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From: email@example.com (john jardine)
Subject: Re: Loudspeaker and microcontroller
Date: 24 Oct 2002 10:18:13 -0700
References: <firstname.lastname@example.org> <3DB65042.66FB6FCB@scazon.com> <email@example.com>
NNTP-Posting-Date: 24 Oct 2002 17:18:13 GMT
firstname.lastname@example.org (jools) wrote in message news:<email@example.com>...
> very helpful guys thanks, just could some one tell me the relevance of
> the speaker impedance, I know higher impedance less current through it
> etc. But which one do I want for the biggest sound output.
> Also the sound is just going to be a two tones, perhaps some wave
> shaping to make it slightly nicer to listen too.
A higher audio voltage ups the watts. A higher audio current ups the
The max audio voltage you can get is set by the power supply voltage.
The max audio current you can get is set by the speaker resistance.
You have choice of control over both. Double the audio voltage and you
get 4 times the watts. Halve the speaker resistance and get only twice
the watts. The magic formula ...
Speaker power is (audiovolts*audiovolts) / SpeakerResistance.
As you are probably stuck with (say) a 9V dc power supply, then this
sets the maximum audio voltage that can be pumped out to the speaker
(using an audio amp)at about 2.5V ac rms. A bigger audio power
therefore can only come from upping the speaker current, which means a
finding a speaker with the least resistance eg, 8 or 4 ohms. Which
would give in turn a max of about 3/4watt and 1.5watts, which is still
not a lot.
The best way to up the power is to go for increasing the amp supply
voltage to say 24V, use a standard 8ohm speaker and feed the audio
amp from the PIC, via a volume control (max 8watts).
Sounds complicated? ... It is!. There is *nothing* simple in
electronics design. Get this Watts/volts/ohms thing, sorted out in
your mind though and (honest!) you are well on you way to designing
100kWatt crowd pleasers.
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