Let's say I've got some 4-ohm speakers that are advertised to take up
to 400W.

Let's also say I want to build a 200W audio amplifier.

Since P=IV, and I=V/R, P=V^2/R, or V=sqrt(PR) (where R=4 ohms).

So if I want a 200W amp, the amp has to put out a peak voltage of
sqrt(200 x 4), or 28V, correct?

Similarly, if I want to build a 400W amp, I'd need the amp to put out
a peak of 40V, right?

Let me know if my calcs are correct, or if I forgot something...

Michael

On Aug 26, 8:58 pm, mrdarr...@gmail.com wrote:
> Let's say I've got some 4-ohm speakers that are advertised to take
up
> to 400W.
>
> Let's also say I want to build a 200W audio amplifier.
>
> Since P=IV, and I=V/R, P=V^2/R, or V=sqrt(PR) (where R=4 ohms).
>
> So if I want a 200W amp, the amp has to put out a peak voltage of
> sqrt(200 x 4), or 28V, correct?
>
> Similarly, if I want to build a 400W amp, I'd need the amp to put
out
> a peak of 40V, right?
>
> Let me know if my calcs are correct, or if I forgot something...
>
> Michael

You forgot peak to peak and the RMS conversion so you need 80 Vp-p
_under load_ meaning you'll need considerably more than that as the
power supply droops under load. To double the power output, the power
supply has to go up 1.41 (square root of 2) which means the load
current also goes up 1.41. Keep in mind the power line voltage wanders
around a bit too. At my house I've measured it between 108 and 122
depending if our A/C is on, how many neighbors A/C is on as 10 houses
share one pole pig here. Point is, unless you use a regulated supply,
your max power capability will vary quite a lot. If you design for 200
watts at 4 ohms and 108 volts, when the power line goes to 122 your
power can go up to 255. Most won't consider this a problem - more
power, right? But you have to be able to handle the peak voltage and
currents at the higher voltages. You get the picture.

Good luck with this

GG

mrdarrett@gmail.com wrote:

> Let's say I've got some 4-ohm speakers that are advertised to take up
> to 400W.
>
> Let's also say I want to build a 200W audio amplifier.
>
> Since P=IV, and I=V/R, P=V^2/R, or V=sqrt(PR) (where R=4 ohms).
>
> So if I want a 200W amp, the amp has to put out a peak voltage of
> sqrt(200 x 4), or 28V, correct?

First off, a 200W (average power) amplifier driven with sinewaves (the normal
way to rate audio amplifiers) puts out a peak of 400W.

From P = V^2/R ..... V = sqrt (P.R) = sqrt (400.4) = 40V


> Similarly, if I want to build a 400W amp, I'd need the amp to put out
> a peak of 40V, right?

56.6 V


> Let me know if my calcs are correct,

They weren't.

Graham

On Aug 26, 9:31 pm, stratu...@yahoo.com wrote:
> On Aug 26, 8:58 pm, mrdarr...@gmail.com wrote:
> > Let's say I've got some 4-ohm speakers that are advertised to take
> up
> > to 400W.
> >
> > Let's also say I want to build a 200W audio amplifier.
> >
> > Since P=IV, and I=V/R, P=V^2/R, or V=sqrt(PR) (where R=4 ohms).
> >
> > So if I want a 200W amp, the amp has to put out a peak voltage of
> > sqrt(200 x 4), or 28V, correct?
> >
> > Similarly, if I want to build a 400W amp, I'd need the amp to put
> out
> > a peak of 40V, right?
> >
> > Let me know if my calcs are correct, or if I forgot something...
> >
> > Michael
>
> You forgot peak to peak and the RMS conversion so you need 80 Vp-p
> _under load_ meaning you'll need considerably more than that as the
> power supply droops under load. To double the power output, the power
> supply has to go up 1.41 (square root of 2) which means the load
> current also goes up 1.41. Keep in mind the power line voltage wanders
> around a bit too. At my house I've measured it between 108 and 122
> depending if our A/C is on, how many neighbors A/C is on as 10 houses
> share one pole pig here. Point is, unless you use a regulated supply,
> your max power capability will vary quite a lot. If you design for 200
> watts at 4 ohms and 108 volts, when the power line goes to 122 your
> power can go up to 255. Most won't consider this a problem - more
> power, right? But you have to be able to handle the peak voltage and
> currents at the higher voltages. You get the picture.
>
> Good luck with this
>
> GG


Ah, ok. I was thinking more along the lines of car audio than home
audio, actually, and was wondering how big of a boost converter I'll
need.

Then again, I saw a 200W car stereo amp from Wal Mart for $50. I
wonder if I can build it any cheaper than that... maybe I'll just buy
it and take it apart to see what toys they put in it.

Michael

On Mon, 27 Aug 2007 08:26:59 -0700, mrdarrett wrote:

> Then again, I saw a 200W car stereo amp from Wal Mart for $50. I
> wonder if I can build it any cheaper than that... maybe I'll just buy
> it and take it apart to see what toys they put in it.

That '200W' was the peak power, sum of channels, with a generous bullshit
factor thrown in.

On Aug 27, 2:22 pm, "Stephen J. Rush" <sjr...@comcast.net> wrote:
> On Mon, 27 Aug 2007 08:26:59 -0700, mrdarrett wrote:
> > Then again, I saw a 200W car stereo amp from Wal Mart for $50. I
> > wonder if I can build it any cheaper than that... maybe I'll just buy
> > it and take it apart to see what toys they put in it.
>
> That '200W' was the peak power, sum of channels, with a generous bullshit
> factor thrown in.


Thanks for that...! My Kenwood already is 45W x 4. Guess I don't
need that amp after all...

Michael