transformer rewinding

[bonjing]

hello every one i am confused of so many formulas about transformer rewinding... can you help with simplified formula of calculating... example 0-220v primary and 50v-0-50v sec at 10ampere ct... thanx.... JINGz

 

[MrAl]

hello every one i am confused of so many formulas about transformer rewinding... can you help with simplified formula of calculating... example 0-220v primary and 50v-0-50v sec at 10ampere ct... thanx.... JINGz

Hi there,

When working with transformer windings you have to exercise some concern
about at least a minimum number of issues or you wont get something
that will work the way you expect it to.

As a minimum, you need to think about these issues:
1. Turns ratio
2. Maximum flux density
3. Primary wire diameter (wire gauge)
4. Secondary wire diameter
5. Primary winding resistance
6. Secondary winding resistance
7. Control of winding to winding isolation
8. Control of layer to layer winding isolation
9. Total power dissipation/heating at full load

Considering all of the above leads to a decent design
which could last a long time.

Skip any *one* of these issues and you could end up with something
that either doesnt work the way you want it too, doesnt work at all,
burns up after a short time of use, or works for a while and then
sometime later shorts out and blows the fuse or burns up.


If you are still interested let me know.

 

[Hero999]

Why not just buy one?

50V-0-50V is a pretty common output voltage, it's not like you require anything specialised.

 

[Nigel Goodwin]

Why not just buy one?

50V-0-50V is a pretty common output voltage, it's not like you require anything specialised.

Presumably it's a homework question?.

 

[Roff]

Hi there,

When working with transformer windings you have to exercise some concern
about at least a minimum number of issues or you wont get something
that will work the way you expect it to.

As a minimum, you need to think about these issues:
1. Turns ratio
2. Maximum flux density
3. Primary wire diameter (wire gauge)
4. Secondary wire diameter
5. Primary winding resistance
6. Secondary winding resistance
7. Control of winding to winding isolation
8. Control of layer to layer winding isolation
9. Total power dissipation/heating at full load

Considering all of the above leads to a decent design
which could last a long time.

Skip any *one* of these issues and you could end up with something
that either doesnt work the way you want it too, doesnt work at all,
burns up after a short time of use, or works for a while and then
sometime later shorts out and blows the fuse or burns up.


If you are still interested let me know.

Shouldn't you include primary inductance? Of course, you did say "As a minimum...".

 

[bonjing]

thanx guys.... but i don't to buy if there is already available 50-0-50 output voltage, i want to learn bout rewinding..... can you giv me an example on how to calculate the turns per volt....

 

[bonjing]

yah am preety much interested.... i hav read books and a lot of references but it does give me or satisfied me.... they should show the systematic formula.... 1.how to find the wattage of that transformer 2.how to calculate the turns per volt 3. how to finf the correct gage of wires used...

 

[MrAl]

Shouldn't you include primary inductance? Of course, you did say "As a minimum...".

Hello,


Roff:
Yes, that's right, i did say 'as a minimum' as that was a feeler to see
if the OP was still interested after seeing what may be involved because
they had mentioned that they wanted something more simple or
straightforward.

Bonjing:
So you think you would be interested in a set of formulas which
relate the core to the windings and you might be willing to do
a few tests before you start?
These formulas/techniques would be based on the core and the
window area of that core, and possibly what kind of bobbin
is on it already.

If so, here are a few questions for you:
What kind of test equipment do you have or can get?
Are the windings still on the old transformer?
Can you purchase any size enamel wire?

 

[Hero999]

An off the shelf transformer with twin 50V secondaries that can be wired in series to give 50V-50V
**broken link removed**

 

[bonjing]

to MraI... the winding are not come from the old windings... ill provide with a new one.... maybe i could purchased if availble....

 

[bonjing]

i want to try this formula by rebotexx but im wondering where does 5.775 came from..
for me this is a simplified example...

Example: Core Size = 1 ¼ X ¾

Formula 1:
Convert first into decimal:
thickness X center tapped = Core Area

1.25 X .75 = 0.9375 (Core Area)

Formula 2:
Compute power watts:
Power=Core Area (5.58 ) 2

0.9375 (5.58 ) =(5.23125) 2
Power =27.365977 Watts

Formula 3:
Compute number of turns:
Turn / Volts = 5.775 / Core Area

5.775 / 0.9375 = 6.16 Turn over volts

Primary winding:
6.16 X 220 Volts = 1,355 Turns

Secondary winding:
6.16 X 6 Volts = 37 + 37 turns for full wave design
6.16 X 6 volts = 37 turns for haft ware design

Formula 4:
Compute for Primary & Secondary Current:
Current = Power / Volts (I= P/V)

For primary current:
I = 27.365977 / 220 Volts = 0.124308 Amp.

For secondary current:
I = 27.365977 / 6 Volts = 4.5609962 Amp. For haft wave design
I = 27.365977 / 6 Volts = 4.5609962 / 2 = 2.2804981 Amp. For full wave design

Formula 5:
Compute for Circular Miles AWG:

Current X 680 or 700 recommended 680 for normal heat

For primary winding:
0.124308 (680) = 84.59 Cir. Mil.
AWG # = 31
(Please refer the nearest value to AWG standard table)

For secondary winding:
2.2804981 (680) = 1551 Cir. Mil.
AWG # = 18
(Please refer the nearest value to AWG standard table)

You have now designed:

220 volts primary
6 volts full wave secondary
2 Amps.
27 Watts

 

[bonjing]

i want to try this formula by rebotexx but im wondering where does 5.775 came from..
for me this is a simplified example...

Example: Core Size = 1 ¼ X ¾

Formula 1:
Convert first into decimal:
thickness X center tapped = Core Area

1.25 X .75 = 0.9375 (Core Area)

Formula 2:
Compute power watts:
Power=Core Area (5.58 ) 2

0.9375 (5.58 ) =(5.23125) 2
Power =27.365977 Watts

Formula 3:
Compute number of turns:
Turn / Volts = 5.775 / Core Area

5.775 / 0.9375 = 6.16 Turn over volts

Primary winding:
6.16 X 220 Volts = 1,355 Turns

Secondary winding:
6.16 X 6 Volts = 37 + 37 turns for full wave design
6.16 X 6 volts = 37 turns for haft ware design

Formula 4:
Compute for Primary & Secondary Current:
Current = Power / Volts (I= P/V)

For primary current:
I = 27.365977 / 220 Volts = 0.124308 Amp.

For secondary current:
I = 27.365977 / 6 Volts = 4.5609962 Amp. For haft wave design
I = 27.365977 / 6 Volts = 4.5609962 / 2 = 2.2804981 Amp. For full wave design

Formula 5:
Compute for Circular Miles AWG:

Current X 680 or 700 recommended 680 for normal heat

For primary winding:
0.124308 (680) = 84.59 Cir. Mil.
AWG # = 31
(Please refer the nearest value to AWG standard table)

For secondary winding:
2.2804981 (680) = 1551 Cir. Mil.
AWG # = 18
(Please refer the nearest value to AWG standard table)

You have now designed:

220 volts primary
6 volts full wave secondary
2 Amps.
27 Watts

 

[ericgibbs]

i want to try this formula by rebotexx but im wondering where does 5.775 came from..
for me this is a simplified example...

hi,
A rule of thumb for the number of turns/volt for a transformer is the cross sectional area [CSA] of the central core, divided into 6.

eg: if the CSA is 1 sq inch its 6t/volt,,, if the CSA is 2 sq inch its 3t/V.

When winding transformer low voltage secondaries this will give a close approximation.

 

[Mikebits]

The formulas is the easy part. The actual fun is when you really try to wind the thing. Do you have tooling for such a project?

 

[ericgibbs]

The formulas is the easy part. The actual fun is when you really try to wind the thing. Do you have tooling for such a project?

hi Mike,
Clamp a small mechanical hand drill in the bench vice and insert a threaded bolt thru the transformer bobbin [ with a couple of flat plates as washers].
Tighten the drill chuck on the threaded bolt and use the drill handle to turn the bobbin in order to wind on the turns.

 

[Mikebits]

I suppose that would work. The last winder I saw looked like a lathe and had auto tension sense along with x-axis feed, and a turns counter all automated, but where there is a will there is a way.

 

[MrAl]

i want to try this formula by rebotexx but im wondering where does 5.775 came from..
for me this is a simplified example...

Hello there,


That constant comes from deriving a turns/volt constant from the well
known transformer equation for max volts:

V=4.44*B*N*A*f*10^-8

where

V is the max rms voltage in volts
B is the max flux density in gauss
A is the area of the core in square centimeters
f is the frequency in Hertz
4.44 and 10^-8 are both constants.

Typical transformer material has a max flux density of 20kG, which
can change a bit with temperature, and taking into account high
line a safe design goal is to stay away from this by 25 percent, which
means 15kG is the target max flux density for a decent design, but
sometimes even a kG lower than that is used.
This is for typical transformer material that is often used for 50 and
60Hz transformer work, but the actual max kG could be lower depending
on material.

Anyway, taking that stuff into account we can come up with an
approximation by lumping several variables into one constant:

First converting square cm to square inches:

V=4.44*B*N*A*6.4516*f*10^-8

and after lumping we get:

TAPV=1/(4.44*B*6.4516*f*10^-8)

which is in units of turns*square_inches per volt.

Substituting B=15kG and f=50Hz, we get:

TAPV=4.655

which is a little less than the 5.775 quoted in your formula, so
that approximation was probably derived using a wider safety
margin with Bmax=12.2kG instead of 15kG:

TAPV=1/(4.44*B*6.4516*f*10^-8)

with B=12.2kG and f=50Hz we get:

TAPV=5.72

which is close to the quoted value.

It's not too hard to apply the original formula directly though,
so perhaps it's better to do that instead:

V=4.44*B*N*A*6.4516*f*10^-8 (with A in square inches again)

Solving for N, we get:

N=V*10^8/(4.44*B*A*6.4516*f)

which after inserting
V=220
B=12200
A=1.25*0.75
f=50

we get:

N=1343 turns.


Of course we could simplify the formula a little to this:


N=V*3.5e6/(B*A*f) (A again in square inches)

and we would get

N=1346 turns.

 

[Mikebits]

Well done MrAl, that looks like you put a lot of time into that post. Kudos

 

[MrAl]

Hi Mike,


Sometimes when someone asks a question i just cant help but give them
an answer

 

[bonjing]

thank you eric and to all who shared their ideas, it really helps me a lot...