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DEVELOPING AN AC CURRENT GENERATOR


Circuit Cellar Online
THE MAGAZINE FOR COMPUTER APPLICATIONS
Circuit Cellar Online offers articles illustrating creative solutions
and unique applications through complete projects, practical
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DEVELOPING AN AC CURRENT GENERATOR

Lessons from the Trenches by Ernesto Gradin & Aubrey Kagan

Start ý Features ý Magnetics ý Primary and Secondary Turns ý Primary and Secondary Winding ý Primary Inductance ý Hardware ý Firmware ý Analog to Digital Conversion ý User Interface ý Sources and PDF

MAGNETICS

Figure 2 shows the equivalent circuit of the current generating transformer. 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. A capacitor can be added to cancel the current required by the inductive component, reducing the current drive requirements of the amplifier.

Figure 2ýNo real electrical component is ideal. The imperfections, as well as impedances reflected from one side of the transformer to the other, are shown in an equivalent circuit. Also indicated is the way the total current splits to pass through individual components.

 

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 load in the unit under test.

ESTABLISH CONSTANTS

To design the transformer, you must introduce values from the actual circuitry that you are going to use. For example, we decided to generate a 10-A output AC current. The load (RL) is 25 milliohms (a shunt internal to the product), and the wiring resistance of the secondary, so Ri is assumed to be 25 milliohms as well.

In addition, a transformer core must be selected. We had several spare toroidal transformers that we disassembled. The measurements of the transformer core were as follows:

ý Cross-sectional height, H = 2.15 cm

ý Cross-sectional width, W = 1.60 cm

ý 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, as shown as follows:

where S is the core cross-sectional area in square centimeters, Cm is the mean circumference of the toroid (2 ý p ý (outside diameterýinside diameter)), and N is the total number of turns. m eff 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 calculated as:

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, as shown above. Alternatively, it can be measured with an LCR bridge. Winding five turns through the core and measuring the inductance gave 40.7 ýH. The inductance is proportional to the square of the number of turns. Therefore, LT is calculated as 40.7/52 = 1.63 ýH/turn2.

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