Impedance matching for DYI RCA attenutators

[Barry]

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

[alexk0il]

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.

[Barry]

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.

[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'.

[alphaGT]

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

Russell

[alexk0il]

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:

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

[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.

[alexk0il]

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?

[Pharos]

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.

[Barry]

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.