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Old 1st April 2009, 02:56 AM   #1
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Default Voltage, power, impedance

I wasn't sure where I should put this, so I guess it ends up here. Anyway, I'm a little confused about the relationship between peak voltage and RMS voltage as it relates to power and impedance. Let's say I have an amp with +/- 30VDC rails and it is driving a power resistor of 4 ohms. The Vp is 30V, so the Vrms = 30/1.414=21.2V. This is the value that I use for power calculations, right?

So P = V^2/R
P = 21.2^2/4
P=112W

Is this correct? If so, then I move on to the following calculation: an amp which needs to provide 100W to a 7-ohm load must have +/- 37.4VDC rails (assuming that it can rail), correct?

Vrms=sqrt(100*7)
Vrms=26.46V
Vp=26.46(1.414)=37.4VDC
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Old 1st April 2009, 03:35 AM   #2
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That's correct.

In reality, your OP trannies can't actually swing rail to rail and then the PSU usually sags a bit at full load hence YMMV.

edit: oh, and a 7ohm speaker isn't really 7 ohms, well maybe at a couple of frequencies
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Old 1st April 2009, 03:47 AM   #3
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Quote:
Originally posted by Iain McNeill
That's correct.

In reality, your OP trannies can't actually swing rail to rail and then the PSU usually sags a bit at full load hence YMMV.

edit: oh, and a 7ohm speaker isn't really 7 ohms, well maybe at a couple of frequencies
Thanks for the help.

I can read impedance charts, though.
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Old 1st April 2009, 11:27 PM   #4
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that sounds about right. a while back i made up a "power guesstimation" chart based on rail voltage, peak voltage and p-p voltage so that techs wouldn't have to mess with a calculator when testing an amp. it's mostly "ballparked" values, but it works and saves a bit of time. as they used to say "it's close enough for rock and roll"......
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