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Thread: Impedance matching for DYI RCA attenutators

  1. #1
    Join Date: Feb 2018

    Location: Bucks

    Posts: 35
    I'm Alex.

    Default Impedance matching for DYI RCA attenutators

    Hi,

    I have a vintage amp with inputs that are too sensitive for my sources. I would like to try a DYI RCA attenuator similar to what is described here:
    http://audiokarma.org/forums/index.p...uators.392186/

    However after reading theory about how pi-attenuators I still don't understand how important is to match the input impedance or if this is not essential. For example, the OP suggests using a symmetrical pi-attenuator to match 47 KOhm both for input and output, while the last post in the thread suggests matching both input and output separately, though this turned out to be impossible and a closely matching attenuator is proposed as a result.

    Since even a close match for 10db, Zin = ~500Ohm and Zout=60kOhm for my system is impossible, I wonder how important is to match the input impedance? Would it make any difference and may be using the Zin to be the same as Zout would still work fine, or shall I try minimizing the mismatch both for Zin and Zout?

    Thanks
    Last edited by alexk0il; 19-06-2018 at 21:00.

  2. #2
    Join Date: Aug 2008

    Location: Suffolk, UK

    Posts: 1,243
    I'm Paul.

    Default

    I am a bit confused by what you have written, but generally you want the source to have an output impedance that is ideally 10 times lower than the imput impedance of the amplifier.

    When using an inline attenuator you design it so it has the same input impedance as the amplifer. This will involve calculating the impedance caused by the attenuators shunt resistor (and any series resistors after) and the input impedance of the amplifier.

    I have found it best not to get caught up with the output impedance of the attenuator. This however does become a problem if your amp has a passive, unbuffered volume control as the first stage. In this case it gets complicated.
    ~Paul~

  3. #3
    Join Date: Oct 2016

    Location: Bolton, England

    Posts: 1,245
    I'm Andrew.

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    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, though it's probably on the low side for some valve gear. Nothing has to "match".
    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, so you want the input impedance as high as it needs to be without going over the top.

    Why a pi attenuator? That isn't necessary.

  4. #4
    Join Date: Jan 2009

    Location: Essex

    Posts: 17,859
    I'm openingabottleofwine.

    Default

    All you need is a simple "L pad" consisting of a 10k series resistor and a 5.1k shunt resistor. The 5.1k resistor will appear in parallel with the nominal 47K of your amplifier to create an effective resistance of 4.625K, which with the 10k series resistor will cause a voltage attenuation ratio of 0.315 (or -10dB).

    The effective input impedance is now 14.6K, which ought not to cause any problem with sources having a source impedance of less than say 1K.
    Have you listened to this month's choice in the Album Club?

    Barry

  5. #5
    Join Date: Feb 2018

    Location: Bucks

    Posts: 35
    I'm Alex.

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    Thank you, everyone who replied.

    I guess I should have provided the background information to my question. Or perhaps I should have asked it in a broader sense.

    Anyway, I am aware of L-pad attenuators; it is a good and a cost efficient solution. However some reviews on the internet suggest that unmatched impedance of RCA attenuators may alter the SQ of the system.

    So in my naive mind, the ideal impedance matching means that the source should see the same impedance with and without the attenuator (output impedance matching). And at the same time, the amp should see the same source impedance with and without the attenuator (input impedance matching). I cuold be wrong, but I think the main difference between the P and L-pads is that the former has a certain flexibility in matching the input impedancewhile the latter only targets the output impedance.

    I guess there is a consensus here that the attenuator does not have to match the impedance of the source, the pi-pad attenuator is an overkill and will not provide any audible difference over a properly matched L-pad amplifier. Is this what you are all saying?


    Thanks

  6. #6
    Join Date: Jan 2009

    Location: Essex

    Posts: 17,859
    I'm openingabottleofwine.

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    As Andrew has explained there is no real matching to be done.

    In audio electronics signal sources are regarded as approximating ideal voltage sources (that is, they have very low output impedance). An ideal or perfect voltage source would have zero output impedance, and would be capable of supplying any amount of current demanded by the following load. However in practice audio signal sources only approximate to an ideal voltage source. For example, most solid-state signal sources have an output impedance of 100Ω or less. Valve sources on the other hand will usually have a higher output impedance of say 1kΩ. Thus in order for the load not to stress or place too much demand on the source, the load impedance needs to be considerably higher than the source impedance. How high this ratio is depends on how much loss can be tolerated. Since the same current will flow through both the source and the load impedance, the two impedances form a voltage divider: so for example, making the load impedance 10x that of the source impedance, the loss is 0.82dB. If the ratio is say 20x, then the loss is reduced to 0.42dB.

    With the simple L-pad attenuator using the resistor values I described, the nominal 47kΩ impedance of your amplifier as seen by the signal source is reduced to 14.6kΩ. Let us assume for example the audio signal source has an output impedance of 100Ω, then with the attenuator in place, the amplifier will 'see' a source impedance of 99.3Ω. The impedance ratio is thus 148x - leading to a loss of only 0.06dB.


    A π-network (or an equivalent T-network) attenuator can be designed to have a prescribed input and output impedance, as well as a certain attenuation value. However when the ratio between these two impedances becomes large (in your quoted case it is 120x), it becomes difficult to achieve the required attenuation value; something that you have already noticed.

    π- and T-network attenuators are used when the input and output impedance are equal and the same as the characteristic impedance of the system. However the concept of characteristic impedance has no meaning in audio circuitry, because the frequencies involved are too low (and the corresponding wavelengths too long). It is only when the length of the signal path approaches 1/6 of the longest wavelength involved, or greater, does the concept of characteristic impedance become relevant; as it does in radio-frequency circuitry. Then both the source (or generator) and the load have the same impedance as the characteristic impedance of the signal path, so as to allow maximum power transfer to take place and to minimise mismatch reflections, which if the signal path is several wavelength long would cause a ‘standing wave’ to occur.
    Last edited by Barry; 22-06-2018 at 21:29. Reason: correction
    Have you listened to this month's choice in the Album Club?

    Barry

  7. #7
    Join Date: Feb 2018

    Location: Bucks

    Posts: 35
    I'm Alex.

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    OK, got it. Thank you, Barry.

  8. #8
    Join Date: Oct 2016

    Location: Bolton, England

    Posts: 1,245
    I'm Andrew.

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    Barry has given a very comprehensive analysis of the situation. Well done.
    I was too lazy to go into that much detail

  9. #9
    Join Date: Jan 2009

    Location: Essex

    Posts: 17,859
    I'm openingabottleofwine.

    Default

    Alex,

    If you really want the 10dB L-pad attenuator to preserve the 47K impedance of your amplifier, make the series resistor 33K and the shunt resistor 22K. The combintion of the 22K resistor in parallel with the 47K of the amplifier is 14.98k (call it 15k), and this forms a potential divider with the 33K resistor leading to a voltage ratio of 15/(15 + 33) = 0.313 (i.e. -10.1dB).

    The input impedance as seen by the source will be 33K + 14.98K = 47.98K.

    If the source has an output impedance of, say, 100 Ohm, this will appear to the amplifier (via the L-pad) as 99.8 Ohm, and so the transfer loss will be 0.02dB.
    Have you listened to this month's choice in the Album Club?

    Barry

  10. #10
    Join Date: Feb 2018

    Location: Bucks

    Posts: 35
    I'm Alex.

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    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

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