Use and misuse of the IR2153 for feeders up to 1,5KW

This item will be considered in the handbook IR2153, or better will be theoretically presented as a basis for the construction of various pulsed power supplies. IR2153 It is a high-voltage driver with an internal oscillator. It allows to realize switching power supplies power up to 1,5 kW based on a semi-bridge circuits with minimum circuitry.

Given that speak in the article IR2153 and in the title are also present models IR2151 and IR2155 ho

made a table which highlights the key differences,

they are interchangeable until the powers involved are not high, but until we are below 300W nothing prevents you indiscriminately use one of the three, if you do not want to risk you should still get the strongest IR2155.

In the article also I explain what these differences and when to use a template in place of another.

 

There are two versions of the same integrated, they differ only in the presence of a diode to the boost voltage:

Block diagram of the IR2153

Functional scheme IR2153D with internal diode D1

The following figure shows the input stage consists of three op amp and the Flip-flop SR:

At first I had not noticed it, ragionandoci some mental fog has disappeared and I realized where I had seen a similar circuit, turns turns even at a distance of 50 years continue to use 555!!

Block diagram of the 555

Initially when you apply voltage C1 is discharged and the inverting input of op-amp is zero, and the non-non-inverting input has a positive voltage provided by the resistive divider. The result is that all three have the output voltage at logic level one.
Because the R input is active with zero level has no influence on the output condition, but at the S input will set the output Q of the flip-flop that C1 starts to charge through the resistor R1.

The course of the voltage on Ct is shown by the blue line, red in the output voltage of OP1, green in the output of Op2, Rose, the Q output of the flip-flop and the Q output mustard denied.

As soon as the voltage on Ct exceeds 5 V the output OP2 It goes to zero, continuing to load C1 the voltage reaches a value slightly higher than in 10 Volt and this time goes to zero the output of OP1, which in turn leads to zero the Q output of flip-flop. From this point C1 begins to discharge through the resistor R1, and as soon as the voltage across it will be a bit 'less than 10V l'uscita Op1 returns to 1. When the voltage on the capacitor Ct becomes less 5V uscita of Op2 will reset the flip flop and restart charging Ct.

In the chip there are two additional modules UV DETECT e LOGIC. The first of them is responsible for enabling the charging process of C1 only above a certain power supply voltage and generates the second delay pulses that are necessary to prevent both endings are conductive at the same time by short-circuiting the power supply through the power stage.Next there is a division of logic levels for the middle and upper deck to the lower one.

Consider the typical simplified scheme of the IR2153:

I pin 8, 7 e 6, respectively, are the output VB, HO and VS, that is, power management upper branch (VB), piloting of the final stage (HO) and the negative of the upper branch of the control module (VS). Attention should be paid to the fact that when the lower branch is active, the diode D1 starts to load C3 fact with T2 into conduction, the capacitor is virtually connected to ground.
Once the outputs change state T2 isolates , and HO via T1 goes into conduction. At this point, the voltage on VS begins to increase up to the level of overall positive power, given that T1 under these conditions has a resistance of tenths of ohms.

It turns out that the maintenance of the conducting transistor requires a gate voltage at least 8 volts higher than the supply voltage, It provides precisely C3 loaded 15 volt, allowing you to keep T1 conductive, thanks to the energy stored in it when T2 was conductive thanks to D1. In this phase, the same diode does not allow capacitors to discharge on the same power supply.
As soon as the pin on impulse control 7 terminates the transistor T1 opens and closes in its place T2, charging again the capacitor C3 to a voltage of 15 V. The C3 value greatly depends on the time in which it is conductive T1. You should avoid using electrolyte for operating frequencies above 10KHz, Whereas such Ic is able to operate from 10Hz a 300KHz.

Working virtually from 40 and 80KHz through use, would be enough power capacity 220nF , to be sure you choose the value of 1uF. The oscillator frequency, It can be determined from this graph in the data of the integrated, for convenience I enclose a copy.

Choosing an appropriate mosfet

In the table below I have summarized the characteristics of the most used mosfet that I found around this integrated. A helpful in case you have to look for the equivalent in case of breakages, sifting in the table we can choose between the models in our possession that although not precisely equivalent goes well anyway.

It can be used for power supply design from scratch.

Calculation of the relative gate resistance

As known, the dynamic properties of a field-effect transistor are not characterized more accurately by the value of its parasitic capacitances, but from the total charge of the gate-Qg. The Qg parameter value is mathematically interconnected by a pulse current with the gate of the transistor switching time, thus allowing the developer to properly calculate the control node.
Take for example the MOSFETs IRF840 very common and present in the table.

With a drain current Id = 8 A, una tensione drain-source Vds= 400 V and a gate-source voltage Vgs = 10 V, is the gate charge Qg = 63 nC.

It should be specified that with the same Vgs, the gate charge decreases with an increase of the drain current Id and with a decrease in the voltage Vds, in the calculations seen that both voltages are constant for good take the value provided by the manufacturer, small changes do not affect the final result of the calculations.
We will calculate the parameters of the control circuit, provided that it is necessary to reach the switch-on time of the transistor ton = 120 ns. To do this, the driver control current must have the value:

Ig= Qg/ ton= 63 x 10-9/ 120 x 10-9= 0.525 (A) (1)

When the amplitude of the control voltage pulses on the gate Vg = 15 V, the sum of the driver output resistance and the resistance of the limiting resistor must not exceed:

Rmax= Vg/ Ig= 15 / 0.525 = 29 (Ohm) (2)

We calculate the output resistance in cascade output of the driver for the chip IR2155:

Ron= Vcc/ Imax= 15V / 210mA = 71,43 ohm
Roff= Vcc/ Imax= 15V / 420mA = 35,71 ohm

Taking into account the value calculated according to the formula (2) Rmax = 29 Ohm, we conclude that the specified speed of the transistor IRF840 It can not be obtained by the driver IR2155. If in the gate circuit, a resistor Rg is installed = 22 ohm, the ignition time of the transistor is defined as follows:

REon= Rg+ Rf, dove

RE = total resistance

Rf = the driver output impedance,

Rg = external resistance in the gate circuit of the power transistor

Reon = + 71,43 = 93,43 ohm;
Ion= Vg/ RE, dove

Ion = the drive current

Vg = value of control gate voltage

Ion= 15 / 93,43 = 160mA;
ton= Qg/ Ion= 63 x 10-9 / 0,16 = 392 nS
The sleep time can be calculated with the above formulas:

REoff= Rf+ Rg= 35,71 + 22 = 57,71 ohm;

Ioff= Vg/ Reoff = 15/58 = 259mA

toff= Qg/ Ioff= 63 x 10-9 / 0,26 = 242nS
To obtain the value of the real time it is necessary to add the time that physically employs the transistor to pass from one stage to another and that is 40ns for the on condition, and 80ns for that will be off the real-time

Your 392 + 40 = 432nS, e Toff 242 + 80 = 322nS.

Now it remains to be determined whether a power transistor will have time to completely close before the second beginning to open. A tal fine, we add Ton and Toff to get 432 + 322 = 754 nS, equivalent to 0,754 μS.

From the data shows that the DEAD TIME the IR2151 can not be used as it is 0,6 μS.

In the datasheet it says that Deadtime (tip.) It is fixed and depends on the model, but there is also a very embarrassing figure from which it emerges that DEAD TIME and the 10% of the control pulse duration:

To dispel doubts, I did some testing with a two-channel oscilloscope on a basic circuit to see what came out, the curiosity of a child for things to me not the new've never lost, this is the result:

The power was 15 V and the frequency was 95 kHz. As you can see from the photograph, with a scan 1 μS, the length of the pause is a bit 'more than one division, that exactly matches 1,2 μS. In addition, reducing the frequency you can be seen as follows:

As you can see from the picture to the frequency of 47 kHz, the pause time is virtually unchanged, then the part that says that Dead time (tip.) 1,2 mS is true.
Since the circuits were already functioning, it was impossible to restrain another experiment, lowering the supply voltage to ensure that the frequency of the generator did not increase. The result is the following image:

However, expectations were not justified, instead of increasing the frequency I witnessed its reduction, Fortunately, however, the variation is less than 2%.

Values ​​negligible considering a change in the supply of more than 30%. It should be also noted that the pause time is slightly increased. This fact is quite good, when the control voltage decreases slightly changing the opening and closing time of the power transistors and increase the break in this case is very useful.
Note that
UV DETECT with its function, blocks the oscillator in the event of excessive lowering of the power supply and then reactivate the chip when it goes beyond the minimum level.
Now back to our example, with the gate resistor 22 Ohm closing and opening is still 0,754 uS con IRF840, this value is less than the pause 1,2 uS, typical of the chip itself.
So, with the IR2155 and IR2153 through resistors 22 Ohm can control IRF840, but surely IR2151 will be discarded, as well as too low dead time because transistors need to be aware of 259 but it is 160 mA, while the IR2151 has the maximum value is 210 but it is 100 ma. obviously, it is possible to increase the resistance installed in the gate of the power transistor, but in this case there is the risk of going beyond Dead time.
To reduce the switching noise of the power transistors in the switching power supply uses a shunt resistor in series with a capacitor in parallel to the winding of the transformer. This node is called snubber. The resistor of the suppression range is selected with an assessment of 5-10 times greater than the drain-source resistance of the MOSFET.

The capacitance is determined by the expression:
C = TDT / 30 x R
TDT is the off time of the upper and lower transistor.

Based on the fact that the duration of the transitional amount to 3RC, it must be 10 times less than the duration of TDT.
The snubber delaying the moments of opening and closing of the FET control voltage fluctuations with respect to its door and reduces the rate of change of voltage between the drain and the gate. Consequently, the peak values ​​of impulsive current pulses are minor and their service life is longer. Almost without changing the turn-on period, the damping circuit significantly reduces the FET off time and limits the interference generated spectrum, you can find it in the position drawn indifferemtemente or directly in parallel to the winding of the trasgormatore, the differences between the two configurations are so marginal as to be considered interchangeable in practice.

Here are some practical patterns seen around.

In almost none of the following schemes the number of coils in transformers is indicated because they are to be calculated according to the characteristics of the transformer itself and also because in most cases the patterns that I found is not specified.
The easiest switching power supply with IR2153 It is an electronic transformer with minimum functions:

in the scheme 1, there is no extra function, and the secondary is formed by two bipolar power rectifiers consisting of a pair of double Schottky diodes. The ability to 220 uF output to the bridge is calculated with the empirical formula of 1 uF per watt on the load. In this case it is used for a stereo amplifier 100W per channel. The two capacitors 2u2 on the primary of the transformer are placed in the range from 1 a 2u2 .

The power depends on the transformer core and the maximum current of the power transistors and in theory can reach 1500 watt. Practically, in this scheme it depends on the maximum current of the transistor temperature STP10NK60Z, the maximum current of 10 A If you have only 25 degrees. When the temperature of the silicon salt to 100 degrees is reduced to 5,7A, and talk about the temperature of the silicon, rather than the temperature of the heatsink.
Thus the maximum power must be chosen according to the divided transistor current 3, if you feed a power amplifier and divided by 4, if one feeds a constant load, such as incandescent lamps.
That said you can theoretically power an amplifier

10/3 = 3,3A 3,3A x 155V = 511W totali.

For a constant load 10/4 = 2,5 A 2,5 A x 155V = 387W.

From calculations reference is made to a fixed voltage 155V, where it comes from that value? It is derived from the effective voltage on the smoothing capacitor at maximum power, the value is empirical but, It is not very different from the real value and allows us to simplify our lives without too great deviations from real.

In both cases, it theorizes a yield of 100%, that can not be reached .

Furthermore, wanting to obtain the maximum power of 1500W given the need to 1 uF of the primary power supply capacity for each watt of power on the load, It needs of one or more capacitors to get to 1500 uF Total and to load them should be a soft-start in order not to jump the counter at each switch.

greater power and protection in current in the following scheme 2:

This is implemented by an overload protection thanks to a current transformer. In most cases you use a ferrite ring with a diameter of 12 a 16 mm, in which are enveloped by 60 a 80 bifilar coils of insulated wire diameter 0,1 mm. To form a winding center tap for the secondary. The primary winding is made by winding from one to two coils, sometimes for convenience also it makes a turn and a half, when it comes into function lowers the power of the integrated circuit causing, thanks to the internal protections stopping the driving of the final. Once scaricatosi electrolyte SCR turns off and reforms the right to return power to run the final regular.
Two resistors 62K in parallel allow to feed the integrated with a good primary power excursion (180 … 240V). In order not to overload the internal zener diode if it uses an external from 1,3 W a 15 V.
An additional circuit based around the bottom transistor allows a gradual start with lower frequency, until the full charge in ± 80 V of capacitors 1000 uF.

With the divider 330K-4k7 and the diodes connected to it is loaded initially from the electrolytic 4u7, such a gate voltage in the transistor which increases the capacity of the oscillator, just enough time to charge the capacitors too without overloading the transformer ferrite.

After this time, the transistor island and the integrated back to running at its operating frequency.

The presence of a snubber network eliminates a good portion of clutter caused feeder.

A further embodiment of the pulsed power supply capable of providing to the load 1500 W contains a soft-start system for the primary power supply while the secondary has an overload protection, also it creates a tension for the fan for forced ventilation of the cooling fin. The problem of the rapid shutdown of the power mosfet is solved by means of two transistors BD138, they discharge the gate capacitance of the mosfet with extreme simplicity.

Such a system allows the use of relatively powerful elements as IRFPS37N50A, SPW35N60C3, not to mention and IRFP360 IRFP460 about.
At the time of the primary ignition voltage on the power diode bridge arrives through the resistor 360 Ohm, since the relay is open. Furthermore, the voltage across the resistor 47k It is applied to the chip, simultaneously via the two resistors 33 e da 360 which refer to the FAN terminal and the relay coil. With them, the capacitor is charged gradually from 100uF Since the second part of the relay coil is part of a zener diode and the scr soon as this voltage reaches 13V It will trigger the scr that will excite the relay. Here you have to remember that the IR2155 already begins to operate with a supply voltage of approximately 9 V, then upon excitation of the relay it already works by generating control pulses for driving the primary.

Piloting that occurs with reduced power seen that with the open relay passes through the resistor 360. It is essential that trick to limit the charging current of the secondary power supply filter capacitors. Once the relay coil is energized by the thyristor his contacts shunt both limiting resistors.
In the transformer there is provided a further winding for supplying the cooling fan (FAN), its resistance with a current limiting.

Lately I was required a low voltage stabilizer but starting from a high starting voltage, the one below is an elegant solution to this problem, the MOSFET T2 is exploited as if it were a diode, when you climb the tensions find fast diodes that withstand high currents are certainly more expensive than a common mosfet.

the scheme 5 using the IR2155 for a voltage booster circuit. In this scheme, the tall driver is connected to the supply voltage:

As in the previous embodiment, the closure of the power transistors is made with the two BD140. Initially part of the edge of the car battery with 12V and then fed voltage stabilized at 15 V via the diodes of the extra-voltage suppression, the limiting resistor and zener stabilize the supply voltage of the integrated.
Not present in the diagram there is a thermal switch to be fixed in the fins, it will stop the REM voltage by switching off the integrated. These diodes must be fast fast SF16 series, HER106, etc.

With this I think I have clarified many aspects of this family of integrated, but as a last treat, place the adapter that I use for my amplifier 200+200W, realized with a transformer recovered from a PC power supply saved from landfill.

Unica addition not present in the diagram and a snubber network composed of resistor 100 Ohm with a series capacitor from 100 pF in parallel to each diode on the secondary.

Such subsequent modification also makes it suitable for classical linear amplifiers.

In it there is a soft start an emi filter and a protection from excessive absorptions, many of passive components are the original power, why look elsewhere for what I had on hand?

I deliberately did not explain this latest scheme to see if what was said in the article really good for something.

Amilcare Greetings

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  1. stony stone
    stony stone says:

    very nice this solution. simple and without many caps around the transformer.
    I continued on the path of the scheme 3. I adjusted the speed to 60khz with about rt 10k and ct 1nf and calculated etd49 with 19 turns on the primary 13+13 for 70v + 70v e 2+2 revs for 21v to be adjusted to 15v with regulator ic. I'd like to show the changes to understand what you think. how can I do?

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    • Amilcare
      Amilcare says:

      At the top you will find the new button, from the drop-down menu select “media”
      From there you can upload the pattern you used, as a final result it gives you the link to the uploaded image which you copy and paste into the message.

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        • Amilcare
          Amilcare says:

          I would raise R16 and R17 so that the transistors intervene more quickly. R22 limits the current peaks too much. The power supply used will hardly exceed the ampere, current that the transistors used calmly withstand, I would eliminate it or at least reduce it by a decade.
          Calmly at the PC I will see if there are other things

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          • stony stone
            stony stone says:

            I want to briefly explain the circuit. at the entrance there is a simple soft start circuit used for large loads. it has a relay with two switches. the first switch with pins 6,7,8 the second with pins 3,4,5.
            when to give power the relay passes the current from the pin 6 al pin 8 where a resistor is connected from 500 ohm 5 watt charging large capacitors. the second switch, on the other hand, is closed. when the soft start is ready the pin 6 from direct current to the caps and the second pin switch 3,4,5 from current to small ac dc 15 was modulo, which starts ir2151d with delay.
            There are 3 protections, one with temperature, one for high current formed by Q6,7,8 and one to block it 0 negative of ir2151 if the R13 resistor jumps. I put 150ohm resistor on the collector head of transistors q2 and q3 to limit the current to 100milli amps
            I inserted resistance

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          • Amilcare
            Amilcare says:

            Your 15V 5W power supply module delivers 300mA in total. Beyond that current, the voltage naturally drops without the need for limitations.
            To have switching edges, peak currents even higher than 1A for 1uS are required, after that time you have no more absorption on the gate. If you put a limiting resistor you will never be able to have steep fronts, effectively nullifies the advantages of added transistors on the gate circuit. They should guarantee the highest possible impulse current to reduce the duration of the edges with the advantage of having greater efficiency and less heat dissipation..

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    • Picmicro675
      Picmicro675 says:

      I would recommend opening a discussion to be able to follow in chronological order and better organized. I do not agree to use the comments for problems other than the published article.

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  2. Amilcare
    Amilcare says:

    Assuming that the transformer is intact, you must try to jot down a schematic of the old power supply.
    It is necessary to understand if it was initially half-deck or full-deck. This is to understand how the transformer was made. Once the network is straightened you have electrolytics in series or in parallel?
    The control unit of the driver side transformer is
    1)connected to a small capacity capacitor
    2) connected 300Vdc
    3) connected to the midpoint of the series of the input smoothing capacitors.
    Depending on the configuration, the driving circuit is different and the transformer is certainly calculated on that configuration that will be replicated.

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    • stony stone
      stony stone says:

      Thanks for your informations, unfortunately I abandoned the idea of ​​wanting to reuse the old smps. I have tried various ICs and changed some values, but it keeps making a bad noise, in addition, the output is not stable and to avoid wasting any more time on them I have dismantled all the cards. But I would like to reuse a lot of components to build your smps in the schematic 3. I think 1500watt maybe too 2000 for small moments it is a good product. Many components are already available, but i would like to know if i use stw45nm50 i have to change some value? and above all I have to redo the windings of the transformer tdk etd49 PC95. For this I need help.

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      • Amilcare
        Amilcare says:

        The primary is calculated on the deliverable power of the transformer. And it's usually supposed to be done right , if you keep the same piloting configuration there is no need to recalculate that section. You just have to determine how the input stage is configured whether half-bridge or full-bridge. Those MOSFETs you intend to use are fine. For the excessive noise it could be due to some component in the secondary or the damping network parallel to the primary winding.
        If you want to rewind the transformer fearing insulation losses, be aware that the primary is divided into two parts. The first innermost attached directly to the spool and the second on top of all other windings. This is the only way to contain the inductances and parasitic capacities that would limit the total efficiency.

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        • stony stone
          stony stone says:

          excuse the Italian, the old one is a whole bridge, it no longer exists, . the control board with l6599 buzzes, it whistles loudly,
          the secondary instead of delivering double 70v goes beyond 100 , while with a load, also a light bulb from 50 watt, the output drops to 5-10 volt.
          the transformer etd 49 it's not like the others, it has two separate offices, one for primary and one for secondary, see type https://it.aliexpress.com/i/32816022767.html?spm=a2g0y.12057483.0.0.29746690QjjiWe
          they are not wrapped, layer upon layer as you say.
          however it is not a problem to build a new transformer, i have a lot of etd media, 34, 44 e 49,, the core is missing but I can buy.
          , I also have many meters of transformer wire 0.07mm x 150 wires or 0.1mm x 120 wires. It's not a problem, but I don't know how many laps for the primary and how many laps for the secondary. if your scheme can give as much power as I can get from it double 70v and double 21v

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          • Amilcare
            Amilcare says:

            Today I started doing some complex calculations and for an ETD59 transformer with N87 ferrite(the most common) I calculated the turns necessary to have 2000W continuous.
            Input filter capacity on the mains voltage 1000uF and a variation of the mains voltage 10%
            number of turns to the primary 12+12
            number of turns to the secondary for dual 70V 6+6
            number of turns on the secondary 21V dual 1,5+1,5
            all this for an operating frequency of 60KHz
            One of these days I will write a tutorial on the necessary calculations that led me to this data.
            A soft start has to be foreseen because upon switching on the capacity would cause the counter to jump.
            Last thing, the primary requires a wire section of at least 2.5mmq, better 3mmq to withstand 13A and the same section should be used for the dual 70V section of the secondary

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          • stony stone
            stony stone says:

            many 2000watt, the nucleus 59 is higher than 49, much bigger
            how come on the primary there are 12+12 rounds? they are with central channel or are two parallel windings, I do not understand.
            for the soft start circuit, that of the diagram 3 it is good if it works as in description

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          • Amilcare
            Amilcare says:

            We are talking about 2000W continuous and to avoid overheating that makes them last longer it is better to abound a little, transformer has 1cm more footprint, it's not all that different from the ETD49 where the number 49 indicates the largest size, 49mm while in what I proposed 59mm
            like all these power supplies they need a minimum load to function properly
            The circuit 3 it is very good so you can take it as a starting point, what changes is only the connection to the primary which is slightly different to take full advantage of all the power without unnecessary complications. as it also greatly simplifies the protection of reverse voltages that flow onto the semi-winding not in use. with these powers at stake, even that aspect is essential to avoid merging the power elements.

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  3. stony stone
    stony stone says:

    good evening, let me introduce myself my name is Pietro Ruta and I found your article very interesting, I state that I don’t have much knowledge of smps, but I like to build audio amplifiers. lately i came across these smps power supplies circuit with ir21531. I have seen that they are very simple to make and very efficient. but lately i would try to build one of these smps to power a powerful subwoofer. some time ago I bought the grande smps llc 3000watt full bridge from China, but due to a terrible storm it stopped working. unfortunately many components have been hidden, the codes have been canceled by the seller and repair operations are very difficult. from a careful study I was able to read the power MOSFETs. they are stw45nm50 and i also believe the l6599d driver. but despite the replacement it does not work properly, it emits a loud buzz …. then here is my question, I want to replace all the control and modify smps, but I need help … meanwhile the amplifier that I have to power works with double 70v, it generates an audio power of 1000watt RMS on a load of 8hom. the power supply fuses are 6ampere per branch, so I think 1kw of smps is enough. my idea is to recover all the components of the old smps and rebuild it using this ir21531 or higher model ir2155 as driver and a pair of stw45nm50 to push the etd49 transformer. in your opinion the ir2155 can drive stw45nm50?

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    • Picmicro675
      Picmicro675 says:

      You probably fall into a failure. In particular case to fight the noise created by high frequencies generated by these SPMS. Your PSU will need a very well engineered filter to suppress all the harmonic frequencies.

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      • stony stone
        stony stone says:

        i understand, but my intent is to completely disassemble the old smps and create a new one but with lower power, and use this little ir2155 driver for a few euros. it bothers me a lot to have to spend a lot of money to buy a new one, when the only burned component is the l6599 driver (but again it’s my thought, because all the components on the board have the codes cleared)

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    • Amilcare
      Amilcare says:

      That type of MOSFET can be driven by the IR2153 but keep in mind that the connections between the power elements on the primary of the transformer and the driver must also be made carefully otherwise the result will be poor.

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    • Amilcare
      Amilcare says:

      If it is hot it means you're using it beyond the limits and the core is going into saturation, therefore I would opt to raise the operating frequency so that the magnetizing current does not rise to levels of saturation. I do not know if you know this but the current in a coil increases linearly with time, the lower the conduction time and the lower the current flowing through the coil and the lower the magnetic flux in the core. Almost always there is the need to leave a gap between the two sections of the ferrite, two overlapping layers of insulating tape are sufficient in most cases.
      I'm trying to avoid complex terms and complicated formulas, you are a hobbyist and do not think you care very much what and from what I understand in the field texts your ideas rather than solving mathematical questions and only after taking tests, therefore I try to give empirical solutions, (Here some I excommunication electronics purist about what I said 😉 ).

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      • maredentro72
        maredentro72 says:

        😭Si in fact I am a hobbyist trying on the field and I follow the theory written by others, the transformer has two windings 20+20 d0.6 coils in the primary isolated with yellow ribbon type to understand, the secondary is a 10+10 with a central outlet and semore intermixed with insulating appisito. Then I gave a slight resin epoxy but maybe this could risparmiarmela.
        I'll try to work on frequency of operation, I now have a resistor of 47k and then I should be on 330pf 40-50 kHz, I'll try to get to 60-80 kHz.

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          • stony stone
            stony stone says:

            I write in Italian, maybe it's better. I introduce myself my name is Pietro Ruta and I found your article very interesting, I state that I do not have much knowledge of smps, but i like to build audio amplifiers. lately i came across these smps power circuits with ir21531. I have seen that they are very simple to make and very efficient. but lately i would try to build one of these smp to power a powerful subwoofer. long ago i bought smps llc 3000watt big bridge from china, but due to a terrible storm it stopped working. unfortunately many components have been hidden, the codes have been cleared by the seller and repairs are very difficult. from a careful study I was able to read the power MOSFETs. i am stw45nm50 and i also believe l6599d driver. but despite the replacement it does not work properly, emits a loud buzz .... so here's my question, i want to replace all control and edit smps, but i need help … meanwhile, the amplifier I have to power works with double 70v, generates an audio power of 1000watt RMS on an 8hom load. the power supply fuses are 6ampere per branch, so i think 1kw of smps is enough. my idea is to recover all the components of the old smps and rebuild it using this model ir21531 or higher ir2155 as driver and a pair of stw45nm50 to push the etd49 transformer. according to you the ir2155 can drive stw45nm50?

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    • Amilcare
      Amilcare says:

      Maybe I explained myself wrong, the ferrite core is made from two pieces joined together, the thickness must be put between the two pieces of the nucleus. For the primary it is fine divided into two parts but the second part of the primary must be placed after wrapping all secondary. You have a higher yield and less malfunctioning.

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      • maredentro72
        maredentro72 says:

        That this is interesting. I am looking for a new core or do I remove one and redo everything. I open a discussion in the forum on the subject elettroamici, maybe back to business also useful to others.

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  4. maredentro72
    maredentro72 says:

    Amilcare Good morning and thank you for this page. Skipping the pleasantries and I come to the point; I do not want to spend € 50-100 for a transformer for making a variable dual power supply with voltages between 40 90Vdc and 500w for branch with which the test audio amplifiers that delight me to realize. So a smps to be right for me and how I did it to me I have to make from scratch. The integrated dedicated are many so I have taken of various types with different types of MOSFETs… Some disassembled by ATX other scratch. Trying fairly simple patterns I saw that those who use IR215x family are perfect for my needs.
    I found that a classic linear power supply is a no brainer to do it, a smps is another world. Luckily I took mosfet, integrated and I made some EE transformer 35.
    Morals have disintegrated it twenty to do various tests. The IR2155 is better, I have to realize a power supply from at least Total 150-200vdc with at least 2 – 3A total so 400 600w total. I used to like the MOSFET IRF740 but in the final version I would use something more robust, magari IGBT.
    Step questions always that you can and have time to respond.
    The resistance to the integrated alimeontazione I found in various values ​​in the schemes but to me it only works a 22k and 10w. (5w too warms up to 150 ° C). And if I change places all burn.
    Once I turned on and running all I tried to connect load as a halogen lamp 400w (approximately 10ohm cold resistance) ma… Bum!
    After going up all over again, Now I tried to do yna change for the variable frequency 40 a 90khz… I hope to do more smoke.
    If you have suggestions to give me for my experiments are all ears.

    Alberto

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    • Amilcare
      Amilcare says:

      A control on the output current is always worth to take into consideration the destruction of the MOSFET at the time of connection of the load power supply. A zener from external to the integrated 15V in case of amateur productions would be a good idea. In industrial ones it should not be proposed either because they tend to save on everything. Another question is the capacity on the 320V line must have a total capacity of 1uF for Watt Output. Finally proper snubber should be considered not to merge the mosfet with overvoltage. The variable version is not so difficult to be used, the protection circuit on the current to turn off the integrated soon as you reach the desired voltage.

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        • stony stone
          stony stone says:

          good evening, I introduce myself my name is Pietro Ruta and I found your article very interesting, I state that I do not have much knowledge of smps, but i like to build audio amplifiers. lately i came across these smps power circuits with ir21531. I have seen that they are very simple to make and very efficient. but lately i would try to build one of these smp to power a powerful subwoofer. long ago i bought smps llc 3000watt big bridge from china, but due to a terrible storm it stopped working. unfortunately many components have been hidden, the codes have been cleared by the seller and repairs are very difficult. from a careful study I was able to read the power MOSFETs. i am stw45nm50 and i also believe l6599d driver. but despite the replacement it does not work properly, emits a loud buzz .... so here's my question, i want to replace all control and edit smps, but i need help … meanwhile, the amplifier I have to power works with double 70v, generates an audio power of 1000watt RMS on an 8hom load. the power supply fuses are 6ampere per branch, so i think 1kw of smps is enough. my idea is to recover all the components of the old smps and rebuild it using this model ir21531 or higher ir2155 as driver and a pair of stw45nm50 to push the etd49 transformer. according to you the ir2155 can drive stw45nm50?

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          • Picmicro675
            Picmicro675 says:

            You make me waste my fingers writing you in English? 😀
            I can't say more about what I know, I am not experienced in SPMS. Wait for Amilcare to give you some indication.
            In the meantime if you already have the components, test and post the results you can get. Even the oscilloscope screens.

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  5. A_X_C
    A_X_C says:

    A valuable aid for the design and even more valuable because it uses simple formulas.
    A pocket calculator just for all calculations.
    Thanks to the simplifications of which you have made us partakers.
    You did seem simple a complex as a switching power supply
    THANK YOU

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