On 10/03/2008, piclist <amerton.piclist@ihug.co.nz> wrote:
> Michael Cunningham wrote:

....
> > "When you convert a AC to DC by a full wave rectifier
> > bridge you have to multiply the output voltage by 1.41
> > to obtain the VA"
> >
> >

Michael,
When measuring AC power levels the problem is that the AC voltage and
current is varying so the instantainious power delivered (the product
of voltage and current at that instant) is not constant. So it is
convienient to convert the AC readings to an equivelent constant
reading that would deliver the same power over an extended period.
(i.e greater than one cycle). This can be done by taking a large
number of readings over the course of one cycle, multiplying the
current and voltage and iintegrating the result.

Or, if we assume we are looking at sinewaves, we can perfom a
mathmatical calculation based on whatever parameter we actually know
or measure.

So for example if we have a sinewave at 325V peak applied to a
resistor, it will produce pretty much the same amount of heat as if a
230V DC voltage was applied.

With a sinewave, the ratio of the peak value  to this "average"
(otherwise known as the rms value)  is root2 = 1.414. That's where the
1.41 factor comes from and mathamatically it appears as a term in the
integration result.

So a "230V" AC supply is actually a calculated value, based on the
waveform being a sinewave with a peak value of just over 325V.

The same goes for current.

So a 30A AC current will have a peak value of ~42.4V.

If you are measuring the current, it is therefore useful to know the
type of measurement you are making - is it an instantainious reding -
or a pek reading. If it is a peak reading, how long is the peak value
held before it decays?

Since you are using a PIC, the readings are likely to be
instantainious, in which case you either need to synchronise the
reading with the incoming signal peak, or take a number of readings
and  process the results. Simply averaging the readings will give you
a value related to the correct value but there will always be some
error as the straightforward avaerage value is NOT the same as the rms
vale..

Best bet / easiest. is to take samples at 1mS intervals  (or less) and
for longer than an AC cycle and use the maximum of these as the peak
value to convert to rms by dividing by 1.414. Or even better, use the
(max-min) result  for a pk-pk reading & divide by 2.818

Always assuming a sine wave.


The complicated way is to take each value, square it, average the
results over a cycle period (17 or 20mS for 60 or 50Hz) and then
square root to get the rms value.
This will produce accurate results in the event the waveform is not a
sine wave but is a lot more calculation intensive.

Richard
-- 
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist