/P impedance it feeds into.

    By merely adding a resistor in series with the I/P this effectively raises the I/P impedance 'seen' by the feeding device, and this lowers the current demand on it, which is good.

    There is no need to consider impedance matching, this is applicable to transmission theory in which maximum power transfer is required, and power transfer is not an issue in voltage signal stages of amplification.

    You could use a pot in series with the I/P to gauge the required attenuation to taste, and then substitute a resistor of that value.

    I agree with Barry's post 6, but not with his earlier assertion that an I/P impedance can be too high, but with the exception from memory that very high I/P impedances can be prone to external noise I/P..

  3. #13
    Join Date: Jan 2009

    Location: Essex

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    I'm openingabottleofwine.

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    Quote Originally Posted by alexk0il View Post
    Hi Barry,

    Can you please clarify how you calculated the 99.8 Ohm to be seen by the amp? Isn't it supposed to be 13.2 KOhm = 1/(1/(33000+100) + 1/22000)?

    Thanks
    Oops - wasn't thinking clearly! Yes Alex you're quite correct, if you include the 47K input impedance of the amplifier the source impedance as seen by the amplifier is 10.3K. So not a good idea after all!

    Ignore what I wrote - one shouldn't have 'bright' ideas so late at night! Stick to a series resistance of 10K with a shunt resistance of 5.1K (placed on the amplifier side of the L-pad), and you will be OK!

    Apologies for misleading you
    Barry
    Last edited by Barry; 23-06-2018 at 12:47. Reason: Appropriate emojis added
    Have you listened to this month's choice in the Album Club?

    Barry

  4. #14
    Join Date: Feb 2018

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    I'm Alex.

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    Hi Barry,

    10k and 5.1k resistors with 47k load and 100Ohm source give me 14.7k and 3.3k at the source and amp sides. I still don't see the 99 Ohm at the source size.

    But let's ignore the 99 Ohm goal. We already know that a passive attenuator with the ratio of 100 and more is not achievable.

    Why do you think your former proposal with 10k and 5.1k resistors is better? If I look at the input/output impedance Ibwould think the one that has 13k/47k is better, as it has a better match on the amp side.

  5. #15
    Join Date: Jan 2009

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    It's simply impossible (and unnecessary) to match both the source impedance and the load impedance - the ratio is too large (47K/100 = 470).

    I think you will find most commercial attenuators use a 10K series resistor. For 10dB attenuation the shunt resistor is 5.1K.

    A quick solution might be for you to buy a pair of 10dB in-line attenuators from Rothwell - they are well made and not expensive.
    Have you listened to this month's choice in the Album Club?

    Barry

  6. #16
    Join Date: Mar 2017

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    There is absolutely no need for anything more complicated than a series resistor which will both attenuate the O/P signal fed to the load amp, and also raise the effective I/P Z of the amp that the source amp 'sees'.

  7. #17
    Join Date: Apr 2015

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    I'm Russell.

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    Iím just ease dropping on this post but thank you for the education! This is valuable information.

    Russell

  8. #18
    Join Date: Feb 2018

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    I'm Alex.

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    Quote Originally Posted by Pharos View Post
    There is absolutely no need for anything more complicated than a series resistor which will both attenuate the O/P signal fed to the load amp, and also raise the effective I/P Z of the amp that the source amp 'sees'.
    OMG, I thought I asked a question that is easy to answer, but it gets more and more confusing, This is how I see the conversation evolved:

    Q: Do I need to match both input and output impedance for pi-attenuator or only one of them?
    • A1: No need to to use pi-attenuator, just use an L-pad with 10K and 5.2K resistors
    • A2: No need to match anything, just buy a commercial RCA
    • A3: No need to use L-pad. A single resistor will do


    I'm sure all three options will work, though I wonder if I would get some clarity if there are any audible differences between all these options. I guess it all boils down to what Rothwell said:
    Quote Originally Posted by RothwellAudio View Post
    The impedance that the source sees needs to be high enough so the source doesn't struggle. 10k should be high enough for most modern sources ... Why not go mad and make the input impedance 1M? Because the output impedance will also be high if you do that, and that isn't good...
    I would definitely want to avoid getting into lengthy debates about which way is better, but I do wonder if someone or may be Rothwell can explain why it is not good to have the high output impedance. That will definitely conclude the lesson.

    Many thanks to all who already contributed to this entertaining thread

  9. #19
    Join Date: Mar 2017

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    Rothwell must be considering the series resistor as a part of the source impedance as far as the load is concerned.

    This is a valid conception in some situations, for example with a loudspeaker where the series resistor would worsen the damping factor, but that is not an issue in purely resisitive signal circuits.

    Versed in Thevenin and Norton equivalent cct. calculations, and Pi and L attenuators years ago at Wood Norton, this is all second nature.

  10. #20
    Join Date: Feb 2018

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    I'm Alex.

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    Quote Originally Posted by Pharos View Post
    Rothwell must be considering the series resistor as a part of the source impedance as far as the load is concerned.

    This is a valid conception in some situations, for example with a loudspeaker where the series resistor would worsen the damping factor, but that is not an issue in purely resisitive signal circuits.

    Versed in Thevenin and Norton equivalent cct. calculations, and Pi and L attenuators years ago at Wood Norton, this is all second nature.
    See your point, however my intuition tells me that having a shunt resistor might help to reduce the input noise on the amp inputs.

    I would expect the amp impedance to be a function of frequency, temperature, current or god know what else. A slight change in amp impedance will proportionally change the attenuation of your single resistor pad, hence you will get the noise coming in. If instead, as an example, you have an an L-pad with a 1000 Ohm shunt resistor in parallel to 10K amp, then the influence of the amp impedance fluctuations is reduced by the factor of 10.

    Or did I get this wrong?

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