Conductor resistance. Resistivity

Ohm's law is the most important in electrical engineering. That is why electricians say: “Whoever does not know Ohm’s Law should sit at home.” According to this law, current is directly proportional to voltage and inversely proportional to resistance (I = U / R), where R is a coefficient that relates voltage and current. The unit of measurement for voltage is Volt, resistance is Ohm, current is Ampere.
To show how Ohm's Law works, let's look at a simple electrical circuit. The circuit is a resistor, which is also a load. A voltmeter is used to record the voltage across it. For load current - ammeter. When the switch is closed, current flows through the load. Let's see how well Ohm's Law is observed. The current in the circuit is equal to: circuit voltage 2 Volts and circuit resistance 2 Ohms (I = 2 V / 2 Ohms = 1 A). The ammeter shows this much. The resistor is a load with a resistance of 2 ohms. When we close switch S1, current flows through the load. Using an ammeter we measure the current in the circuit. Using a voltmeter, measure the voltage at the load terminals. The current in the circuit is: 2 Volts / 2 Ohms = 1 A. As you can see, this is observed.

Now let's figure out what needs to be done to increase the current in the circuit. First, increase the voltage. Let's make the battery not 2 V, but 12 V. The voltmeter will show 12 V. What will the ammeter show? 12 V/ 2 Ohm = 6 A. That is, by increasing the voltage across the load by 6 times, we obtained an increase in current strength by 6 times.

Let's consider another way to increase the current in a circuit. You can reduce the resistance - instead of a 2 Ohm load, take 1 Ohm. What we get: 2 Volts / 1 Ohm = 2 A. That is, by reducing the load resistance by 2 times, we increased the current by 2 times.
In order to easily remember the formula of Ohm's Law, they came up with the Ohm triangle:
How can you determine the current using this triangle? I = U / R. Everything looks quite clear. Using a triangle, you can also write formulas derived from Ohm's Law: R = U / I; U = I * R. The main thing to remember is that the voltage is at the vertex of the triangle.

In the 18th century, when the law was discovered, atomic physics was in its infancy. Therefore, Georg Ohm believed that the conductor is something similar to a pipe in which liquid flows. Only liquid in the form of electric current.
At the same time, he discovered a pattern that the resistance of a conductor becomes greater as its length increases and less as its diameter increases. Based on this, Georg Ohm derived the formula: R = p * l / S, where p is a certain coefficient multiplied by the length of the conductor and divided by the cross-sectional area. This coefficient was called resistivity, which characterizes the ability to create an obstacle to the flow of electric current, and depends on what material the conductor is made of. Moreover, the greater the resistivity, the greater the resistance of the conductor. To increase the resistance, it is necessary to increase the length of the conductor, or reduce its diameter, or select a material with a higher value of this parameter. Specifically, for copper the resistivity is 0.017 (Ohm * mm2/m).

Conductors

Let's look at what types of conductors there are. Today, the most common conductor is copper. Due to its low resistivity and high resistance to oxidation, with fairly low fragility, this conductor is increasingly being used in electrical applications. Gradually, the copper conductor is replacing the aluminum one. Copper is used in the production of wires (cores in cables) and in the manufacture of electrical products.

The second most commonly used material is aluminum. It is often used in older wiring that is being replaced by copper. Also used in the production of wires and electrical products.
The next material is iron. It has a resistivity much greater than copper and aluminum (6 times more than copper and 4 times more than aluminum). Therefore, as a rule, it is not used in the production of wires. But it is used in the manufacture of shields and tires, which, due to their large cross-section, have low resistance. Just like a fastener.

Gold is not used in electrical applications because it is quite expensive. Due to its low resistivity and high oxidation protection, it is used in space technology.

Brass is not used in electrical applications.

Tin and lead are commonly used in alloying as solder. They are not used as conductors for the manufacture of any devices.

Silver is most often used in military equipment high frequency devices. Rarely used in electrical applications.

Tungsten is used in incandescent lamps. Due to the fact that it does not collapse at high temperatures, it is used as filaments for lamps.

It is used in heating devices, as it has a high resistivity with a large cross-section. A small amount of its length is needed to make a heating element.

Coal and graphite are used in electric brushes in electric motors.
Conductors are used to pass current through themselves. In this case, the current does useful work.

Dielectrics

Dielectrics have great importance specific resistance, which is much higher in comparison with conductors.

Porcelain is used, as a rule, in the manufacture of insulators. Glass is also used to produce insulators.

Ebonite is most often used in transformers. It is used to make the frame of the coils on which the wire is wound.

Also often used as dielectrics different types plastics Dielectrics include the material from which the insulating tape is made.

The material from which the insulation in the wires is made is also a dielectric.

The main purpose of a dielectric is to protect people from electric shock and to insulate current-carrying conductors among themselves.

It happens that when assembling a particular device, you need to decide on the choice of power source. This is extremely important when devices need powerful block nutrition. Today it is not difficult to purchase iron transformers with the necessary characteristics. But they are quite expensive, and their large size and weight are their main disadvantages. And assembling and setting up good switching power supplies is a very complicated procedure. And many people don’t take it up.

Next, you will learn how to assemble a powerful and yet simple power supply, using an electronic transformer as the basis for the design. By and large, the conversation will be about increasing the power of such transformers.

A 50-watt transformer was taken for the conversion.

It was planned to increase its power to 300 W. This transformer was purchased at a nearby store and cost about 100 rubles.

A standard transformer circuit looks like this:

The transformer is a conventional push-pull half-bridge self-generating inverter. The symmetrical dinistor is the main component that triggers the circuit, since it supplies the initial impulse.

The circuit uses 2 high-voltage transistors with reverse conductivity.

The transformer circuit before modification contains the following components:

1. Transistors MJE13003.
2. Capacitors 0.1 µF, 400 V.
3. A transformer with 3 windings, two of which are master windings and have 3 turns of wire with a cross-section of 0.5 square meters. mm. One more as current feedback.
4. The input resistor (1 ohm) is used as a fuse.
5. Diode bridge.

Despite the lack of short-circuit protection in this option, the electronic transformer operates without failure. The purpose of the device is to work with a passive load (for example, office halogen lights), so there is no output voltage stabilization.

As for the main power transformer, its secondary winding produces about 12 V.

Now take a look at the transformer circuit with increased power:

There are even fewer components in it. A feedback transformer, resistor, dynistor and capacitor were taken from the original circuit.

The remaining parts were taken from old computer power supplies, and these are 2 transistors, a diode bridge and a power transformer. Capacitors were purchased separately.

It wouldn’t hurt to replace the transistors with more powerful ones (MJE13009 in a TO220 package).

The diodes were replaced with a ready-made assembly (4 A, 600 V).

Diode bridges from 3 A, 400 V are also suitable. The capacitance should be 2.2 μF, but 1.5 μF is also possible.

The power transformer was removed from the 450 W ATX format power supply. All standard windings were removed from it and new ones were wound. The primary winding has been wound triple wire 0.5 sq. mm in 3 layers. The total number of turns is 55. It is necessary to monitor the accuracy of the winding, as well as its density. Each layer was insulated with blue electrical tape. The calculation of the transformer was carried out experimentally, and a golden mean was found.

The secondary winding is wound at the rate of 1 turn - 2 V, but this is only if the core is the same as in the example.

When you first turn it on, be sure to use a 40-60 W incandescent safety lamp.

It is worth noting that at the moment of startup the lamp will not flash, since there are no smoothing electrolytes after the rectifier. The output frequency is high, so in order to make specific measurements, you must first rectify the voltage. For these purposes, a powerful dual diode bridge assembled from KD2997 diodes was used. The bridge can withstand currents of up to 30 A if a radiator is attached to it.

The secondary winding was supposed to be 15 V, although in reality it turned out to be a little more.

Everything that was at hand was taken as a load. This powerful lamp from a 400 W film projector at a voltage of 30 V and 5 20-watt 12 V lamps. All loads were connected in parallel.

Biometric lock - LCD display diagram and assembly

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Probably, the problem we’ll talk about today is familiar to many. I think everyone has had the need to increase the output current of the power supply. Let's look at a specific example, you have a 19-volt power adapter from a laptop, which provides an output current of, well, let's say, around 5A, and you need a 12-volt power supply with a current of 8-10A. So the author (YouTube channel “AKA KASYAN”) once needed a power supply with a voltage of 5V and a current of 20A, and had a 12-volt power supply at hand for LED strips with an output current of 10A. And so the author decided to remake it.

Yes, assemble the required power source from scratch or use the 5-volt bus of any cheap computer unit power supply is of course possible, but many homemade electronics engineers will find it useful to know how to increase the output current (or in common parlance amperage) of almost any pulse block nutrition.

As a rule, power supplies for laptops, printers, all kinds of monitor power adapters, and so on, are made according to single-ended circuits; most often they are flyback and the construction is no different from each other. There may be a different configuration, a different PWM controller, but the circuit diagram is the same.

A single-cycle PWM controller is most often from the UC38 family, a high-voltage field-effect transistor that pumps a transformer, and at the output a half-wave rectifier in the form of a single or dual Schottky diode.

After that there is a choke, storage capacitors, and a voltage feedback system.

Thanks to feedback, the output voltage is stabilized and strictly kept within the specified limit. Feedback usually built on the basis of an optocoupler and a tl431 reference voltage source.

Changing the resistance of the divider resistors in its wiring leads to a change in the output voltage.

This was a general introduction, and now about what we have to do. It should be noted right away that we are not increasing the power. This power supply has an output power of about 120W.

We are going to reduce the output voltage to 5V, but in return we will increase the output current by 2 times. We multiply the voltage (5V) by the current (20A) and as a result we get design power about 100W. We will not touch the input (high-voltage) part of the power supply. All alterations will affect only the output part and the transformer itself.

But later, after checking, it turned out that the original capacitors are also quite good and have a fairly low internal resistance. Therefore, in the end the author soldered them back.

Next, we unsolder the throttle, and then pulse transformer.

The diode rectifier is quite good - 20 ampere. The best thing is that the board has a seat for a second diode of the same type.

As a result, the author did not find a second such diode, but since he recently received exactly the same diodes from China only in a slightly different package, he plugged a couple of them into the board, added a jumper and strengthened the tracks.

As a result, we get a 40A rectifier, that is, with a double current reserve. The author installed diodes at 200V, but this makes no sense, he just has a lot of them.

You can install regular Schottky diode assemblies from a computer power supply with a reverse voltage of 30-45V or less.
We're done with the rectifier, let's move on. The choke is wound with this wire.

We throw it away and take this wire.

We wind about 5 turns. You can use a native ferrite rod, but the author had a thicker one lying around nearby, on which the turns were wound. True, the rod turned out to be slightly long, but later we will break off all the excess.

The transformer is the most important and responsible part. Remove the tape, heat the core with a soldering iron on all sides for 15-20 minutes to loosen the glue, and carefully remove the core halves.

Leave the whole thing for ten minutes to cool. Next, remove the yellow tape and unwind the first winding, remembering the direction of winding (or just take a couple of photos before disassembling, in which case they will help you). Leave the other end of the wire on the pin. Next, unwind the second winding. Also, we do not solder the second end.

After this, we have before us the secondary (or power) winding of our own person, which is exactly what we were looking for. This winding is completely removed.

It consists of 4 turns, wound with a bundle of 8 wires, each with a diameter of 0.55 mm.

The new secondary winding we will wind contains only one and a half turns, since we only need 5V of output voltage. We will wind it in the same way, we will take a wire with a diameter of 0.35 mm, but the number of cores is already 40 pieces.

This is much more than is needed, but, however, you can compare it yourself with the factory winding. Now we wind all the windings in the same order. Be sure to follow the winding direction of all windings, otherwise nothing will work.

It is advisable to tin the cores of the secondary winding before winding begins. For convenience, we divide each end of the winding into 2 groups so as not to drill giant holes on the board for installation.

After the transformer is installed, we find the tl431 chip. As mentioned earlier, it is this that sets the output voltage.

We find a divider in its harness. In this case, 1 of the resistors of this divider is a pair of smd resistors connected in series.

The second divider resistor is located closer to the output. In this case, its resistance is 20 kOhm.

We unsolder this resistor and replace it with a 10 kOhm trimmer.

We connect the power supply to the network (necessarily through a safety incandescent network lamp with a power of 40-60W). We connect a multimeter and preferably a small load to the output of the power supply. In this case, these are low-power 28V incandescent lamps. Then, very carefully, without touching the board, we rotate the trimming resistor until the desired output voltage is obtained.

Next, we turn everything off and wait 5 minutes so that the high-voltage capacitor on the unit is completely discharged. Then we unsolder the trimming resistor and measure its resistance. Then we replace it with a permanent one, or leave it. In this case, we will also have the ability to adjust the output.

Occasionally need to increase force happening in an electrical circuit current. This article will discuss the basic methods of increasing current without the use of difficult devices.

You will need

• Ammeter

Instructions

1. According to Ohm's law for continuous current electrical circuits: U = IR, where: U is the magnitude of the voltage supplied to the electrical circuit, R is the total resistance of the electrical circuit, I is the magnitude of the current occurring in the electrical circuit, to determine the current strength, it is necessary to divide the voltage supplied to circuit to its total resistance. I=U/RAccordingly, in order to increase the current, it is possible to increase the voltage supplied to the input of the electrical circuit or reduce its resistance. The current will increase if the voltage is increased. The increase in current will be proportional to the increase in voltage. Let's say, if a circuit with a resistance of 10 Ohms was connected to a standard battery with a voltage of 1.5 Volts, then the current flowing through it was: 1.5/10 = 0.15 A (Ampere). When another 1.5 V battery is connected to this circuit, the total voltage will become 3 V, and the current flowing through the electrical circuit will increase to 0.3 A. The connection is made in stages, that is, the plus of one battery is connected to the minus of the other. Thus, by combining a sufficient number of power sources in steps, it is possible to obtain the required voltage and ensure the flow of current of the required strength. Several voltage sources combined into one circuit are called a battery of elements. In everyday life, such designs are usually called “batteries” (even if the power source consists of each of one element). However, in practice, the increase in current strength may differ slightly from the calculated one (proportional to the increase in voltage). This is mainly due to the additional heating of the circuit conductors, which occurs with an increase in the current passing through them. In this case, as usual, the resistance of the circuit increases, which leads to a decrease in current strength. In addition, an increase in the load on the electrical circuit can lead to its burnout or even fire. You must be extremely careful when operating electrical appliances, which can only work at a fixed voltage.

2. If you reduce the total resistance of an electrical circuit, the current will also increase. According to Ohm's law, the increase in current will be proportional to the decrease in resistance. Say, if the voltage of the power source was 1.5 V, and the circuit resistance was 10 Ohms, then an electric current of 0.15 A passed through such a circuit. If after this the circuit resistance is halved (made equal to 5 Ohms), then the resulting along the circuit, the current will double and amount to 0.3 Amperes. An extreme case of decreasing load resistance is a short circuit, in which the load resistance is actually zero. In this case, of course, an immense current does not appear, because there is an internal resistance of the power source in the circuit. A more significant reduction in resistance can be achieved if the conductor is cooled tightly. The acquisition of high currents is based on this result of superconductivity.

3. To increase the strength of alternating current, all kinds of electronic devices are used, mainly current transformers, used, say, in welding units. The strength of the alternating current also increases as the frequency decreases (because the net result is that the energetic resistance of the circuit decreases). If there are energetic resistances in the alternating current circuit, the current will increase as the capacitance of the capacitors increases and the inductance of the coils (solenoids) decreases. If the circuit contains only capacitors (capacitors), the current will increase as the frequency increases. If the circuit consists of inductors, then the current strength will increase as the frequency of the current decreases.

According to Ohm's law, increasing current in a circuit is permissible if one of two conditions is fulfilled: an increase in voltage in the circuit or a decrease in its resistance. In the first case, change the source current on another, with greater electromotive force; in the second, select conductors with lower resistance.

You will need

• a regular tester and tables for determining the resistivity of substances.

Instructions

1. According to Ohm's law, on a section of the chain the force current depends on 2 quantities. It is directly proportional to the voltage in this area and inversely proportional to its resistance. Universal connectedness is described by an equation that can be easily derived from Ohm’s law I=U*S/(?*l).

2. Assemble an electrical circuit that contains a source current, wires and electricity buyer. As a source current use a rectifier with the possibility of adjusting the EMF. Connect the circuit to such a source, having previously installed a tester into it in stages for the buyer, configured to measure force current. Increasing the emf of the source current, take readings from the tester, from which it can be concluded that as the voltage on a section of the circuit increases, the force current it will increase proportionally.

3. 2nd method to increase strength current– reduction of resistance in a section of the circuit. To do this, use a special table to determine the resistivity of this section. To do this, find out in advance what material the conductors are made of. In order to increase force current, install conductors with lower resistivity. The smaller this value, the greater the force. current in this area.

4. If there are no other conductors, resize the ones that are available. Increase their cross-sectional areas and install the same conductors parallel to them. If current flows through one core of the wire, install several wires in parallel. By how many times the cross-sectional area of ​​the wire increases, the current will increase by how many times. If possible, shorten the wires used. By how many times the length of the conductors decreases, by how many times the force increases current .

5. Methods for increasing strength current allowed to combine. Say, if you increase the cross-sectional area by 2 times, reduce the length of the conductors by 1.5 times, and the emf of the source current increase by 3 times, get an increase in strength current you 9 times.

Tracking shows that if a current-carrying conductor is placed in a magnetic field, it will begin to move. This means that some force is acting on it. This is the Ampere force. Because its appearance requires the presence of a conductor, magnetic field And electric current, the metamorphosis of the parameters of these quantities will allow the Ampere force to increase.

You will need

• - conductor;
• – current source;
• – magnet (continuous or electro).

Instructions

1. A conductor carrying current in a magnetic field is acted upon by a force equal to the product of the magnetic induction of the magnetic field B, the strength of the current flowing through the conductor I, its length l and the sine of the angle? between the magnetic field induction vector and the direction of the current in the conductor F=B?I?l?sin(?).

2. If the angle between the magnetic induction lines and the direction of the current in the conductor is acute or obtuse, orient the conductor or field in such a way that this angle becomes right, that is, there should be a right angle of 90? between the magnetic induction vector and the current. Then sin(?)=1, and this is the highest value for this function.

3. Enlarge force Ampere, acting on the conductor, increasing the value of the magnetic induction of the field in which it is placed. To do this, take a stronger magnet. Use an electromagnet, one that allows you to get a magnetic field of different intensities. Increase the current in its winding, and the inductance of the magnetic field will begin to increase. Force Ampere will increase in proportion to the magnetic induction of the magnetic field, say, increasing it 2 times, you will also get an increase in strength by 2 times.

4. Force Ampere depends on the current strength in the conductor. Connect the conductor to a current source with variable emf. Enlarge force current in the conductor by increasing the voltage at the current source, or replace the conductor with another one, with the same geometric dimensions, but with lower resistivity. Let's say replace an aluminum conductor with a copper one. Moreover, it must have the same cross-sectional area and length. Increased Strength Ampere will be directly proportional to the increase in current strength in the conductor.

5. To increase the force value Ampere increase the length of the conductor, the one in the magnetic field. At the same time, strictly consider that the current strength will decrease proportionally; therefore, a primitive lengthening will not give results; at the same time, bring the value of the current strength in the conductor to the initial value, increasing the voltage at the source.

Video on the topic

Video on the topic

Instructions

According to Ohm's law for direct current electrical circuits: U = IR, where: U is the value supplied to the electrical circuit,
R is the total resistance of the electrical circuit,
I is the amount of current flowing through an electrical circuit; to determine the current strength, you need to divide the voltage supplied to the circuit by its total resistance. I=U/RAccordingly, in order to increase the current, you can increase the voltage supplied to the input of the electrical circuit or reduce its resistance. The current will increase if you increase the voltage. An increase in current will result in an increase in voltage. For example, if a circuit with a resistance of 10 Ohms was connected to a standard 1.5 Volt battery, then the current flowing through it was:
1.5/10=0.15 A (Ampere). When another 1.5 V battery is connected to this circuit, the total voltage will become 3 V, and the current flowing through the electrical circuit will increase to 0.3 A.
The connection is made “in series,” that is, the plus of one battery is connected to the minus of the other. Thus, by connecting a sufficient number of power sources in series, you can obtain the required voltage and ensure the flow of current of the required strength. Several voltage sources are combined into one circuit by a battery of cells. In everyday life, such designs are usually called “batteries” (even if the power supply consists of only one element). However, in practice, the increase in current strength may differ slightly from the calculated one (proportional to the increase in voltage). This is mainly due to the additional heating of the circuit conductors, which occurs with an increase in the current passing through them. In this case, as a rule, there is an increase in the resistance of the circuit, which leads to a decrease in current strength. In addition, an increase in the load on the electrical circuit can lead to its burnout or even fire. You need to be especially careful when using electrical appliances that can only operate at a fixed voltage.

If you reduce the total resistance of an electrical circuit, the current will also increase. According to Ohm's law, the increase in current will be proportional to the decrease in resistance. For example, if the voltage of the power source was 1.5 V, and the circuit resistance was 10 Ohms, then an electric current of 0.15 A passed through such a circuit. If then the circuit resistance is halved (made equal to 5 Ohms), then the current flowing through the circuit current will double and amount to 0.3 Amperes. An extreme case of a decrease in load resistance is a short circuit, in which the load resistance is practically zero. In this case, of course, infinite current does not arise, since the circuit has internal resistance of the power source. A more significant reduction in resistance can be achieved by greatly cooling the conductor. The production of enormous currents is based on this effect of superconductivity.

To increase the power of alternating current, all kinds of electronic devices are used, mainly current transformers, used, for example, in welding machines. The strength of the alternating current also increases as the frequency decreases (since, due to the surface effect, the active resistance of the circuit decreases). If there are active resistances in the alternating current circuit, the current strength will increase as the capacitance of the capacitors increases and the inductance of the coils (solenoids) decreases. If the circuit contains only capacitors (capacitors), the current will increase as the frequency increases. If the circuit consists of inductors, then the current strength will increase as the frequency of the current decreases.