# What's the minimum electrical requirements to ac-ac conversion

I bought a 220VAC -> 12VAC (0.3A 3.6VA) cheap power source, and tried to do a simple conversion of the output voltage, with two coils, around a ferrite core, with different length of turns.

simulate this circuit – Schematic created using CircuitLab

I also tried with a resistance load:

simulate this circuit

I got 0V AC. I also tried with some coils, that I've got from some old electronics hardware like:

but also got 0V AC.

In both cases, I could get some voltage in the voltmeter, when I disconnected, or connected the coils, sometimes bigger than 12V AC if I did connect/disconnect many times.

My guess is that the inductance of those inductors, was not enough to store all the electromagnetic energy for each voltage cycle. All of the inductors were in the micro henry range.

So, my question is: what's the inductance both inductors have to have to hold this together, and if I should add some resistances to the circuit, or any electronic component, or is any other thing wrong.

Also, does the frequency of the output of the charger makes any difference?

I've ordered some cable of insulated copper wire, (0.6mm), and will try again with more inductance.

Edit: I've put "50/60 Hz" for the 12VAC signal but, in the power source stick, it says that frequency value for the MAINS input signal. It doesn't mention the frequency of the output signal.

• Are you sure that the output of your "cheap power source" is really AC? Most "wall-wart" power supplies output DC, while transformers require AC to operate. Jan 23, 2015 at 1:23
• My guess is that your "transformer" primary does not have enough inductance (turns) and you're effectively shorting the supply. (This appears to be what Dave is saying as well.) Jan 23, 2015 at 1:25
• (For what you're attempting it might be wise to start with an audio output transformer.) Jan 23, 2015 at 1:27
• It's labeled AC, priceminister.com/offer/buy/209043199/…, and my multimeter got 12.5VAC from it. About the shorting of the supply, I think it makes sense because it got too hot and it died. I'm buying another one. Jan 23, 2015 at 1:34

An inductor with inductance $L$, operating at frequency $f$, has an impedance $Z = 2 \pi j f L$. For $\mu$H inductors at 50 Hz, the impedance would be in the m$\Omega$ range. But your power supply can't source enough current to maintain 12 V over such an inductor. Try measuring the voltage on the power supply side when the circuit is running - I bet it is close to 0 V.
To get the voltage step-down/up you are expecting, you therefore need the inductance of the primary to be much larger than $V_\mathrm{PSU}/(2 \pi f I_\mathrm{PSU})$, in your case approximately 100 mH.