We looked at what action to take if we have a short circuited ATX power supply fuse. This means that the problem is somewhere in the high-voltage part, and we need to check the diode bridge, output transistors, power transistor or mosfet, depending on the model of the power supply. If the fuse is intact, we can try to connect the power cord to the power supply, and turn it on with the power switch located on the back of the power supply.

And here a surprise may await us, as soon as we flip the switch, we can hear a high-frequency whistle, sometimes loud, sometimes quiet. So, if you hear this whistle, do not even try to connect the power supply for tests to the motherboard, assembly, or install such a power supply in the system unit!

The fact is that in the standby voltage circuits there are the same electrolytic capacitors familiar to us from the last article, which lose capacity when heated, and from old age, their ESR increases, (abbreviated in Russian as ESR) equivalent series resistance . At the same time, visually, these capacitors may not differ in any way from working ones, especially for small values.

The fact is that on small denominations, manufacturers very rarely make notches in the upper part of the electrolytic capacitor, and they do not swell or open. Without measuring such a capacitor with a special device, it is impossible to determine its suitability for operation in the circuit. Although sometimes, after desoldering, we see that the gray stripe on the capacitor, which marks the minus on the capacitor body, becomes dark, almost black from heating. As repair statistics show, next to such a capacitor there is always a power semiconductor, or an output transistor, or a duty diode, or a mosfet. All these parts emit heat during operation, which has a detrimental effect on the life of electrolytic capacitors. I think it would be superfluous to explain further about the performance of such a darkened capacitor.

If the cooler of the power supply has stopped due to grease drying out and being clogged with dust, such a power supply will most likely require replacing almost ALL electrolytic capacitors with new ones due to the increased temperature inside the power supply. Repairs will be quite tedious and not always advisable. Below is one of the common schemes on which Powerman 300-350 watt power supplies are based, it is clickable:

ATX Powerman power supply circuit

Let's look at which capacitors need to be changed in this circuit in case of problems with the duty room:

So why can't we plug the power supply whizzing into the assembly for testing? The fact is that in the standby circuits there is one electrolytic capacitor, (highlighted in blue) with an increase in the ESR of which, we have an increase in the standby voltage supplied by the power supply to motherboard, even before we press the power button system unit. In other words, as soon as we clicked the key switch on the back wall of the power supply, this voltage, which should be equal to +5 volts, goes to our power supply connector, the purple wire of the 20 Pin connector, and from there to the computer’s motherboard.

In my practice, there were cases when the standby voltage was equal (after removing the protective zener diode, which was in the short circuit) to +8 volts, and at the same time the PWM controller was alive. Fortunately, the power supply was of high quality, Powerman brand, and there was a 6.2 volt protective zener diode on the +5VSB line (as the duty room output is indicated in the diagrams).

Why is the zener diode protective, how does it work in our case? When our voltage is less than 6.2 volts, the zener diode does not affect the operation of the circuit, but if the voltage becomes higher than 6.2 volts, our zener diode goes into a short circuit (short circuit) and connects the duty circuit to the ground. What does this give us? The fact is that by connecting the control panel to ground, we thereby save our motherboard from supplying it with the same 8 volts, or another high voltage rating, through the control panel line to the motherboard, and protect the motherboard from burnout.

But this is not a 100% probability that in case of problems with the capacitors the zener diode will burn out; there is a possibility, although not very high, that it will go into a break and thereby not protect our motherboard. In cheap power supplies, this zener diode is usually simply not installed. By the way, if you see traces of burnt PCB on the board, you should know that most likely some semiconductor went into a short circuit, and a very large current flowed through it, such a detail is very often the cause (although sometimes it also happens to be the effect) breakdowns.

After the voltage at the control room returns to normal, be sure to change both capacitors at the output of the control room. They can become unusable due to the supply of excessive voltage to them, exceeding their rated voltage. Usually there are capacitors with a nominal value of 470-1000 microfarads. If, after replacing the capacitors, a voltage of +5 volts appears on the purple wire relative to ground, you can short the green wire with the black one, PS-ON and GND, starting the power supply, without the motherboard.

If the cooler starts to rotate, this means with a high degree of probability that all voltages are within normal limits, because our power supply has started. The next step is to verify this by measuring the voltage on the gray wire, Power Good (PG), relative to ground. If +5 volts are present there, you are in luck, and all that remains is to measure the voltage at the 20 Pin power supply connector with a multimeter to make sure that none of them are too low.

As can be seen from the table, the tolerance for +3.3, +5, +12 volts is 5%, for -5, -12 volts - 10%. If the control panel is normal, but the power supply does not start, we do not have Power Good (PG) +5 volts, and there is zero volt on the gray wire relative to ground, then the problem was deeper than just with the control panel. We will consider various options for breakdowns and diagnostics in such cases in the following articles. Happy repairs everyone! AKV was with you.

One of the most serious problems that both beginners and professional radio amateurs periodically face is heating of the elements. Almost all medium and high power devices get hot. In this case, it is not the heating itself that is dangerous (many devices, for example an electric kettle, are designed specifically for this purpose), but the overheating of the device - when its temperature rises above a certain maximum permissible level. At the same time, some other non-semiconductors become charred (i.e., literally “burn out”), and in semiconductors, breakdown of p-n junctions occurs, and these junctions, instead of passing current in only one direction, begin to pass it in both (i.e., they “turn” into ordinary conductors with little resistance) or do not pass it at all, either in the forward or reverse direction. About such devices, by analogy with resistors, they also say that they “burnt out”, although this is not entirely correct, especially since modern semiconductors (,) are produced in sealed cases, due to which it is impossible to determine whether this device has “burnt out” or not.

The reason for heating is the power released by the element, or, in scientific terms, the power dissipated by the element. Power dissipation, like any other power, depends on the voltage drop across the element and the current flowing through it:

where Rras is power dissipation, W; U - voltage drop. IN; I - flowing current. A; R - element, Ohm.

For example, let's collect the simplest scheme(Fig. 1.42): high-voltage (relatively!) voltage to power a low-voltage light bulb. Supply voltage - 15 V, zener diode stabilization voltage - 3.6 V, current in the circuit - 0.2 A. Since it is connected according to the circuit (the pin to which power is supplied is considered common), the voltage at its emitter (and, accordingly , on the light bulb) is 0.6 V less than the voltage at the base - i.e. 3.0 V. The power dissipated on the light bulb is 3 V · 0.2 A = 0.6 W.

Since only 3 V is supplied to the light bulb, the remaining 15 - 3 = 12 (V) fall on the transistor - after all, they must go somewhere, and the supply voltage (15 V) is constant, and reduce it. Let's assume it's impossible. Therefore, the transistor dissipates a power of 12 V · 0.2 A = 2.4 W - 4 times more than a light bulb.

The simplest analogue of a switching step-down power supply is shown in Fig. 1.43. It is advisable to choose a more powerful light bulb (more than 10...20 W), and use two wires rubbing against each other as the S1 button.

When two wires are connected to each other, the contact between them is not broken and the light bulb burns fully. But when you start rubbing the wires against each other, the contact between them will periodically begin to break down and the brightness of the light bulb will decrease; If you practice, the brightness can be reduced by 5...10 times, and the light bulb will barely glow.

The explanation for this effect is very simple. The fact is that all incandescent lamps have significant thermal inertia (and the greater the power of the lamp, the greater the thermal inertia - this is why I advise you to choose a more powerful light bulb), i.e. their spiral heats up very slowly and cools down just as slowly, and The hotter the spiral is, the brighter it shines. When wires rub against each other, it is because their surface is partially oxidized (the oxide layer does not conduct electricity), and also due to their imperfectly smooth surface, the contact between them is chaotically broken and restored again. When there is no contact, it is infinite; when there is contact, it is close to zero. Therefore, the light bulb does not receive D.C. amplitude 12 V, and pulsed, with the same amplitude. The spiral of the light bulb, due to thermal inertia, smoothes out these pulses, and since the constant component of the pulse current is always less than the amplitude of the pulse, the light bulb glows as if its supply voltage has decreased, and the shorter the duration of the current pulse, compared to the duration of the pause between pulses , the weaker the light bulb glows.

performance is maximum (since the transistor is “helped” by the output of the op-amp - until due to inertia it has time to open completely, the current from the output of the op-amp through the base-emitter junction flows into its load), and also, unlike, it consumes from the source The signal current is not very high, i.e. it minimally loads the op-amp output. But the powerful one is switched on according to the circuit: although this consumes much more current than, the voltage drop at the collector-emitter junction of the open transistor is less (no more than 0.2...0.5 V), i.e. we lose in terms of control current , but overall (in terms of efficiency) we win. If VT2 is turned on according to the circuit, then even with a load current of more than 200 mA it becomes quite hot; The cascade with OE at this current is practically cold.

Pulses from the collector of transistor VT2 through L1 enter the load. The voltage on capacitor C2 depends on the current consumed by the load - the higher the current, the lower the voltage. This can be compensated by increasing resistor R5. IN modern schemes such compensation works automatically: another op-amp is connected to capacitor C2, which automatically changes the duty cycle of the signal at output DA1 so that the output voltage always remains unchanged, i.e. it functions in the same way as the AGC system. We will look at this scheme a little later.

The main parameter of inductors is their. In our circuit, L1 should be larger, so it needs to be wound on some kind of core: when winding a coil on a magnetic core, it increases by a certain number of times, which is called the magnetic permeability of the core. The magnetic permeability of even the worst cores exceeds 50, i.e., a coil with a certain given inductance, when using a core, has 50 times fewer turns than the same coil, but without a core. At the same time, you save both the wire and the space occupied by the coil, and also significantly reduce the coil windings. , which have a magnetic core, are called “choke”.

As cores, they usually use either iron plates (for example, transformers) or rings made of so-called “ferrite”: iron plates are good only when used in low-frequency devices (up to 400 Hz) - at higher frequencies they begin to heat up and the efficiency of the device sharply decreases . This is due to the emerging Foucault currents (eddy currents), the cause of which is the non-zero thickness of the plates and their low density. In an ideal core, current should flow only along the plates (perpendicular to the coil), but since the plates have a certain thickness, part of the current flows across the plates, causing only harm. Therefore, modern iron cores are composed of many plates insulated with a varnish coating, the thickness of one plate is much less than its length, and only an insignificant part of the energy is spent on it. But still, the iron core works well only at frequencies up to 400 Hz - at high frequencies the thickness of the plates should be very small, and it will be difficult to work with such plates.

At frequencies above 400 Hz, cores are usually used. Ferrite is a ceramic rather than a metal and does not conduct electricity. Therefore, no electric current arises inside it, i.e., there are no eddy currents, regardless of the thickness of the core. Ferrites operate normally at frequencies up to tens of megahertz; at high frequencies, too much is not needed, and a regular coil without a core is quite sufficient.

To work in this scheme, it is best to use standard size Κ20χ10χ5, i.e. its outer (total) diameter is 20 mm, internal (hole diameter) is 10 mm, thickness is 5 mm. The number of turns of inductor L1 is about 50...100 with a wire of diameters 0.5...0.8 mm in varnish insulation (transformers, electric motors and other “pieces of hardware” in which electric current is converted into a magnetic field and (or) vice versa are wound with such wire). The coil is wound across the ring, that is, the wire is threaded into the ring, pulled out from the opposite side, wrapped around the outside of the ring and threaded into it again. And so - 50... 100 times. It is advisable to place the turns side by side (each subsequent one next to the previous one); if the length of the inner surface of the ring is “not enough” to place the entire coil in one layer, wind the second (and so on) layer, but the winding direction of each subsequent layer must coincide with the winding direction of the previous one!

The ring can be taken with either a larger or smaller diameter, while in the first case you need to slightly increase the number of turns and reduce the diameter of the wire (the load current will decrease), and in the second case you need to reduce the number of turns, and if you increase the diameter of the wire, then by selecting VT2, it will be possible to increase the load current. It makes sense to use rings with an outer diameter of less than 10 mm only with a load current of no more than 100 mA, although, in principle, you can increase the operating frequency and replace VT1 and VT2 with higher frequency ones - then the number of inductor turns will need to be reduced, i.e. it can be reduced will be wound with a thicker wire, due to which the maximum permissible load current will increase.

It is advisable to connect a film or ceramic capacitance of 0.047...0.22 µF in parallel with capacitor C2. Simply electrolytic, due to the peculiarities of the internal structure, are inertial and react poorly to impulses arriving through the L1 coil. Because of this, the ripple of the output voltage sharply increases and the efficiency of the device decreases slightly. A “fast-acting” small capacitance (it is called “blocking” - do not confuse it with the “filtering” capacitor C2!) blocks the passage of pulses to the output, charging itself, and during the pause between pulses it transfers its charge (very small, but the duration pulse is small) to capacitor C2 and to the load.

One of the features of such a power supply is that, when properly assembled and configured, the current in the load can exceed the current consumed from the power source! This is due to the fact that it transforms voltage and current, and

where U n „ T and 1 power supply are, respectively, the supply voltage and the current consumed from the power source; U H and 1 n - voltage and current in the load.

That is, in the ideal case, if the supply voltage is 10 times less, then this () from the power source (mains rectifier, batteries) consumes a current that is 10 times less than the load current. The linear stabilizer discussed above (Fig. 1.42) at any load voltage consumes from the power source a current equal to and even slightly greater than the load current.

But this is only in the ideal case, when the efficiency is 100%. In real circuits, due to the inertia of work powerful transistors and diodes, and also due to the imperfectly selected inductance of inductor L1 (in this circuit it is better to change not the inductor, but the frequency of the generator - by selecting the capacitance of capacitor C1), the efficiency is rarely higher than 80...90%. But this is also a lot, especially if there is a large difference between the input and output voltages: after all, the efficiency of a linear stabilizer in this case tends to zero. For a pulse stabilizer, the efficiency is practically independent of the voltage difference and is always maximum.

The higher the efficiency of the device, the less you pay for the electricity it consumes. In addition, with an increase in efficiency, the heating of the power elements (i.e., the power transistor and diode) sharply decreases. Mine, assembled using a powerful output stage field effect transistor, with a load power of 40 W (the electric soldering iron) practically does not heat up - a little more than 1 W is released on the transistor, and it is able to dissipate such an insignificant power independently, without a radiator. But before that, I used the “services” of a linear stabilizer, which, with the same load power and the same difference between the input and output voltages, overheated even when using a radiator the size of this book. But heating also requires energy!

The only drawback of a switching stabilizer is the very high level of noise both in the load and in the stabilizer power supply. In addition, the magnetic field around the coil L1 of the stabilizer operating at a certain load is variable, i.e. it emits powerful electromagnetic interference. This interference is capable of drowning out all low-frequency long-wave radio stations within a radius of tens of meters from the throttle.

It is possible to fight these “misfortunes,” although it is very difficult. You can reduce the level of noise in the wires by increasing the capacitance of capacitors C2 and SZ (SZ should be located in close proximity to the emitter terminal of transistor VT2 and the anode of diode VD3 - it is advisable to solder it directly to the terminals of these elements), as well as by soldering blocking low-inertia small capacitances parallel to them. But dealing with electromagnetic interference is more difficult. In principle, if you are not going to operate it together with a long-wave radio, then you don’t have to fight them - they don’t affect anything else -1 ·. But if they need to be eliminated, L1 should be screened, i.e. “hidden” in. any completely closed metal box (take care of reliable electrical insulation!), and the thickness of its walls should not be less than 0.5...1.0 mm. To ensure that the power lines around the throttle do not close on the screen, the distance from any point on the surface of the throttle to the screen should not be less than half its diameter.

Because of this power supply feature, they are mainly used only in conjunction with powerful digital circuits - supply voltage ripple “to the light bulb”. To power low-power analog circuits, you only need to use: analog circuits, especially those with a significant gain, are extremely sensitive to interference, so it is better to immediately sacrifice efficiency than to try to eliminate interference later. But in some cases, when the range of analog operating frequencies does not come into contact with the operating frequency of the power supply (for example, it operates in the range of 20...20000 Hz, and either in terms of efficiency they were even worse than linear ones, or they distorted the signal very much. And in the output stage of the linear one it is subject to those the same laws as in Fig. 1.42 Unfortunately, nothing can correct the situation yet, so here I will only talk about how you can indirectly reduce the heating of the output transistors.

First, the amplifier's supply voltage must be matched to the load resistance. For example, it will be used with a speaker with a resistance of 4 Ohms and should produce power up to 50 W. With such power, the voltage on the column should be (amplitude and alternating voltage). Taking into account the small voltage drop on the power (output) transistors (after all, under no circumstances should they be brought to saturation!), the amplifier supply voltage should be equal to ±17...20 V. If the supply voltage is lower, with a small voltage at the base (gate), they need to be opened a little - then they simply will not “enter” the nonlinear mode. And since the current-voltage characteristic of the transistor is very weak from the supply voltage, the quiescent current of both high-voltage and low-voltage amplifiers is almost the same. Therefore, the “rest power” is less for a low-voltage amplifier, i.e., it heats up less than a high-voltage amplifier.

Oddly enough, it heats up the most at “average” output power (volume), and at minimum and maximum sound volumes it heats up much less. But there is nothing strange here. It’s just that at a minimum sound volume, although the voltage on the output transistors is quite significant, the current flowing through them is negligible, and the power P = I U released on them is also minimal. With maximum output power flowing through ultra-high requirements, it is best assembled - at the same time you will save on parts.

What is desirable to have for checking the power supply.
A. - any tester (multimeter).
b. - light bulbs: 220 volts 60 - 100 watts and 6.3 volts 0.3 amperes.
V. - soldering iron, oscilloscope, solder suction.
g. - magnifying glass, toothpicks, cotton swabs, industrial alcohol.

It is safest and most convenient to connect the unit being repaired to the network through a 220v - 220v isolation transformer.
Such a transformer is easy to make from 2 TAN55 or TS-180 (from tube b/w TVs). The anode secondary windings are simply connected accordingly, there is no need to rewind anything. The remaining filament windings can be used to build an adjustable power supply.
The power of such a source is quite sufficient for debugging and initial testing and provides a lot of convenience:
- electrical safety
— the ability to connect the grounds of the hot and cold parts of the unit with a single wire, which is convenient for taking oscillograms.
— we install a biscuit switch — we get the ability to change the voltage stepwise.

Also, for convenience, you can bypass the +310V circuits with a 75K-100K resistor with a power of 2 - 4W - when turned off, the input capacitors discharge faster.

If the board is removed from the unit, check for any metal objects of any kind underneath it. Under no circumstances DO NOT reach into the board with your HANDS or TOUCH the radiators while the unit is running, and after turning off, wait about a minute for the capacitors to discharge. There can be 300 or more volts on the power transistor radiator; it is not always isolated from the block circuit!

Principles of measuring voltages inside a block.
Please note that ground is supplied to the power supply housing from the board through conductors near the holes for the mounting screws.
To measure voltages in the high-voltage (“hot”) part of the unit (on power transistors, in the control room), a common wire is required - this is the minus of the diode bridge and input capacitors. Everything relative to this wire is measured only in the hot part, where the maximum voltage is 300 volts. It is advisable to take measurements with one hand.
In the low-voltage (“cold”) part of the power supply, everything is simpler, the maximum voltage does not exceed 25 volts. For convenience, you can solder wires into the control points; it is especially convenient to solder the wire to the ground.

Checking resistors.
If the nominal value (colored stripes) is still readable, we replace it with new ones with a deviation no worse than the original (for most - 5%, for low-resistance current sensor circuits it can be 0.25%). If the marked coating has darkened or crumbled due to overheating, we measure the resistance with a multimeter. If the resistance is zero or infinity, the resistor is most likely faulty and will need to be determined to determine its value. circuit diagram power supply or study standard schemes inclusions.

Checking diodes.
If the multimeter has a mode for measuring the voltage drop across the diode, you can check without desoldering. The drop should be from 0.02 to 0.7 V. If the drop is zero or so (up to 0.005), unsolder the assembly and check. If the readings are the same, the diode is broken. If the device does not have such a function, set the device to measure resistance (usually the limit is 20 kOhm). Then, in the forward direction, a serviceable Schottky diode will have a resistance of about one to two kilo-ohms, and a regular silicon one will have a resistance of about three to six. In the opposite direction, the resistance is infinity.

Checking the field effect transistor

To check the power supply, you can and should collect a load.
See an example of successful execution here.
We take the soldered one from the unnecessary ATX boards connector and solder wires with a cross-section of at least 18 AWG to it, trying to use all contacts along the +5 volt, +12 and +3.3 volt lines.
The load must be calculated at 100 watts across all channels (it can be increased to test more powerful units). To do this, we take powerful resistors or nichrome. You can also use powerful lamps with caution (for example, 12V halogen lamps), but it should be taken into account that the resistance of the filament in a cold state is much less than in a heated state. Therefore, when starting with a seemingly normal load of lamps, the unit may go into protection.
You can connect light bulbs or LEDs in parallel to the loads to see the presence of voltage at the outputs. Between the PS_ON and GND pins we connect a toggle switch to turn on the block. For ease of operation, the entire structure can be placed in a power supply case with a fan for cooling.

Block check:

You can first turn on the power supply to the network to determine the diagnosis: there is no duty (problem with the duty, or a short circuit in the power section), there is a duty, but there is no startup (problem with swing or PWM), the power supply goes into protection (most often - the problem is in output circuits or capacitors), excessive standby voltage (90% - swollen capacitors, and often as a result - dead PWM).

Initial block check
We remove the cover and begin checking, paying special attention to damaged, discolored, darkened or burnt parts.
Fuse. As a rule, burnout is clearly visible visually, but sometimes it is covered with heat-shrinkable cambric - then we check the resistance with an ohmmeter. A blown fuse may indicate, for example, a malfunction of the input rectifier diodes, key transistors, or the standby circuit.
Disc thermistor. It rarely fails. We check the resistance - it should be no more than 10 ohms. In the event of a malfunction, it is not advisable to replace it with a jumper - when the unit is turned on, the impulse current charge of the input capacitors, which can lead to breakdown of the input rectifier diodes.
Diodes or diode assembly of the input rectifier. We check each diode with a multimeter (in voltage drop measurement mode) for opens and short circuits; you don’t have to unsolder them from the board. If a short circuit is detected in at least one diode, it is also recommended to check the input electrolytic capacitors to which alternating voltage was applied, as well as the power transistors, since there is a very high probability of their breakdown. Depending on the power of the power supply, the diodes must be designed for a current of at least 4...8 amperes. We immediately replace two-ampere diodes, often found in cheap units, with more powerful ones.
Input electrolytic capacitors. Checking external inspection for swelling (a noticeable change in the upper plane of the capacitor from a flat surface to a convex one), we also check the capacitance - it should not be lower than indicated on the marking and differ between two capacitors by more than 5%. We also check varistors that are parallel to the capacitors (usually they clearly burn into charcoal) and equalizing resistors (the resistance of one should not differ from the resistance of the other by more than 5%).
Key (also known as power) transistors. For bipolar ones, use a multimeter to check the voltage drop at the base-collector and base-emitter junctions in both directions. In a working bipolar transistor, the junctions should behave like diodes. If a transistor malfunction is detected, it is also necessary to check its entire “piping”: diodes, low-resistance resistors and electrolytic capacitors in the base circuit (it is better to immediately replace the capacitors with new ones of higher capacity, for example, instead of 2.2 µF * 50V we set 10.0 µF * 50V). It is also advisable to bypass these capacitors with ceramic capacitors of 1.0...2.2 µF.
Output diode assemblies. We check them with a multimeter, the most common fault is a short circuit. It is better to install a replacement in the TO-247 housing. In TO-220 they die more often... Usually for 300-350 W blocks of diode assemblies like MBR3045 or similar 30A - with the head.
Output electrolytic capacitors. The malfunction manifests itself in the form of swelling, traces of brown fluff or streaks on the board (when electrolyte is released). We replace them with capacitors of normal capacity, from 1500 µF to 2200...3300 µF, working temperature— 105° C. It is advisable to use the LowESR series.
We also measure the output resistance between the common wire and the block outputs. For +5V and +12V volts - usually around 100-250 ohms (the same for -5V and -12V), +3.3V - about 5...15 ohms.

Darkening or fading printed circuit board under resistors and diodes indicates that the circuit components were operating abnormally and requires analysis of the circuit to determine the cause. Finding such a place near the PWM means that the 22 Ohm PWM power resistor is heating up due to exceeding the standby voltage and, as a rule, it is the one that burns out first. Often the PWM is also dead in this case, so we check the microcircuit (see below). Such a malfunction is a consequence of the operation of the “on duty” in abnormal mode; you should definitely check the standby mode circuit.

Checking the high-voltage part of the unit for a short circuit.

We take a light bulb from 40 to 100 watts and solder it instead of a fuse or into a break in the network wire.
If, when the unit is connected to the network, the lamp flashes and goes out - everything is in order, there is no short circuit in the “hot” part - remove the lamp and continue to work without it (replace the fuse or splice the network wire).
If, when the unit is plugged in, the lamp lights up and does not go out, there is a short circuit in the unit in the “hot” part. To detect and eliminate it, do the following:
We unsolder the radiator with power transistors and turn on the power supply through the lamp without shorting the PS-ON.
If it is short (the lamp is on, but did not light up and go out), we are looking for the reason in the diode bridge, varistors, capacitors, 110/220V switch (if there is one, it is better to remove it altogether).
If there is no short, we solder the duty transistor and repeat the switching procedure.
If there is a short one, we look for a fault in the control room.
Attention! It is possible to turn on the unit (via PS_ON) with a small load while the light is not turned off, but firstly, unstable operation of the power supply cannot be ruled out, and secondly, the lamp will light up when the power supply with the APFC circuit is turned on.

Checking the standby mode (duty) circuit.

A quick guide: we check the key transistor and all its wiring (resistors, zener diodes, diodes around). We check the zener diode located in the base circuit (gate circuit) of the transistor (in the circuits on bipolar transistors nominal from 6V to 6.8V, in the field, as a rule, 18V). If everything is normal, pay attention to the low-resistance resistor (about 4.7 Ohms) - power supply to the standby transformer winding from +310V (used as a fuse, but sometimes the standby transformer burns out) and 150k~450k (from there to the base of the standby key transistor mode) - offset to start. High-resistance ones often break, while low-resistance ones also “successfully” burn out from current overload. We measure resistance primary winding standby trance - should be about 3 or 7 ohms. If the transformer winding is broken (infinity), we change or rewind the trance. There are cases when, with normal resistance of the primary winding, the transformer turns out to be inoperative (there are short-circuited turns). This conclusion can be made if you are confident in the serviceability of all other elements of the duty room.
We check the output diodes and capacitors. If available, be sure to replace the electrolyte in the hot part of the control room with a new one, solder a ceramic or film capacitor of 0.15...1.0 μF in parallel to it (an important modification to prevent it from “drying out”). We unsolder the resistor leading to the PWM power supply. Next, we attach a load in the form of a 0.3Ax6.3 volt light bulb to the +5VSB (purple) output, connect the unit to the network and check the output voltages of the duty room. One of the outputs should have +12...30 volts, the second - +5 volts. If everything is in order, solder the resistor in place.

Checking the PWM chip TL494 and similar (KA7500).
More information will be written about the remaining PWMs.
We connect the block to the network. On the 12th leg there should be about 12-30V.
If not, check the duty desk. If there is, check the voltage on leg 14 - it should be +5V (±5%).
If not, change the microcircuit. If so, check the behavior of the 4th leg when the PS-ON is shorted to ground. Before the circuit there should be about 3...5V, after - about 0.
We install the jumper from the 16th leg (current protection) to the ground (if not used, it is already sitting on the ground). Thus, we temporarily disable the MS current protection.
We close PS-ON to ground and observe pulses on the 8th and 11th legs of the PWM and then on the bases of the key transistors.
If there are no pulses on 8 or 11 legs or the PWM gets hot, we change the microcircuit. It is advisable to use microcircuits from well-known manufacturers (Texas Instruments, Fairchild Semiconductor, etc.).
If the picture is beautiful, the PWM and drive cascade can be considered live.
If there are no pulses on the key transistors, we check the intermediate stage (drive) - usually 2 pieces of C945 with collectors on the drive transistor, two 1N4148 and capacitances of 1...10 μF at 50V, diodes in their wiring, the key transistors themselves, soldering of the legs of the power transformer and the separating capacitor .

Checking the power supply under load:

We measure the voltage of the standby source, first loaded onto the light bulb, and then with a current of up to two amperes. If the duty station voltage does not sag, turn on the power supply, shorting PS-ON (green) to ground, measure the voltages at all outputs of the power supply and on the power capacitors at 30-50% load for a short time. If all voltages are within tolerance, we assemble the unit into the housing and check the power supply at full load. Let's look at the pulsations. The output PG (gray) during normal operation of the unit should be from +3.5 to +5V.

After the repair, especially if there are complaints about unstable operation, we measure the voltages on the input electrolytic capacitors for 10-15 minutes (preferably with a 40% load of the unit) - often one “dries out” or the resistance of the equalizing resistors “floats away” (they stand parallel to the capacitors) - here and glitchy... The spread in the resistance of the equalizing resistors should be no more than 5%. The capacitor capacity must be at least 90% of the nominal value. It is also advisable to check the output capacitances on the +3.3V, +5V, +12V channels for “drying out” (see above), and if possible and desire to improve the power supply, replace them with 2200 µF or better, 3300 µF and from trusted manufacturers. We replace power transistors “prone” to self-destruction (type D209) with MJE13009 or other normal ones, see the topic Power transistors used in power supplies. Selection and replacement... Feel free to replace the output diode assemblies on the +3.3V, +5V channels with more powerful ones (such as STPS4045) with no less permissible voltage. If in the +12V channel you notice two soldered diodes instead of a diode assembly, you need to replace them with a diode assembly of the MBR20100 type (20A 100V). If you don’t find one hundred volts, it’s not a big deal, but you need to set it to at least 80V (MBR2080). Replace electrolytes 1.0 μFx50V in the base circuits of powerful transistors with 4.7-10.0 μFx50V. You can adjust the output voltages at the load. In the absence of a trimming resistor, use resistor dividers that are installed from the 1st leg of the PWM to the +5V and +12V outputs (after replacing the transformer or diode assemblies, it is MANDATORY to check and set the output voltages).

Repair recipes from ezhik97:

I will describe the complete procedure of how I repair and check the blocks.
The actual repair of the unit is the replacement of everything that was burned out and that was revealed by a regular test
We modify the duty room to operate on low voltage. Takes 2-5 minutes.
We solder a 30V variable from the isolation transformer to the input. This gives us such advantages as: the possibility of burning something expensive from the parts is eliminated, and you can fearlessly poke at the primary with an oscilloscope.
We turn on the system and check that the voltage on duty is correct and that there is no pulsation. Why check for ripple? To make sure that the unit will work on the computer and there will be no “glitches”. Takes 1-2 minutes. Immediately we MUST check the equality of voltages on the network filter capacitors. It’s also a moment, not everyone knows. The difference should be small. Let's say up to about 5 percent.
If it is more, there is a very high probability that the unit will not start under load, or will turn off during operation, or start the tenth time, etc.. Usually the difference is either small or very large. It will take 10 seconds.
We close PS_ON to ground (GND).
Using an oscilloscope, we look at the pulses on the secondary of the power trance. They must be normal. What should they look like? This must be seen, because without load they are not rectangular. Here you will immediately see if something is wrong. If the pulses are not normal, there is a malfunction in the secondary circuits or in the primary circuits. If the pulses are good, we check (for formality) the pulses at the outputs of the diode assemblies. All this takes 1-2 minutes.
All! The unit will 99% start and work perfectly!
If there are no pulses in point 5, there is a need to troubleshoot. But where is she? Let's start from the top
We turn everything off. Using suction we unsolder the three legs of the transition trance from the cold side. Next, take the trans with your finger and simply warp it, lifting the cold side above the board, i.e. stretching his legs out from the board. We don’t touch the hot side at all! ALL! 2-3 minutes.
We turn everything on. We take the wiring. We short-circuit the area where the middle point of the cold winding of the separating trance was with one of the extreme terminals of this same winding and watch the pulses on the same wire, as I wrote above. And the same on the second shoulder. 1 minute
Based on the results, we conclude where the problem is. It often happens that the picture is perfect, but the amplitude of volts is only 5-6 (should be around 15-20). Then either the transistor in this arm is dead, or the diode from its collector to the emitter. When you make sure that the impulses in this mode are beautiful, even, and with a large amplitude, solder the transition trance back and look at the outer legs with an oscil again. The signals will no longer be square, but they should be identical. If they are not identical, but slightly different, this is 100% a mistake.

Maybe it will work, but it won’t add reliability, and I won’t say anything about any incomprehensible glitches that might crop up.
I always strive for identity of impulses. And there can’t be any dispersion of parameters there (the same swing arms are there), except in the half-dead C945 or their protective diodes. Just now I made a block - I restored the entire primary, but the pulses on the equivalent of the transition transformer were slightly different in amplitude. On one arm there is 10.5V, on the other 9V. The block worked. After replacing the C945 in the arm with an amplitude of 9V, everything became normal - both arms are 10.5V. And this often happens, mainly after a breakdown of power switches from a short circuit to the base.
Looks like a leak strong K-E at 945 due to a partial breakdown (or whatever happens) of the crystal. Which, together with a resistor connected in series with the build-up trans, leads to a decrease in the amplitude of the pulses.
If the pulses are correct, we are looking for a jamb on the hot side of the inverter. If not - with a cold one, in swinging chains. If there are no pulses at all, we dig PWM.
That's all. In my experience, this is the fastest reliable verification method.
Some people immediately supply 220V after repairs. I gave up such masochism. It’s good if it just doesn’t work, but maybe it will bomb, simultaneously taking out everything that you managed to solder.