• 01-17-2009, 10:09 AM
    Mikey F-14
    Guitar Cabinet Wattage Input Question.
    Just say that I have two bass guitar cabinets.

    4x10neo Avatar:
    1000 watts RMS Total
    250 watts per speaker @ 8 Ohms - Parallel

    2x12neo Avatar
    500 watts RMS total
    250 watts per speaker @8 Ohm - Parallel


    The question is, say that i have a 1200 watt amplifier and I had them daisy chained (bridged mono = 1200 watts) with the 4x10s first, then the 2x12. They are both wired parallel. Would the 4x10 suck the entire 1000 watts out of the current before it got to the 2x12 and leaving only 200 watts for it, or would the 1200 watts always automatically get dispersed equally to each speaker? 200 watts to each?
  • 01-17-2009, 05:44 PM
    Mr Peabody
    I'm not sure I understand your set up. Is your amp a head like for a piggy back or a pro amp? Really either way, if both cabinets are chained to a single output they both will see the same power. And, if no crossovers are used in the cabinet all the speakers will see the same power.
  • 01-17-2009, 08:27 PM
    Mikey F-14
    Its just an amplifier. i've got it on bridged mono running one signal to my first cab (the 4x10s) then, through and extension out of the first cabinet, (daisy chained) i got a chord running from my 4x10s to my second cab (2x12s). I have cross overs, but they are disabled when running bridged mode so the whole 1200 watts are going daisy chained to both cabs. but i think you've answered my question, thanks a lot.


    -Mikey F-14
  • 02-11-2009, 05:20 PM
    kainbrake2
    the thing is that nothing '' puts out wattage'' at any given time the entire system can be at 2 watts or 1200 watts no mater the amplifier if an amp is ratted at 100 watts it means that you can put a load on it that will dissipate 100 watts before overheating. if a speaker is ratted at 100 watts it means that the speaker can handle enough current to dissipate 100 watts before over heating. so amps don't put out watts they put out voltage and current. volts times current = watts. its all in ohms law. if really thing about it an amps watt ratting is not its output but its power handling. for instance the normal home has a 200 amp breaker panel that is fed with 240 volts. so volts times amp ( 240x200= 48,000) so the home has a 48,000 watt power source. if the only thing in the house on is one 100 watt light bulb does that mean your pushing the light with 48,000 watts. no it just means the light only dissipates 100 watts. if you have a 200 watt amp and a 50 watts speaker and you turn it up and the speaker blows, the amp didnt blow the speaker the speaker blew itself by drawing to much current .
  • 02-11-2009, 07:21 PM
    Luvin Da Blues
    Quote:

    Originally Posted by kainbrake2
    .....so amps don't put out watts they put out voltage and current. volts times current = watts. its all in ohms law........

    Close, but with inductive circuits you need to use I^2*R due to phase shift not E*I which works for purely resistive circuits. But we get da drift.:smile5:
  • 02-11-2009, 07:49 PM
    kainbrake2
    yes that formula would give you a true power but i think the guy was looking for apparent or average power . but like i said if your were to used an oscilloscope at any wave crests between E and I you could use the DC ohms law to find your power. i wasn't looking for a debate of electronics i was just trying to help the guy understand what he is dealing with. besides it would have been a very long message if i were to get into reactive power and phase angles.
  • 02-11-2009, 07:56 PM
    Luvin Da Blues
    A very long topic indeed. Sorry, the electrician in me couldn't resist. Didn't mean to step on your toes. :thumbsup:

    Cheers,

    LDB
  • 02-11-2009, 08:06 PM
    kainbrake2
    im an electrician myself but you know whats funny, when i was in class we spent alot of time on this for service and feeder calculations, but it wasn't until i was out of class that i learned these things we are talking about. the NEC completely walks around phase shift and other things like this. i didn't know what reactive resistance was until a few years after the state exams. i never pursued it but i would imagine this type of calculation is already figured in to stuff like FLC for motors and stuff.
  • 02-11-2009, 08:18 PM
    Luvin Da Blues
    Yep, totally understand. Our CEC (Canada) basically only covers installation, protection and control. The way I understand it, nameplate ratings on equipment factor in all parameters, that's the load you need to calc. demands and conductor/brkr sizing, etc. Design and manufacturing is governed by engineering codes and the CSA (UL).
  • 02-11-2009, 08:36 PM
    kainbrake2
    i guess we are on a need to know basis! i guess it would also be a nightmare to figure start current, run current, and in rush or FLC and then figure for over current and overload device if all you had was volts and henries. hey i got a good one for you that i cant get a good answer for not even from engineers. ground fault and arc fault protection. i understand an arc fault as far as i know you can tell me if im off. an arc fault sees the difference in a working and nonworking arc by measuring current in line and grounded conductor, if the sum of the currents doesn't equal 0 it trips? now maybe you can help me understand the difference in working and non working arc better than the NEC. also GFCI i cant find a good diagram or explanation on it and one of my helpers asked me about it and i had to change the subject because i felt like a real tool.
  • 02-11-2009, 08:39 PM
    Kevio
    Since all speakers are 8 ohms and all speakers are wired it parallel (the daisy chain connection is another parallel connection), the available power will be split evenly between the 4 speakers.

    1200 total over four speakers is 300 watts per speaker. That's more than the 250 watts the speakers are rated for which means that if you turn it up and push the amp to full capacity, you may damage one or more of the speakers.

    Also with 4 8-ohm speakers wired in parallel, the amplifier will see a 2-ohm load. Make sure the amplifier is rated to handle such a load. If it isn't, smoke may come out of it.
  • 02-11-2009, 08:49 PM
    Luvin Da Blues
    Not sure what you mean by arc fault? Are you talking high voltage switchgear? HV gear that I've installed only used arc quenching.

    A GFCI works basically the same way. There's a trip mechanism that's shunted between the H & N. when there's a difference of 5 milliamps (safe human limit) or more between the two legs it will trip a switch and disconnect the power. Basically, if there's a difference, that voltage has to be going somewhere, either directly to ground or thru a person or other object to ground.

    Hope this helps.
  • 02-11-2009, 09:01 PM
    kainbrake2
    ok ark fault is kinda new here its now required in any bedroom of any dwelling unit. its abbrev. is AFCI and this is why i had to ask because for years i understood the GFCI to work this way and now they give AFCI and im thinking was i wrong all that time. i trying to figure out what the diff is between them. anyway weve got way off subject this stuff belongs in another forum all togethern its been a pleasure my friend.

    for the guy with the speaker problem if you will give the ohms of each speaker in the 2 cabs someone on this forum can give you a diagram to get your ohms up and get the power handling up.
  • 02-11-2009, 09:04 PM
    kainbrake2
    if you have 2 200 watt drivers in parallel you have a 400 watt speaker circuit, but if you have 2 200 watt drivers in parallel and 1 200 watt in series then depending on were the speaker is in the circuit you can reduce the total power handling.
  • 02-13-2009, 11:23 AM
    hermanv
    This isn't all that obscure. An amplifier operated within it's ratings is a voltage source. It will deliver the same voltage to any number of speakers wired in parallel (up to it's current rating limit).

    So power in each speaker is calculated as if the other speakers didn't exist. If an amplifier was adjusted to an output of 2V RMS then an eight ohm speaker would dissipate 1/2 watt and any number of additional eight ohm speakers would also dissipate 1/2 watt each. If a four ohm speaker was also wired in parallel it would dissipate 1 watt, but have no effect on the power dissipated by the existing eight ohm speakers.

    Because the amplifier output impedance is not zero ohms there will be a very small drop in output voltage as more speakers are added, but this confuses the question. In a parallel circuit all elements receive the same voltage and each added element has no effect on the dissipation of prior elements up to the rating of the amplifier.

    So in the original question, if the volume was adjusted to deliver 200 watts to one eight ohm driver, every other driver would also dissipate 200 watts. Of course eight ohms or four ohms is just a nominal rating, few if any speakers will measure either 4 or 8 ohms exactly, actually they will vary from this rating by quite a lot. Also note that musical instrument speakers are one of only a few places where 16 ohm drivers are common.
  • 02-13-2009, 04:36 PM
    Kevio
    Quote:

    Originally Posted by hermanv
    This isn't all that obscure. An amplifier operated within it's ratings is a voltage source. It will deliver the same voltage to any number of speakers wired in parallel (up to it's current rating limit).

    Most amplifiers will have different power specifications for different load impedances. Ohm's law says you get twice the power when you halve the impedance. The amplifier may have other limitations that would say otherwise. Read the specs.
  • 02-13-2009, 05:08 PM
    hermanv
    Quote:

    Originally Posted by Kevio
    Most amplifiers will have different power specifications for different load impedances. Ohm's law says you get twice the power when you halve the impedance. The amplifier may have other limitations that would say otherwise. Read the specs.

    I did say current limit. The equation is I squared R, halving the impedance will double the current, so my post is correct for a constant output voltage. Since I is doubled I squared is four times, but R is 1/2 so the end result is a doubling of power. This completely consistent, P = E x I, double I will also double power. Since the volume will also double, many users would lower the volume setting.

    The original post asked if power was distributed to multiple speakers based on their power rating. The answer is, no, it is distributed based on their impedance.
  • 02-18-2009, 02:39 PM
    kainbrake2
    again i keep seeing the words deliver or push watts. nothing delivers,pushes,outputs,or supplies anything with watts. amplifiers or any other sorce simply supply the means ( volts and current) needed to dissipate power (watts). now if out put was 2vrms and the speaker was eight ohms then you would dissipate 1/2 watt, this is true but if you connected another 8 ohm speaker in parallel then you have a 1 watt circuit. everytime you add an 8 ohm load in parallel you will up the power by 1/2 watt at a time.

    now lets picture a schematic of 3 1/2 watt speakers in parallel, with this you will have a 1.5 watt circuit. now place one more 1/2 watt speaker in series between the amplifier and the other speakers on the + line. now you have a circuit that will fail at just over 1/2 watt. now the only thing that fails is the speaker in series, the rest of them will live but the ciruit will now work unless you replace the speaker in series or remove it.