You will need

- Active LC meter or multimeter with a function for measuring inductance

Instruction

1

Get an LC meter. In most cases, they are similar to conventional multimeters. There are also multimeters with a function for measuring inductance - such a device you will also work. Any of these devices can be purchased in specialized shops selling electronic components.

2

Power off the fee, which is a coil. If necessary, discharge the capacitors on the Board. Wipaire coil,

**the inductance of**which you want to measure from the circuit Board (if this is not done, the measurement will be made visible error), and then connect to the input sockets of the device (to which it is specified in his instructions). Switch off the device at the exact limit, usually designated as "2 mH". If**the inductance**of the coil is less than two, MH, it will be defined and shown on the display, after which the measurement can be considered complete. If it is greater than this value, the meter will show overload - in the senior category will appear one and in the other spaces.3

If the meter showed overload, switch off the device at the next, coarser limit - "20 mH. Note that the decimal point on the display shifted to change the scale. If the measurement this time is not successful, continue to shift the limits towards more rough as long as the overload disappears. Then read the result. Looking then to the switch, you will know in what units the result is expressed in Henry's or the MH.

4

Disconnect the coil from the input jacks of the device, and then solder back to Board.

5

If the instrument shows zero even at the exact limit, the coil either has a very small

**inductance**, or contains the short-circuited coils. If even at the coarsest limit is indicated overload, the coil is either broken or has too much**inductance**, the measurement of which the device is not designed.Note

Never connect the LC-meter to the circuit, under voltage.

Useful advice

Some LC-meters have a special knob to adjust. Check the instructions for the instrument, how to use it. Without adjusting the readings will be inaccurate.

# Advice 2 : How to make an inductor

Coil

**inductance**is a rolled into a spiral conductor, storing magnetic energy in a magnetic field. Without this element it is impossible to build neither a radio transmitter nor a radio receiver, for use of the device wired connection. And TV, to which many of us are so accustomed to, without coil**inductance**unthinkable.You will need

- Wires of various sections, paper, glue, plastic cylinder, knife, scissors

Instruction

1

The basis of the coil

**inductance**is a conductor. Around the conductor passing through it a current the magnetic field is always present. The strength of this field depends on the strength of current in the conductor. Another way to strengthen the magnetic field to collapse the conductor into a spiral. This is nothing more than a coil**of inductance**. The smaller the coil diameter, the more turns the stronger the magnetic field generated by the coil. Hams usually wound these coils on their own.2

Inductance is the ability of the coil to create a magnetic field. Inductance is measured in Henry (H).

3

Coil

**inductance**is not produced in the form of standard parts with standard specifications, as calculated and manufactured for each specific device separately. Therefore, in the manufacture of coil you will need to first consider the characteristics of input and output signals of your radio.4

For VHF and short-wave oscillatory circuits are made coils with few turns of thick wire. Some of these coils don't have a frame.

5

For transmission and reception of radio signals at medium and long wavelengths are used multiturn coil (single-layer and multi-layer). To make the frame for these coils you will need paper or plastic.

6

The number of turns of the coil when you configure the radios and other equipment will have to pick up experimentally, changing the inductance of the coil. You can do it otmuchivaniem and Domitiana of turns of the coil, but this method is not quite useful in practice. This files most often placed inside the coil pull out a core of special magnetic material. This can be alsever (an alloy of aluminum, iron and silicon).

7

Magnetic cores concentrate the magnetic field of the coil, thus increase its inductance. With this you can reduce the number of turns of the coil, which leads to the decrease of its sizes and dimensions disturbed.

# Advice 3 : How to calculate the inductance of the coil

An inductor can store magnetic energy during the flow of electric current. Its main characteristic is its

**inductance**, which is denoted by the letter L and is measured in Henry (H).**The inductance**depends on its characteristics.*of the coil*You will need

- the coil material and its geometrical parameters

Instruction

1

**The inductance**proportional to the linear dimensions

*of the coil is**of the coil*, magnetic permeability of the core and the square of the number of turns of winding.

**The inductance of a**wound on a toroidal core is: L = ?0*?r*s*(N^2)/l. In this formula ?0 — magnetic constant, which is approximately equal to 1.26*(10^-6) Ng/m ?r — relative magnetic permeability of core material, which depends on the frequency), s is the cross — sectional area of the core, l is the length of the middle line of the core, N is the number of turns of the

*coil**coil*.

Relative magnetic permeability and material, as well as the number of turns N are dimensionless quantities.

2

Thus,

**the inductance***of the coil*is greater, the larger the area of its cross section. This condition increases the magnetic flux through the coil when the same current.**The inductance**inductance in µh can be calculated also by the formula: L = L0*(N^2)*D*(10^-3). Here N is the number of turns, D is the diameter of*of the coil**coil*in inches. The coefficient of L0 depends on the ratio of the length*of the coil*to its diameter. For single-layer*coil*it is equal to: L0 = 1/(0,1*((l/D)+0,45)).3

If circuit

If

*coil*are connected in series, their total**inductance**equals the sum of the inductances of all coils: L = (L1+L2+...+Ln)If

*coils*are connected in parallel, the total**inductance**equal to: L = 1/((1/L1)+(1/L2)+...+(1/Ln))Note

The main parameter characterizing the properties of inductors and chokes is the inductance. The inductance of the coil depends on its size and shape, number of turns and the magnetic permeability of the medium. . Characterizes the energy losses in the coil and is determined by the ratio of its inductive reactance to active resistance

Useful advice

The physical nature of inductance. Inductors have the property to render the reactance to alternating current with little resistance to direct current. Together with the capacitors they are used to create filters performing the frequency selection of electrical signals and for generating delay elements of signals and storage elements...

# Advice 4 : How to measure inductance

To measure

**the inductance**of the coil, use an ammeter, voltmeter and frequency counter (if not known, the frequency of the AC source, then take readings and calculate the**inductance**. In the case of the solenoid (coil whose length is much greater its diameter) to determine the inductance required to measure the length of the solenoid, the area of its cross section and number of turns of the conductor.You will need

- an inductor tester

Instruction

1

Measurement of inductance by the method of voltmeter-ammeter.

To find the

To find the

**inductance**of a conductor by this method, use the AC source with a known frequency. If the frequency is not known, measure its customera joining it to the source. Connect to the power supply coil,**the inductance**of which is measured. Then in series turn on the ammeter, and to the ends of the coil in parallel is a voltmeter. Passing the current through the coil, clear the readings. Accordingly, the current in amperes and voltage in volts.2

According to this data, calculate the value of inductance of the coil. For this value of the voltage divide successively by 2, the number 3.14, the values of current frequency and current. The result is the value of inductance for the coil in Henry (H). Important: the coil connect only to AC power source. The resistance of the conductor used in the coil should be negligible.

3

The measurement of the inductance of the solenoid.

To measure inductance of the solenoid take a ruler or other instrument for determining lengths and distances, and determine the length and diameter of solenoid in meters. Then count the number of its coils.

To measure inductance of the solenoid take a ruler or other instrument for determining lengths and distances, and determine the length and diameter of solenoid in meters. Then count the number of its coils.

4

Then find

**the inductance**of the solenoid. To do this, take the number of turns in the second degree, the result and multiply by 3.14, the diameter of the second degree, and divide the result by 4. The number you divide by the length of the solenoid and multiply by 0,0000012566 (1,2566*10-6). This will be the value of the inductance of the solenoid.5

If possible, to determine the inductance of this conductor use a special device. It is based on scheme, called bridge AC.

# Advice 5 : How to find the inductance of the coil

An electric current flowing through a conductor creates magnetic field around it. The coefficient of proportionality between the current in the circuit and the magnetic flux created by this current is called the inductance of the coil.

Instruction

1

Based on the definition of inductance, it is easy to guess about the calculation of this value. The simplest formula to calculate the inductance of the solenoid is: L=f/I where L is inductance, f is the magnetic flux embracing the coil of the magnetic field, I is the current in the coil. This formula is defining the unit of inductance: 1 Weber / 1 Ampere = 1 Henry or, for short, 1 WB, 1 And = 1 GN.

Example 1. Through the coil flows a current of 2 A, formed around it a magnetic field, the magnetic flux which is 0,012 WB. Determine the inductance of this coil. Solution: L= 0,012 WB / 2 (A = 0,006 H = 6 mH.

Example 1. Through the coil flows a current of 2 A, formed around it a magnetic field, the magnetic flux which is 0,012 WB. Determine the inductance of this coil. Solution: L= 0,012 WB / 2 (A = 0,006 H = 6 mH.

2

Inductance (L) depends on the size and shape of the coils, from the magnetic properties of the environment in which the current-carrying conductor. Accordingly, the inductance of a long coil (solenoid) can be determined according to the formula given in figure 1, where µ0 – magnetic constant, equal to 12.6*(10) in -7 degree GN/m; µ - relative magnetic permeability of the medium, which is a coil current (tabular value specified in physical books); N is the number of turns in the coil, lкат – coil length, S – area of one loop.

Example 2. Find the inductance of the coil having the characteristics: the length is 0.02 m, the area round to 0.02 sq. m. number of turns = 200. Solution: If the environment in which the solenoid is not specified, the default is taken to air, the magnetic permeability of air is unity. Therefore, L = 12,6*(10) in -7 degree *1*(40000/0,02)*0,02=50,4*(10) in -3 degree H = 50,4 mH.

Example 2. Find the inductance of the coil having the characteristics: the length is 0.02 m, the area round to 0.02 sq. m. number of turns = 200. Solution: If the environment in which the solenoid is not specified, the default is taken to air, the magnetic permeability of air is unity. Therefore, L = 12,6*(10) in -7 degree *1*(40000/0,02)*0,02=50,4*(10) in -3 degree H = 50,4 mH.

3

Also calculate the magnetic induction of the solenoid is possible, based on the formula of magnetic field energy current (see figure 2). It can be seen that the induction can be calculated by knowing the energy of the field and the current in the coil is: L = 2W/(I) squared.

Example 3. Coil in which is flowing a current of 1 A, creates a magnetic field around itself with the energy of 5 joules. Determine the inductance of a coil. Solution: L = 2* 5/1 = 10 GN.

Example 3. Coil in which is flowing a current of 1 A, creates a magnetic field around itself with the energy of 5 joules. Determine the inductance of a coil. Solution: L = 2* 5/1 = 10 GN.

# Advice 6 : How to increase the inductance of the coil

**The inductance**

**of the coil**depends on a number of its design features. These include number of turns, diameter, type of core, its location etc. in order that the inductance has changed, sufficient to modify at least one of these parameters.

Instruction

1

Domotica to the coil additional coils. This will increase the inductance of

**coils**with constant parameters the remaining structural elements, and the variometer (**coil**with a movable core) - will shift both limit changes in inductance (top and bottom) upwards. When winding additional turns may be that they do not fit on the frame. Do not be tempted to use a finer wire than that used in the original coil, so as not to cause heating of the windings flowing through it current.2

To the coil not having a core, add such. But remember that it needs to be made of such a material, in which the operating frequency

**of the coil**does not occur losses in eddy currents. For an electromagnet operating on direct current, suitable solid steel core, for 50-Hz transformer - core, recruited from the blue plate in the higher frequency coils will have to use cores made of ferrites of different grades.3

Remember that even with the same number of turns and other parameters of the coil of greater diameter will have a higher inductance. It is clear, however, that the wire for its production will need more.

4

The ferrite is manufactured with different magnetic permeability. Replace one ferrite core in the coil to another, which is the value of this parameter above, and its inductance will increase. But it will decrease the marginal rate at which this coil can operate without any appreciable loss in the core.

5

There are

**coils**equipped with special mechanisms to move the core. In order to increase the inductance in this case, slide the core into the frame.6

A closed magnetic circuit, ceteris paribus provides a higher inductance than open. But try not to apply this decision in transformers and chokes operating in the presence of a DC component. It is able to podmanivaya and saturate a closed core, thereby, on the contrary, causing a decrease in inductance

**of the coil**.# Advice 7 : How to determine the inductance of the coil

The inductance of the coil can be measured directly or indirectly. In the first case, you will need active or bridge device, and the second will have to use the generator, voltmeter and milliammeter, and then to carry out a series of calculations.

You will need

- - active or bridge meter inductance;
- - generator of sinusoidal voltage;
- - voltmeter and milliammeter AC;
- - frequency;
- - scientific calculator.

Instruction

1

To measure the inductance of the active device, connect the coil and then successively selecting the measurement range switch, select such of them to the result was approximately in the middle of the range. Read the result. If the meter has an analog scale, when reading the results, take into account the price of division, and the coefficient shown next to the corresponding switch position.

2

On the bridge device after each selector turn the knob balancing of the bridge at any of the end positions, and then rotate it all the way in the opposite direction. Find a range in which the handle is possible to balance the bridge. Having achieved the disappearance of the sound in the speaker or headphones or decreasing readings dial indicator to zero, read the indications on the scale of the regulator (but not the gauge). In this case, as in the previous case, consider the division and the factor that should be multiplied by the range statement.

3

To measure inductance indirectly assemble the measuring circuit. The AC voltmeter switched to the limit at which the upper boundary of the range corresponds to a voltage of several volts, connect the parallel output of the generator. There connect and a frequency counter. Also in parallel, they connect the series circuit consisting of the test inductors as well as milliamperemeter AC. Both devices should show the current, not the amplitude value measured value, and can be calculated for sinusoidal oscillations.

4

The generator will enable the generation of the voltage sine wave. Ensure that the voltmeter showed about two volts. Increase the frequency until then, until the readings on the milliammeter begins to decrease. Make sure that they reduce to about half the original value. Select frequency limit corresponding to the measured frequency. Read the testimony of all three devices, and then disconnect the generator and disassemble the measuring circuit.

5

Put readings in SI units. Divide the voltage across the current. Will the inductive reactance of the coil at the frequency at which the measurement was carried out. It will be expressed in ohms.

6

Calculate the inductance according to the formula: L=X/(2πF), where L is the frequency, G (Henry), X is the inductive reactance in Ohms, F is frequency, Hz. If necessary, move the result of the calculation in derived units (e.g., MH, microgenre).

Note

Do not touch the elements of the measurement circuit when it is energized.