# An AC Current Generator, Part 2

In An AC Current Generator, Part 1, I talked about the overall principle in creating an AC current generator. Today I am going to start to describe how to create a transformer, specifically to generate 10 amps AC. It is rather detailed and lengthy, so I am going to have to split it into multiple parts. The process is similar for any desired output current.

When testing, a 10 amp AC current is fed into the module being tested (DUT) and the current is measured across a 25 mΩ shunt internal to the DUT. The arrangement can be represented by the circuit in Figure 2-1.

Figure 2-1

This is the equivalent circuit of the current generating transformer and its connections.

The output current from the power amplifier has three components of three different phases. IL is the current through the inductive component; IR is the current through the reflected resistive component (from the secondary); and IC is the current through the capacitance of the circuit. An additional capacitor can be added to the parasitic already present to cancel the current required by the inductive component, reducing the current drive requirements of the amplifier.

The internal resistance of the secondary Ri is made up of wire resistance, plus the transferred impedance of the primary, plus the resistance of the measurement shunt and the transferred impedance from the secondary of the current measuring transformer. RL is the shunt in our DUT.

Step 1: Establish constants
In order to design the transformer, we must introduce values from the actual circuitry that we are going to use. For this example we decided to generate an output current of 10 amps AC. The load RL is 25 mΩ (the shunt internal to the product), and the wiring resistance of the secondary, Ri is assumed to be 25 mΩ as well.

The Bicron B5303 current transformer (BICRON_Current_Sensors) has a current turns ratio of 5,000 and a wire resistance of 207 ohms. The output is connected in series with a 10 ohm resistor in its secondary (internal to the measurement device). The reflected impedance is 217/50002 = 9 μ Ω so this should not pose a problem and is included in the nominal 25 mΩ.

A transformer core must be selected. We had several spare toroidal transformers, which we disassembled. The measurements of the transformer core selected were as follows:

Cross–sectional height H = 2.15 cm
Cross-section width W = 1.60 cm
Inner diameter D = 3.93 cm

Prior to the other calculations the inductance per turn LT must also be known. This can be calculated from the magnetic constants if the core data is available using the following formula:

Inductance of the winding, L (in Henry) = (1.256*S*N2 * μ e *10-8 )/Cm ………(1)

Where S is the core cross section in square cm, Cm is the mean circumference of the toroid (2* Π*(outside diameter-inside diameter)/2), and N is the total number of turns. μ e is the effective permeability of the material. In a toroid this is essentially the permeability of the material. Once L has been calculated from the above formula, LT can be simply calculated as follows:

LT = L/ N2 …….(2)

However, if you are like us, you don’t have those details, so alternatively it can be measured with an LCR bridge. Winding 5 turns through the core and measuring the inductance gave 40.7 μ H. Using (2) above LT is calculated as 40.7/52 = 1.63 μ H/turn2 .

Step 2: Primary turns
Just to be sure that our numbers will work I remind you that the maximum allowable flux density before saturation of an iron core is 12,000 to 15,000 gauss. 8,000 to 10,000 gauss is a safe value. Iron cores can be used up to a frequency of 400 Hz. For higher frequencies a ferrite core must be used where the flux density must be reduced to no more than 2,500 to 3,000 gauss to prevent saturation of the core. So there!

Magnetic theory predicts the following relationship for the primary of the transformer:

E=4.44*f*NP *S*B*10-8 …….(3)

Where E is the RMS voltage, f is the frequency, NP is the number of turns on the primary, S is the cross sectional area of the core, and B is the flux density in the core (in gauss).

We can re-arrange this to find the number of turns:

NP = (E*108 )/(4.44*f*S*B) …….(4)

In our setup E=14Vrms. This is derived from the supplies being +/-24V. Allowing for 4V “headroom” for the power amplifier, this gives Vpp of 40V.

Vrms = Vpp /(2*1.414)=E …….(5)

The frequency, f, is nominally 60Hz.

The cross sectional area, S is equal to H*W=2.15*1.60=3.44 cm2

The constant B is our flux density of 8,000 gauss.

Substituting these values

NP = (14*108 )/(4.44*60*3.44*8000) ≅190 turns …….(6)

This is the number of turns required to maintain the core below saturation at the maximum input voltage.

This number of coils can be quite difficult to wind and you can see how to do it in Hint No 3 in my blog, “Top 17 Helpful Hints for Constructing Electronic Systems.”

Step 3: Secondary turns
The next step is to calculate the number of turns on the secondary, Ns . As can be seen from Figure 2-1 and based on Ohm’s Law:

Vout = Iout*(RL+Ri) …….(7)

Since we have chosen the output to be 10Amps and RL=25 mΩ and Ri=25 mΩ, Vout=10*0.05=0.5Vac.

The turns ratio is the proportion of the input voltage to the output voltage n= E/Vout=14/0.5 = 28.

Therefore the number of turns on the secondary Ns :

Ns = Np /n = 190/28 = 7.

We have calculated the primary and secondary turns, but we still have to figure out what gauge of wire to use on the primary and secondary, to say nothing of reflected impedances. I will leave you on tenterhooks till Part 3.

## 3 comments on “An AC Current Generator, Part 2”

1. xiaonanhai
December 12, 2014

If you are interesting in laser drivers,please visit our website to see our products. This series laser driver is designed for driving diode lasers with low noise current at high efficiency, > 76%. The output current of laser drivers can be set either by an analog voltage ranges from 0 to 2.5V or by a potentiometer to from 0 to 20A, 30A, 40,45A.        analogtechnologies.com/High_Efficiency_AC_Input_Laser_Drivers

2. Davidled
December 12, 2014

Coli winding design could be a crucial in terms of power efficiency. Coil winding machine could be used for mass production in manufacture and decide the transformer durability depending on primary winding and secondary winding. A few engineers might check actual transformer design.

3. Victor Lorenzo
December 14, 2014

Thanks for the series, @Aubrey, and for pointing us to your top 17 helpful hints post.

Toroid transformers perform great as magnetic flux can stay more confined than in E section transformers, that makes them attractive in many applications. But winding a high number of turns into a toroid transformer can be time consuming and laborious.

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