Impedance matching for DYI RCA attenutators

[alexk0il]

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.

[Pharos]

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

[Barry]

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.

[Pharos]

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.

[Pharos]

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.