Help needed! Building/winding 3000VA transformer!

transformer secondary sheet winding

I am trying to rewind an old transformer (EI type) taken from a 3000W voltage stabilizer.. i will use it in dc-ac inverter 12VDC to 220-240VAC (50Hz).. Please tell me all the specifications i.e. no. of turns in primary and secondary, the dimensions of the core (if relevant!! current dimesions are given below), diameter of the winding coil etc.. it will be a center tapped transformer with 2 primary windings for switching purpose.. and 1 secondary winding.. however.. it might not be used at full 100% load i.e. 3000W loads.. loads will be around 1500W.. and i'll be using 12V 200AH battery.. or 2 batteries of same specifications i.e. 12V 200AH each!

Any ideas for making an efficient transformer which could work properly and not heat up?? Should i use 3000VA transformer? or would it be more efficient to use 1500VA transformer?? however, need might arise to use it for loads of upto 3000W!

Dimensions
length of 'I' = 10.5 cm
width of 'I' = 1.7 cm
Thickness of the pile/core = 5.5 cm
length of 'E' = 10.5 cm
width of 'E' = 7.7 cm
width of the middle section of E on which coils are wound = 3.5 cm

3000va transformer

Hello.
It seems there is something wrong with the sizes you gave.
These sizes are from a transformer that can handle roughly 400W as maximum power....
Are you sure it is coming from a 3000W equipment????

S.

winding 230 volt pulse transformer

well actually yes.. i took it out of an old 3000W voltage stabilizer.. or at least thats whats written on its case.... only 400W?? no wonder why its all blown up lolz.. when i took it out all its windings are burnt black.. please give me details as to how long, and wide should the transformer be if i have to use it for 3000W?? and is it ok to use 1500W load on a 3000W transformer? or will it lead to power losses??

waiting for a reply..

3000w transformer winding for inverter

have you thought of the circuit of 3000W inverter i hope you know current will be upto 250A

Added after 14 minutes:

Transformer:

We can make the transformer by changing the windings of an old transformer.



fig. 4: shell-type transformer core transformer

A transformer will provide best characteristics when the primary coil, that takes over magnetization of the iron core, fits closely around the core. For industrial transformers this would be the 230 Volts coil, on our inverters however it will be the 12 volt coil.



fig. 5: "EI"-sheet metals of a 850 VA-shell-type transformer

For the computation of the numbers of turns the following consideration applies:

The peak value of the primarily generated alternating voltage is given by the battery voltage. This determines the number of the primary windings of the transformer. On the secondary side of the transformer likewise the peak value must be taken also for computation, i.e. 325 Volt. In the case of a fully loaded battery the supply voltage of the inverter amounts to 13.8 Volts. The peak value of the 230 Volts output voltage may not exceed that of the usual supply networks, even if the rms could be held on 230 Volts by reduction of the duty cycle. The following table for the output voltage results (without pulse width regulation):


Battery voltage
Upeak (secondary)
Urms (secondary)

11,8 Volt
297 Volt
210 Volt

12,35 Volt
311 Volt
220 Volt

12,7 Volt (accord. 9 Vrms
320 Volt
227 Volt

12,9 Volt
325 Volt
230 Volt

13,5 Volt
340 Volt
240 Volt



The table applies to fixed duty cycle of 25% and/or sine-wave voltage and without consideration of the magnetization energy. Our inverter will keep the output voltage constant on an rms of 230 Volts, due to its pulse width regulation, even if the peak value will drop or rise, due to the battery voltage. The peak value will not exceed 350 Volts (247 Vrms for sine-wave voltage), critical for electronics, even in case of fully loaded battery. Theoretical, without pulse width regulation, the rms could rise again up to the theoretical factor of 2, according to a duty cycle of 50%, because of the magnetization energy. The recirculating magnetization energy already forms the beginning of the next half wave of the output voltage (see fig. 6). But without load there is no rms by definition, so this consideration is only of theoretical nature, with one exception: A measuring instrument, calibrated on rms would indicate a wrong output voltage and small consumers, who need less than the magnetizing energy of the transformer, could get damaged.



fig. 6: output voltage with no load or inductive load

The table shows, that the transformer needs a ratio of windings of 1 : 25. The schematic diagram shows, that it has two primary windings and one secondary. Both primary windings have the same number of turns and the secondary winding must have by factor 25 more turns (110 Volts: factor 13).

Here a selection of used transformers:

Length
Width
Deep
Power
Primarily
Secondary
idle current

150 mm
125 mm
50 mm
460 VA
2 x 13 W
325 W
1,4 Ampere

150 mm
125 mm
50 mm
460 VA
2 x 14 W
350 W
1,2 Ampere

150 mm
125 mm
67 mm
600 VA
2 x 10 W
250 W
2,2 Ampere

150 mm
125 mm
67 mm
600 VA
2 x 11 W
275 W
1,6 Ampere

150 mm
125 mm
67 mm
600 VA
2 x 12 W
300 W
1,4 Ampere

150 mm
125 mm
95 mm
1000 VA
2 x 9 W
225 W
1,4 Ampere


-
-
2000 VA
2 x 11 W
275 W
2,2 Ampere

170 mm
140 mm
80 mm
850 VA
2 x 12 W
300 W
1,5 Ampere

170 mm
140 mm
75 mm
850 VA
2 x 13 W
325 W
1,3 Ampere

175 mm
140 mm
60 mm
750 VA
2 x 13 W
325 W
1,2 Ampere


Length: length of the "I" from fig. 5

Deep: thickness of pile of all iron sheet metals

Power: rated output

The table shows, that the number of turns is not particularly critical. Only the ratio of primary windings to secondary must be correct. The rms of the output voltage will be finally adjusted by the automatic controller with R16 to the value of 220 or 230 Volts. It is of great importance however, that both primary coils are absolutely symmetrical. They must be wound bifilar, so that they are very close to each other. While one winding will magnetize the core, the corresponding winding will return the magnetizing energy. If there is no close coupling of both primary coils, energy losses will result by overvoltage, causing avalanche effects on the transistors. Despite completely symmetrical structure of the windings, the transformer will show a small magnetical bias (DC biasing), recognizable from the asymmetrical magnetizing currents, which can be watched on R20 with an oscilloscope. This biasing will change on every change of load, in particular with strong inductive loads. This effect is completely normal for square wave voltages at inductances and is connected with the heavy non-linearity of ferro-magnetical materials. The second half wave of the output voltage applies other magnetizing conditions to the ferro-magnetic transformer core due to the remaining remanence. (only with sine-wave voltages an equilibrium can adjust itself after several oscillations, due to hysteresis losses, see "Rush effect"). Critical unbalances, which develop e.g. after an impact short-circuit, are eliminated surely by the electronic shutdown system.



Wire strength:

Current densities from 3.5 A/mm2 to 4 A/mm2 are used on industrial transformers. If our inverter is not beeing used excessivly, current densities may even be higher. A transformer with 1000 VA needs approx. 84 ampere from the 12 Volt battery on nominal load. Since the two primary coils alternate mutually, we may count from 42 amperes. (This is strictly not correct, since the acceptance applies only if both windings would exhibit double surface for heat emission). For a round wire this would mean a diameter of 4 mm. Such wire is hardly to wind, also automats can't do it perfectly. A solution may be wires with rectangular cross section or several smaller wires in parallel.

After winding the transformer, the sheet metals must be inserted again. With each layer we change the direction of the sheet metals, while in the original condition several sheet metals were probably summarized into packages, in order to increase the air gap and linearize magnetizing currents. This effect isn't needed for our inverter. Magnetizing currents are always extremely nonlinear in square wave transformers, and they are asymmetrical also. This has no effect on the performance of the inverter and the output voltage.

After the transformer has been built, it should first be tested. Therefore we attach its 230 Volts windings to public electricity mains or any other 230 Volts source. Each low-voltage coil should now show 9 Volts. Now we can connect the beginning of one "primary" coil with the end of the other. At the free ends a voltage of 18 V now should appear. If this voltage would be 0 V, the windings have been connected the wrong way.



The making of a transformer is a very laborious work. Nobody likes to take a transformer apart for a second time to correct the windings. With unknown transformers it is advisable to apply first a sample coil of thin and easy to handle wire and test the power input on idle. The windings of the sample coil can be changed without dividing the transformer. For this test the transformer does not need the secondary 230 Volts coil. Only the electronics must be adjusted correctly (tested with another, correct transformer or an oscilloscope: duty cycle 25%).





Transforer computation:

For the first regard, it appears difficult to seize the obscure and for precisive computation not accessible magnetization procedures in the magnetic core of a transformer. I want to show, that in our case this is not necessary. As the table with the numbers of turns shows, a tranformer may be built on different numbers of turns, only the relation to each other must be exact.

We specify the maximum magnetic induction on a value of 1.5 Tesla. For computation now only two simple equations are necessary:


Uind = n x F /t converted: 1') n=Uind x t/F


F = B x A


Uind=induced voltage
n=number of turns

F = magnetic flux
t=transistor switch-on time

B= magnetic induction
A=cross-section area of transformer core



For power electronics resistive load shall not calculate on energy conversion. Thus the whole battery voltage will apply on the transformer coil for the whole switch-on time of the transistor. The switch-on time results in 5 milliseconds, dependend on the period of the 50 cycles / second oscillation and a duty-cycle of 25% (period of a 50 cycle oscillation is 1 / 50 Hz = 20 milliseconds).



Sample calculation for the above described 850 VA transformer:

The cross-section area of the transformer calculates to A = 60 mm x 80 mm = 4,8 x 10-3 m2



Uind = 12.7 Volt
B= 1.5 Tesla = 1.5 Vs/ m2

t= 5 ms
A= 4.8 x 10-3 m2




With equation 2) the magnetic flux calculates to F = B x A = 1.5 Vs/ m2 x 4.8 x 10-3 m2 = 7.2 x 10-3 Vs

set in equation 1') results

Number of turns n=Uind x t/F = 12.7 V x 5 x 10-3 s / (7.2 x 10-3 Vs) = 8.82 (rounded up 9 turns).

By trying I built the transformer with 2 x 12 turns. The losses were clearly smaller thereby. The calculated flux in this case was only 1.1 Tesla

effect of diameter of winding in transformer

with a ordinary transformer of 400w core you can go upto 3kva to stepup from 200v to 240v. but for normal use you can use only for 400va