3

So I've been into sound for a while, but there's just one thing that goes over my head. I'll use an example. Say you have an amplifier, in which datasheet it is specified that it outputs 300 watts per channel at 8 ohms . Now my question is: Does this mean that the amp will put out 300 watts when there is a total 8 ohm load present or when there is an 8 ohm load present at each channel effectively making it a 4 ohm load?

I do have an understanding of voltage, current and Ohm's law, however I am sorry if I don't understand the answers posted.

Sorry if this has been asked before or is not on topic.

Regards.

4
  • I think it could likely get confused between the engineers & the marketing department, but essentially "300W per channel at 8Ω" is 300W into each 8Ω speaker. A stereo amp is effectively 2 amps in this respect, one for each channel [though tied as a pair]
    – Tetsujin
    Commented Mar 7, 2015 at 14:22
  • Ah, so basically, 300 watt per channel into 4 ohm total load, assuming 1 8ohm speaker per channel. Thank you man. Still, I'd like someone else to answer, so I can be safe
    – Matthew
    Commented Mar 7, 2015 at 15:45
  • it's not really into a 4Ω load, it's into 2 8Ω loads. They're not in parallel, they're independent loads.
    – Tetsujin
    Commented Mar 7, 2015 at 15:47
  • Oh, okay, thank you man, just like I said to the other guy.
    – Matthew
    Commented Mar 7, 2015 at 16:21

1 Answer 1

0

Just so you have an answer and not only comments: Each channel is an independent amp and so a spec that says "300W at 8 ohms" applies separately to each 'half' of the stereo. There is no 4 ohm load (the 8 ohms aren't in parallel) there are two loads of 8 ohms each, one for each channel.

2
  • Oh... So it turns out I had forgotten how an amp works. Thank you both for clearing it up to me, and sorry for asking such a stupid question.
    – Matthew
    Commented Mar 7, 2015 at 16:21
  • There are no stupid questions [well, within limits] only stupid answers - of which this isn't one ;)
    – Tetsujin
    Commented Mar 7, 2015 at 16:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.