How to hook up several monitor systems to one monitor mix?

Discussion in 'Pro Audio' started by Shawn, May 11, 2004.

1. ShawnGuest

Hi,
I searched for an hour and did not find a direct answer to this
question.
We are playing a larger gig than normal.
I have an extra power amp and will probably buy two more wedge stage
monitors.
My question is how do I take the monitor mix output from the foh board
and split it into the two monitor power amps?
I've heard that y-ing at line level is not a problem. I've also heard
that there is some signal loss.
What say you all?

Thanks
Shawn

Shawn, May 11, 2004

2. Scott DorseyGuest

If the output impedance of the device going into the amps is half the
input impedance of the amps, or lower, there will be no problem. If it
is not, there will be a problem.

This means if you're working with old 600 ohm constant current stuff, you
cannot freely Y it off. If you are working with modern gear where the
input impedances are very high and the output impedances are very low, you
can run dozens of amps off of one output.
--scott

Scott Dorsey, May 11, 2004

3. Michael R. KestiGuest

If we assume that you're using voltage transfer (I.e., a low source impedance
driving a high load impedance.) and concern ourselves only with resistive
losses (I.e., ignore the low pass filter formed by the source impedance
and the cable capacitance.), I say that there will be some signal loss but
it will very likely not be a problem. How much loss occurs depends on the
source and load impedance of the gear being used and whether it is a problem
depends on how much loss your system can tolerate.

I'll demonstrate with an example. Let's say that we have an output with a
source impedance of 100 ohms that generates one volt of signal into an open
circuit. If we connect that output to an input with a load impedance of
10,000 ohms, then the voltage divider rule tells us that the signal voltage

10,000
1 volt X ------------ = 0.99 volt
10,000 + 100

If we use a wye connection to connect a second 10,000 ohm load in parallel
with the first, then the load impedance becomes 5000 ohms (The total
impedance of N equal loads connected in parallel is the impedance of one
load divided by N.) and the signal voltage applied to the load is:

5,000
1 volt X ----------- = 0.98 volt
5,000 + 100

The loss is:

0.99 volt
20 X log(10)--------- = 0.09 dB
0.98 volt

So there is a loss but it's a small enough to probably be negligible.

As an exercise, Shawn, determine the load impedance that would result in
an entire dB of loss with a source impedance of 100 ohms. How many 10,000
ohm loads connected in parallel are required to achieve that load impedance?

A 1 dB loss occurs when a 100 ohm source drives an 810 ohm load.

First, determine the voltage ratio that results in 1 dB of loss:

-1 = 20 X log(10)N
N = 10**(-1/20)
N = 0.89

Then determine the impedance that results in this voltage ratio:

Z
0.89 = -------
Z + 100

Z = 810

An 810 ohm load occurs when 12.3 10,000 ohm loads are connected in
parallel.

10,000
N = ------ = 12.3
810

Michael R. Kesti, May 11, 2004
4. ShawnGuest

The main monitor amp (oxymoron?) is a qsc, 150 watts per side, not
sure of model number. The new one is an Alesis M100(?) I swiped from
my home studio in order to get a couple more, small, monitors.
Thanks again,
Shawn

Shawn, May 11, 2004
5. Laurence PayneGuest

Correct, with modern equipment.
Correct. But very little.

CubaseFAQ www.laurencepayne.co.uk/CubaseFAQ.htm
"Possibly the world's least impressive web site": George Perfect

Laurence Payne, May 13, 2004