That depends on how you define the VA rating.
If you use the common convention of VA rating = apparent power deliverable to the load in the intended connection of the transformer then it is indeed double for a 2:1 autotransformer of the same size as an isolating transformer. That is, you see the transformer as a black box with an input and an output and do not care about what is inside.
If you buy a 100VA 2:1 autotransformer it will usually mean that it was made to deliver 100VA to the load. You could do the same with a 50VA rated 1:1 isolating transformer of suitable voltage externally connected as an autotransformer.
If you use the common convention of VA rating = apparent power deliverable to the load in the intended connection of the transformer then it is indeed double for a 2:1 autotransformer of the same size as an isolating transformer. That is, you see the transformer as a black box with an input and an output and do not care about what is inside.
If you buy a 100VA 2:1 autotransformer it will usually mean that it was made to deliver 100VA to the load. You could do the same with a 50VA rated 1:1 isolating transformer of suitable voltage externally connected as an autotransformer.
I still can't follow that explanation, on the one hand you say it is the same VA rating and then state is doubles and can use a half VA transformer.
Let's see it the way I have come to understand it.
Buy a 230:10Vac transformer.
Wire the 10Vac secondary in series with the 230Vac primary.
Connect the 230Vac primary to the 230Vac mains power.
The output voltage across the primary is 230Vac, the output voltage across the series pair is 241Vac to 242Vac when open circuit and when fully loaded will drop to 240Vac.
The maximum rated 240Vac output current is the current rating of the 10Vac secondary.
If the isolating transformer were a 50VA 230:10Vac, then when used as an autotransformer it will give ~1200VA when delivering 5Aac.
I cannot see how any explanation that says "it depends" or it's "the same VA" can resolve my understanding of the autotransformer operation.
Let's see it the way I have come to understand it.
Buy a 230:10Vac transformer.
Wire the 10Vac secondary in series with the 230Vac primary.
Connect the 230Vac primary to the 230Vac mains power.
The output voltage across the primary is 230Vac, the output voltage across the series pair is 241Vac to 242Vac when open circuit and when fully loaded will drop to 240Vac.
The maximum rated 240Vac output current is the current rating of the 10Vac secondary.
If the isolating transformer were a 50VA 230:10Vac, then when used as an autotransformer it will give ~1200VA when delivering 5Aac.
I cannot see how any explanation that says "it depends" or it's "the same VA" can resolve my understanding of the autotransformer operation.
The rated VA from the manufacturer depends on the intended application.
A transformer that is internally an autotransformer will be rated by how much it can deliver to the load.
If your 230V:10V 50VA isolating transformer instead had come connected as an autotransformer from the factory it would have been rated 1200VA by the manufacturer if it was intended to transform between 230V and 240V.
A transformer that is internally an autotransformer will be rated by how much it can deliver to the load.
If your 230V:10V 50VA isolating transformer instead had come connected as an autotransformer from the factory it would have been rated 1200VA by the manufacturer if it was intended to transform between 230V and 240V.
Andrew:
Using your example, you put in 1200VA and get out 1200VA. The auto may be nameplated as a 1200VA transformer. This is deceptive, IMO, but I won't complain.
If you weigh the xfmr, analyze wire size, and get some dimensions of the core, you will find the unit is curiously similar to a 230V:10V isolation transformer carrying a 50VA rating. This is no coincidence. I can take that little 50VA xfmr, connect it as an auto, and get 1200VA worth of input/output performance. It's not magic.
The key is that your 1200VA auto has only transformed (by magnetic flux) 50VA worth of 'power'. The only transformation occurring is that boosted by the 10V winding. I don't care for the 1200VA rating because you haven't actually transformed 1200VA; but megajocke is correct that all the user cares about is input/output ratings, not what is actually occurring internally.
Using your example, you put in 1200VA and get out 1200VA. The auto may be nameplated as a 1200VA transformer. This is deceptive, IMO, but I won't complain.
If you weigh the xfmr, analyze wire size, and get some dimensions of the core, you will find the unit is curiously similar to a 230V:10V isolation transformer carrying a 50VA rating. This is no coincidence. I can take that little 50VA xfmr, connect it as an auto, and get 1200VA worth of input/output performance. It's not magic.
The key is that your 1200VA auto has only transformed (by magnetic flux) 50VA worth of 'power'. The only transformation occurring is that boosted by the 10V winding. I don't care for the 1200VA rating because you haven't actually transformed 1200VA; but megajocke is correct that all the user cares about is input/output ratings, not what is actually occurring internally.
Ziga,
I find your last paragraph quite confusing.
It does not make clear anything that we do agree on.
That is the same with post19.
I can see what is meant, but it is not what you are saying.
I find your last paragraph quite confusing.
It does not make clear anything that we do agree on.
That is the same with post19.
I can see what is meant, but it is not what you are saying.
Maybe this thread would help, especially the pdf drawings.
http://www.diyaudio.com/forums/tubes-valves/133843-how-wire-buck-transformer.html
Back to the example, when connected as an auto, all the transformer has to do is boost a current of 5A through a potential rise of 10V. The rest of the 230 V is already provided by the source, and goes through no transformation to the 240V output. The reason this is possible is there is a direct electron connection from source to load (i.e. no galvanic isolation).
So on the input side, there is some magnetizing current drawn to energize the 230V winding, establishing flux in the core. This effectively does nothing other than providing the 10V potential across the secondary winding. Since the phase relationship of primary and secondary are essentially identical, polarities connected for boost operation, the source has 10V added to it on the output side, hence 240V output.
With load added, the source is providing all of the 230V portion of the output direct, without needing the transformer at all. In order to pass current through the 10V winding, flux linkage is required, and this is where the transformer is actually doing its work.
Consider the pdf attachment below and ask yourself three questions:
What voltage is provided to the load ?
How much current is flowing in primary and secondary ?
How much has the transformer 'transformed', based on volts*amps ?
Now take it to the next step by connecting to the 10V winding.
http://www.diyaudio.com/forums/tubes-valves/133843-how-wire-buck-transformer.html
Back to the example, when connected as an auto, all the transformer has to do is boost a current of 5A through a potential rise of 10V. The rest of the 230 V is already provided by the source, and goes through no transformation to the 240V output. The reason this is possible is there is a direct electron connection from source to load (i.e. no galvanic isolation).
So on the input side, there is some magnetizing current drawn to energize the 230V winding, establishing flux in the core. This effectively does nothing other than providing the 10V potential across the secondary winding. Since the phase relationship of primary and secondary are essentially identical, polarities connected for boost operation, the source has 10V added to it on the output side, hence 240V output.
With load added, the source is providing all of the 230V portion of the output direct, without needing the transformer at all. In order to pass current through the 10V winding, flux linkage is required, and this is where the transformer is actually doing its work.
Consider the pdf attachment below and ask yourself three questions:
What voltage is provided to the load ?
How much current is flowing in primary and secondary ?
How much has the transformer 'transformed', based on volts*amps ?
Now take it to the next step by connecting to the 10V winding.
Yup,
You can either look at how much the transformer has "transformed" magnetically (50VA) or what is delivered to the load (1200VA). The first one is what determines the size of the unit but the second one is usually what is stated on the nameplate of a factory-made autotransformer because this is what you are interested in when connecting a load.
You can either look at how much the transformer has "transformed" magnetically (50VA) or what is delivered to the load (1200VA). The first one is what determines the size of the unit but the second one is usually what is stated on the nameplate of a factory-made autotransformer because this is what you are interested in when connecting a load.
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Good explanantion, pretty clear.
For "In order to pass current through the 10V winding...", I am not saying this is wrong, but I would prefer to state it in this way, hopefully a little bit clear:
Since there is a 10V emf of the upper winding, if it's connected to load, current will flow "naturally". The consequence of this current is to decrease the flux of the winding (for both upper and lower since they are coupled), which means the emf provided by the lower windings will be lower, i.e., will be less than the input voltage...but the emf of the lower winding has to be equal to the input voltage, the only way is to increase the flux upto the orignal value (the value at unloaded condition), and to this end, the current in the lower winding has to be increased to balance the flux decrease caused by the 10V windings----"this is where the transformer is actually doing its work".....
Sure, but the OP request was surplus autos that had no nameplate. A little reverse engineering will be needed. Without consideration of the "actual flux transforming" capabilities of the unit, you will have a difficult time determining the overall input/output ratings.
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