**Introduction**

Inverter system is one of today alternative to power supply. During the recent years, a lot of research as been carried out to improvement the efficiency performance of inverter system, hence different types of inverter as been designed and improved on such include square wave inverter, modified square wave, sine wave, modified sine wave e.t.c using different techniques commonly used push pull.

For all inverter been design a common and important component, device must been designed which must suit the rating of the inverter to be design, which is a TRANSFORMER.

In this post , practical step taking in designing and implementing a 500VA Transformer will be discuss

**Require components and tools for transformer design**

- Iron core stamping E/I or U/T)
- Bobbin made from either plastic,ceramic, ceiling board.
- Two winding (electrically isolated and magnetically coupled) using super enameled copper wire made
- Masking tape
- Bot and nut
- Hammer

**Determining the transformer rating for 12V, 500VA Inverter System**

After reading through different source on inverter design, most designer uses 12V/220V transformer for a 12V inverter system. According to Lab researching and experiment carried out, a 12V/220V is not appropriate for such 12V inverter system especially when the same transformer will be used for charging purpose . Below is some theory calculations

From Theory

“Using Vpp = Vrms * root of 2,

hence Vrms = Vpp/root of 2″.

Therefore: Voltage output from FET= Voltage supply by battery / 1.4142

At battery voltage of 12V,

Vrms= 12/1.4142 = 8.48V

At battery voltage of 10.5V

Vrms = 10.5/1.4142= 7.42V

from calculation above an inverter system of 12 voltage will have an output of 8.48V RMS from the H-Bridge,

so using a 12v / 220v or rather 12v /230v.

using transformation ratio

Transformer AC output = (220 x 8.48)/12

= 155V

using a transformer of 6V/220v

Transformer AC output = (220 x 8.48)/6

= 310.9V

to adjust my output to 220, the duty cycle of the FET will be reduce.

**Calculating the require duty cycle for the FET**

At 12V battery voltage

require RMS voltage at transformer primary terminal is 6V RMS

Vpp= 6 x root of 2 = 8.46

duty cycle at 12V battery voltage = (8.46/12) x 100 = 70%

Hence FET is pulsed at 70% to have 6V RMS at the transformer terminal to achieve 220 AC voltage output

At 10.5V minimum battery voltage usage

duty cycle at 12V battery voltage = (8.46/10.5) x 100 = 80%

Hence FET is pulsed at 80% to have 6V RMS at the transformer terminal to achieve 220 AC voltage output

from the calculation it is shown that , at any point of the battery usage the battery voltage is not drawn to the maximum, this help for better utilization of the battery and a constant AC output is maintained

For better battery usage in inverter mode to have a constant output AC voltage of 220/230v and to have an efficient charging capability the primary side of the transformer voltage value range from 6V to 7.42V. With this, as the battery voltage level drop battery,the output of the FET is been compensate so to have a constant AC output power.

i.e voltage drop to 10.5V , with a 6V/220V give a transformer output of 272 AC Voltage, to achieve the 220 V the FET is rate pulse is reduce ans shown in the calculation above.

Hence, appropriate transformer primary voltage should lie between 6V to 7.42V..

For our design , since the same Transformer will perform both inverter mode and charging mode 6V/220V Transformer will be design.

using the same FET pulse changing the battery can be charge in different mode of operations at at different value of supplied AC voltage as low as 100 AC such as:

- Bulk charge mode
- Absorption mode
- Float mode

More details is explained in 500VA Inverter design.

**For designing a transformer**, we need:

- Power rating
- Voltage levels (primary and secondary)
- Currents on both sides
- Primary and secondary coils wire diameter/size
- Iron Core area
- Numbers of turns (primary and secondary)

We are going to design a 500 VA step up transformer of 6V to 220V. Necessary calculations along with formulae are given below in details:

**Step 1: power rating**

power= 500VA,

Primary current

Voltage=6V

Ip = 500/6 = 83.33A

where Ip is the transformer primary current

Assuming at 90% efficient transformer

Ip = 500/6×0.9= 92.59A

Secondary current

Vs = 220

Is = P/Vs

Is = 500/220 = 2/27

Is = 2.3A

Bobin size (Area) : recall that

A = 1.152 sqrt (Power(VA))

= 1.152 sqrt (500)

=25.759

=25.76cm^2

Turns per Volt from this equation

Ai= 1/4.44 x F x Bm x T

T= 1/4.44x F x Bm x Ai

Tpv = 1/4.44×10^-4 x25.76 x1 x 50

= 1.748643053 (where Bm = 1.0)

Number of turns in primary

Np =Tpv x Vp

= 1.75 x 6 = 10.5

Number of turns in secondary

Ns =1.75 x 220 = 385 turns

Primary wire size

Aprm = Ip/ d = 92.59/2.3

= 40.26mm^2

from

A = πr^2

r^2 = A/π r = sqrt(Area/π)

d = 2 x sqrt (Area/ π)

= 2x sqrt(40.26/ π)

= 7.16mm

primary diameter copper size = 7.16mm

use the SWG Table from here

primary cable gauge is SWG is 1

Secondary wire size

Asec = Is/d = 2.3A/2.3A/mm^2

d = 2 x sqrt 1mm^2 /π = 1.1283

=1.13mm

Copper size = 17SWG

Weight of copper (primary)

Bobbin perimeter = 0?

Area = 25.76

Perimeter = 5 x 2 + (5.152 x 2)

= 20.304

Length of one turn is given by = 20.301

= 0.21m

Total length = 0.21 x Np

Lprm = 0.21 x 11 = 2.31m

Volume V1 = Lprm x Aprm

= 2.31m x 40.26 x 10^-6 m^2

= 9.3 x 10^-5 m^3

Weight = 8960 x 9.3 x 10^-5

8.333Kg

Weight of copper secondary

Lsec = o.21 x Ns = 0.21 x 385

=80.85m

Volume (V) = Lsec x Asec

=80.85m x 1 x 10 ^-6 m^2

=8.085 x 10^-5

Weight = 8960 x 8.085 x 10^-5

= 0.7244kg

=724.4gram

**Design implementation of the Transformer**

Making the Bobbin

Using a plywood the shape of the bobbin is measure and cut out and attached together using gum.

Winding the transformer

Placing the iron core

Tightening the transformer

Testing the Transformer

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