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Power amp wattage distribution to two cabinets?

Discussion in 'Amps and Cabs [BG]' started by BassWombat, Dec 26, 2006.


  1. First up, apologies for not being able to find a thread that covered this subject even though I'm sure it has been discussed.

    Here's the story: I have a Focus SA that delivers 800w @ 4 ohms and 1000w @ 2 ohms through a single path. I am interested in what happens with the following two cabinets - a 4ohm Schroeder 1212 (rated at 1000w) and an 8ohm Epifani T210 (rated at 500w).

    If I ran it all together, what do I need to be aware of in terms of not overpowering the Epifani? Is there any way i can approximately tell how much power is going to each cabinets? And (as much out of interest as anything) is the power output of a power amp linear in the sense that when I turn it to halfway is it roughly putting out half rated power output?

    Cheers - and happy holidays to all.
     
  2. BassyBill

    BassyBill The smooth moderator... Gold Supporting Member

    Mar 12, 2005
    West Midlands UK
    If you're talking about running these in parallel, you'll have an overall impedance of around 2.7 ohms, so your power amp will deliver maybe 900W. About 600W of this will be dissipated by the Schroeder and the remaining 300W by the Epifani based on the impedances you quote. Sounds like you shouldn't expect any problems at all with that setup - as always, use your ears as a guide to whether any of your gear is in distress, but you won't be overpowering either cab according to these specs.
     
  3. Thanks bassybill - much appreciated.
     
  4. Jim Carr

    Jim Carr Dr. Jim Gold Supporting Member

    Jan 21, 2006
    Denton, TX or Kailua, HI
    fEARful Kool-Aid dispensing liberal academic card-carrying union member Musicians Local 72-147

    Interesting. How did you calculate this? I assume the difference in power dissipated is because of the impedence of the cabinets. I don't doubt your assertions, I'm just curious. Thanks! :confused:
     

  5. (cab1 x cab2)/(cab1+cab2). In this case, (4x8)/(4+8)=32/12=2.67
     
  6. BassyBill

    BassyBill The smooth moderator... Gold Supporting Member

    Mar 12, 2005
    West Midlands UK
    Your assumption is correct.

    Same pd (voltage) across each of the two cabs in parallel. Twice as much current will flow through the lower impedance 4 ohm cab as the 8 ohm (I=V/R). Therefore, as power=volts * current, the total power of about 900 watts will be split two thirds/one third in favour of the lower impedance cab.

    EDIT I'm simplifying here as I'm treating the cabs' impedances as simple resistances rather than "reactive" loads, but as long as one cab is roughly half the impedance of the other most of the time then this reasoning still works.
     
  7. BassyBill

    BassyBill The smooth moderator... Gold Supporting Member

    Mar 12, 2005
    West Midlands UK
    Happy to help! :)
     

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