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).
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.