Daisy Chain with Speakon

Discussion in 'Amps and Cabs [BG]' started by atldeadhead, Feb 26, 2004.

  1. atldeadhead


    Jun 17, 2002
    I've never used mutiple cabs before. Can I do this?

    I've got an SWR SM500 that I'd like to run bridged mono into 2 Aguilar GS 112's. They are both 8 ohms so by daisy chaining them I'd have a total load of 4 ohms. This is what I want, that way I can get the total amount of power the 500 will produce.

    My question is about the speaker cables I have. Tell me if this is an acceptable way to daisy chain the cabainets.

    I have one cable (A) that has a speakon on one end and a 1/4" plug on the other end. I had been using this cable to go from my SWR to my Hartke 4.5xl. I have another cable (B) that has a speakon on both ends. It's the speakon cable that came with my SWR.

    I'd like to run cable A from the SWR into the 1/4" input of GS112 (A). From there I'd like to use cable B, the speakon cable, to go from GS112 (A) into GS112 (B).

    I don't *think* there's anything wrong with doing it this way but I want to make sure so I don't damage any of my equipment.

    Can I do this?

  2. the only thing that would hold this setup back is if the speakon connectors are not wired in parallel with the .250" jacks. this is highly unlikely, however. you won't damage any of your equipment this way. the worst that could happen is that the second GS-112 cabinet doesn't produce any sound. but that is highly unlikely.

  3. atldeadhead


    Jun 17, 2002
    Anybody know if the jack's are all wired in parallel?
  4. reddavid


    Oct 11, 2001
    Wayne, PA
    not sure about the wiring, but i would think again about the impedence...from another thread:

    If the cabs are in series, you just add the impedances. Eg 8 + 8 =
    16. That's why most amps/cabs don't have the speakers in series,
    adding a second cab makes the setup quieter!

    So, most cabs are in parallel. The general case for parallel cabs of
    impedance A and B: (A x B) / (A + B)

    Eg (8 * 4) / (8 + 4) = 32 / 12 = 2.7

    so, if you run daisy chain (series) i would think your impedence was 16 ohms.

    i'm sure someone else more expert than i can clarify.

  5. Conventional speaker jacks are in parallel, so daisy chaining them like this will produce a 4 ohm load (if both cabs are 8 ohms). Should work fine.
  6. atldeadhead


    Jun 17, 2002
    Nashville Bill is correct. The following excerpt comes straight outta the SM500 the manual:

    Speaker 1 = 8 Ohms
    Speaker 2 = 8 Ohms

    This example shows two 8 ohm speaker enclosures connected in parallel for a total load of 4ohms. This is the minimum impedance that the SM-500 is designed to drive safely in the Bridge Mode.

    Connecting multiple enclosures that have a combined total impedance of less than 4ohms may result in damage to your amplifier.

    The following cabinet configurations may be used:
    • one 4 ohm enclosure
    • one 8 ohm enclosure
    • two 8 ohm enclosures <--This would be me with 2 GS112's
    • four 16 ohm enclosures

    SM-500 Power Output Ratings:
    Bridge/Mono Mode

    500 Watts @ 4 Ohms
    400 Watts @ 8 Ohms
    250 Watts @ 16 Ohms

    (minimum load = 4 Ohms)
  7. I build all my cabs with dual Speakon, two-pair type in parallel. All four wires are intact for bi-amping purposes.

    As long as the amp will stand the load, you can daisy chain from Speakon to Speakon. With the Lows in Pair #1 and the Highs in Pair #2, all the intelligence is in the jack dish. This allows the use of standard Speakon two-pair cables.

    Each cabinet is internally wired to use Pair 1 or Pair 2, depending on if the cabinet is a Low or High type for bi-amping. Full range typically uses Pair 1.

    The beauty of Speakon...

  8. Can I ask for a little clarification to see if I really understand what is being said.

    I have a Clarus which can run at 8, 4, or 2 Ohms.

    Currently I have 1 x 8 Ohm speaker which gives me a load of 8 Ohms

    I want to run 2 x 8 Ohm speakers, one out of each of the outputs on the amp, giving me a load of 4 Ohms.

    Now if I get a further 2 x 8 Ohm speakers and plug one into the back of each of the 2 x 8 Ohm speakers attached to the Clarus would the load then reduce to 2 Ohms?

    Thanks from a confused Brit.
    [no change there you all say!]

  9. that is correct.

    if each cabinet you own has an impedance of Z -- in your case you have two cabinets where Z = 8 -- the math looks like this:

    1/Z + 1/Z = 1/Zsum

    for you that's:

    1/8 + 1/8 = 1/Zsum


    2/8 = 1/4 = 1/Zsum


    Zsum = 4/1 = 4Ohms.

    if you have three 8Ohm cabinets:

    1/8 + 1/8 + 1/8 = 1/Zsum


    3/8 = 1/Zsum


    Zsum = 8/3 = 2.67Ohms.

    and if you have four 8Ohm cabinets:

    1/8 + 1/8 + 1/8 + 1/8 = 1/Zsum


    4/8 = 1/Zsum

    so Zsum = 8/4 = 2Ohms.

    i hope that makes sense to people.

  10. rcz

    Many thanks for the clarification.


  11. de nada. knowledge is power.

  12. To answer the original question this is what you need to do:

    Neutrik to Neutrik from head to cab #1. or Neutrik to 1/4" from head to cab #1 plugging the 1/4" into the cab.

    And 1/4" to 1/4" from cab #1 to cab #2.

    When I owned 2 GS112's this is how I cabled-up to run my poweramp bridged into these cabs.

  13. mr e

    mr e

    Nov 17, 2003
    be sure your speakon connectors are wired properly
  14. atldeadhead


    Jun 17, 2002
    Man this stuff can get confusing.

    I'd like to use the speakon cable that SWR supplied with my SM 500 to run from one cabinet to the other. I didn't know there where different kinds of speakon cables.

    Would this cable be wired correctly?
  15. i would be shocked if it isn't. i think your original idea was the best for your situation. if it doesn't produce sound from both cabinets, then it's time to think about it.

  16. atldeadhead


    Jun 17, 2002
    Well I hooked everything up last night and it all worked so I have to assume that everything is copacetic.

    I'm sure ya'll can appreciate my caution. I have a lot 'o money tied up in all this stuff and don't want to blow anything up. Everytime it seemed like I had my answer someone else would chime in with another twist and all it's done is get me more and more confused :confused:

    I think I've got it all straight now.
  17. seanm

    seanm I'd kill for a Nobel Peace Prize! Supporting Member

    Feb 19, 2004
    Ottawa, Canada
    Note that in the general case where all cabs have the same impedence: total impedence = impedence of one cab / number of cabs.

    For an example of four 8 ohm cabs: 8 / 4 = 2