Calculation of inductors. Encyclopedia of radio electronics and electrical engineering

Encyclopedia of radio electronics and electrical engineering / Beginner radio amateur

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Any current-carrying conductor creates a magnetic field around itself. The ratio of the magnetic flux of this field to the current that generates it is called inductance. The inductance of a straight piece of conductor is small and amounts to 1 ... 2 μH per meter of length, depending on the diameter of the wire (thin conductors have a large inductance). More accurate results are given by the formula

where is the length of the wire; d is its diameter. Both sizes must be taken in meters (under the sign of the logarithm it is permissible in any, but the same units), the inductance will turn out in microhenry. To facilitate calculations, we recall that the natural logarithm of any number is 2,3 times the decimal logarithm (which can be found using tables, a slide rule or a calculator), i.e. Inx \u2,3d XNUMXlgx.

Why did we give this formula? Let's explain with an example.

Let the conclusions of some radio element have a length of 4 cm with a diameter of 0,4 mm. Let's calculate their inductance:

2,3lg100 = 4,6 and 0,2-0,04-3,6 = 0,03 (round off).

So, the inductance of each pin is close to 0,03 uH, and the inductance of the two pins is 0,06 uH. With a capacitance of only 4,5 pF (and the mounting capacitance can be more), this inductance forms an oscillating circuit tuned to a frequency of 300 MHz - remember Thomson's formula:

f = 1/2π√LC.

That is why on VHF it is impossible to carry out installation with long wires and leave long leads of parts.

To increase the inductance, the conductor is folded into a ring. The magnetic flux inside the ring increases, and the inductance becomes about three times larger:

L = 0,27πD(ln8D/d-2).

Here D is the diameter of the ring, the dimensions are the same. A further increase in inductance occurs with an increase in the number of turns, while the magnetic fluxes of individual turns not only add up, but also affect all other turns. Therefore, the inductance increases with the square of the number of turns. If there are N turns in the coil, the inductance obtained for one turn must be multiplied by N².

For a single-layer cylindrical coil with a length much greater than the diameter D (Fig. 23), the inductance is calculated quite accurately by the formula

strictly derived for a very long solenoid or torus. All dimensions here are in the SI system (meters, Henry), μ0 = 4π 10-7 H/m - magnetic constant; S = πD2/4 - cross-sectional area of ​​the coil; μ - effective magnetic permeability of the magnetic circuit. For open magnetic circuits, it is much less than the permeability of the material itself. For example, for a magnetic antenna rod made of 600NN grade ferrite (magnetic permeability 600) and barely reaches 150. If there is no magnetic circuit, μ = 1.

This formula gives very accurate results for toroidal coils, and corresponds to the circumference of the annular magnetic circuit, measured along its center line. The formula is also suitable for low-frequency transformers wound on a W-shaped magnetic core (Fig. 24).

In this case, S = ab is the cross-sectional area of ​​the magnetic circuit, and - this is the average length of the magnetic field line, shown in the figure by the dotted line. For closed magnetic circuits assembled without a gap, as for ferrite rings, and is taken equal to the magnetic permeability of the material. A small gap slightly reduces μ. Its influence can be taken into account by increasing the length of the magnetic field line by δμ, where δ is the gap width, μ is the magnetic permeability of the core material.

As you can see, the inductance practically does not depend on the diameter of the wire. For low-frequency coils, the wire diameter is selected based on the permissible current density, for copper conductors 2 ... 3 amperes per mm2 of the conductor section. In other cases, especially with RF coils, the goal is to achieve a minimum conductor resistance in order to increase the quality factor (the ratio of inductive to active resistance).

To this end, it would seem that the diameter of the wire should be increased, but then the length of the winding increases, which reduces the inductance, and with a close, multilayer arrangement of turns, the effect of "displacing" the current from the winding is observed, which increases the resistance. The effect is similar to current displacement at high frequencies in any conductors, whereby the current flows only in a thin skin layer near the surface of the conductor. The skin layer thickness decreases and the wire resistance increases in proportion to the square root of the frequency.

Thus, to obtain the desired inductance and quality factor, it is not at all necessary to choose the thickest wire. For example, if a single-layer coil (see Fig. 23) is wound with a thick wire turn to turn or twice as thin as a wire, but with a step equal to the diameter of the wire, the inductance will remain the same and the quality factor will hardly decrease. The quality factor increases with the increase along with the diameter of the wire of all sizes of the coil, mainly its diameter.

To obtain the maximum quality factor and inductance, it is more advantageous to make the coil short, but large in diameter, with the ratio D/ about 2,5. The inductance of such coils is more accurately calculated by the empirical (selected empirically) formula

, where the dimensions are taken in centimeters, and the inductance is obtained in microhenries. It is curious that the same formula is applicable to a spiral or basket flat coil (Fig. 25).

As D take the average diameter:

D = (Dmax + Dmin)/2

but as - winding width,

= (Dmax - Dmin)/2.

The inductance of a multilayer coreless coil (Fig. 26) is calculated by the formula

where the dimensions are substituted in centimeters, and the inductance is obtained in microhenries. With a dense ordinary winding, the quality factor does not exceed 30 ... 50, "loose" winding (bulk, universal) gives high values ​​of the quality factor. Even better is the "cellular" winding, now almost forgotten. At frequencies up to 10 MHz, the quality factor increases when using a litz wire - a wire twisted from many thin insulated veins. The litz wire has a larger total surface of the wire, through which, in fact, the current flows due to the skin effect, and therefore, there is less resistance at high frequency.

A magnetodielectric trimmer increases the inductance up to 2-3 times, depending on the size of the trimmer. An even greater increase in inductance is provided by closed or partially closed magnetic circuits, for example, pot-shaped ones. In this case, it is better to use the strict formula for the solenoid or torus (see above). The quality factor of a coil on a closed magnetic circuit is determined not so much by the wire as by losses in the core material.

At the end of the chapter, we present some useful formulas for calculating the active resistance of wires. The linear resistance (per meter of length) of a copper wire at direct current and low frequencies (Ohm / m) is easy to find by the formula

FL = 0,0223/d2,

where d is the wire diameter, mm. Skin thickness for copper (mm) is approximately 1/15√f (MHz). Please note: already at a frequency of 1 MHz, the current penetrates the wire to a depth of only 0,07 mm! In the case when the wire diameter is larger than the skin layer thickness, the resistance increases in comparison with the DC resistance. The linear resistance of the wire at high frequency is estimated by the formula

R = √f/12d (mm).

Unfortunately, these formulas cannot be used to determine the active resistance of coils, because due to the proximity effect of the turns, it turns out to be even greater.

It is time to give answers to the first tasks given in the previous sections. Problem from introduction of ("Radio", 2002, No. 9, p. 52): what is the duration of unit pulses (with respect to the period) at the output of the logic element (Fig. 2), if it switches at a voltage of 2 V, and a sinusoidal signal with amplitude 4 V?

It is easier and clearer to solve this problem graphically - it is necessary to draw a sinusoid with an amplitude of 4 V as accurately as possible and draw a straight horizontal line at the level of the switching threshold of the element, i.e. 2 V (Fig. 27).

The element will switch at the times corresponding to the points of intersection of the sinusoid with this line. The duration of the resulting pulses (marked with thick lines) can now be measured with a ruler - it will be 1/3 of the period.

On the horizontal axis of the graph, it is advisable to postpone not the time, but the phase of the oscillation φ. The full period will be 360°, and the switching times are found from the equation 4sinφ = 2 or sinφ =1/2 (it equates the instantaneous voltage value to the switching threshold). Equation solutions: φ = 30°, 150°, etc. The phase difference between the switching points is 150 - 30 = 120°, the pulse duration with respect to the period will be 120/360 = 1/3. Thus, the problem can be solved algebraically, but it is easy to get confused in the multivalued solution of the equation for φ, so drawing a graph turned out to be very useful. Even if you do not try to draw the graph accurately, we will get an approximate estimate from it, and from the solution of an algebraic equation - an exact result.

Now the second problem suggested at the end of the first section: Battery measurements showed an EMF of 12 V and a short circuit current of 0,4 A. Which light bulb should I take so that the light is as bright as possible? Determine the internal resistance of the battery:

r \u3d E / lK12 \u0,4d 30 / XNUMX \uXNUMXd XNUMX Ohms.

In order for the light to be as bright as possible, maximum power must be released on the lamp bulb (not voltage, and not current, but power, which is then converted into heat: Q \u6d P t). This happens when the load resistance is equal to the internal resistance of the source: R \u0,2d g. Of all the listed bulbs, only one satisfies this condition - we find its resistance according to Ohm's law: 30 V / 6 A \u0,2d XNUMX Ohm. She will be the brightest. Note also that a voltage of XNUMX V will be released on it and a current of XNUMX A will flow, i.e., the lamp will shine in the mode recommended for it.

Author: V.Polyakov, Moscow

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