Hello All,

I have been trying to determine the right size gauge for plugging wire into an industrial transformer (Brand: Hevi-duty, Company: Sola, ID: E100 I attached the datasheet below). The primary winding is 480V AC but I will only be using half of it (240V AC). The secondary winding produces 120VAC. According to the manual the larger of the voltages is considered primary while the the smaller is secondary. However, I need to convert 120VAC to 480VAC, so I figured I can use the secondary winding as a primary winding and the primary winding as a secondary winding. I'm not sure if this is okay or not. Also, I can find the Volt-Amps value for the inductor and since only one conductor is attached, the power factor must be about 90 degrees, so power is being transferred through the magnetic field of the coil and the resistance of the wire used to transfer the power is small, so the real power is small as well. Then, how would I calculate the current going through the wire assuming I know resistivity and the other factors I just mentioned? I'm planning on using a 12 gauge wire to connect a transformer from an outlet, transform it to 240V, connect it to a power module, which converts single phase to 3 phase system 230VAC and attach that to a motor. This brings me to another question. The label of the power module says "Motor: 0.75hp" but the motor I'm using is only about 1/5 of that (120W). I read through the manual for hours and could not find anything useful. I'm going to ask technical support, but if you know right away, I would appreciate your response. I have attached the manual of the power module as well. How do I figure out the amount of current that would flow into the wire from this previous information? Am I missing any information? Am I overthinking it?

Best,
Adanovinivici

 

Theoretically, the input is applied to the primary winding and the output is extracted from the secondary winding.
The ratio of the input voltage to the output voltage is equal to the number of turns on the primary divided by the number of turns on the secondary.
For a transformer of 100% efficiency, the volts x amps on the primary equals the volts x amps on the secondary.

There might be other physical limitations and considerations which I am not aware of.

If the transformer is designed to transform 480VAC to 120VAC, then the turns ratio would be 4:1. This is called a step-down transformer.
If you turn the transformer around, then it would transform 120VAC to 480VAC. As one might guess, this would be called a step-up transformer.

The current in the primary (120VAC) will be 4 times the current in the secondary (480VAC).
For example, if the load takes 1A @ 480VAC, the current in the primary will be 4A @ 120VAC.

 

Alright that helps a bunch! Thank you MrChips.

 

usually you can't just turn around the TF-s because their primary is considered to be the primary e.g has too high resistance for secondary
if you somehow overcome that -- the core will saturate . . . that's why they use inverters . . . i guess

 

On that note, be aware that the winding that was intended to be driven was designed to drive the magnetizing current as well as the load on the other winding. When you turn the transformer around and drive the load side you will probably see increased harmonic distortion and less output voltage than expected for a given load because of IR drop in the driven winding because of the magnetizing current.

Depending upon the application the magnetizing current can be nearly as high as the intended load current.

 

Just curious. Why would there be harmonic distortion, and how would it affect a power supply?

 

I am confused. What is the voltage of your available power source? I take it that it's single phase.

And you need 240 V, 3 phase for a constant load?

If you have 120 V........then you have 240 V. Single phase.

 

Just curious. Why would there be harmonic distortion, and how would it affect a power supply?

To expand the explanation just a little, in the general case the magnetizing current is a nonlinear function of voltage and time. In the photo below the yellow trace is current through the driven winding of a lightly loaded power transformer, so most of the current is magnetizing current. (This is a case in which a low voltage winding "secondary"is driven to produce 240 VAC on what was intended to be the "primary" winding.



The current in and the resistance of the winding cause an IR drop within the winding. A result of the IR drop is that the waveform on the output is distorted (blue trace) and the output voltage is less. In the case of an impedance protected transformers (short-proof) in which the resistance of the secondary driven as a primary was so high that 15% of the P-P voltage was lost to magnetizing current related IR drop.

Just something to be aware of.

 

I appreciate it guys! I wasn't aware of these nuances, so I will definitely be aware of it next time around. Thank you DickCappels and everyone who contributed to the post.

 

To expand the explanation just a little, in the general case the magnetizing current is a nonlinear function of voltage and time. In the photo below the yellow trace is current through the driven winding of a lightly loaded power transformer, so most of the current is magnetizing current. (This is a case in which a low voltage winding "secondary"is driven to produce 240 VAC on what was intended to be the "primary" winding.

View attachment 113491

The current in and the resistance of the winding cause an IR drop within the winding. A result of the IR drop is that the waveform on the output is distorted (blue trace) and the output voltage is less. In the case of an impedance protected transformers (short-proof) in which the resistance of the secondary driven as a primary was so high that 15% of the P-P voltage was lost to magnetizing current related IR drop.

Just something to be aware of.

Could you explain to me what you mean by IR drop? I'm not familiar with that term.

 

I is current
R is resistance

The voltage drop, E across a resistance (in this case the resistance the driven winding) is equal to current through the winding times the resistance of the winding, conventionally this is written as E= IR. IR refers to the voltage drop.

 

Are we talking constant dc current here?
I was taught that reactance controls current and therefore the voltage drop across the coil.
The IR drop is the heat of the coil. Not the waveform.

 

If you do not mind a question or two, what is the purpose of this scheme? That is, what is the small three phase motor going to be doing?

 

Are we talking constant dc current here?
I was taught that reactance controls current and therefore the voltage drop across the coil.
The IR drop is the heat of the coil. Not the waveform.

It has nothing to do with DC current.
It's the transformer impedance, not just the reactance that determines the current.
Transformer AC primary current is a combination of magnetizing current from the transformer inductive reactance in series with the winding resistance, and the load current (normally the inductive reactance is much higher than the winding resistance).
These currents cause an IR drop equal to the sum total of these currents times the winding resistance.