Outlined here are the various causes of inefficiency, methods of estimating them and techniques for minimizing them.
Your life as a designer of transformers, for example, would be far simpler if you did not need to worry about power losses; you would simply select a part having the turns ratio needed - and that would be that. It's the requirements to keep the losses under control that largely determines whether your transformer weighs one gram and costs one dollar or weighs one ton and costs ten thousand dollars.
Power is lost in an inductor through several different mechanisms:
Only the first of these, the copper loss, operates even when the current in the winding is not changing. The other losses all fall to zero with frequency.
Hysteresis and eddy current losses are collectively known as iron loss or core loss (even if a ferrite core is used). The figures for common core materials are tabulated in Kaye & Laby.
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See also ...
[↑
Producing wound components]
[A guide to the terminology used in the science of magnetism]
[ Air coils]
[The force produced by a magnetic field]
[ Faraday's law]
[Guide to
unit systems in electromagnetism]
[ AWG explained]
Copper loss or winding loss is the term used to describe the energy dissipated by resistance in the wire used to wind a coil.
Quantity name | Current density |
---|---|
Quantity symbol | J |
Unit name | amperes per metre squared |
Unit symbols | A m^{-2} |
The question of current density is a critical one for the designer of wound components. It appears in formulae which relate the gauge of wire needed to the size of coil former required to hold the wire. In simplistic terms, a high value for J leads to compact, cheap and inefficient inductors, whilst a low value implies bulky, expensive, more efficient and, possibly, more reliable designs. Generally -
where A is the cross sectional area, in metres squared, of some part of
the winding and I the total current flow, in amperes, in that area. In
data sheets the value will often be quoted in amps per square
millimetre. In that case you will neeed to multiply by 10^{6}
before substitution in the formulae given here.
In these pages you will meet two kinds of current density. The first is what you might term the 'raw' current desnsity: J_{C}. Here the area under consideration is A_{C}: exactly that part of the cross section of one of the turns which is composed entirely of copper conductor. Surrounding insulation space is ignored. For small transformers the value of this density will be in the order of 3.5 A mm^{-2}.
The second variation is J_{W} where the area considered, A_{W}, is the entire winding aperture of the coil former - including insulation around the wire itself, air between the wires and any tape used to separate one layer of turns from the one above. J_{W} will be considereably lower than J_{C}. The (dimensionless) ratio of these two is the winding copper factor, F_{W} -
where A_{C} is the the cross sectional area of the copper in a
single turn. Practical values of F_{W} for the type of
inductors most used in electronics will be about 0.5. In other words,
only about half the winding space will be filled with copper.
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In most cases your wire will be made of copper, whose resistivity at 20 °C is about 1.73×10^{-8} ohm metres. Real world coils often operate above room temperature. You can then multiply by a 'fiddle factor', m:
Temperature (°C) | 20 | 40 | 60 | 80 | 100 |
---|---|---|---|---|---|
m (Mullard, 1979) | 1.0 | 1.079 | 1.157 | 1.236 | 1.314 |
You can conveniently estimate the resistance of a wire as
Material | Resitivity(20°C)/ Ω m ×10^{-8} |
---|---|
Aluminium | 2.65 |
Copper | 1.73 |
Graphite | 1375 (0°C) |
Iron (pure) | 10 |
Iron (3% Si) | 47 |
Silver | 1.59 |
where D_{W} is the diameter of the conductor in millimetres. Usually, you will know the length of a winding from
l_{w} = N π D_{av} metres
where N is the number of turns and D_{av} is the average diameter of the winding in metres. Your copper losses will then be
P_{W} = 0.022 l_{w} (I / D_{W})^{2} watts
where I is the RMS current in the winding.
The usual range of current densities seen in power transformers lies between about 1.5 to 5 amps per square millimetre. A good starting point is 3.5A mm^{-2}.
BS Wire Gauge | Diameter / mm | Current / A |
---|---|---|
10 | 3.24 | 2.9E+01 |
12 | 2.64 | 1.9E+01 |
14 | 2.03 | 1.1E+01 |
16 | 1.62 | 7.2E+00 |
18 | 1.22 | 4.1E+00 |
20 | 0.91 | 2.3E+00 |
22 | 0.71 | 1.4E+00 |
24 | 0.56 | 8.6E-01 |
26 | 0.46 | 5.8E-01 |
28 | 0.376 | 3.9E-01 |
30 | 0.304 | 2.5E-01 |
32 | 0.274 | 2.1E-01 |
34 | 0.234 | 1.5E-01 |
36 | 0.193 | 1.0E-01 |
38 | 0.152 | 6.4E-02 |
40 | 0.122 | 4.1E-02 |
42 | 0.102 | 2.9E-02 |
44 | 0.081 | 1.8E-02 |
AWG Wire Gauge | Diameter / mm | Current / A |
---|---|---|
0 | 8.25E+00 | 1.87E+02 |
2 | 6.54E+00 | 1.18E+02 |
4 | 5.19E+00 | 7.40E+01 |
6 | 4.11E+00 | 4.65E+01 |
8 | 3.26E+00 | 2.93E+01 |
10 | 2.59E+00 | 1.84E+01 |
12 | 2.05E+00 | 1.16E+01 |
14 | 1.63E+00 | 7.28E+00 |
16 | 1.29E+00 | 4.58E+00 |
18 | 1.02E+00 | 2.88E+00 |
20 | 8.12E-01 | 1.81E+00 |
22 | 6.44E-01 | 1.14E+00 |
24 | 5.11E-01 | 7.17E-01 |
26 | 4.05E-01 | 4.51E-01 |
28 | 3.21E-01 | 2.83E-01 |
30 | 2.55E-01 | 1.78E-01 |
32 | 2.02E-01 | 1.12E-01 |
34 | 1.60E-01 | 7.05E-02 |
36 | 1.27E-01 | 4.43E-02 |
38 | 1.01E-01 | 2.79E-02 |
40 | 7.99E-02 | 1.75E-02 |
You can look up the diameter corresponding to this density in table
PLA. If efficiency, long life for the insulation and good voltage
regulation are important then choose a lower current density. If size,
weight, cost or frequency range are important then choose a higher density.
Unfortunately the above analysis holds only for low frequencies. Copper losses increase with frequency due to the skin effect.
Skin effect refers to the tendency of current flow in a conductor to be confined to a layer in the conductor close to its outer surface. At low frequencies the skin effect is negligible and the distribution of current across the conductor is uniform. As frequency is increased the depth to which the flow can penetrate, D_{S} / metres, is reduced.
where ρ is the conductor resistivity (ohm
metres), μ_{r} is the relative permeability and f
is the frequency (Hz).
50Hz | 1kHz | 100kHz | 1MHz | 10MHz | |
---|---|---|---|---|---|
Copper | 9.36E-03 | 2.09E-03 | 2.09E-04 | 6.62E-05 | 2.09E-05 |
Aluminium | 1.16E-02 | 2.59E-03 | 2.59E-04 | 8.19E-05 | 2.59E-05 |
Iron (3% Si) μ_{r}=300 | 2.82E-03 | 6.30E-04 | 6.30E-05 | 1.99E-05 | 6.30E-06 |
Graphite | 2.64E-01 | 5.90E-02 | 5.90E-03 | 1.87E-03 | 5.90E-04 |
Skin effect causes the apparent resistance of a coil to increase above that suggested by the DC value, thus lowering the Q-factor in resonant circuits and reducing the efficiency in switching supplies. Skin effect occurs because current flow moves away from those regions of the conductor having the strongest magnetic field. A consequence of this is that the number of flux linkages between turns will be reduced. Therefore skin effect produces a decrease in inductance; of about 2%, though more if the wire is short.
Having chosen a diameter of wire that can cope with the current at zero Hz you should check that skin effect is not going to be a problem at the frequency at which you actually need to work. The formula below (Terman) gives the diameter of wire that will suffer a 10% increase in resistance at the frequency of operation, f (in Hz) -
D_{W} = 200 / √f millimetres
This formula applies for isolated conductors. In a coil surrounded by other turns the actual resistance will be higher because of the proximity effect. Taking the example of a coil working at 10 kHz you will see that a wire diameter of 2 mm is about as thick as you can go without skin effect increasing its AC resistance. Incidentally, if you are designing a switch mode supply don't forget to take harmonics of the switching frequency into account.
If you find that the diameter determined for zero Hz is greater than the diameter at which skin effect will take place at your operating frequency then what options do you have?
R = 8.32 10^{-5} √f / D_{W} ohms per metre
Where f is frequency (Hz), and D_{W} the wire diameter (millimeters).
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Litz (a contraction of Litzendraht) wire consists of thin strands of enamel insulated wire held together by an outer covering of silk or cotton. There will usually be at least 5 strands but perhaps as many as 1000 or more. Its purpose is to minimize the resistive losses that arise at high frequencies due to skin effect. Litz wire is mostly used within the frequency range 10kHz to 1MHz. Because it is expensive and difficult to obtain you should usually only consider this type of wire if you are certain that the increased performance is essential. A typical application is long wave rod antennas. A more detailed paper analyzing the skin and proximity effects in litz wire is available at Dartmouth College.
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The Q factor of a coil is important when it is used in a resonant circuit because it affects the 'sharpness' of the response curve.
The Q or quality factor of a resonant circuit is a measure of the ratio of the energy stored in it to the energy lost during one cycle of operation. All practical inductors exhibit losses due to the resistance of the wire or absorption by materials within the magnetic field surrounding it. It is possible to model these losses as a resistance, R, in series with a perfect or loss free inductance L.
The value of R will be greater than that of the DC resistance of the wire due to skin effect. The above formula suggests that the Q of any given inductor will increase indefinitely with frequency. This is never the case because of an effect known as self resonance.
Practical values of Q range from around 10 for a high loss circuit through about 100 for a reasonable one up to around 1000 with careful design and favourable conditions.
Modern ferrite materials have low hysteresis and eddy current losses but they can still be significant. Cores with an air gap are needed to achieve the highest Q factor and temperature stability.
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Quantity name | coercivity, alias coercive force |
---|---|
Quantity symbol | H_{c} |
Unit name | ampere per metre |
Unit symbols | A m^{-1} |
Coercivity occurs within ferromagnetic materials and is the field strength which must be applied in order to reduce (or coerce) a remnant flux to zero.
Materials with high coercivity (such as those used for permanent magnets) are sometimes called hard. A good example of a hard material is Alnico, which may have an H_{c} of about 80,000 A m^{-1}. Conversely, materials with low coercivity (such as those used for transformers) are called soft. Supermalloy has an H_{c} of about 0.5 A m^{-1}.
Those materials with a low coercivity tend to have a high permeabilty and vice versa. See the paragraph on hysteresis for more details.
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Quantity name | remnance, alias remanence or residual induction |
---|---|
Quantity symbol | B_{r} |
Unit name | tesla |
Unit symbol | T |
Base units | kg s^{-2} A^{-1} |
Remnance (or remanence) occurs within ferromagnetic materials and is the flux density which remains in a magnetic material when any externally applied field is removed (H = 0). For materials used in permanent magnets you usually need a high value of remnance. For transformers you need low remnance. See the paragraph on hysteresis for more details.
Remnance is also what makes possible all magnetic recording technologies; including the hard disk drive on which this text was stored until you loaded it into your browser.
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Figure PHY shows the effect within ferromagnetic materials known as hysteresis.
We start with an unmagnetized sample at the origin (P1) where both field strength and flux density are zero. The field strength is increased in the positive direction and the flux begins to grow along the dotted path until we reach P2. This is called the initial magnetization curve.
If the field strength is now relaxed then some curious behavior occurs. Instead of retracing the initial magnetization curve the flux falls more slowly. In fact, even when the applied field is returned to zero there will still be a remaining (remnant or remanent) flux density at P3. It is this phenomenon which makes permanent magnets possible.
To force the flux to go back to zero we have to reverse the applied field (P4). The field strength here is called the coercivity. We can then continue reversing the field to get to P5, and so on round this type of magnetization curve called (by J. A. Ewing) a hysteresis loop.
What is the significance of this from the point of power losses? It is that we have had to expend energy in order to set up the remnant flux. To show this more clearly we'll look at separate figures below. The area (shown shaded) between the B-H curve and the B axis represents the work done (per unit volume of material).
W = _{0}∫^{B}H dB J m^{-3}
You can see that the energy required to 'pump up' the core by moving from P1 to P2 is more than that which it returns when going from P2 to P3. This is evocatively termed inelastic behavior. You could look at this another way by thinking about the 'back emf' which opposed the initial increase in coil current. The emf generated is always proportional to the change in flux; but the flux changes less on the 'way down' than it does going up.
We can go a stage further and deduce that the total power lost over one complete cycle is proportional to the area within the hysteresis loop. Because this effect is related to an area, hysteresis loss is roughly proportional to the square of the working flux density. In fact, the non-linearities will, for transformer iron, reduce this to about B^{1.6}. The particular value for a given material is called the Steinmetz exponent, n. Unfortunately, it is very rare to see a data sheet which states n directly. For an iron core device it is sometimes assumed to be 1.6. For ferrite grade 3C8 it is 2.5. Data sheets sometimes have graphs of loss versus flux density on a log scale. These can be used to estimate n.
Because a hysteresis loss is incurred each time the core cycles, from positive to negative values of B, the loss rate (watts) is directly proportional to the frequency of operation, f (Hz). We can combine these proportionalities in a single formula for the hysteresis loss:
P_{h} = K_{h}×f×B^{n} watts m^{-3}
Where K_{h} also depends on the particular core material. Significantly, hysteresis loss in a transformer is essentially independent of load current; it depends just on flux and hence voltage. Consequently the efficiency of a transformer drops towards zero with the load current.
You can build a simple circuit to display hysteresis. By the way, this tells us why it really isn't advisable to subject a transformer to a voltage (and hence flux) overload - it's going to get hot before you go very far :-(
The hysteresis of a material strongly affects its suitability for a
particular application.
The left hand curve shows a 'soft' magnetic material such as iron
alloyed with silicon. Its area is small so it's ideal for a low loss
transformer core. The addition of 3 % of silicon to iron reduces the
hysteresis loss at 1 tesla from about 250 to 163 J m^{-3}. The
material on the right hand curve is 'hard' magnetic. Its large area is
commonly seen in materials such as Alnico (an
iron/cobalt/nickel/aluminium alloy) used for permanent magnets.
A word of warning about the terms 'hard' and 'soft': many writers use it
to denote only that the value of coercive force is high or low respectively. Others use
them to say that the shape of the loop is 'wide' or 'thin'. That
is ambiguous. Does it just refer to H_{c} or to the ratio of H_{c}
to B_{r}? Other
writers say that hard or soft means the area of the loop is large or small. Yet
others (e.g. Duffin)
use hard or soft to denote the value of the remnant field! Groan :-(
In practise, high
H_{c} often goes together with high B_{r} and everyone is happy. If
you want a winner then go for straight coercivity. Hard means
H_{c} above 10 kAm^{-1}. Soft means H_{c} below
1 kAm^{-1}.
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Eddy current losses occur whenever the core material is electrically conductive. Most ferromagnetic materials contain iron: a metal that has fairly low resistivity (roughly 10^{-7} Ω m). The problem is intuitively obvious if you consider that the magnetic field is contained within a 'circuit' or loop formed by the periphery of the core in the same way as it is contained within a turn on the windings. Around that periphery a current will be induced in the same way as it is in an ordinary turn which is shorted at its ends.
What is needed, then, is some method of increasing the resistance of the core to current flow without inhibiting the flow of magnetic flux. In mains transformers this is achieved by alloying the iron with about 3% of silicon. This lifts the resistivity to 4.5×10^{-7} Ω m. Depending upon the amount of silicon this material is called 'transformer iron', 'electrical iron' or 'armature iron'. The alternative name 'silicon steel' is a misnomer because steel is iron alloyed with carbon; and carbon does no good in a transformer core. The silicon does, though, increase the mechanical hardness of iron in the same way as carbon - try sawing up a transformer core and you'll discover this quickly.
In any resistive circuit the power is proportional to the square of the applied voltage. The induced voltage is itself proportional to f×B and so the eddy losses are proportional to f^{2}B^{2}. The flux is also related to the size of the loop. Figure PLM shows how the idea of lamination is used to reduce the power losses caused by eddy currents in mains transformers. The same principle applies to motors and generators too. Using a solid iron core (as in cross-section B) results in a large circulating current. So, instead, the core is made up of a stack of thin (~0.5 millimetre) sheets (cross section C). Here I have shown only four laminations but there will normally be many more. The lines of magnetic flux can still run around the core within the plane of the laminations. The situation for the eddy currents is different. The surface of each sheet carries an insulating oxide layer formed during heat treatment. This prevents current from circulating from one lamination across to its neighbors.
Clearly, the current in each lamination will be less than the very large current we had with the solid core; but there are more of these small currents. So have we really won? The answer is yes, for two reasons.
Power loss (the reduction of which is our aim) is proportional to the square of induced voltage. Induced voltage is proportional to the rate of change of flux, and each of our laminations carries one quarter of the flux. So, if the voltage in each of our four laminations is one quarter of what it was in the solid core then the power dissipated in each lamination is one sixteenth the previous value. Hurrah!
But wait; it gets better. Look at the long thin path that the eddy current takes to travel round the lamination. Suppose we made the laminations twice as thin (we halved d_{1}). The path length of the current isn't much changed; it's still about 2×d_{2}. However, the width of the path has halved and therefore its resistance will double and so the current will be halved. The bottom line is that eddy current loss is inversely proportional to the square of the number of laminations. Iron losses should be between about 1 and eight watts per kilogram at 50Hz and 1.5T for good transformer steel.
This idea of dividing up the iron into thin sections is carried a stage further in the iron dust cores. Here the iron is ground into a powder, mixed with some insulating binder or matrix material and then fired to produce whatever shape of core is required. These cores can function at several megahertz but their permeability is lower than solid iron.
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In general, the cross sectional area of a transformer core, A_{e}, determines how much flux the core can carry and therefore how much voltage is induced in each turn. The cross sectional area of the winding aperture, A_{w}, determines how thick the wire can be and therefore how much current it can handle. From these you can estimate the VA rating.
The peak core flux is
φ_{pk} = B_{pk} × A_{e} webers | Equation PLG |
For most laminated iron cores B_{pk} is kept below about 1.2 teslas. The transformer equation relates this flux to induced RMS voltage
V = 4.44 N × f φ_{pk} volts | Equation PLB |
where N is the number of turns on the primary and f is the frequency of operation in hertz.
Now, what about the current capacity of the primary? Unfortunately, the raw figure for the area of the primary winding aperture isn't directly useful here because, in a practical coil, much of this area is occupied not by copper (which can carry current) but by insulation and other gaps (which can't) between the turns. We represent this difference by a parameter called the winding factor or the space utilization factor, F_{w}
A_{Cu} = A_{w} × F_{w} m^{2} | Equation PLC |
Where A_{Cu} is the cross sectional area of actual copper. For many designs F_{w} is about 0.5. That is, half of the primary cross section is actually copper.
Even less certain is the working current density in the wire, j. This is about 3 amps per millimetre squared (3×10^{6} A m^{-2}) for most small mains transformers. Whatever value is pulled out of the hat for j you then get the winding current by
I = j × A_{Cu} / N amps | Equation PLD |
Putting this together
V × I = 4.44 j × A_{Cu} × f φ_{pk} volt-amps | Equation PLE |
Take as an example the 50 VA transformer kit from RS, stock no. 182-9919, working at 50 Hz. This has a rectangular section central core for the lamination stack 28 x 43 mm so A_{e} = 1.16×10^{-3} m^{2} and so φ_{pk} = 1.2 × 1.16×10^{-3} = 1.39×10^{-3} weber.
The primary winding is 37.5 mm high by 3 mm deep so A_{w} = 1.13×10^{-4} m^{2}, giving an A_{Cu} = 0.5 × 1.13×10^{-4} = 5.63×10^{-5} m^{2}.
V × I = 4.44 × 3×10^{6} × 50 × 5.63×10^{-5} × 1.39×10^{-3} = 52 volt-amps | Equation PLF |
This is a closer result than you will normally get.
Fundamentally, the rate at which a transformer looses heat depends upon its surface area - which increases in proportion to the square of the transformer's linear dimension. How fast heat is generated inside a transformer depends upon the volume of iron and copper - which increases as the cube of the linear dimension. Large transformers tend to be run at lower flux and current ratings which improves efficiency (99% is common). This is partly for reasons of cost savings through reduced energy wastage but they also have to be more efficient than small transformers because they have a harder job getting rid of heat. This explains why the largest transformers are fitted with oil-filled cooling loops: they effectively increase surface area.
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By various means, such as the use of air gaps, or by selecting a lower permeability material, it's possible to trade off copper losses against iron losses. This raises the question of how this should be done to obtain the most efficient operation. The answer, as described by Jansson, is succinct:
P_{W} = n×P_{C} / 2
where P_{W} is the copper loss, P_{C} is the iron loss and n is the Steinmetz exponent for the core material.
It is also useful during the design of a transformer to be able to relate the power losses (P_{L} in watts) to its temperature rise (Δ_{T} in °C). Some data sheets will give this but, if not, then this empirical formula (SEI Ltd.) should give an indication at least.
A_{S} = 10^{-4}(59(1000/(T_{A}+273))^{1.69} (P_{L}^{0.82} /Δ_{T}))^{1.22} m^{2} | Equation PTR |
Where A_{S} is the exposed surface area of the core (metres squared), and T_{A} is the ambient temperature (°C). For example, if the ambient is 25°C, the total power loss is 13.7 watts and the temperature rise is 58 °C then an area of 1.7×10^{-2} m^{2} is required. One manufacturer of the FX3750 core quotes this as its area when fully wound, but says that only 9 watts are required to give this temperature rise in free air. Discrepancies of this magnitude must be expected.
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E-mail:
R.Clarke@surrey.ac.uk
Last revised: 2008 August 2nd.