CONUNDRUM

Handymann: In the event of signal overload, I sometimes use interconnect cables that have a built-in two-resistor network that attenuate the signal by a fixed amount. Since only two resistors are used (one series, one shunt), and the attenuator is built into the RCA connector plug right before the line stage input, there isn't much degradation in sound.

You need to determine by how many decibels you need to attenuate the signal from the PH-7, figure out (or ask someone) a resistor network pair that attenuates the signal by the desired amount without stressing the PH-7, then build (or have someone build) that resistor pair into the RCA connector of a decent-quality interconnect cable.

Since the modification will be to the interconnect and not to the phono stage or line preamp, the work should be fairly simple, and won't affect the resale value of the ARC or Aesthetix.

hth, jonathan carr

The input impedance of the Calypso is 20kohm if unbalanced, 40kohm if balanced.

The output impedance of the PH-7 is 200 ohms unbalanced. According to ARC, the output load should be a minimum of 10kohm, so let's assume that value as the total load, and estimate the values of some resistor networks that could be built into the interconnect between the PH-7 and Aesthetix.

If you want to reduce the PH-7's output to half (-6dB), you would need two 4.99kohm resistors per signal polarity. One resistor pair if you use the unbalanced outputs from the PH-7, two resistor pairs if you use the balanced outputs.

One of the 4.99k resistors should be soldered in series with the "hot" signal wire of the interconnect cable, the other resistor is soldered from the output side of the first 4.99kohm resistor to ground (outer shell of the RCA connector).

However, we must keep in mind that the Calypso already has a 20kohm input resistor, so for the second resistor (from output to ground), we want a value that forms 4.99kohm when placed in parallel with the Calypso's 20kohm resistor.

One such value for the second resistor would be 6.8kohm, which in parallel with 20kohm, will form 5075 ohms, which is a +1.7% error.

Another possible value for the second resistor would be 6.65kohm, which in parallel with 20kohm, forms 4.991kohm, for an error of +0.01%.

In either case, the first resistor remains 4.99kohm.

For a reduction to one-quarter (-12dB), the series resistor should be 7.5kohm and the ground-side resistance should be 2.49kohm.

Again, since the Calypso already has 20kohm at its inputs, we want a value for the second resistor that forms 2.49kohm when placed in parallel with the Calypso's 20kohm resistor.

One possible value is 2.7kohm, which in parallel with 20kohm, forms 2.379kohm, which is an error of -4.46%.

A more suitable value would be 2.87kohm, which in parallel with 20kohm, forms 2.51kohm, which is an error of +0.8%.

If your ears suggest that the total 10kohm load is too heavy for the PH-7 (slightly muted dynamics, slight loss in resolution), generally you would double all of the resistance values. 10kohm becomes 20kohm, 4.99kohm becomes 10 kohms, 7.5kohm becomes 15kohms, 2.49kohm becomes 5kohm. But all of the second resistors will need to be re-calculated, I suggest that you try the 10kohm setting first, and if that isn't to your liking, let us know and I or someone else will do the math for 20kohm instead.

Please keep in mind that the lower resistance, the lower the noise, so I would not increase the resistor values unless I thought that the sound lacked life.

Theoretically, the resistor network could be placed either at the output of the PH-7, or the input of the Calypso, but if you place it at the output of the PH-7, you will get a roll-off in the high frequencies, due to the capacitance of the interconnect. Better to place the resistor network at the input of the Calypso.

I believe that the above will be the easiest and most economical solution to your troubles.

kind regards, jonathan carr