I want to generate a magnetic field from a coil made of copper. Let's consider a simple circuit with a power source (12VAC), a transformer and a coil.


simulate this circuit – Schematic created using CircuitLab

I always see that step-up transformer increases the voltage and reduces the current while a step-down transformer increases the current and reduces the voltage because the power is conserved (P = VI). Since the magnetic field depends on current, I guess I should use a step-down transformer.

However, the coil has some resistance. So if I use a step-up transformer, I could generate a really large voltage across the coil and the current (I = RV) would also be large.

From my understanding, the current is dependant on the load following Ohm's law. If I use a step-down transformer, the voltage across the coil would lower and the current would lower as well. So I don't get why we say that step-up transformer (or a buck converter) would increase the current if it's decided by the load and the voltage across it and the power is conserved.

The way I see it is that the 12V is fixed at the primary. I feel like I'm missing something, please correct me if my understanding is wrong.

  • 1
    \$\begingroup\$ Is your 12 V AC or DC? \$\endgroup\$
    – winny
    Apr 4 at 17:37
  • 1
    \$\begingroup\$ You can't use a transformer with a DC voltage source, so the question doesn't really work as stated. Also, the "right" choice depends on the construction of your coil and what you're trying to do with it. For a wire-wrapped-around-a-nail electromagnet, 1 volt or so is sufficient. If you're using a thousand turns of magnet wire, that's different. \$\endgroup\$
    – hobbs
    Apr 4 at 17:38
  • \$\begingroup\$ I think you mean boost converter (not buck). You don't often need a transformer to drive a coil. You can design the coil to suit the AC voltage available. \$\endgroup\$
    – Andy aka
    Apr 4 at 18:01
  • \$\begingroup\$ My goal is to generate a stable magnetic field of approximately 1 mT. If possible, I would also like to make an oscillating magnetic field at 100kHz. This will be harder since the impedance of the coil at that frequency will be high but will a resonant capacitor, it might be possible. I still have more research to do about that part. \$\endgroup\$ Apr 4 at 22:08
  • \$\begingroup\$ I saw this on wikipedia and in many videos. Buck converter Wikipedia: A buck converter or step-down converter is a DC-to-DC converter which steps down voltage (while stepping up current) from its input (supply) to its output (load). But with user 253751's answer, it makes more sense now. \$\endgroup\$ Apr 4 at 22:11

3 Answers 3


The magnetic field is proportional to the current multiplied by the number of turns of wire. By using a different number of turns, you can make coils with any current/voltage ratio you like, for the same magnetic field strength, within reason. Of course, you should use smaller wire for the coils with more turns - otherwise they'll get too big - and larger wire for the coils with fewer turns - otherwise they'll have too much resistance.

If you can change the number of turns to get the voltage and current you want, you should do that. A transformer has two coils in it - why use three coils when one is enough? And if you don't use a transformer it won't block DC. And if you want a steady magnetic field you have to use DC.

You seem to be a bit confused about transformers, as well. Ohm's law applies to the load on the transformer - the electromagnet. As you increase the voltage, the current increases proportionally. More voltage, more current.

But you also know that a step-up transformer increases voltage and decreases current. How does that work? Well, a transformer has two different currents and two different voltages and you are confusing them with each other. It's true that the voltage on the secondary side is higher than the voltage on the primary side, and it's true that the current on the secondary side is lower than the current on the primary side. But that doesn't mean the current is lower than if the transformer wasn't there!

Indeed, if you add a 1:4 transformer in front of a resistive load, the voltage across the resistor increases 4 times, the current through the resistor increases 4 times, and the current from the power supply increases 16 times. The resistor current is lower than the power supply current - in accordance with the transformer law - but both of them are higher than they were before you added the transformer.

The equations you need are:

  • Transformer increases voltage: \$V_{sec} = V_{pri}\cdot N\$
  • Transformer decreases current: \$I_{sec} = \frac{I_{pri}}N\$
  • Ohm's law: \$V_{sec} = R_{load}\cdot I_{sec}\$
  • \$\begingroup\$ Thanks for your detailed answer. I was missing the part where the current in the primary would also increase. I did not consider the coil directly in the source which could also work! \$\endgroup\$ Apr 4 at 22:24

Note that transformers don't work on DC, which is what you have drawn, so I will assume AC.

If you have a 12 VAC power source that is able to provide an AC current \$I_{AC}\$ and a resistive coil with impedance \$Z\$, you use the transformer to match impedances for maximum power transfer. The impedance ratio is:

$$\frac{I_{AC} Z}{V}=\sqrt n$$

If you are free to choose the power supply and/or wind the coil, you can just tune the impedances to match without a transformer.


Here's a simple example using 60 W lamps.

  • At 120 V the lamp resistance (when hot) will be given by
    \$ R = \frac {V^2}P = \frac {120^2}{60} = 240 \ \Omega \$.
  • At 12 V the lamp resistance will be
    \$ R = \frac {V^2}P = \frac {12^2}{60} = 2.4 \ \Omega \$.

For a given power the resistance must change by the square of the ration of the change in voltage.


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