How to raise the current in the power supply. How to increase the strength of the electric current. conductor resistance. Resistivity. Coal, graphite are used in electric brushes in electric motors. Conductors are used to pass forces through them.

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Probably, the problem we will talk about today is familiar to many. I think everyone had a need to increase the output current of the power supply. Let's look at a specific example, you have a 19 volt laptop power adapter that provides an output current, 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 at hand there was a 12-volt power supply for LED strips with an output current of 10A. And so the author decided to remake it.

Yes, it is certainly possible to assemble the necessary power supply from scratch or use the 5-volt bus of any cheap computer power supply, but it will be useful for many home-made electronics engineers to know how to increase the output current (or amperage in common people) of almost any switching power supply.

As a rule, power supplies for laptops, printers, all kinds of monitor power adapters, and so on, are made according to single-cycle circuits, most often they are flyback and the construction is no different from each other. There may be a different equipment, a different PWM controller, but the schematics are 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 it, a choke, storage capacitors, and, well, a voltage feedback system.

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

A change in the resistance of the resistors of the divider in its binding 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 do not increase the power. This power supply has an output power of about 120W.

We are going to reduce the output voltage to 5V, but instead double the output current. We multiply the voltage (5V) by the current (20A) and as a result we get the estimated power of 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 native capacitors are also not bad and have a rather low internal resistance. Therefore, in the end, the author soldered them back.

Next, we solder the inductor, well, and a pulse transformer.

The diode rectifier is pretty good - 20 amps. The best thing is that the board has a seat for a second diode of the same kind.

As a result, the author did not find a second such diode, but since recently exactly the same diodes came to him from China only in a slightly different case, he stuck a couple of pieces into the board, added a jumper and strengthened the tracks.

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

You can also supply ordinary Schottky diode assemblies from a computer power supply with a reverse voltage of 30-45V or less.
With the rectifier finished, let's move on. The throttle is wound with such a wire.

We throw it out and take this wire.

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

The transformer is the most important and critical part. We remove the adhesive tape, heat the core with a soldering iron from all sides for 15-20 minutes to loosen the glue and carefully remove the halves of the core.

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

After that, we have a 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 0.55 mm in diameter.

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

This is much more than necessary, well, however, you yourself can compare 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 even before the start of winding. For convenience, we divide each end of the winding into 2 groups so that we do not drill giant holes for installation on the board.

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

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

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

Solder this resistor and replace it with a 10 kOhm trimmer.

We connect the power supply to the network (necessarily through a safety network incandescent lamp with a power of 40-60W). We connect a multimeter to the output of the power supply and preferably not a large load. In this case, these are low-power 28V incandescent lamps. Then, very carefully, without touching the board, we rotate the tuning resistor until the desired output voltage is obtained.

Then we cut everything down, wait 5 minutes, so that the high-voltage capacitor on the block is completely discharged. Then we solder the tuning resistor and measure its resistance. Then we replace it with a constant, or leave it. In this case, we will also have the opportunity to adjust the output.

Instruction

According to Ohm's law for DC electrical circuits: U \u003d IR, where: U is the value supplied to the electrical circuit,
R is the total resistance of the electrical circuit,
I - the value of the current flowing through the electrical circuit, to determine the current strength, it is necessary to divide the voltage supplied to the circuit by its impedance. I \u003d U / R Accordingly, 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 increase the 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 \u003d 0.15 A (Amp). 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 carried out “in series, that is, the plus of one battery is connected to the minus of the other. Thus, by connecting in series a sufficient number of power sources, 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 by a battery of elements. In everyday life, such designs are usually called “batteries (even if the power is from 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. Particular care must be taken when operating household appliances that can only operate at a fixed voltage.

If you reduce the impedance of the electrical circuit, then the current will also increase. According to Ohm's law, an increase in current will be proportional to a 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 (make it equal to 5 ohms), then the current flowing through the circuit current will double and amount to 0.3 Ampere. An extreme case of a decrease in load resistance is a short circuit, in which the load resistance is almost zero. In this case, of course, there is no infinite current, since 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 strongly cooled. This effect of superconductivity is based on obtaining currents of enormous strength.

To increase the strength of the alternating current, various electronic devices are used, mainly current transformers, used, for example, in welding machines. The strength of the alternating current also increases with decreasing frequency (since the active resistance of the circuit decreases due to the skin effect). If there are active resistances in the alternating current circuit, then the current strength will increase with an increase in the capacitance of the capacitors and a decrease in the inductance of the coils (solenoids). If there are only capacitances (capacitors) in the circuit, then the current strength will increase with increasing frequency. If the circuit consists of inductors, then the current strength will increase with decreasing current frequency.

The article will discuss how to increase the current in the charger circuit, in the power supply, transformer, generator, in the USB ports of the computer without changing the voltage.

What is current strength?

Electric current is an ordered movement of charged particles inside a conductor with the obligatory presence of a closed circuit.

The appearance of current is due to the movement of electrons and free ions with a positive charge.

In the process of moving, charged particles can heat the conductor and have a chemical effect on its composition. In addition, the current can affect neighboring currents and magnetized bodies.

Current strength is an electrical parameter that is a scalar quantity. Formula:

I=q/t where I is current, t is time and q is charge.

Ohm's law is also worth knowing, according to which current is directly proportional to U (voltage) and inversely proportional to R (resistance).

There are two types of current - positive and negative.

Below we consider what this parameter depends on, how to increase the current strength in the circuit, in the generator, in the power supply and in the transformer.

What does the strength of the current depend on?

To increase I in a circuit, it is important to understand what factors can affect this parameter. Here you can highlight the dependence on:

resistance. The smaller the parameter R (Ohm), the higher the current strength in the circuit.
Voltages. According to the same Ohm's law, we can conclude that as U increases, the current strength also increases.
Magnetic field strength. The larger it is, the higher the voltage.
The number of turns of the coil. The larger this indicator, the larger U and, accordingly, the higher I.
The power of the force that is transmitted to the rotor.
Conductor diameters. The smaller it is, the higher the risk of heating and burnout of the supply wire.
Power supply designs.
The diameter of the wires of the stator and armature, the number of ampere-turns.
Generator parameters - operating current, voltage, frequency and speed.

How to increase the current in the circuit?

There are situations when it is necessary to increase the I that flows in the circuit, but it is important to understand that you need to take action on, this can be done using special devices.

Consider how to increase the current strength using simple devices.

You will need an ammeter to do the job.

Option 1.

According to Ohm's law, current is equal to voltage (U) divided by resistance (R). The simplest way to increase the force I, which suggests itself, is to increase the voltage that is supplied to the input of the circuit, or to reduce the resistance. In this case, I will increase in direct proportion to U.

For example, when connecting a 20 ohm circuit to a power source with U = 3 Volts, the current will be 0.15 A.

If you add another 3V power supply to the circuit, the total value of U can be increased to 6 Volts. Accordingly, the current will also double and reach a limit of 0.3 Amperes.

The power sources must be connected in series, that is, the plus of one element is connected to the minus of the first.

To obtain the required voltage, it is enough to connect several power supplies into one group.

In everyday life, constant U sources combined into one group are called batteries.

Despite the obviousness of the formula, practical results may differ from theoretical calculations, which is associated with additional factors - the heating of the conductor, its cross section, the material used, and so on.

As a result, R changes in the direction of increase, which leads to a decrease in the force I.

Increasing the load in the electrical circuit can cause overheating of the conductors, burnout or even a fire.

That is why it is important to be careful when operating devices and take into account their power when choosing a section.

The value of I can be increased in another way by reducing the resistance. For example, if the input voltage is 3 volts, and R is 30 ohms, then a current equal to 0.1 amperes passes through the circuit.

If you reduce the resistance to 15 ohms, the current strength, on the contrary, will double and reach 0.2 amperes. The load decreases almost to zero during a short circuit near the power source, in this case I increase to the maximum possible value (taking into account the power of the product).

You can further reduce the resistance by cooling the wire. Such an effect of superconductivity has long been known and is actively used in practice.

To increase the current strength in the circuit, electronic devices are often used, for example, current transformers (as in welders). The strength of the variable I in this case increases with decreasing frequency.

If there is active resistance in the AC circuit, I increases with an increase in the capacitance of the capacitor and a decrease in the inductance of the coil.

In a situation where the load is purely capacitive, the current increases with increasing frequency. If the circuit includes inductors, the force I will increase simultaneously with the decrease in frequency.

Option 2.

To increase the current strength, you can focus on another formula, which looks like this:

I = U*S/(ρ*l). Here we know only three parameters:

S - wire section;
l - its length;
ρ is the specific electrical resistance of the conductor.

To increase the current, assemble a chain in which there will be a current source, a consumer and wires.

The role of the current source will be performed by a rectifier, which allows you to regulate the EMF.

Connect the circuit to the source, and the tester to the consumer (pre-set the device to measure the current strength). Increase the EMF and control the performance on the device.

As noted above, as U increases, the current can also be increased. A similar experiment can be done for resistance.

To do this, find out what material the wires are made of and install products that have a lower resistivity. If you cannot find other conductors, shorten those that are already installed.

Another way is to increase the cross section, for which parallel to the installed wires it is worth mounting similar conductors. In this case, the cross-sectional area of the wire increases and the current increases.

If we shorten the conductors, the parameter (I) of interest to us will increase. If desired, options for increasing the current strength can be combined. For example, if the conductors in the circuit are shortened by 50%, and U is raised by 300%, then the force I will increase by 9 times.

How to increase the current in the power supply?

On the Internet, you can often find the question of how to increase I in the power supply without changing the voltage. Consider the main options.

Situation #1.

The 12 volt power supply operates with a current of 0.5 amperes. How to raise I to the limit value? To do this, a transistor is placed in parallel with the PSU. In addition, a resistor and a stabilizer are installed at the input.

When the voltage across the resistance drops to the desired value, the transistor opens, and the rest of the current flows not through the stabilizer, but through the transistor.

The latter, by the way, must be selected according to the rated current and a radiator should be installed.

In addition, the following options are available:

Increase the power of all elements of the device. Install a stabilizer, a diode bridge and a higher power transformer.
If there is current protection, reduce the value of the resistor in the control circuit.

Situation #2.

There is a power supply for U \u003d 220-240 Volts (at the input), and at the output a constant U \u003d 12 Volts and I \u003d 5 Amperes. The task is to increase the current to 10 amperes. At the same time, the PSU should remain approximately the same size and not overheat.

Here, to increase the output power, it is necessary to use another transformer, which is recalculated for 12 Volts and 10 Amps. Otherwise, the product will have to be rewound on its own.

In the absence of the necessary experience, it is better not to take risks, because there is a high probability of a short circuit or burnout of expensive circuit elements.

The transformer will have to be changed to a larger product, as well as recalculate the damper chain located on the DRAIN of the key.

The next point is the replacement of the electrolytic capacitor, because when choosing a capacity, you need to focus on the power of the device. So, for 1 W of power, there are 1-2 microfarads.

After such an alteration, the device will heat up more strongly, so you cannot do without installing a fan.

How to increase the current in the charger?

In the process of using chargers, you may notice that the chargers for a tablet, phone or laptop have a number of differences. In addition, the speed at which the device is charged may also vary.

Here a lot depends on whether the original or non-original device is used.

To measure the current that comes to the tablet or phone from the charger, you can use not only the ammeter, but also the Ampere application.

With the help of software, it is possible to find out the rate of charge and discharge of the battery, as well as its condition. The application is free to use. The only downside is the ads (the paid version doesn't have any).

The main problem with charging batteries is the low current of the charger, which makes the capacity build-up time too long. In practice, the current flowing in the circuit directly depends on the power of the charger, as well as other parameters - the length of the cable, its thickness and resistance.

With the help of the Ampere app, you can see at what current the device is charging, and also check if the product can be charged at a faster speed.

To use the capabilities of the application, just download it, install and run it.

After that, the phone, tablet or other device is connected to the charger. That's all - it remains to pay attention to the parameters of current and voltage.

In addition, information about the type of battery, U level, battery status, and temperature conditions will be available to you. You can also see the maximum and minimum I occurring during the period of the cycle.

If you have several memory devices at your disposal, you can run the program and try to charge each of them. Based on the test results, it is easier to make a choice of a memory that provides maximum current. The higher this parameter is, the faster the device will charge.

Measuring current is not the only thing the Ampere app can do. With it, you can check how much I is consumed in standby mode or when you turn on various games (applications).

For example, after turning off the brightness of the display, deactivating GPS or transferring data, it is easy to notice a decrease in load. Against this background, it is easier to conclude which options drain the battery to a greater extent.

What else is worth noting? All manufacturers recommend charging devices with "native" chargers that deliver a certain current.

But during operation, there are situations when you have to charge your phone or tablet with other chargers that have more power. As a result, the charging speed may be higher. But not always.

Few people know, but some manufacturers limit the current limit that the battery of the device can accept.

For example, the Samsung Galaxy Alpha device comes with a 1.35 Amp charger.

When a 2-amp charger is connected, nothing changes - the charging speed remains the same. This is due to the limitation that is set by the manufacturer. A similar test was made with a number of other phones, which only confirmed the guess.

Given the above, we can conclude that "non-native" memory is unlikely to harm the battery, but can sometimes help in faster charging.

Let's consider one more situation. When charging the device via a USB connector, the battery gains capacity more slowly than if you charge the device from a conventional charger.

This is due to the limitation of the current strength that the USB port is capable of delivering (no more than 0.5 Amperes for USB 2.0). In the case of using USB3.0, the current strength increases to the level of 0.9 Amperes.

In addition, there is a special utility that allows the “troika” to pass a larger I through itself.

For Apple devices, the program is called ASUS Ai Charger, and for other devices, ASUS USB Charger Plus.

How to increase the current in a transformer?

Another question that worries electronics lovers is how to increase the current strength in relation to a transformer.

Here are the following options:

Install a second transformer;
Increase the diameter of the conductor. The main thing is to allow the section of the "iron".
Raise U;
Increase the cross section of the core;
If the transformer works through a rectifier, it is worth using a product with a voltage multiplier. In this case, U increases, and with it, the load current also increases;
Buy a new transformer with suitable current;
Replace the core with a ferromagnetic version of the product (if possible).

A transformer has a pair of windings (primary and secondary). Many output parameters depend on the wire cross section and the number of turns. For example, on the high side there are X turns, and on the other side there are 2X.

This means that the voltage on the secondary winding will be lower, as well as the power. The output parameter also depends on the efficiency of the transformer. If it is less than 100%, U and the current in the secondary circuit decrease.

Taking into account the above, the following conclusions can be drawn:

The power of the transformer depends on the width of the permanent magnet.
To increase the current in the transformer, a decrease in R load is required.
The current (A) depends on the diameter of the winding and the power of the device.
In case of rewinding, it is recommended to use thicker wire. In this case, the ratio of the wire by weight on the primary and secondary windings is approximately identical. If 0.2 kg of iron is wound on the primary winding, and 0.5 kg on the secondary, the primary will burn out.

How to increase the current in the generator?

The current in the generator directly depends on the load resistance parameter. The lower this setting, the higher the current.

If I is higher than the nominal parameter, this indicates the presence of an emergency mode - a decrease in frequency, overheating of the generator and other problems.

For such cases, protection or disconnection of the device (part of the load) must be provided.

In addition, with increased resistance, the voltage decreases, U is added at the output of the generator.

To keep the parameter at an optimal level, the excitation current is regulated. In this case, an increase in the excitation current leads to an increase in the generator voltage.

The mains frequency must be at the same level (be a constant value).

Consider an example. In a car alternator, it is necessary to increase the current from 80 to 90 amperes.

To solve this problem, it is required to disassemble the generator, separate the winding and solder the output to it, followed by connecting the diode bridge.

In addition, the diode bridge itself is changed to a part of higher performance.

After that, it is required to remove the winding and a piece of insulation in the place where the wire should be soldered.

If there is a faulty generator, the output is bitten off from it, after which legs of the same thickness are built up with the help of copper wire.

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, let him sit at home." According to this law, current is directly proportional to voltage and inversely proportional to resistance (I = U / R), where R is the coefficient that relates voltage and current. The unit of measurement for voltage is Volt, resistance is Ohm, current strength is Ampere.
To show how Ohm's Law works, let's look at a simple electrical circuit. The circuit is a resistor, it is also a load. A voltmeter is used to measure voltage. For load current - ammeter. When the switch is closed, current flows through the load. We look at how Ohm's Law is respected. The current in the circuit is equal to: the voltage of the circuit is 2 Volts and the resistance of the circuit is 2 ohms (I \u003d 2 V / 2 Ohms \u003d 1 A). The ammeter shows that much. The resistor is a load, 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. With the help of a voltmeter - 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 raise the current in the circuit. First, increase the voltage. Let's make a battery not 2 V, but 12 V. The voltmeter will show 12 V. What will the ammeter show? 12 V / 2 ohms \u003d 6 A. That is, by increasing the voltage at the load by 6 times, we got an increase in current strength by 6 times.

Consider another way to raise the current in the circuit. You can reduce the resistance - instead of a load of 2 ohms, 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 Ohm's triangle:
How can the current be determined from this triangle? I = U / R. Everything looks quite clear. Using a triangle, you can also write formulas derived from Ohm's Law: R \u003d U / I; U = I * R. The main thing to remember is that the voltage is at the top 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 like a pipe in which a 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 more significant with an increase in its length and less with an increase in diameter. Based on this, Georg Om derived the formula: R \u003d p * l / S, where p is some 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 large value of this parameter. Specifically, for copper, the resistivity is 0.017 (Ω*mm2/m).

conductors

Consider what conductors are. Today, the most common conductor is made of copper. Due to the low resistivity and high resistance to oxidation, with a rather low brittleness, this conductor is increasingly being used in electrical applications. Gradually, the copper conductor displaces the aluminum. Copper is used in the production of wires (cores in cables) and in the manufacture of electrical products.

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

Gold is not used in electrics, as it is quite expensive. Due to its low resistivity and high protection against oxidation, it is used in space technology.

Brass is not used in electrical.

Tin and lead are commonly used in the alloy as solder. As conductors, they are not used for the manufacture of any devices.

Silver is most often used in military technology for high-frequency devices. Rarely used in electrical applications.

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

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

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

Dielectrics

Dielectrics have a high resistivity value, which is much higher in comparison with conductors.

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

Ebonite is most often used in transformers. A frame of coils is made from it, on which the wire is wound.

Also, various types of plastics are often used as dielectrics. Dielectrics refers to 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, to isolate conductive wires from each other.

Progress does not stand still. Computer performance is growing rapidly. And as performance increases, so does power consumption. If earlier almost no attention was paid to the power supply, now, after nVidia's statement about the recommended power supply for their top-end solutions of 480 W, everything has changed a bit. Yes, and processors consume more and more, and if all this is still properly overclocked ...

With the annual upgrade of the processor, motherboard, memory, video, I have long resigned myself, as with the inevitable. But for some reason the upgrade of the power supply makes me really nervous. If the hardware progresses dramatically, then there are practically no such fundamental changes in the circuitry of the power supply. Well, the trance is bigger, the wires on the chokes are thicker, the diode assemblies are more powerful, the capacitors ... Is it really impossible to buy a more powerful power supply, so to speak, for growth, and live at least a couple of years in peace. Without thinking about such a relatively simple thing as high-quality power supply.

It would seem that it would be easier, buy the largest power supply you can find, and enjoy a quiet life. But it was not there. For some reason, all employees of computer companies are sure that a 250-watt power supply will be enough for you in excess. And, what infuriates most of all, they begin to lecture categorically and groundlessly prove their case. Then you reasonably notice that you know what you want and are ready to pay for it, and you need to quickly get what they ask for and earn a legitimate profit, and not anger a stranger with your senseless, unsupported persuasion. But this is only the first hurdle. Move on.

Let's say you still found a powerful power supply, and here you see, for example, such an entry in the price list

Power Man PRO HPC 420W - 59 u
Power Man PRO HPC 520W - 123 u

With a difference of 100 watts, the price has doubled. And if you take it with a margin, then you need 650 or more. How much is it? And that is not all!

The vast majority of modern power supplies use the SG6105 chip. And its switching circuit has one very unpleasant feature - it does not stabilize voltages of 5 and 12 volts, and the average value of these two voltages obtained from a resistor divider is applied to its input. And it stabilizes this average value. Because of this feature, a phenomenon such as "voltage distortion" often occurs. Previously used chips TL494, MB3759, KA7500. They have the same feature. I will quote from the article Mr. Korobeinikov .

"... The voltage imbalance occurs due to the uneven distribution of the load on the +12 and +5 Volt buses. For example, the processor is powered from the + 5V bus, and the hard disk and CD drive hang on the +12 bus. The load on + 5V is many times exceeds the load by +12V. 5 volts fails. The microcircuit increases the duty cycle and +5V rises, but +12 increases even more - there is less load. We get a typical voltage imbalance..."

On many modern motherboards, the processor is powered by 12 volts, then the opposite occurs, 12 volts goes down, and 5 goes up.

And if the computer works normally in the nominal mode, then during overclocking the power consumed by the processor increases, the distortion increases, the voltage decreases, the protection of the power supply against undervoltage is activated and the computer turns off. If there is no shutdown, then the lower voltage is still not conducive to good overclocking.

So, for example, it was with me. I even wrote a note on this topic - "Overclocker's lamp" Then two power supplies worked in my system unit - Samsung 250 W, Power Master 350 W. And I naively believed that 600 watts was more than enough. Enough may be enough, but due to skew, all these watts are useless. I unknowingly strengthened this effect by connecting the motherboard from Power Master, and from Samsung a screw, disk drives, etc. That is, it turned out - mainly 5 volts are taken from one power supply, 12 from the other. And the other lines are "in the air", which increased the effect of "skew".

After that, I purchased a 480 watt Euro case power supply. Because of his addiction to silence, he remade it into a fanless one, which he also wrote about on the pages of the site. But even in this block there was SG6105. When testing it, I also encountered the phenomenon of "voltage skew". The newly purchased power supply is not suitable for overclocking!

And that is not all! I kept wanting to buy a second computer, and leave the old one "for experiments", but it was elementary "toad crushed". Recently, I still persuaded this beast and purchased iron for the second computer. This is of course a separate issue, but I bought a power supply for it - PowerMan Pro 420 W. I decided to check it for "skew". And since the new mother feeds the processor via the 12 volt bus, I checked it using it. How? Find out if you read the article to the end. In the meantime, I will say that with a load of 10 amperes, twelve volts failed to 11.55. The standard allows a voltage deviation of plus or minus 5 percent. Five percent of 12 is 0.6 volts. In other words, at a current of 10 amperes, the voltage dropped almost to the maximum allowable mark! And 10 amps corresponds to 120 watts of processor consumption, which is quite real during overclocking. In the passport for this unit, a current of 18 amperes is declared on the 12 volt bus. I think I won’t be able to see these amps, since the power supply will turn off much earlier from the “skew”.

Total - four power supplies in two years. And you have to take the fifth, sixth, seventh? No, enough. Tired of paying for things you don't like. What prevents me from making a kilowatt power supply myself and living quietly for a couple of years, with confidence in the quality and quantity of my pet's food. In addition, I started the manufacture of a new case. I started making a huge case and a power supply, a non-standard size, should fit there without problems. But owners of standard cases may also need such a solution. You can always make an external power supply, especially since there are already precedents. It seems Zalman has released an external power supply.

Of course, making a power supply of such power from scratch is difficult, long, and troublesome. Therefore, the idea arose to assemble one block from two factory ones. Moreover, they already exist and, as it turned out, in their current form are unsuitable for overclocking. This thought prompted me all the same.

"... To introduce separate stabilization, you need a second transformer and a second PWM chip, and this is done in serious and expensive server blocks ..."

In a computer power supply, there are three high-current lines with a voltage of 5, 12 and 3.3 volts. I have two standard power supplies, let one of them produce 5 volts, and the other, more powerful, 12 and all the rest. The voltage of 3.3 volts is stabilized separately and does not cause distortion. Lines producing -5, -12, etc. - low-power and these voltages can be taken from any unit. And to carry out this event, use the principle set forth in the same article by Mr. Korobeinikov - turn off unnecessary voltage from the microcircuit, and adjust the necessary one. That is, now the SG6105 will stabilize only one voltage and, therefore, there will be no "voltage skew" phenomenon.

The mode of operation of each power supply is also facilitated. If you look at the power section, a typical power supply circuit (Fig. 2), you can see that the 12, 5 and 3.3 volt windings are one common winding with taps. And if from such a trance you do not take all three at once, but only one voltage, then the power of the transformer will remain the same, but for one voltage, and not for three.

For example, a block on the lines of 12, 5, 3.3 volts gave out 250 watts, now we will get almost the same 250 watts on the line, for example, 5 volts. If earlier the total power was divided between three lines, now all the power can be obtained on one line. But in practice, for this it is necessary to replace the diode assemblies on the used line with more powerful ones. Or include in parallel additional assemblies taken from another block on which this line will not be used. Also, the maximum current will limit the cross section of the inductor wire. The protection of the power supply against power overload may also work (although this parameter can be adjusted). So we won’t get a completely tripled power, but there will be an increase, and the blocks will heat up much less. You can, of course, rewind the inductor with a larger wire. But more on that later.

Before proceeding to the description of the modification, a few words must be said. It is very difficult to write about the alterations of electronic equipment. Not all readers understand electronics, not everyone reads circuit diagrams. But at the same time, there are readers who deal with electronics professionally. No matter how you write it, it turns out that for some it is incomprehensible, but for others it is annoyingly primitive. However, I will try to write in a way that would be understandable to the vast majority. And experts, I think, will forgive me.

It is also necessary to say that you make all alterations to the equipment at your own peril and risk. Any modification will void your warranty. And of course, the author is not responsible for any consequences. It would not be superfluous to say that a person who undertakes such a modification must be confident in his abilities and have the appropriate tool. This modification is feasible on power supplies assembled on the basis of the SG6105 chip and slightly outdated TL494, MB3759, KA7500.

First, I had to look for a datasheet for the SG6105 chip - it turned out to be not so difficult. I quote from the datasheet the numbering of the legs of the microcircuit and a typical switching circuit.

Figure 1. SG6105

Rice. 2. Typical switching circuit.

Rice. 3. Wiring diagram SG6105

I will first describe the general principle of modernization. First upgrade units on SG6105. We are interested in pins 17(IN) and 16(COMP). A resistor divider R91, R94, R97 and a tuning resistor VR3 are connected to these pins of the microcircuit. On one block, turn off the voltage of 5 volts, for this we solder the resistor R91. Now we adjust the voltage value of 12 volts with a resistor R94 roughly, and with a variable resistor VR3 exactly. On the other block, on the contrary, we turn off 12 volts, for this we solder the resistor R94. And we adjust the voltage value of 5 volts with a resistor R91 roughly, and with a variable resistor VR3 exactly.

The PC-ON wires of all power supplies are connected to each other and soldered to a 20-pin connector, which is then connected to the motherboard. The PG wire is more difficult. I took this signal from a more powerful power supply. In the future, several more complex options can be implemented.

Rice. 4. Connector wiring diagram

Now about the features of upgrading blocks based on the TL494, MB3759, KA7500 microcircuit. In this case, the feedback signal from the output voltage rectifiers of 5 and 12 volts is fed to pin 1 of the microcircuit. We act a little differently - we cut the PCB track near pin 1. In other words, we disconnect pin 1 from the rest of the circuit. And we apply the voltage we need to this output through a resistor divider.

Fig 5. Schematic for TL494, MB3759, KA7500 chips

In this case, the resistor values are the same for stabilizing 5 volts and for 12. If you decide to use the power supply to get 5 volts, then connect the resistor divider to the 5V output. If for 12, then to 12.

Probably enough theory and it's time to get down to business. First you need to decide on the measuring instruments. To measure voltages, I will use one of the cheapest multimeters DT838. Their voltage measurement accuracy is 0.5 percent, which is quite acceptable. I use an ammeter to measure current. Currents need to be measured large, so you will have to make an ammeter yourself from a pointer measuring head and a home-made shunt. I could not find a ready-made ammeter with a factory shunt of an acceptable size. I found a 3 amp ammeter, disassembled it. Pulled the shunt out of it. Got a microammeter. Then there was a little difficulty. To manufacture a shunt and calibrate an ammeter made from a microammeter, a standard ammeter was needed that could measure current in the range of 15-20 amperes. For these purposes, current clamps could be used, but I did not have any. I had to find a way out. I found the easiest way out, of course, not very accurate, but quite. I cut the shunt from a steel sheet 1mm thick, 4mm wide and 150mm long. I connected 6 bulbs 12V, 20W to the power supply through this shunt. According to Ohm's law, a current equal to 10 amperes flowed through them.

P(Wt)/U(V)=I(A), 120/12=10A

One wire from the microammeter was connected to the end of the shunt, and the second was moved along the shunt until the arrow of the device showed 7 divisions. Up to 10 divisions was not enough length of the shunt. It was possible to cut the shunt thinner, but due to lack of time, I decided to leave it as it is. Now 7 divisions of this scale correspond to 10 amperes.

Photo 1 Budget stand for shunt selection.

Photo 2. Stand with 6 bulbs on 12 volts 20 watts.

The last photo shows how the voltage of 12 volts sank at a current of 10 amperes. PowerMan Pro 420 W power supply. Minus 11.55 shows due to the fact that I mixed up the polarity of the probes. In fact, of course, plus 11.55. I will use the same stand as a load to adjust the finished power supply.

I will make a new power supply based on the PowerMaster 350 W, it will produce 5 volts. According to the sticker on it, it should give 35 amps along this line. And PowerMan Pro 420 W. I will take all other voltages from it.

In this article I will show the general principle of modernization. In the future, I plan to convert the resulting power supply into a passive one. Perhaps I will rewind the chokes with a larger wire. I will finalize the connecting cables to reduce pickups and ripples. I will monitor currents and voltages. And much more is possible. But that's in the future. I will not describe all this in this article. The purpose of the article is to prove the possibility of obtaining a powerful power supply by upgrading two or three units of lower power.

A little about safety. All soldering is done, of course, with the unit turned off. After each shutdown of the unit, before further work, discharge the large capacitors. They have a voltage of 220 volts, and they accumulate a very decent charge. Not fatal, but extremely annoying. Electrical burns take a long time to heal.

I'll start with PowerMaster. I disassemble the block, take out the board, cut off the extra wires ...

Photo 3. PowerMaster 350 W unit

I find a PWM chip, it turned out to be TL494. I find pin 1, carefully cut the printed conductor and solder a new resistor divider to pin 1 (see Fig.5). I solder the input of the resistor divider to the five-volt output of the power supply (usually red wires). Once again I check the correct installation, it is never superfluous. I connect the upgraded unit to my budget stand. Just in case, hiding behind a chair, I turn it on. The explosion did not happen and it even caused a slight disappointment. To start the unit, I connect the PS ON wire to a common wire. The unit turns on and the lights come on. First victory.

With a variable resistor R1 on a small load of the power supply (two bulbs of 12V, 20W and a spot of 35W), I set the output voltage to 5 volts. I measure the voltage directly at the output connector.

My camera is not the best, it does not see small details, so I apologize for the quality of the pictures.

The power supply can be turned on without a fan for a short period of time. But you need to monitor the temperature of the radiators. Be careful, there is voltage on the heatsinks of some models of power supplies, sometimes high.

Without turning off the unit, I begin to connect an additional load - light bulbs. The voltage does not change. The block stabilizes well.

In this photo, I connected all the bulbs that were available to the block - 6 lamps of 20w, two of 75w, and a spot of 35w. The current flowing through them according to the readings of the ammeter is within 20 amperes. No "sagging", no "distortions"! Half done.

Now I'm taking PowerMan Pro 420 W. I'm also disassembling it.

I find the SG6105 chip on the board. Then I look for the necessary conclusions.

The circuit diagram given in the article by Mr. Korobeinikov corresponds to my block, the numbering and resistor values are the same. To turn off the 5 volts, I solder the resistor R40 and R41. Instead of R41, I solder two variable resistors connected in series. Nominal 47 kOhm. This is for a coarse voltage adjustment of 12 volts. For fine adjustment, resistor VR1 is used on the power supply board.

Fig 6. A fragment of the PowerMan power supply circuit

Again I take out my primitive stand and connect the power supply to it. First, I connect the minimum load - spot 35W.

I turn it on and adjust the voltage. Then, without turning off the power supply, I connect additional light bulbs. The voltage does not change. The block works great. According to the ammeter readings, the current reaches 18 amperes and there is no "sag" voltage.

The second stage is over. Now it remains to check how the blocks will work in pairs. I bite the red wires going from PowerMan to the connector and molex, isolate them. And I solder a five-volt wire from the PowerMaster 350 W to the connector and molex, and also connect the common wires of both blocks. I combine the Power On wires of the power supplies. I take PG with PowerMan. And I connect this hybrid to my system unit. In appearance, it is somewhat strange, and if someone wants to know more about him, I ask for a PS.

The configuration is like this:

Mother Epox KDA-J
Athlon 64 3000 processor
Memory Digma DDR500, two sticks of 512Mb
Screw Samsung 160Gb
Video GeForce 5950
DVD RW NEC 3500

Turn it on, everything works great.

The experience went well. Now you can proceed to further modernization of the "combined power supply". Transferring it to passive cooling. The photo shows a panel with instruments - all this will be connected to this unit. Pointer devices - current monitoring, digital devices in round holes under the pointers - voltage monitoring. Well, a tachometer, and all that, I already wrote about this on my personal computer. But this is in the future.

I did not check the influence of the "combined power supply" on further overclocking. I'll finish it, then I'll check. The processor has already been overclocked to 2.6 gigahertz on the bus, with a voltage of 1.7 volts on the processor. I drove it on a fanless power supply, but with such overclocking, 12 volts on it sagged to 11.6 volts. And the hybrid puts out exactly 12. So maybe I'll squeeze a few more megahertz out of it. But that will be another story.

List of used literature:

Magazine "Radio". - 2002.-№ 5, 6, 7. "Circuit engineering of power supplies for personal computers" author. R. Alexandrov

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