When the right off-the-shelf part just can=E2=80=99t be found, you = may want to=20 consider designing a custom transformer to fit your specific=20 requirements.
Thousands of ready-made off-the-shelf transformers out in the market = are=20 available to be designed into electronic or electrical circuits. = However, when=20 the right part just can=E2=80=99t be found, you may want to consider = designing a custom=20 transformer to fit your specific requirements.
Download=20
this article in .PDF format = This=20 file type includes high resolution graphics and schematics when=20 applicable. |
You need to determine a few things before you can start to design a=20 transformer, though. At the minimum, these include the input voltage(s) = and=20 frequency, and the output voltage(s) and current(s). There may very well = be=20 other parameters to consider, both physical and electrical, such as = available=20 space for mounting, mounting style, isolation requirements, leakage = currents,=20 etc. Environmental conditions may also be a consideration.
Note: At this step, it=E2=80=99s important to point out = that all of the=20 decisions made concerning the design will be verified through = calculations that=20 can only be made after many =E2=80=9Crules of thumbs=E2=80=99=E2=80=9D = and =E2=80=9Ceducated guesses=E2=80=9D are used=20 to get us to that point.
Selecting the Core
The first step is to determine the type of core for the design. You = should=20 consult with a core manufacturer to obtain the specific characteristics = and=20 power-handling capabilities for each type and size of core. However, a = general=20 starting point is:
=E2=80=A2 When less than 400 Hz, a silicon steel lamination is = generally used.
=E2=80=A2 At 400 to 2000 Hz, consider a tape-wound or nickel-alloy = core.
=E2=80=A2 Above 2000 Hz, look at ferrite.
Remember this is only a guideline; it=E2=80=99s not uncommon to go = outside of these=20 ranges (e.g., audio transformers can use silicon steel laminations and = operate=20 from 20 to 20,000 Hz). There are many other core types, and many sizes, = shapes,=20 and material grades within the cores listed above. The exact core chosen = may=20 depend on board spacing, location, mounting style, or any of a number of = physical and electrical parameters that only you can decide.
Most core types will also need a winding bobbin to fit the core that = you=20 choose, and possibly assist in the mounting of the finished product. = Make sure a=20 sufficient bobbin style and material is available, and that you have all = of the=20 mechanical measurements to determine winding details later on in the = design.=20 Certain cores don=E2=80=99t require a bobbin, but we=E2=80=99ll save = those for another=20 discussion.
After selecting a core and bobbin, you need to calculate the correct = number=20 of primary turns needed using formula 1 or 2 (see =E2=80=9CBasic = Design Formulas,=E2=80=9D=20 below). There are several variables to consider when using these = formulas;=20 you will need to consult the core manufacturer=E2=80=99s data for = answers to=20 specifications such as flux density and stacking factor.
1) N(p) =3D (V x 108) / (4.44 B A f K) sine wave
2) N(p) =3D (V x 108) / (4 B A f K) square wave
3) N(s) =3D V(s) / V(p) x N(p)
4) I(p) =3D (VA(s) + losses) / V(p)
5) I(s) =3D P(out) / V(out)
6) Open circuit voltage (Voc) =3D N(s)/N(p) x V(p)
7) Loaded voltage (Vld) =3D Voc-[IR(sec) + (IR(pri) x = (N(s)/N(p))]
8) Temp rise (T(C)) =3D (losses/(0.008 x surface area))
N(p) =3D Primary turns
N(s) =3D Secondary turns
B =3D Flux density in gauss
A =3D Core area in centimeters squared
f =3D Frequency
K =3D Stacking factor
L =3D Inductance
A(l) =3D Inductance per turn squared for a given core
Windings and Wire
The primary winding current and wire size needs to be determined. The = primary=20 current will be equal to the total output power plus transformer power = losses,=20 divided by the primary voltage. For power losses, I start at a 10% = increase in=20 the input power, assuming a 90% efficient transformer. For example, a=20 transformer with a 12-V, 2-A output at 120 V input would be:
12 V x 2 A =3D 24 VA; 24 VA x 1.10 (110%) =3D 26.4 VA needed in the =
primary=20
winding;
26.4 VA/120 V =3D 0.22 A in =
the primary=20
winding
The next step will be a subject for debate and adjustment depending = on the=20 transformer characteristics: I generally start at approximately 500 = circular=20 mills (cm) per amp to choose the starting wire gauge. This number may be = smaller=20 for small transformers, and larger for large power transformers; that = decision=20 is again up to the designer. Using the example above, 0.22 A x 500cm/A = =3D 110cm;=20 I would start with a 29 gauge wire (127.7cm) for the primary.
You now need to determine the number of turns that will be required = for each=20 secondary winding. The first step is to use formula 3 (N(s) =3D V(s) / = V(p) x=20 N(p)) to determine the turns for a perfect transformer. This number then = needs=20 to be increased to account for the losses in the coils. As a rule of = thumb=20 again, I start with a 10% increase in the number of turns, assuming a = 90%=20 efficient transformer: N(s) x 1.10 =3D N Turns. This percentage will = vary=20 depending on the characteristics of your design. Use the same method to=20 determine the secondary wire gauge(s) that you used for the primary.
Now you need to see if the windings fit into your winding area and = determine=20 the actual losses of the coils. Start by determining the winding width = (ww) of=20 the bobbin, the winding height (wh) of the bobbin, and the mean length = of turns=20 (mlt) from the mechanical drawing (see the figure).
With this information, you can now calculate how many turns of wire = can fit=20 in each layer of winding for each different winding. After that, = determine how=20 many layers of each wire gauge used it will take to wind the calculated = number=20 of turns for each given winding. Finally, calculate the total height of = all of=20 the windings wound concentrically. Remember to include the insulation = used=20 between the windings in figuring your total winding height.
Verification
The next step is to verify your design. Will it fit in the allowable = winding=20 height with enough =E2=80=9Cplay=E2=80=9D to allow for error? Keep in = mind that the windings=20 will not be perfectly layered, so you must allow for some bowing of the = copper=20 wire as it bends around each corner of the bobbin. I use a maximum of = 85% fill=20 (or build) when comparing the actual winding height to the allowable = bobbin=20 winding height:
(Actual winding height/Allowable winding height) =C3=97 100 =3D < = 85%
A build of around 75% is more desirable, and will make the = transformer easier=20 to manufacture in the long run.
Next, you need to calculate the resistance of each winding, and the = loaded=20 voltage losses in the winding to determine loaded output voltages. = First, the=20 resistance of each winding is calculated by multiplying the mean length = of turn=20 by the number of turns. This gives the total length of the wire = conductor.=20 Whatever unit of measure you are using, convert this length to thousands = of feet=20 (K ft); for example, a length of 400 in./12000 =3D 0.03 K ft. Now from a = magnet=20 wire chart, you can get resistance of each gauge of wire in thousands of = feet.=20 For our example, 29 gauge has a resistance of 81.21 =CE=A9/K ft, so a = 400-in. wire=20 would have a resistance of 0.03 K ft =C3=97 81.21 =CE=A9 /K ft =3D 2.44 = =CE=A9.
Once you have the resistance of a winding, you simply calculate the = voltage=20 drop across that winding by multiplying the resistance by the current in = the=20 winding. In this primary it would be 2.44 =CE=A9 x 0.22 A =3D 0.540 V. = Calculate the=20 voltage drop across each winding in the same manner.
The next step is to calculate the open-circuit voltage of each = secondary=20 using formula 6 and the loaded voltage using formula 7 of (see = =E2=80=9CBasic Design=20 Formulas=E2=80=9D). I put these together because I usually do both = calculations on=20 each secondary winding before moving on to multiple windings if I have = them.
The key thing to note here is that the primary voltage drop is = reflected into=20 the secondary loaded voltage by the ratio of the turns. After that, = it=E2=80=99s added=20 to the secondary voltage drop. The sum of the two is then subtracted = from the=20 open circuit voltage of the given secondary. This gives you the loaded = output=20 voltage of that secondary (I realize the formula could be mathematically = simplified, but this format allows me easier use with a calculator; = again, my=20 opinion).
If the voltage is not what you want, divide the desired output = voltage by the=20 calculated output for an error ratio. Use this ratio multiplied by the = number of=20 turns on the given secondary and recalculate everything starting at the = number=20 of layers of winding.
Temp Calculations
After calculating the turns, you need to know the calculated = temperature=20 rise. There are two main causes of temperature rise in a transformer: = core power=20 losses and winding power losses. To determine the core power losses, = refer to=20 manufacturer=E2=80=99s datasheets and the flux density used in your = design. Winding=20 power losses are easily calculated by multiplying the voltage drop = across the=20 winding again by the current in the winding (I2R). The sum of = the=20 power losses in the windings, primary(s), and secondary(s) is multiplied = by 1.33=20 (I=E2=80=99ll be honest, I don=E2=80=99t fully know where this factor = came from, but it has to=20 do with heat transfer and concentric windings, and it works) to give the = total=20 effective heating losses of the windings. Add the core losses to the = winding=20 losses for total power dissipation.
To determine how well the transformer dissipates power losses, we = need to=20 calculate the surface area of the completed device. This simply requires = looking=20 at all of the surfaces that will be exposed to air and adding them up in = inches=20 squared. Now use formula 8 from the design formulas to calculate the = estimated=20 temperature rise of the transformer.
Download=20
this article in .PDF format = This=20 file type includes high resolution graphics and schematics when=20 applicable. |
Once again, what constitutes an acceptable temperature rise depends = on the=20 application and the designer. I always use 50=C2=BAC as my maximum = rise=20 allowed. Keep in mind that forced air cooling or heat sinks may be used = in the=20 end product, which could push that number higher.
Now that you=E2=80=99ve made all of your decisions on the design = aspects, based on=20 =E2=80=9Crules of thumb=E2=80=9D and =E2=80=9Ceducated guesses,=E2=80=9D = you should be able to confirm the=20 choices made using the formulas and calculations discussed in the = article. If=20 things anywhere along the way don=E2=80=99t work, you have to go back, = make the=20 appropriate changes, and start over. This includes the temperature-rise=20 calculation.
You haven't included anything about determining the correct amount of = air gap=20 in the core to allow, if you don't have one for a power transformer you = may get=20 saturation and unacceptable losses, for a signal transformer this will = manifest=20 itself as unacceptable distortion at certain frequency and power levels. = I think=20 you're trying to vastly oversimplify a field of design that it takes = most people=20 an entire career to master.
But if you include an air-gap, inductance will drop ... requiring = more turns=20 to restore a reasonable "excitation current" in the primary. It's = generally=20 better to just use more core material to avoid saturation. The only = transformers=20 that normally use air gaps are those that have substantial DC current in = primary=20 or secondary ... so they're generally used only in switch-mode power = supplies or=20 single-ended signal (audio) transformers. All else being equal, they'll = have=20 poorer low-frequency response. Also, any transformer that operates near = core=20 saturation (as do most 50/60 Hz power transformers are to make them = cheap) will=20 also radiate a substantial external field ... making them troublesome = neighbors=20 for nearby equipment. Designing transformers is truly "a world of = trade-offs." =20 Bill Whitlock, Technology Manager, Jensen Transformers, Inc.
I have a line isolation TRF (<500W with @16cm2 core) and wanna = modify it=20 to fuse rods of small diameter (max. 10mm2). Do you know how to workout = the=20 current to fuse metals and at what voltage?
I'd love to meet you guys for lunch once a week, to kick this subject = around.
" I think you're trying to vastly oversimplify a field of design that = it=20 takes most people an entire career to master."
Profoundly true, and it takes some courage to write in one article a=20 transformer design guide. A few additional hand-waves:
1. To optimize transfer power, full core utilization is the goal. The = core=20 should be run at both its power-loss and saturation limits (with design = margins). Then the smallest volume core can be used. Transfer power = applies to=20 coupled inductors, but the same basic utilization criteria apply to=20 transformers.
2. To optimize power transfer, the winding and core power losses = should be=20 about the same. The goal is to maximize power transfer through the = transductor=20 (transformer or coupled inductor) by adjusting the design for winding to = core=20 power loss such that output power / input power is maximized. This = happens=20 neither at equal core and winding loss nor at minimum loss but = in-between the=20 two.
Core and winding losses can be traded off by adjusting turns and wire = size.=20 However, circuit design constraints also affect both. Wire size is = fairly easy=20 to choose from a wire table and circuit current requirements. The = optimal turns=20 is the geometric mean between the max-power-loss minimum and the = saturation=20 maximum.
the pdf available for download seems to be corrupted.
The start of =
each=20
page has omissions.
It stops after the table on page 4.
Sponsored Introduction Continue on to=20 (or wait seconds)=20 =C3= =97