An AC Current Generator, Part 3

In An AC Current Generator, Part 1, I talked about the overall principle in creating an AC Current Generator. In An AC Current Generator, Part 2 I moved on to the initial calculations for the primary and the secondary. Today we will complete the transformer design, but you will need to refer back for some of the details.

Step 4- Primary Winding

The length of wire required for the primary is given by the perimeter of the cross section multiplied by the number of turns. For our case the perimeter is (2×2.15)+(2×1.60)=7.5 cm. The length of wire is 190×7.5= 1425cm ≅ 15 meters.

In order to choose the correct wire for the primary we need to consider the inside circumference of the toroid as in Figure 3-1. The interior diameter of the core is 39.3 mm. The internal circumference is ΠD = 3.14×39.3 ≅ 123 mm. The maximum diameter of a wire that will fit into this circumference is 123/190 ≅ 0.6 mm with one layer. More than one layer is permitted, if necessary. Using wire tables magnetic wire of size AWG22 is used (0.71mm with insulation).

Figure 3-1

Primary winding, wire thickness calculation.

Primary winding, wire thickness calculation.

The resistance of the primary Rp is the product of the wire resistivity (0.53 mΩ /cm for AWG22 copper) and the wire length Rp = 0.00053 x 1500 ≅ 0.8 Ω. The reflected resistance to the secondary is Rp ’= Rp /n2 = 0.8/282 ≅ 1.0 mΩ

Step 5- Secondary Winding

Based on wire tables we need a wire gauge of at least AWG#12 for 10Amps. It has a resistivity of 0.052 mΩ /cm. Allowing for 100cm of wire (only about 52 cm is needed for the actual loops) the wire resistance Rs , is 5.2 mΩ. The total internal resistance Ri is the sum of Rp ’ and Rs (the shunt) i.e. 5.2+1 = 6.2 mΩ.

If Ri plus the sum of the addition resistance of the wire and contact resistance of the module under test exceed the 25 mΩ that we selected at the start, the 14 Vrms that we use to drive the primary would have been insufficient. Increasing the primary voltage would increase the flux density and the core would saturate, so if these calculations need to be redone if Ri is excessive.

Step 6- Primary Inductance

The primary inductance is the inductance per turn multiplied by the (number of turns)2 .

Lm = Ns 2 x LT = 1902 x 1.63 μH/turn2 = 58.8mH ≅ 60mH. The magnetic (reactive) current IL (see Figure 2-1) flows through this. The impedance is given by XL = 2x ΠxfxLm = 2×3.14x60x60mH ≅ 22 Ω

Step 7- Primary Reflected Impedance

The secondary load of nominally 50 mΩis reflected to the primary Rs ’= Rs x n2 = 50 mΩx 272 ≅36 Ω. The current IR (see Figure 2-1) flows through this resistive load.

Step 8- Amplifier Drive Current

If the capacitance Xc is purely parasitic, the current through the capacitance is negligible so the total current is the vector sum of IL and IR . This is

I= √ (IL 2 + IR 2 ) = √ (E2 /XL 2 + E2 /RL 2 ) = E x √ (1/XL 2 +1/RL 2 ) = = 14x √ (1/222 + 1/362 ) ≅ 750mA. The amplifier must be capable of delivering this. In addition the power supplies must be capable of sourcing and sinking this current as well as supplying the other needs of the circuit.

If a capacitor is added it should be equal to the inductive component

XC = XL = 1/(2 x Π x f x C) where f is the frequency and C is the capacitance. From above this is 22 ohm

C= 1 / (2 x Π x f x XC ) =1 / (2 x 3.14 x 60 x 22) ≅ 120 μ F.

This capacitor must be non-polar. It can be achieved by back to back electrolytic capacitors of double the value. Since the amplifier that we chose was not overtaxed by the uncompensated requirements, we chose not to use the capacitor.

Step 9- Maximum Possible Current

Based on Ohm’s Law the secondary current is be expressed as follows:

Is = Ep /(nxRs + Rp /n)

Where Ep is the maximum voltage across the primary, n the turns ratio Np /Ns , Rs is the secondary resistance and Rp the primary resistance. Through some painful calculus it is possible that to show that a maximum current (for given resistances) is achieved when nxRs = Rp /n. Solving this for n we get

n = √ (Rp /Rs )

Using our calculations to date, n= √ (0.8/0.05) = √ (16) =4. With an Np of 190 Ns becomes 47. From above

Is max = Ep /(4 x Rs + Rp /4)= 14 / (0.2 + 0.2) = 35 amps!

However with the maximum number of turns in the secondary, the secondary resistance will go up, but we can see that there is still quite a bit of “headroom” and it is possible to get a higher current simply by increasing the number of windings.

Phew. I hope I hope I got all of parts 2 and 3 right. I can’t promise to answer your questions- I may have to defer to Ernesto, who I mentioned in Part 1.

10 comments on “An AC Current Generator, Part 3

  1. uchiha
    December 30, 2014

    why electricity generators have to work harder for higher loads..?

  2. samicksha
    December 30, 2014

    I faced similar kind of question during my engineering classes, and one of the standard answer for this comes with an example, i.e.

    Let's say I have a 120VAC generator and either a 1A or a 10A load, and assume it can handle 1200W without issue. For 10A, the generator uses more gas and works harder compared to 1A.

    Is this because the 10A is passively causing a magnetic field that makes the generator shaft physically more difficult to spin, Or is it because the circuitry in the generator is actively sensing voltage drop .

  3. uchiha
    December 30, 2014

    @samicksha : Your example is very clear. Thanks a lot for this clarification & got an idea how it works.

    December 30, 2014

    @uchiha: Are you doing any research on this module ? 

  5. vasanjk
    January 2, 2015

    Hi AK

    Is there some limitation based on the start up load current?



  6. antedeluvian
    January 3, 2015

    Hi JK

    Is there some limitation based on the start up load current?

    The transformer in all our cases has been designed to work into really low loads, so even a dead short will work, with some possible saturation of the core. I have never seen an issue. at the other end of the scale we almost often start up in an open circuit with no adverse

    I have to think (and ask) the implication if you designed this for several tens of ohms of load (or more), but I don't think normal startup conditions will have a huge influence.

     I look at your question again and I think perhaps that is not the question I attempted to answer. Are you asking- given a particular starting current, is there a limiitation on what you can use to drive the primary side?

  7. vasanjk
    January 3, 2015


    Are you asking- given a particular starting current, is there a limiitation on what you can use to drive the primary side?

    “Are you asking- given a particular starting current, is there a limiitation on what you can use to drive the primary side?”

    Yes.I was asking the same.


  8. antedeluvian
    January 3, 2015


    The limiting factor of the whole arrangement is the magnetic characteristics of the core. I believe that if you work following the formulas that I have included you should not run into any limitation. However I have seen instances where we have cobbled together a quick circuit to test (as I describe in part 1- with just a variac and a transformer that is hanging around) where it does not work at all or as expected although there has never been any disastrous consequences like setting the lab on fire.

    During developement you may have issues (at least I did) driving the primary of the transformer with a power amplifier- I can't remember the issues that I believe caused my first attempt to blow up (it was an expensive lesson) but I did find a very resilient power op-amp, the LM12, for the job. Unfortunatley it has been discontinued so a new part will have to be found. I believe I discuss it in part 4, but it may have been cut.

    And of course you have to be aware of all kinds of parasitic resitances like the contact of connectors, self heating of wires and the secondary side which all contribute to the overall votl drop- if you transformer can't suppy the voltage the output current will drop.


  9. vasanjk
    January 3, 2015


    Got it.

    “parasitic resitances like the contact of connectors, self heating of wires”.

    I was told by a colleague who worked on a similar project that the self heating of wires is his worst concern. He further explained that the material of the wires has a lot to contribute and choosing the material/quality of material is an art rather than engineering some times. I have seen one of his transformers heat up so much that we used to sit next to it in the winter to keep warm. He solved the problem somehow.

  10. antedeluvian
    January 4, 2015


    I forgot to mention that prmary side of the transformer must be AC coupled (also to be mentioned in part 4).

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