Is there a rule of thumb for calculating the transormer's VA rating?
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 17th February 2003, 03:27 PM #1 diyAudio Member     Join Date: Nov 2002 Location: Grenoble, FR Is there a rule of thumb for calculating the transormer's VA rating? For a class AB amp, according to the output power?
diyAudio Member

Join Date: Apr 2002
Location: Llanddewi Brefi, NJ
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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 17th February 2003, 04:26 PM #3 diyAudio Member     Join Date: Nov 2002 Location: Grenoble, FR 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
 17th February 2003, 04:29 PM #4 diyAudio Member   Join Date: Sep 2002 Location: India 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...
 17th February 2003, 04:36 PM #5 diyAudio Member     Join Date: Nov 2002 Location: Grenoble, FR 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?
diyAudio Member

Join Date: Oct 2002
Location: Cape Town
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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 17th February 2003, 05:23 PM #7 diyAudio Member     Join Date: Jun 2002 Location: Left Coast 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.
 17th February 2003, 07:09 PM #8 diyAudio Member   Join Date: Sep 2002 Location: Sweden 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??
 20th February 2003, 08:50 AM #9 Account Disabled   Join Date: May 2002 check out databook DL151/D from www.onsemi.com for more precise info..
 20th February 2003, 09:04 AM #10 Electrons are yellow and more is better! diyAudio Member     Join Date: Apr 2002 Location: Göteborg, Sweden Blog Entries: 4 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!) __________________ /Per-Anders (my first name) or P-A as my friends call me Tube Buffered Gainclone in work |Thread

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