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Old 17th February 2003, 03:27 PM   #1
Bricolo is offline Bricolo  France
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Default Is there a rule of thumb for calculating the transormer's VA rating?

For a class AB amp, according to the output power?
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Old 17th February 2003, 04:21 PM   #2
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Default here's some information

This table appeared in an article entitled: "Rewinding Transformers with CAD" from "Ham Radio", December 1986 -- interesting to have a good memory, huh?
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Old 17th February 2003, 04:26 PM   #3
Bricolo is offline Bricolo  France
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thanks, i'll archive this on my computer


but I wasn't asking how to make a trafo, but how to calculate the trafo's power I need, accrding to the amp's output power
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Old 17th February 2003, 04:29 PM   #4
sangram is offline sangram  India
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I had asked the same Q earlier, and I got answers ranging from 1.4 to 1.8, give or take.

I also got someone who claimed to be running two LM3875 off a 50VA transformer, or something similar.

I also managed to run a TDA 2030 off an 18VA transformer. Not very loud or clean, but it did run, so go figure...
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Old 17th February 2003, 04:36 PM   #5
Bricolo is offline Bricolo  France
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It's for 2 lm3875 at +-18V AC in 4 ohms, so, say 50W output power each, 100W total out of the system


is 225VA good? too much, not enough?
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Old 17th February 2003, 05:19 PM   #6
JDeV is offline JDeV  South Africa
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Quote:
Originally posted by Bricolo
It's for 2 lm3875 at +-18V AC in 4 ohms, so, say 50W output power each, 100W total out of the system


is 225VA good? too much, not enough?

225VA is good. 18V AC will give you about 24V DC and that will give you about 45Watt.

This is not just another gainclone
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Old 17th February 2003, 05:23 PM   #7
sam9 is offline sam9  United States
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Default How simple do you want it?

A more presice calculation is possible but 3:1 will be pretty close of a stereo pair in to 8-ohms. I.e., if you want to build a 2-channel amp of 100W/ch into 8-ohms use a 300VA transformer. (100x3=300). If you want 4-ohm capability, double that. If you only want one channel halve it. Etc.

This is crude and inexact but it will get you close. Treat the value as a minimum an choose a transformer at least equal to the figure.

I assume you knoe how to determine the required voltages of the secondaries. Oddly enough, I found this easier to make an error on than the VA rating.
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Old 17th February 2003, 07:09 PM   #8
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To dig a little deeper into the matter than just a rule of thumb,
I have been wondering about what the relevant factors are for
calculating an appropriate VA rating. The simplest approach is
to say that the average power consumed by the amplifier should
match the VA rating of the transformer. Most commercial amps
will use a lower value than this, relying on the capacitors to
store enough energy for occasional peaks. In this case
temperature is an issue, since the transformer may overheat if
we draw more power than it is rated for.

Most DIYers will not be satisfied by this solution so
let's assume we use the max. average consumed power as
a minimum for the VA rating. At a first glance, this should also
be sufficient and there should be no risk of overheating.

However, what we ignore here is the conduction angle.
Since the capacitors are not being charged during the whole
cycle, the instantaneous current drawn from the transformer
is much higher than the average current. Due to winding
resistance the transformer will not be able to deliver its
specified voltage, so it seems we must use a transformer
with higher rating for this reason, but how much higher rating?
An obvious answer would be to use a transformer that can
deliver the require voltage at an average load current equal to
the actual instantaneous charge current. That would however
force us to use extremely large transformers, so it seems we
must compromise here, and accept that we lose some voltage.

That is about how far my own thinking has got me, although
it does not provide an answer to how much the voltage will
drop, unless we know the winding resistance. However, I
suspect that also magnetic phenomenae may come into play
here. I am very bad at magnetics theory, but I suspect the
winding inductances must also be taken into consideration.
Furthermore, I also suspect that we might get problem with the
core saturating from the high instantaneous currents. Ihave no
clue myself, however, how to take these things into account.

Anyway, it seems to me that the conduction angle matters, so
the choice of VA rating is partially dependent on the choice of
capacitors. Can somebody fill in some of the missing things
in my reasoning, or is it rather that I am wandering off in the
wrong direction??
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Old 20th February 2003, 08:50 AM   #9
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check out databook DL151/D from
www.onsemi.com

for more precise info..
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Old 20th February 2003, 09:04 AM   #10
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The only thing that really matters in a transformer is the temperature inside. Too high temperature distroys the isolation.

Little smoothing, = very high ripple = DC power out = VA rating
Medium smoothing, = high ripple = DC power out = 0.7 * VA rating
Heavy smoothing, = low ripple = DC power out = 0.5 * VA rating

300 VA transformer with a huge smoothing package = 150 W DC power = 70-90 audio-W out to load, 100% duty cycle (not very likely!)
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