EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Throy on July 03, 2015, 06:53:39 pm
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In the datasheet here, it says the rated output current is 7.14A for the VTX-146-500-135. Can anyone tell me if that is per secondary winding or total (3.57A per winding)?
http://www.farnell.com/datasheets/1909346.pdf (http://www.farnell.com/datasheets/1909346.pdf)
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The rating is 500 VA. 500 VA / 7.14 A = 70 V. Since there are two 35 V secondaries, it is 7.14 A per secondary.
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7.14A * 35V * 2 windings = 500 VA.
Note that you can't draw 500W if you are intent on using a capacitor input rectifier. The power factor will be about 0.5, so that the maximum continuous power handing will be closer to 250W.
Tim
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Cool thanks for the help! :-+
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7.14A * 35V * 2 windings = 500 VA.
Note that you can't draw 500W if you are intent on using a capacitor input rectifier. The power factor will be about 0.5, so that the maximum continuous power handing will be closer to 250W.
Tim
I recently did some load test on my 300VA toroidal transformer, 2x 30V secondaries. After bridge rectifier and 10,000 uF filter cap, I measured 31VAC and 4.76A on one loaded secondary side (AC measurement), which gives 147W. The second half of the secondary was disconnected. This gives me PF of 0.98 ?? How could it be when you say it will be around 0.5 ? I must have done something wrong.. :o ?
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You have DC voltage after the bridge rectifier and capacitors, so what made you think to measure AC voltage?
For bridge rectifer+capacitor, I usually use Idc = 0.62 x Iac and Vdc peak = 1.414 x Vac [ - 2 x Vdiode (which varies with current) ] and Capacitance = Current / [2 x Frequency AC x (Vpeak - Vmin)] (most people use 2200-3300uF for 1A) to approximate voltage and current and capacitance required.
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To understand the problem, you must remember that heating of the windings depends of the effective value (root mean square = rms) of the alternating current.
When we say that secondary current is 10A max, it is 10Arms
Current waveform with rectifier + capacitor is very bad, it is short pulses with high peak value.
It has a high rms value for a low average one. (average = dc current)
PF is indeed high, but waveform is bad.
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I recently did some load test on my 300VA toroidal transformer, 2x 30V secondaries. After bridge rectifier and 10,000 uF filter cap, I measured 31VAC and 4.76A on one loaded secondary side (AC measurement), which gives 147W. The second half of the secondary was disconnected. This gives me PF of 0.98 ?? How could it be when you say it will be around 0.5 ? I must have done something wrong.. :o ?
According to your numbers you had a load of 147 VA on a 300 VA transformer. That seems to be about 0.5.
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@IvoS:
After bridge rectifier and 10,000 uF filter cap, I measured 31VAC and 4.76A on one loaded secondary side (AC measurement), which gives 147W.
With what kind of multimeter did you measure those values ? True rms or not ?
If not, those values are useless because secondary ac current is absolutely not sinusoidal and secundary ac voltage is highly distorted by the leakage inductance of the transformer.
If you have measured with a true rms multimeter, than 31Vrms and 4.76Arms are 147.56 VA, and not 147.56 W.
To know the power in W, you must use a Wattmeter if you want to measure on the ac side or measure output dc current and voltage. P(W) = Vdc x Idc
P(W) = P(VA) x PF
PF = P(W)/P(VA)
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I recently did some load test on my 300VA toroidal transformer, 2x 30V secondaries. After bridge rectifier and 10,000 uF filter cap, I measured 31VAC and 4.76A on one loaded secondary side (AC measurement), which gives 147W. The second half of the secondary was disconnected. This gives me PF of 0.98 ?? How could it be when you say it will be around 0.5 ? I must have done something wrong.. :o ?
According to your numbers you had a load of 147 VA on a 300 VA transformer. That seems to be about 0.5.
Yes, but using only one side of secondaries.
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@IvoS:
After bridge rectifier and 10,000 uF filter cap, I measured 31VAC and 4.76A on one loaded secondary side (AC measurement), which gives 147W.
With what kind of multimeter did you measure those values ? True rms or not ?
If not, those values are useless because secondary ac current is absolutely not sinusoidal and secundary ac voltage is highly distorted by the leakage inductance of the transformer.
If you have measured with a true rms multimeter, than 31Vrms and 4.76Arms are 147.56 VA, and not 147.56 W.
To know the power in W, you must use a Wattmeter if you want to measure on the ac side or measure output dc current and voltage. P(W) = Vdc x Idc
P(W) = P(VA) x PF
PF = P(W)/P(VA)
Absolutely true RMS, Agilent and Fluke.
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And BTW, it was PLITRON trafo.
http://shop.plitron.com/shopexd.asp?id=167 (http://shop.plitron.com/shopexd.asp?id=167)
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According to your numbers you had a load of 147 VA on a 300 VA transformer. That seems to be about 0.5.
Yes, but using only one side of secondaries.
That's not (that) important. The VA rating of a transformer is for the transformer as a whole, not for the secondaries individually. It is mostly determined by the size of the iron core.
Whether you take 150 VA from one secondary alone, or 75 VA from one secondary and 75 VA from another, it is still 150 VA in total.
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Beware: VA rating is total, and individual windings (primary and secondary) have their own ratings.
In this case, the transformer was operated properly: each winding is rated for about 5A, and about 5A RMS was drawn from the winding used. (So all we're missing is what the DC load was, which would tell power factor, but oh well.)
One secondary is certainly not rated for the full VA load, although you could probably get away with up to 40% higher load with just one winding. The toroid's construction will have a thinner layer of windings than a shell type, leading to a lower peak internal temperature despite the greater demands on the wire.
Some transformers are rated for full VA on partial windings; there are tapped 120/240 primaries for instance, where either winding is rated for full VA. There are transformers designed for FWCT rectifiers that have to handle more than normal rated VA because each winding is only being used half the time (same 40% figure at work). Doing that on more than a few windings requires more winding area, and you get into some funky core cuts/shapes when you need a lot of that (typical for transformers with weak coupling, magnetic shunts, current limited outputs, ferroresonant transformers, etc.).
Tim