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Thread: Impedance matching for DYI RCA attenutators

  1. #21
    Join Date: Mar 2017

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    From a pre to power, for the audio bandwidth and the temp there will IMO be a totally insignificant effect without a shunt resistor. I am sure it is independent of current. I think it is standard to assume that an amps I/P Z is constant and almost purely resistive over the audio range. I cannot see how a change in attenuation can increase S/N ratio.

    The BBC designed and amp which was for vinyl cartridge use, which actually had a noise figure within half a dB of the maximum theoretical figure, but that of course was for very low signal levels, and power amps are typically at a sensitivity of 0.5 to 1V.

    The advantages of using more complex pads is that they can deal with ccts in which it is crucial to maintain constant Zs, this especially true with maximum power transfer theory in transmission lines. Of course certain configs of L or Pi pads can be used to improve the ratio of I/P to O/P Z.

    I had absolutely masses of calcs to do for the attenuator networks for the Heils on my own Heil - Rogers mid/woof design, trying to maintain a relatively constant I/P Z and O/P Z with pre-determined attenuation levels.
    This involved pages of tables, but of course it involved the Heil driver (3.6 ohms), and the near zero O/P Z of the power amp.

    It actually worked out rather well with a -5dB pot range, and a switch giving another -5dB, a total of 10 dB, in 1/3 dB steps of resolution.

  2. #22
    Join Date: Jan 2009

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    True, one could use a single series resistor of 100k to form a potential divider with the 47k input impedance of the amplifier, to give an attenuation of 10dB, however that would make the effective source impedance > 100k.

    It is preferable to include a shunt resistor in the network.


    Out of curiosity, I measured the component values for some commercial in-line attenuators/attenuating cables.

    The first was a modified cable to allow a CD player to be used into the ‘Radio’ input of a Quad 33 preamp. It provides 14dB attenuation and uses an 8.2k series resistor and a 2k shunt resistor. The impedance of the ‘Radio’ input of the Quad 33 is 100k, so the attenuation factor is 14.3dB. The source will ‘see’ a load of 10.16k, that is ~ 100x that of the source impedance. The preamp will see a source impedance of 1.61k.

    The second example was with a switched in-line attenuator, offering 10, 15 or 20 dB attenuation. For all settings the series resistor was measured to be 10.13k (i.e. a standard 10k resistor) with measured shunt resistor values of 4.68k, 2.17k and 0.99k respectively (i.e. 4.7k, 2.2k and 1k standard values). Assuming the load impedance is 47k, and then the respective attenuation values are 10.64dB, 15.46dB and 21.25dB.

    The source impedance as seen by the amplifier will be 3k, 1.72k and 0.89k respectively. The load impedance as seen by the source will be 14.36k, 12.2k and 11.1k respectively.

    In all of the above I have assumed a source output impedance of 100 Ohm.
    Have you listened to this month's choice in the Album Club?

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  3. #23
    Join Date: Oct 2016

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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'.
    Quote Originally Posted by alexk0il View Post
    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.
    Quote Originally Posted by Barry View Post
    True, one could use a single series resistor of 100k to form a potential divider with the 47k input impedance of the amplifier, to give an attenuation of 10dB, however that would make the effective source impedance > 100k.

    It is preferable to include a shunt resistor in the network
    I agree that it is preferable to have a shunt resistor in the network. There are two problems with using only a series resistor.

    The first problem is that you need to know the input impedance of the power amp (or whatever it is you're feeding the signal into) to know what series resistor is required, and you will only be able to use the attenuator with that power amp (or one with the same input impedance) to get the specified attenuation.

    The second problem is that the input impedance of the power amp (or whatever) may be listed as, say, 47k in the spec sheet, but is it really 47k? Probably not, at high frequencies anyway. More likely than not there will be an RF filter at the input, and raising the source impedance may turn the RF filter into a "top end of the audio band" filter.

    Lets take an extreme case as an example - a valve power amp with an input impedance of 1M which you would like to attenuate by 20dB. If you try to use just a series resistor instead of an L pad you would need a 10M resistor. Lets say there's an RF filter at the input of the amp and it's formed by a 5k resistor and a 270pF capacitor. Assuming a low source impedance from the preamp, that gives you an RF filter at about 118kHz.
    However, once you have raised the source impedance to 10M the RF filter will roll off at only 59Hz. Yes, that's 59Hz
    Ok, so take out the RF filter cap. What now? You still have a problem because of something called Miller capacitance. Basically, the input valve acts as if it has a small value of capacitance built-in to it. In fact, parasitic capacitances are shown on the spec sheets for valves. Anyway, you will still have a rolled-off top end even if you remove the RF filter. That's why it's preferable to have an L pad attenuator rather than simply a series resistor.

    Actually, there's a third problem with the series resistor - Johnson noise. All resistors have a noise voltage called Johnson noise - and the bigger the resistor, the greater the noise. It may not be significant with 10k resistors and signals at line level, but get up to 10M and the noise can't be ignored.

  4. #24
    Join Date: Feb 2018

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    Quote Originally Posted by RothwellAudio View Post
    I agree that it is preferable to have a shunt resistor in the network. There are two problems with using only a series resistor.

    The first problem is that you need to know the input impedance of the power amp (or whatever it is you're feeding the signal into) to know what series resistor is required, and you will only be able to use the attenuator with that power amp (or one with the same input impedance) to get the specified attenuation.

    The second problem is that the input impedance of the power amp (or whatever) may be listed as, say, 47k in the spec sheet, but is it really 47k? Probably not, at high frequencies anyway. More likely than not there will be an RF filter at the input, and raising the source impedance may turn the RF filter into a "top end of the audio band" filter.
    ...
    Actually, there's a third problem with the series resistor - Johnson noise. All resistors have a noise voltage called Johnson noise - and the bigger the resistor, the greater the noise. It may not be significant with 10k resistors and signals at line level, but get up to 10M and the noise can't be ignored.
    Crystal clear, brilliant explanation. Thank you.

  5. #25
    Join Date: Mar 2017

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    Doing some calcs but the hay fever is so bad I cannot concentrate or see properly.
    Last edited by Pharos; 26-06-2018 at 13:41.

  6. #26
    Join Date: Oct 2016

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    Quote Originally Posted by Pharos View Post
    Doing some calcs but the hay fever is so bad I cannot concentrate or see properly.
    If you look closely at my post #23 you'll find errors. Firstly, for 20dB attenuation you would need a 9M resistor, not 10M. That doesn't make a great deal of difference. However, the roll-off won't be that of a 10M/270pF RC filter. My mistake. The 9M series resistor and the 1M load resistor of the power amp will become effectively a 0.9M source impedance and the roll-off will be at more like 600Hz. Still not an impressive bandwidth

    One more thing to consider: if the attenuator (or simple series resistor) is at the output of the preamp it will have the capacitance of the interconnect cable to deal with. If the attenuator is at the input to the power amp it won't be affected by cable capacitance.
    Anyway, using an L pad instead of a just a series resistor solves a lot of problems and makes sense to me.

  7. #27
    Join Date: Jan 2009

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    I wouldn't worry about the mistakes Andrew - we all make them, as I did in one of my earlier posts. Anyway, the points you made were correct, even though the example you used was a bit extreme.

    One can readily buy in-line attenuators fitted with RCA phono connectors. They use a simple L-pad and do the job very well. They are not expensive either.
    Have you listened to this month's choice in the Album Club?

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  8. #28
    Join Date: Mar 2017

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    OK, amazing how out of practice I have become, and I also had to fix my 43 year old calculator.

    I would tend to ignore the frequency dependant stuff because I don' think that filters have been present on many of the I/Ps to amps I have dealt with, but it may of course be a reality in any one case.
    I agree, placing one series R at the power amp end does dispense with cable capacitance effects.

    If we take two values of resistor w.r.t. Johnsons or thermal noise, say 1K and 1M, we gat the following;

    Vn = root (4KBTR) volts, where K = Boltzmanns constant, 1.38 x (10 -23) J/K, B = bandwidth, T = temp Kelvin, and R = resistance.

    (10-23) is my way of representing indices, 10 to the power of -23.

    Assuming an ambient temp of 20C which is 293K.

    For 1K Vn = root [(4 x 1.38 x (10-23) x 2 x (10-4) x 293 x (10-3)]

    = root [8 x 1.38 x (10-16) x 293]

    = root [3.2 x (10-13)]

    = 5.6 x (10-7)V = 0.5 microvolts

    This is -124dB w.r.t. 1V, and -118dB w.r.t. 0.5V.

    For 1M Vn = root [4 x 1.38 x (10-23) x 2 x (10+4) x 293 x (10+6)]

    = root [8 x 1.38 x (10-13) x 293]

    = root [3.235 x (10-10)]

    = 0.18 microV

    This is -94.9dB w.r.t. 1V and -89dB w.r.t. 0.5V.

    These figures are not at all bad even for the worst case of a 1Mohm resistor, and the rather unlikely low value perhaps of the 1K resistor.

  9. #29
    Join Date: Oct 2016

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    Quote Originally Posted by Pharos View Post
    I would tend to ignore the frequency dependant stuff because I don' think that filters have been present on many of the I/Ps to amps I have dealt with, but it may of course be a reality in any one case.
    RF filters at the inputs to power amps are pretty standard in my experience.

  10. #30
    Join Date: Mar 2017

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    You've probably dealt with many more than I have, and I accept your word on that.

    The Nelson Jones 10 + 10 Class A went up to 1MHz, and didn't suffer from oscillation problems.

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