8
\$\begingroup\$

I’m trying to figure out how a step-up transformer works. A step-down transformer is simple and logical enough; you start out with a higher voltage and end with less, the remainder being wasted as heat. But with a step-up transformer, you end up with more voltage than you start with.

I tried looking it up, but all I can find (whether online or even in some electronics texts) is general information on how transformers work (induction, Faraday’s law, construction, etc.) and explanations of the difference between step-ups and step-downs in terms of the number of turns, but not specifically how step-ups result in more voltage.

Where does that extra voltage come from? Not magic…

\$\endgroup\$
12
  • 26
    \$\begingroup\$ You are wrong about "wasted as heat." \$\endgroup\$
    – markrages
    Mar 3, 2011 at 16:46
  • 20
    \$\begingroup\$ Yes, it does matter where it goes. You are confusing voltage with power, which is why your question is wrong. \$\endgroup\$
    – markrages
    Mar 3, 2011 at 16:55
  • 1
    \$\begingroup\$ The heat in a transformer is due to copper loss and core loss, neither of which is due to voltage. Copper loss is the I^2R loss due to the current and resistance, and core loss is due to the magnetic material and magnetic field which is also current derived. \$\endgroup\$
    – Martin
    Mar 3, 2011 at 17:50
  • 3
    \$\begingroup\$ @synetech, think of voltage as potential energy, just like gravitational. you can get higher voltage by using one mass(current) on one side to counterweight your mass(current) on the other. As long as P in on the left side is equal to the P out on the right you are golden. \$\endgroup\$
    – Kortuk
    Mar 3, 2011 at 19:53
  • 1
    \$\begingroup\$ Using the proposed logic, a step-up transformer would get cold :) \$\endgroup\$
    – gbarry
    Oct 22, 2012 at 16:51

4 Answers 4

34
\$\begingroup\$

I think what you're missing is the current...

  • Step down transformers change a high voltage/low current, to low voltage/high current.

  • Step up transformers change a low voltage/high current, to high voltage/low current.

So, in an ideal 100% efficient transformer, the power doesn't change and no heat will be generated by the transformer, i.e. the power in = the power out, because Power = Volts x Amps.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Aha, that makes sense. The explanations I could find all made it seem like a step-up was giving something for nothing without explaining where it comes from. Adding current to the query, finds a proper explanation much easier now: powertransformer.us/stepuptransformers.htm Thank you! \$\endgroup\$
    – Synetech
    Mar 3, 2011 at 17:02
  • \$\begingroup\$ @Joby: Thanks... I was going to tidy it up later anyway! \$\endgroup\$
    – BG100
    Mar 3, 2011 at 17:10
  • \$\begingroup\$ @Synetech inc.: No problem. \$\endgroup\$
    – BG100
    Mar 3, 2011 at 17:11
27
\$\begingroup\$

Think of a transformer as being a like gearbox (or pulley system, or lever, or other such machine). A 10:1 gearbox can turn a 60rpm rotation into 600rpm, but if the output requires a certain amount of torque to rotate, the input will require at least ten times that much (slightly more, in practice, because of friction in the gearbox itself).

\$\endgroup\$
1
  • \$\begingroup\$ nice analogy! \$\endgroup\$
    – BG100
    Mar 3, 2011 at 18:15
9
\$\begingroup\$

Think about it this way: magnetic field have no idea how many loops you have on the second half of transformer.

So, each loop on second side works like a small 'bettery' connected in series, the more loops - the more batteries - the more voltage.

But as there is the same amount of magnetic field is divided on all loops, achivable current is less.

Same works another way: Less loops - less voltage, but more current as more magnetic field is left for 1 loop.


In ideal situation no heat is generated. Heat is generated the way you say only in linear regulators.

\$\endgroup\$
3
\$\begingroup\$

"you start out with a higher voltage and end with less, the remainder being wasted as heat"

That's absolutely false. In an ideal transformer no heat would be generated, no matter how much the voltage difference is. A transformer transforms the input voltage (actually input power) into a variable magnetic field. That magnetic field generates a voltage in the secondary winding, and the voltage ratio between primary and secondary is equal to the ratio of the number of turns. So you can get a higher output voltage by giving the secondary winding more turns that the primary's.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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