# 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

 VOTE
Tags:
59 replies
1. GDChristoff says:

Hello Amilcare,
I noticed some errors in schema2 and schema3. Check please connection of secondary side of current transformers. In schema2 output of 1N4148 rectifier is connected to midpoint of the half bridge capacitors; this connection should be removed! In schema 3 the midpoint of secondary of CT is connected to midpoint of half bridge capacitors, instead to primary side ground.
Also current peak of MCR100 (15V, 220uF/ESR=0.2Ohm, rd of thyristor = 0.1Ohm) will reach value several times more than peak non-repetitive surge current, permissible to this device!
Also worth to mention, that ir 2153 has third comparator in its timing circuit, dedicated to current protection or shut down (by pulling CT pin below 1/6 Vcc). IR2151 and IR2155 do not have this comparator, so they are not fully analogous to IR2153.

 Approvazioni
2. Gabriel says:

Gabriel, Gabriel, Gabriel. Gabriel. Gabriel?. Gabriel

Gabriel, Gabriel, Gabriel. Gabriel. Gabriel?. Thank you

 Approvals
• Amilcare says:

Gabriel.
Initially wrap the primary half then all the secondary ones and at the end the other half of the primary, remembering to separate each winding with at least two layers of insulation.

 Approvals
3. Elisha Lezz says:

Hi, I seem to have a problem with the last circuit diagram of a SMPS you posted, could help by giving the value of R12 which is right next to the transformer.
Your help will be highly appreciated.

I am in the process of constructing this very circuit.
Thank you so much.

 Approvals
4. yuriy says:

Dear Amilcare, before calculation of REon, you wrote: “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”.

 Approvazioni
• yuriy says:

So, why we still need the external resistor Rg? Could you please explain why you select Rg=22Ω. In some datasheets of ST MOSFET I found additional parameter, called ‘Intrinsic gate resistance’ (Rig). This Rig=6.6Ω for STF13N60M2. Thahk you!

 Approvazioni
• Amilcare says:

The driver can only supply a limited current which can be of 210 or 420mA depending on the output state well below 500 and beyond that they would be needed for that MOSFET.
The use of limiting resistors serves precisely to not exceed those current limits and to safeguard the driver itself. When switching begins, the gate capacitance acts as a short circuit for the switching current, then as it charges, its resistance decreases until it tends to zero..

 Approvals
• yuriy says:

As I know, for a bipolar transistor a current must be applied between the base and emitter terminals to produce a flow of current in the collector.
Opposite, a MOSFET, produces a flow of current in the drain when a VOLTAGE is applied between the gate and source terminals.
The gate of a MOSFET is composed of a silicon oxide layer.
Since the gate is insulated from the source, an application of a DC voltage to the gate terminal does not theoretically cause a current to flow in the gate, except in transient periods during which the gate is charged and discharged. In practice, the gate has a tiny current on the order of a few nanoamperes. When there is no voltage between the gate and source terminals, no current flows in the drain except leakage current, because of a very high drain-source impedance.
You can find and download Toshiba MOSFET Gate Drive Circuit Application Note.

 Approvazioni
• Amilcare says:

The gate is similar to a capacitor and the voltage across its plates determines the conduction state of the MOSFET.
It has a linear zone of operation and a saturation zone. Our intent is to reduce linear operation to nanoseconds.
Therefore, this virtual capacitor is charged as quickly as possible.
To charge a capacitor, the only way is to supply electrons to one of the two plates. The potential difference in fact depends on the number of electrons supplied as a function of the physical dimensions of the capacitance. With high current flow it is possible to contain the time necessary to get the MOSFET out of the linear zone. There is no mention of constant current from peaks for very short times, beyond which the current is practically zero if the loss in the dielectric is excluded.

 Approvals
• Amilcare says:

In circuits with high-power MOSFETs with large gate capacities, a pair of additional transistors is often used to overcome the current limit that can be waived by the driver..

 Approvals
5. boredtc says:

Hello,
Ik have created a PCB based on schema1. But the voltage on the ir2153 keeps decreasing over time untill it comes under the 10v limit so the chip stops working. Any ideas? Have build the actual schema or is it pure theory? Thanks!

 Approvazioni
• Amilcare says:

The first scheme is more or less the principle scheme of the date provided by the manufacturer. If the voltage drops, the limiting resistances to the internal zener or an excessive absorption of the output MOSFETs may not be correct..
The pattern is so small that there can't be many elements creating excessive absorption

 Approvals
6. tricon says:

Amilcare: Thanks so much for the advice.

 Approvals