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Speaker & Amp Impedance Question

Discussion in 'General Electronics Discussion' started by john2k, Nov 17, 2020.

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  1. john2k

    john2k

    185
    3
    Jun 13, 2012
    I have some 8ohm impedance speakers and I have an amp that has output of either 50W at 4ohm or 40W at 8ohm. If connect two of the 8ohm speakers to the single output in parallel that would make the total impedance 4ohm. 4ohm is something the amp supports but question is will the speaker support it? because each speaker is an 8ohm speaker so by wiring up in parallel and make it into 4ohms will that work? I'm assuming if i measure resistance across two terminals of the speaker it will read 8ohms and if i measured it after hooking it up in parallel then it would read 4ohms right?
     
  2. bertus

    bertus Moderator

    1,883
    709
    Nov 8, 2019
    Hello,

    The resistance will measure an other value than the impedance.
    You are right about putting 2 times an 8 Ohm speaker will result in a 4 ohm load for the amplifier.
    Some bigger speaker boxes can have 2 times 8 Ohms speakers parallel to create a 4 ohms box.

    Bertus
     
    Cannonball likes this.
  3. Harald Kapp

    Harald Kapp Moderator Moderator

    11,522
    2,654
    Nov 17, 2011
    For the speaker there is no problem at all.
    The limit is on the amplifier's side:
    From this data it is (more or less) obvious, that the amplifier' output voltage is limited.
    To drive 50 W into 4 Ω the required voltage is V = sqrt(50 W × 4 Ω) = 14.4 V.
    To drive 40 W into 8 Ω the required voltage is V = 17.8 V.
    If you were to drive 50 W into 8 Ω you'd need V = 20 V.
    The amplifier cannot output more than 17.8 V (at least within the limits of the acceptable distortion).

    Now, if you connect 2 × 8 Ω speakers in parallel the output voltage will be the same as for a single 4 Ω speaker, namely 14.4 V (the amplifier doesn't "know" that there are actually 2 × 8 Ω speakers). This is less than a single 8 Ω speaker would have to support, therefore there is no danger for the speakers - if they are rated at 25 W each at minimum (the 50 W from the amplifier are split between the two speakers).
     
  4. ratstar

    ratstar

    486
    20
    Aug 20, 2018
    Can u add a 4 ohm resistor to the 4 ohm speaker and turn it into an 8 ohm?
     
  5. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    You can but you lose 50% of audio output as dissipated power in the resistor.
     
    ratstar likes this.
  6. ratstar

    ratstar

    486
    20
    Aug 20, 2018
    Its not dissipated - isnt that when it turns into heat? But the power probably wouldnt get to the speaker, thats for sure.
     
  7. bertus

    bertus Moderator

    1,883
    709
    Nov 8, 2019
    Hello,

    When you put a 4 Ω resistor in series with a 4 Ω speaker, half the power will be dissipated (turned into heat) in the resistor as @Harald Kapp said.
    Where would the power that goes into the resistor otherwise go?

    Bertus
     
    davenn and Martaine2005 like this.
  8. ratstar

    ratstar

    486
    20
    Aug 20, 2018
    It stays in the battery.
     
  9. Audioguru

    Audioguru

    3,317
    705
    Sep 24, 2016
    A speaker and a resistor in series both conduct exactly the same current.
    2A in a 4 ohm resistor and in a series 4 ohm speaker drives and heats the resistor with (2A squared) s 4 ohms= 16W and heats (dissipates in) the speaker also with (2A squared) x 4 ohms= 16W.

    Since a speaker has a strong resonance you do not want to connect a speaker in series with a resistance or in series with an identical speaker because then the speaker will resonate like a bongo drum.
    any half-decent amplifier has an extremely low output impedance of 0.04 ohms or less that damps the resonance of a single speaker.
     
    Martaine2005 and ratstar like this.
  10. ratstar

    ratstar

    486
    20
    Aug 20, 2018
    Is that the electrical delay caused by the coil in the previous speaker?!?!?
    Nah I doubt that, that would only be in nanoseconds...
     
  11. Audioguru

    Audioguru

    3,317
    705
    Sep 24, 2016
    Impedance is a resistance to AC.
    A speaker with an 8 ohms impedance has a resonant impedance of about 70 or 80 ohms at a low frequency. When it is resonating then its coil is moving back and forth with its magnet and is becoming an AC generator. The extremely low output impedance of a modern amplifier shorts the generated power which stops the resonation when the amplifier "tells" it to stop moving.
     
    ratstar likes this.
  12. ratstar

    ratstar

    486
    20
    Aug 20, 2018
    So thats like back emf in motors. Wouldnt the charge generated only be attracted back the opposite end of the coil of the speaker?
     
  13. Audioguru

    Audioguru

    3,317
    705
    Sep 24, 2016
    With a motor, if you short it then its generated power quickly stops it. The same with a speaker, an amplifier with a very low impedance shorts the speaker coil (a speaker is a motor that moves its cone and the air around it) when the signal stops, then it quickly stops vibrating.
     
    ratstar likes this.
  14. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Sometimes I think you haven't understoof the most basic principles of electronics. You do know Ohm's law, do you?
    Charge is not generated. What one does when "producing" electricity is to separate positive and negative charges to create a potential difference which we call voltage.
     
  15. ratstar

    ratstar

    486
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    Aug 20, 2018
    I didnt understood it.
     
  16. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Try this page for an explanation. Ohm's law is truly one of the most basic concepts. You need to understand it.
     
  17. ratstar

    ratstar

    486
    20
    Aug 20, 2018
    Yeh ok. then how does it apply to t=rc? how come volts doesnt appear in this equation?
     
  18. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    That is not Ohm's law. This equation describes the time constant of an RC filter.
    But it does: C = V/(i × t)
     
    Martaine2005 likes this.
  19. johnshaw

    johnshaw

    5
    2
    Nov 5, 2020
    Agreed
     
    ratstar likes this.
  20. Harald Kapp

    Harald Kapp Moderator Moderator

    11,522
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    Nov 17, 2011
    Disagreed.
    P = I² × R
    As soon as there is current through the resistor, power will be dissipated in the form of heat.
     
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