Nissan 

How to increase the output voltage current. What is voltage, how to lower and increase voltage. Overclocking the power supply

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.

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 equal to: 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 relatively 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 electrics, as 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 for 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 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 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, different types of plastics are often used as dielectrics. 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.

Overclocking the power supply.

The author is not responsible for the failure of any components resulting from overclocking. By using these materials for any purpose, the end user assumes all responsibility. The site materials are presented "as is"."

Introduction.

I started this experiment with frequency due to the lack of power in the power supply.

When the computer was purchased, its power was quite sufficient for this configuration:

AMD Duron 750Mhz / RAM DIMM 128 mb / PC Partner KT133 / HDD Samsung 20Gb / S3 Trio 3D/2X 8Mb AGP

For example, two diagrams:

Frequency f for this circuit it turned out to be 57 kHz.


And for this frequency f equal to 40 kHz.

Practice.

The frequency can be changed by replacing the capacitor C or/and resistor R to a different denomination.

It would be correct to install a capacitor with a smaller capacitance, and replace the resistor with a series-connected constant resistor and a variable type SP5 with flexible leads.

Then, decreasing its resistance, measure the voltage until the voltage reaches 5.0 volts. Then solder a constant resistor in place of the variable one, rounding the value up.

I took a more dangerous path - I sharply changed the frequency by soldering in a capacitor of smaller capacity.

I have had:

R 1 =12kOm
C 1 =1.5nF

According to the formula we get

f=61.1 kHz

After replacing the capacitor

R 2 =12kOm
C 2 =1.0nF

f =91.6 kHz

According to the formula:

the frequency increased by 50% and the power increased accordingly.

If we do not change R, then the formula simplifies:

Or if we don’t change C, then the formula is:

Trace the capacitor and resistor connected to pins 5 and 6 of the microcircuit. and replace the capacitor with a capacitor with a smaller capacity.


Result

After overclocking the power supply, the voltage became exactly 5.00 (the multimeter can sometimes show 5.01, which is most likely an error), almost without reacting to the tasks being performed - with a heavy load on the +12 volt bus (simultaneous operation of two CDs and two screws) - the voltage on the bus is + 5V may drop briefly to 4.98.

The key transistors began to heat up more. Those. If before the radiator was slightly warm, now it is very warm, but not hot. The radiator with rectifier half-bridges did not heat up any more. The transformer also does not heat up. From 09/18/2004 to this day (01/15/05) there are no questions about the power supply. Currently the following configuration:

Links

  1. PARAMETERS OF THE MOST COMMON POWER TRANSISTORS USED IN PUSH-CYCLE UPS CIRCUITS MANUFACTURED FOREIGN.
  2. Capacitors. (Note: C = 0.77 ۰ Nom ۰SQRT(0.001۰f), where Nom is the rated capacitance of the capacitor.)

Rennie's comments: The fact that you increased the frequency, you increased the number of sawtooth pulses over a certain period of time, and as a result, the frequency with which power instabilities are monitored increased, since power instabilities are monitored more often, the pulses for closing and opening of transistors in a half-bridge switch occur at double frequency . Your transistors have characteristics, specifically their speed: By increasing the frequency, you have thereby reduced the size of the dead zone. Since you say that the transistors do not heat up, it means that they are in that frequency range, which means that everything seems to be fine here. But there are also pitfalls. Do you have an electrical circuit diagram in front of you? I’ll explain it to you now using the diagram. There in the circuit, look where the key transistors are, diodes are connected to the collector and emitter. They serve to dissolve the residual charge in the transistors and transfer the charge to the other arm (to the capacitor). Now, if these comrades have a low switching speed, through currents are possible - this is a direct breakdown of your transistors. Perhaps this will cause them to heat up. Now further, this is not the case, the point is that after the direct current that passed through the diode. It has inertia and when a reverse current appears: for some time the value of its resistance is not restored and therefore they are characterized not by the frequency of operation, but by the recovery time of the parameters. If this time is longer than possible, then you will experience partial through currents, which is why surges in both voltage and current are possible. In the secondary it’s not so scary, but in the power department it’s just fucked up: to put it mildly. So let's continue. In the secondary circuit, these switchings are not desirable, namely: There, Schottky diodes are used for stabilization, so at 12 volts they are supported with a voltage of -5 volts (approx. I have silicon ones at 12 volts), so at 12 volts that If only they (Schottky diodes) could be used with a voltage of -5 volts. (Due to the low reverse voltage, it is impossible to simply put Schottky diodes on the 12 volt bus, so they are distorted this way). But silicon diodes have more losses than Schottky diodes and the reaction is less, unless they are one of the fast-recovery diodes. So, if the frequency is high, then the Schottky diodes have almost the same effect as in the power section + the inertia of the winding at -5 volts relative to +12 volts makes it impossible to use Schottky diodes, so an increase in frequency can eventually lead to failure of them. I'm considering the general case. So let's move on. Next is another joke, finally connected directly with the feedback circuit. When you create negative feedback, you have such a thing as the resonant frequency of this feedback loop. If you reach resonance, then your entire scheme will be screwed. Sorry for the rude expression. Because this PWM chip controls everything and requires its operation in mode. And finally, a “dark horse” ;) Do you understand what I mean? It's a transformer, so this bitch also has a resonant frequency. So this crap is not a standardized part, the transformer winding product is manufactured individually in each case - for this simple reason you do not know the characteristics of it. What if you introduce your frequency into resonance? You burn your trance and you can safely throw away the power supply. Externally, two absolutely identical transformers can have completely different parameters. Well, the fact is that by choosing the wrong frequency you could easily burn out the power supply. Under all other conditions, how can you still increase the power of the power supply? We increase the power of the power supply. First of all, we need to understand what power is. The formula is extremely simple - current to voltage. The voltage in the power section is 310 volts constant. So, we cannot influence the voltage in any way. We have only one trans. We can only increase the current. The amount of current is dictated to us by two things - transistors in the half-bridge and buffer capacitors. The conductors are larger, the transistors are more powerful, so you need to increase the capacitance rating and change the transistors to ones that have a higher current in the collector-emitter circuit or just a collector current, if you don’t mind, you can plug in 1000 uF there and not strain yourself with calculations. So in this circuit we did everything we could, here, in principle, nothing more can be done, except perhaps taking into account the voltage and current of the base of these new transistors. If the transformer is small, this will not help. You also need to regulate such crap as the voltage and current at which your transistors will open and close. Now it seems like everything is here. Let's go to the secondary circuit. Now we have a lot of current at the output windings....... We need to slightly correct our filtering, stabilization and rectification circuits. For this, we take, depending on the implementation of our power supply, and change the diode assemblies first of all, so that we can ensure the flow of our current. In principle, everything else can be left as is. That's all, it seems, well, at the moment there should be a margin of safety. The point here is that the technique is impulsive - this is its bad side. Here almost everything is built on the frequency response and phase response, on t reaction.: that’s all

!
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 at hand 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’s certainly possible to assemble the required power source from scratch or use the 5-volt bus of any cheap computer power supply, but it will be useful for many DIY electronics engineers to know how to increase the output current (or in common parlance, the amperage) 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-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 is usually built on the basis of an optocoupler and a reference voltage source tl431.




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 in the end we get a calculated 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 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 inductor and the 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.