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

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.

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.

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

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.

OK, got it. Thank you, Barry.

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.

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

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 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..

Quote:

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Originally Posted by alexk0il

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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Oops - wasn't thinking clearly! :o 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! :doh: 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

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.

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.

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'.

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

Russell

Quote:

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Originally Posted by Pharos

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'.

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OMG, I thought I asked a question that is easy to answer, but it gets more and more confusing,:lol::lol::lol: 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:

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Originally Posted by RothwellAudio

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...

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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 :)

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.

Quote:

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Originally Posted by Pharos

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.

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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?

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.

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.

Quote:

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Originally Posted by RothwellAudio

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.

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Crystal clear, brilliant explanation. Thank you.

Doing some calcs but the hay fever is so bad I cannot concentrate or see properly.

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.

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.

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.