0
\$\begingroup\$

What I would consider two basic facts about electricity are conflicting (to my understanding, obviously not in real life). I know it's said that a transformer provides isolation. I think I remember correctly, reading that the load on a circuit determines how much current is drawn from the source. So can a load on the secondary side determine current drawn from the primary side? It would seem not from the word "isolation", but I don't understand induction well enough to see why not I guess. Thank you for any answers.

\$\endgroup\$
0
\$\begingroup\$

When they talk about isolation in the context of a transformer they are usually referring to what is called Galvanic Isolation, not that the input power is independent of the load.

Galvanic Isolation is when there is no direct connection between the two circuits. In the case of a transformer the only connection is through magnetic flux linkage.

Transformers are often used like this for safety reasons to minimize the chances of electrocution.

\$\endgroup\$
5
\$\begingroup\$

Isolation is often misunderstood for safety or for somehow limiting the power that can pass through a transformer. Although a transformer CAN increase safety, just the presence of a transformer in no way guarantees it.

Isolation transformers provide Galvanic isolation. The principle of isolating functional sections of electrical systems to prevent current flow; no direct conduction path is permitted between them.

schematic

simulate this circuit – Schematic created using CircuitLab

The circuit above contains an isolation transformer. However as you can see, if you grab both sides of the lamp you will be electrocuted just the same as if the transformer was not there.

What about this circuit? Is it isolated?

schematic

simulate this circuit

If you answer yes it is an isolated circuit you would be correct. Some of you will say, "But they are both connected to the same point!", You are right, however that does not matter, there is no current loop between the left side and the right side, so no current can pass between each ground point. They are simply held at the same voltage.

The isolation part comes into it's own by the fact that you can use different references on each side of the transformer.

schematic

simulate this circuit

Notice in this version, though the transformers are basically the same, we can use a different point as a reference to change how the voltage is biased around ground. This is because the circuit, before we grounded it, was actually floating.

You can do much more interesting things with this too, like the following.

schematic

simulate this circuit

In short, an isolation transformer de-couples your signal from whatever zero potential reference you are using on the primary side so you can attach it to a different reference on the secondary side.

Conservation of Energy

The other critical thing about transformers is whatever power goes out one side must come in the other. Whether it be a 1:1 transformer like I used here or a step-down or step-up transformer that rule always holds true. (Within the capability of the windings and core anyway.)

In reality you have to put a little more in than you get out because of losses due to heating and vibration of the transformer. But it can never be less.

That is why isolation transforms give you no guarantee of safety. They may reduce the voltage or current to a level that will not kill you, but the transformer itself can still be capable of supplying the same raw power.

Indeed in the case of a step-up transformer, it is usually MORE likely to kill you.

\$\endgroup\$
2
  • \$\begingroup\$ Pictures and commentary. Gotta love it. \$\endgroup\$ – jonk Apr 6 '17 at 5:49
  • \$\begingroup\$ @jonk... ;)..... \$\endgroup\$ – Trevor_G Apr 6 '17 at 14:20
0
\$\begingroup\$

Power transformers often use 10% of rated real current using phase shifted reactive current to couple the magnetic flux between primary and secondary. This 10% input power is mostly reactive power or VAR not real power such that it does not cause much heat for the rated voltage.

But if one applied over-rated voltage, then the voltage can "saturate" the transformer magnetic flux in the core and then the inductance goes away and it draws excessive current like a wirewound resistor even with no load. (bad, really bad...)

The turns ratio for voltage is such that if n=Vout/Vin then current is the inverse n=Iin/Iout thus the output load current is "transformed" to the input such that

V * A in = V * A out + <10% loss at rated load with defined temp rise..

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.