The present article explains through a practical example the process of applying the various formulas for making an inverter transformer. The various formulas required for designing a transformer has been already discussed in one my previous articles.
An inverter is your personal power house, able to transform any high current DC source into readily usable AC power, quite similar to the power received from your house outlets. Although inverters are extensively available in the market today, but designing your own customized inverter unit can make you overwhelmingly satisfied and moreover it's great fun.
At Bright Hub I have already published many inverter circuit diagram, ranging from simple to sophisticated sine wave and modified sine wave designs. However folks keep on asking me regarding formulas that can be easily used for designing a inverter transformer. The popular demand inspired me to publish one such article dealing comprehensively with transformer design calculations.
Although the explanation and the content was up to the mark, quite disappointingly many of you just failed to grasp the procedure. This prompted me to write this article which includes one example thoroughly illustrating how to use and apply the various steps and formulas while designing your own transformer. Let’s quickly study the following attached example:
Suppose you want to design an inverter transformer for a 120 VA inverter using a 12 Volt automobile battery as the input and need 230 Volts as the output. Now, simply dividing 120 by 12 gives 10 Amps, this becomes the required secondary current.
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The data in hand are:
Primary Voltage = 230 Volts,
Secondary Current (Output Current) = 10 Amps.
Secondary Voltage (Output Voltage) = 12-0-12 volts, that is equal to 24 volts.
Output Frequency = 50 Hz
First we need to find the core area CA = 1.152 ×√ 24 × 10 = 18 sq.cm
We select CRGO as the core material.
Calculating Turns per Volt TPV = 1 / (4.44 × 10^-4 ×18 × 1.3 × 50) = 1.96
Calculating Primary Current = 24 × 10 / 230 × 0.9 (assumed efficiency) = 1.15 Amps,
By matching the above current in Table A we get the approximate Primary copper wire thickness = 21 SWG.
Therefore the Number of Turns for the primary winding is calculated as = 1.96 × 230 = 450
Next, Primary Winding Area becomes = 450 / 137 (from Table A) = 3.27 sq.cm.
Now, the required secondary current is 10 Amps, therefore from Table A we match an equivalent thickness of copper wire = 12 SWG.
Calculating Secondary Number of Turns = 1.04 (1.96 × 24) = 49.
Calculating Secondary Winding Area = 49 / 12.8 (From Table A) = 3.8 Sq.cm.
Therefore, the Total Winding Area Comes to = (3.27 + 3.8) × 1.3 (insulation area added 30%) = 9 sq.cm.
Calculating Gross Area we get = 18 / 0.9 = 20 sq.cm.
Next, the Tongue Width becomes = √20 = 4.47 cm.
Consulting Table B yet again through the above value we finalize the core type to be 6 (E/I) approximately.
Finally the Stack is calculated as = 20 / 4.47 = 4.47 cm