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

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

  1. Shawn

    Shawn Guest

    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
    #1
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  2. Shawn

    Scott Dorsey Guest

    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
    #2
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  3. 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
    applied to the load is:

    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?

    The answers are below.











































    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
    #3
  4. Shawn

    Shawn Guest

    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
    #4
  5. 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
    #5
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