The number of splits that can be done without degradation, or at least without degradation that can be technically predicted and explained, depends on:

1)The output impedance of the preamp (the lower the better).

2)The input impedances of the amps (the higher the better).

3)The TOTAL of the lengths of all of the interconnect cables that are used to connect each channel between the preamp and all of the amps (the shorter the better).

4)The capacitance per unit length of the cables (the lower the better).

There are two goals that should be met:

1)The parallel combination (see below) of the input impedances of all of the amps should ideally be at least ten times greater than the output impedance of the preamp, at the frequency for which the preamp's output impedance is highest (which will usually be 20 Hz if the preamp has an output coupling capacitor, as most tube preamps and some solid state preamps do).

2)The total capacitance of all of the cables on a given channel should be such that the corresponding capacitive reactance (see below) at 20 kHz (20,000 Hz) is ideally at least around five times greater, and preferably ten or more times greater, than the preamp's output impedance at high frequencies (which is usually, but not always, similar to the specified nominal output impedance of the preamp).

If the amps all have the same input impedance, their combined parallel input impedance equals the input impedance of one amp divided by the number of amps. If they have different (or the same) input impedances, their combined parallel input impedance is the reciprocal of the sum of the reciprocals of the input impedances of each of the amps. For example if there are three amps, and they have input impedances of 25K, 50K, and 75K, their combined input impedance is:

1/(1/25 + 1/50 + 1/75) = 13.6K

Capacitive reactance, Xc, is measured in ohms and is calculated as follows:

Xc = 1/(2 x pi x F x C), where:

pi = 3.14

F = 20,000 Hz for purposes of this calculation

C = The total capacitance of all of the interconnect cables (for one channel), in farads. One farad = 1 trillion pf (picofarads).

For example, if you split four ways into cables that are each 15 feet long, and the cables have capacitances of 25 pf/foot, the total capacitance (ignoring the capacitance of the splitters, which will be small in comparison to the capacitance of the cables) would be:

C = 4 x 15 x 25 = 1500 pf

The capacitive reactance at 20 kHz would be:

1/(2 x 3.14 x 20000 x 1500Exp-12) = 5308 ohms, or about 5.3K

Regards,

-- Al

1)The output impedance of the preamp (the lower the better).

2)The input impedances of the amps (the higher the better).

3)The TOTAL of the lengths of all of the interconnect cables that are used to connect each channel between the preamp and all of the amps (the shorter the better).

4)The capacitance per unit length of the cables (the lower the better).

There are two goals that should be met:

1)The parallel combination (see below) of the input impedances of all of the amps should ideally be at least ten times greater than the output impedance of the preamp, at the frequency for which the preamp's output impedance is highest (which will usually be 20 Hz if the preamp has an output coupling capacitor, as most tube preamps and some solid state preamps do).

2)The total capacitance of all of the cables on a given channel should be such that the corresponding capacitive reactance (see below) at 20 kHz (20,000 Hz) is ideally at least around five times greater, and preferably ten or more times greater, than the preamp's output impedance at high frequencies (which is usually, but not always, similar to the specified nominal output impedance of the preamp).

If the amps all have the same input impedance, their combined parallel input impedance equals the input impedance of one amp divided by the number of amps. If they have different (or the same) input impedances, their combined parallel input impedance is the reciprocal of the sum of the reciprocals of the input impedances of each of the amps. For example if there are three amps, and they have input impedances of 25K, 50K, and 75K, their combined input impedance is:

1/(1/25 + 1/50 + 1/75) = 13.6K

Capacitive reactance, Xc, is measured in ohms and is calculated as follows:

Xc = 1/(2 x pi x F x C), where:

pi = 3.14

F = 20,000 Hz for purposes of this calculation

C = The total capacitance of all of the interconnect cables (for one channel), in farads. One farad = 1 trillion pf (picofarads).

For example, if you split four ways into cables that are each 15 feet long, and the cables have capacitances of 25 pf/foot, the total capacitance (ignoring the capacitance of the splitters, which will be small in comparison to the capacitance of the cables) would be:

C = 4 x 15 x 25 = 1500 pf

The capacitive reactance at 20 kHz would be:

1/(2 x 3.14 x 20000 x 1500Exp-12) = 5308 ohms, or about 5.3K

Regards,

-- Al