Inverter using salvaged transformer

[buick51]

Available sources for ferrite transformers are CFL,CDROM supply(IDE power supply) or ATX PC power supply.
The CDROM adapter is cheap it has 220 ac input ,12v 2A 5v 2A output.
I think its ferrite transformer can handle the low power (less than 5 watt) 12 volt to 300v dc /dc inverter i want to make.
To deal with this transformer without the trial and error method i have to figure out its size and shape code for ex. ee16 ,ee35,etd,ei,...and its ferrite material type.
According to its size and shape it is an EI as in the attached figure but is it custum
and does not belong to the standard series ?
It is gapped so what is the most efficient inverter circuit that i should use?

 

[moffy]

The easiest way to determine the transformer is to take some measurements.
1. A sine wave input into one winding and measuring the outputs will give you the ratios.
2. Applying a square wave voltage to the winding you want to use and observing the current ramp, will give you the inductance. If the current rises quickly at the end then you have the saturation current. The current needs to be of the order that you intend to use to get realistic readings. Also because transformers can look a bit like a short circuit if they saturate limit the maximum current.

 

[buick51]

Thank you for your reply.
I know the number of turns for each winding (in its original circuit) can this be a guide in making the inverter circuit ? , i have dismantled it and if it is not a custom transformer it is much easier to know its power limit from datasheet . is this a standard size transformer?

 

[moffy]

The most direct way to find what you need is to do the measurements. Sorry the diagram doesn't provide any detail of the transformer apart from the number of windings.

 

[buick51]

I agree that measurement is needed but i have dismantled it ,unwind it, now i am to rewind it.

Now before rewinding it that is the data i had :

1-winding (7,8) has 32 turns,winding (1,3) has 4 turns ,winding (2,4) 4 turns and winding (5,6) 6 turns (known by dismantling ).

2-the supply 308v (220*1.4) applied to pin 8 of winding 7,8 as there is a mains bridge rectifier followed by 22UF 400v ,capacitor partially shown in the schematic .

That was the original circuit data.

Now i have made a simple driver circuit (555 followed by a power transistor 2sd401). The new circuit will have a supply of 12v and i am going to rewind the trasnsformer.

Am i to wind 32 turns for the secondary and 32*12/300= 1.3 turn for the primary?
or 4 turns primary and 4*300/12 =100 turn for secondary,the problem is that the original circuit has feedback as winding7,8 and 1,3 have a ratio of 32/4=8 and the amplitude of the signal is controlled by this feedback.

To make it short !!! if i am to do measurement i have to rewind the transformer ,better for the new application(12 to 300v),assuming almost 12 v squarewave input (neglecting VCEsat ).what number of turns should i use to do the measurement?

 

[moffy]

Just put say 4 turns on the transformer and you can do all the measurements. Once that is done you can design and then wind your final transformer. Lp = k*N*N where:
Lp = primary inductance
k = constant associated with that transformer core
N = number of turns.
t is time
I is current

Lp = V*t/I for a step input.(as long as it doesn't saturate)

The other important constant is the maximum V*t which is equivalent to L*I. This determines the saturation point of the core.
Vo/Vi = Vi*No/Ni.
That's the essential formulas for designing a transformer. Also allow about a 30% drop in saturation level for temperature effects on the core.

 

[buick51]

Thank you.

 

[moffy]

Sorry but I forgot the important formula that saturation is proportional to N*I (N= no of turns, I=current).e.g. if you have 5turns at 10A and 10turns at 5A they have the same saturation or magnetizing value even though the inductor with 10turns is 4 times larger than the one with 5.