Audio smps 700w (IR2153)

Respecto al transformador no sé qué número de vueltas usaste en el primario. Si tiene un espacio en el transformador, la inductancia para un número determinado de vueltas será menor. El número de vueltas se calcula según la frecuencia de conmutación. Cuanto menor sea la inductancia, mayor debe ser la frecuencia de conmutación. Esta topología HB no necesita un espacio en el núcleo central del transformador. El número de vueltas que se muestra en el esquema es para un transformador ETD39 sin espacio. Su transformador tiene un espacio y la inductancia se vuelve bastante baja para el número de vueltas. Le estoy publicando un enlace para un video ruso para que vea cómo llenar el vacío en su transformador.
ver de 2:10 a 4:10. Enlace [MEDIA = youtube] RH6axUpVDKM [/ MEDIA]
Si el relleno del núcleo central aumenta, puede lijarlo con un trozo de papel de lija colocado sobre una superficie plana (un trozo de vidrio es mejor)
Notarás que solo la mitad del núcleo tiene una pierna central corta y la otra mitad tiene sus 3 piernas iguales. Llenarás la mitad con la pierna central corta.
El uso de una frecuencia más alta puede causarle problemas con el calentamiento de los mosfets y también del transformador.

Saludos Silvio
Excelente aporte, muy buen video explicativo. Yo uso EE42.
primario: 28T (4x0.45 mm) trenzado. secundario: 11T + 11T (6x0.45mm) en paralelo.
 

Silvio

Active member
Which EE42 are you using 42/21/15 or 42/21/20? And what is the frequency of operation? The primary turns differ from one core to another.
 
¿Qué EE42 estás usando 42/21/15 o 42/21/20? ¿Y cuál es la frecuencia de funcionamiento? Los giros primarios difieren de un núcleo a otro.
¿Qué EE42 estás usando 42/21/15 o 42/21/20? ¿Y cuál es la frecuencia de funcionamiento? Los giros primarios difieren de un núcleo a otro.
I am using 42/21/15 ..... based on this core and the 28T that I put in the 4x0.45mm primary is that with a software I assumed it would work better at 50kHz so I adjusted it to that frequency and gave it a try again with the load having the same results. At AC input (265W) at DC output (165W) ... since I don't have a wattmeter, I decided to test by measuring at DC input after rectification and there it was around 163W.
Measures:
295 V DC * 0.7 A = 206 W
135 V DC * 1.2 A = 162 W
Efficiency: 78%
There it looks prettier, but I still wonder how often it should be working and also why such a difference with the AC input, does it mean that it has a very ugly cosine? (reagent)



I continued the tests with different loads always taking the DC prim and DC sec:
300Vdc * 1.7A = 510W
135Vdc * 3.5A = 473W
Efficiency: 92%

another purge (another load):
294Vdc * 2.55A = 750W
132Vdc * 5.5A = 726W
Efficiency: 97%

I notice that the greater the load, the greater its efficiency ........... ok? Or shouldn't efficiency be maintained regardless of load?
 
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Silvio

Active member
I am using 42/21/15 ..... based on this core and the 28T that I put in the 4x0.45mm primary is that with a software I assumed it would work better at 50kHz so I adjusted it to that frequency and gave it a try again with the load having the same results. At AC input (265W) at DC output (165W) ... since I don't have a wattmeter, I decided to test by measuring at DC input after rectification and there it was around 163W.
Measures:
295 V DC * 0.7 A = 206 W
135 V DC * 1.2 A = 162 W
Efficiency: 78%
There it looks prettier, but I still wonder how often it should be working and also why such a difference with the AC input, does it mean that it has a very ugly cosine? (reagent)



I continued the tests with different loads always taking the DC prim and DC sec:
300Vdc * 1.7A = 510W
135Vdc * 3.5A = 473W
Efficiency: 92%

another purge (another load):
294Vdc * 2.55A = 750W
132Vdc * 5.5A = 726W
Efficiency: 97%

I notice that the greater the load, the greater its efficiency ........... ok? Or shouldn't efficiency be maintained
Hi first of all it should be noted that the peak voltage within a capacitor charged from a sinewave cannot hold much as if you consider the sine shape it is rather narrow at the top. So the peak voltage will drop very quickly due to this. The DC voltage will get stronger as it lowers down due to the sinewave is more wider in the middle. It is important to have adequate capacitance in the input. As a rule of thumb you should have 1uF per watt of power consumed at 320vdc. This means that if you have 2 capacitors rated 200v and have a capacitance of 500uF each then total capacitance across the 320v is 250uF and the voltage rating is 400v because they are wired in series.

Regarding your turns for the transformer you may have a little more turns than needed in the primary and for EE42 42/21/15 should be 24 turns primary for a flux density of 0.16 tesla and a peak input voltage of 245vac. I guess the smps will give you better efficiency at 50Khz because at this frequency you have the correct number of turns in the primary. (Lower frequency more turns)

Power factor of 1 means that the voltage wave and current wave are in phase. Usually an smps present a power factor of 0.5 when not loaded. This means that the voltage wave is leading the current wave. Due to this the smps shows that it is drawing more current than it is consuming. That is why I told you to use a watt meter.

When an smps is loaded the power factor will get closer to 1 (unity) That is why you seem to get more efficiency. That is why nowadays they are using Power factor correction in smps. Some also use boost PFC to hold the primary DC voltage stable.

I hope I made things clear enough for you Regards Silvio
 
Hi first of all it should be noted that the peak voltage within a capacitor charged from a sinewave cannot hold much as if you consider the sine shape it is rather narrow at the top. So the peak voltage will drop very quickly due to this. The DC voltage will get stronger as it lowers down due to the sinewave is more wider in the middle. It is important to have adequate capacitance in the input. As a rule of thumb you should have 1uF per watt of power consumed at 320vdc. This means that if you have 2 capacitors rated 200v and have a capacitance of 500uF each then total capacitance across the 320v is 250uF and the voltage rating is 400v because they are wired in series.

Regarding your turns for the transformer you may have a little more turns than needed in the primary and for EE42 42/21/15 should be 24 turns primary for a flux density of 0.16 tesla and a peak input voltage of 245vac. I guess the smps will give you better efficiency at 50Khz because at this frequency you have the correct number of turns in the primary. (Lower frequency more turns)

Power factor of 1 means that the voltage wave and current wave are in phase. Usually an smps present a power factor of 0.5 when not loaded. This means that the voltage wave is leading the current wave. Due to this the smps shows that it is drawing more current than it is consuming. That is why I told you to use a watt meter.

When an smps is loaded the power factor will get closer to 1 (unity) That is why you seem to get more efficiency. That is why nowadays they are using Power factor correction in smps. Some also use boost PFC to hold the primary DC voltage stable.

I hope I made things clear enough for you Regards Silvio
I was thinking of using the EE55 core (horizontal) and feeding the IC externally or with another method, I only have 0.45mm / 0.70mm / 1mm wire, do you have any projects or have you seen one in the forum to read?
Thank you
IvanElectric
 
An EE55 can give you 2KW of power. what are your requirements in this supply?
The truth is that I carry out my own projects and those of others, analyzing functionality, protections, IC or transformer driver, complexity, etc ...... all as a hobby and I have a box of those horizontal EE55s that I bought and I still haven't left the classics. IE33, EE42.
 

Silvio

Active member
As far as gating the mosfets or IGBTs I found best to use a mosfet driver such as IR2110 etc. With a GDT you get more isolation especially not to mix Low and High voltage grounds . Opto couplers have their limits when it comes to high output voltages such as trying to regulate the full voltage of a dual supply (60-0-60). GDT tend to spike during switching like all transformers do and suppression may be needed at times. However mosfet drivers give a good clean waveform with little complications and the one mentioned above is capable of delivering 2 amps of current.

Considering protection we usually use a voltage comparator such as the LM393 or an LM358 followed by an NE555 to latch on the shutdown pin of the switching IC if using one with only voltage control such as the SG3525. Personally I like to use this chip as it has a totem pole in its outputs. However a good chip with 2 opamps for regulation of voltage and current such as the SG3524 or TL494 are also to be considered but these are best considered for a regulated output and perhaps fold back current protection. These chips have only one transistor on the output and a further transistor has to be wired to them to give a proper pulse.
A current transformer or a sense resistor has to be fitted to limit current or short circuit protection, however using a current transformer can protect both high and low pulses while a sense resistor only protects the low side mosfet.

Considering high power smps such as 2 or 3Kw more care has to be taken into account as stray inductances and capacitance may disturb the switching to the gates. I like to use a pin header to get the drive circuit as close as possible to the switching transistors this helps a lot and save space on the pcb.
More robust mosfets or IGBTs and output diodes have to be put in high power smps. As a rule of thumb the switching devices have to take 3 times the load current for audio purpose and 4 times when used for continuous rating. It is not wise to operate mosfets to their limits as stress may blow them prematurely. A good heatsink is also necessary and maybe also a simple overheating protection done by a thermal switch.

It is a good thing to experiment with protection circuits and using a simulator will help you out considerably and save you a lot of time

If you look around on this site you will find the tread by Ludo3232 and also my version based on this circuit with a different arrangement of the pcb for a 1000W smps. Using better mosfets and diodes will give you 2Kw with a EE55

Regards Silvio
 
I have only one problem now with his smps, my IC burns out for no reason and suddenly. When I go to turn it on (even without load) it doesn't work out of nowhere, I start to check and it turns out to be the IC and sometimes I change ZD2 and ZD3, which by the way I made a few changes here since it did not work at first:
ZD2 = 15V
ZD3 = 13V
R7 = 100 ohms / 0.25W
R8 = 100 ohms / 1W
What I can do? there's a solution ? or should i feed externally?
Thanks and regards!
 

Silvio

Active member
Leave the original values of the resistors and zener diodes. Use a genuine IR2153 and the circuit will work. ICs from China are all fakes and do not start as they draw more current at start up. I encountered that problem myself and I know what I am telling you. Make sure that you are tapping in the center tap of the auxiliary winding feeding pin 1(at 4 turns). This should give you around 20v which then will be stabilized to 13v, a 200R resistor is the right value to pass the right current 35mA to the Zener diode, This will be 0.245w across the zener. Any more than this will surely burn your 13v zener. If the supply voltage to the IC exceed 15v the internal zener of the IC will burn and it will be damaged.
 
Leave the original values of the resistors and zener diodes. Use a genuine IR2153 and the circuit will work. ICs from China are all fakes and do not start as they draw more current at start up. I encountered that problem myself and I know what I am telling you. Make sure that you are tapping in the center tap of the auxiliary winding feeding pin 1(at 4 turns). This should give you around 20v which then will be stabilized to 13v, a 200R resistor is the right value to pass the right current 35mA to the Zener diode, This will be 0.245w across the zener. Any more than this will surely burn your 13v zener. If the supply voltage to the IC exceed 15v the internal zener of the IC will burn and it will be damaged.
Don't forget that I use different number of turns of wire, frequency and core.
EE42 / 21/15 at 50kHz
prim: 28T (4 * 0.45mm)
sec: 11T + 11T (6 * 0.45mm)
aux IC: 5T (after the diode I have 17Vdc). I had 4T in that auxiliary and the source did not start, I changed to 6T and it was at 20V so I removed 1 turn and I left it in 5T, I am sorry to tell you that if I leave that 10V ZD2 at the base of the MPSA my ZD3 does not reach 13V and remains at 8.9V approx. That is why I changed those values of zeners and resistors a little, so that a correct supply reaches the IC.
Currently as I modified it I have about 14V at the base of TR2 and 94-96V at the base of TR1 (since it has two zeners in series of 47V). And my IC gets almost 13V.
 

Silvio

Active member
If you look at the datasheet of the IR2153 you will find that the rising voltage lockout has a minimum voltage of 8.1 a typical voltage of 9v and a maximum of 10v. This means that the IC starts oscillation at this voltage hence the 10v Zener. However as the auxiliary voltage kicks in then the voltage to pin 1 will rise to 13v which is clamped by the 13v zener. You can change R6 to say 39K to enhance a little more the collector current of TR2 thus pushing a little more current on startup. You can exchange R7 with a small diode (1n4148) You need also to change ZD2 from 10 to 12v to compensate for the voltage drop in the diode and the base emitter junction of TR2. This will get the start up voltage around 11v or so. It is important that the voltage at pin 1 will be higher than the startup voltage so that the startup circuit around TR1 and TR2 will be switched off during operation.

Some calculations
Primary 28 turns, so 155v / 28 turns = 5.5v per turn. So having 5 turns you get a peak auxiliary voltage of 27.5v ( I hope you are using a fast diode here).
So now for the current in the 13v zener. Let us say the rectified voltage drops to 27v due to the drop in the diode.
Then 27v-13v = 14v so assuming a 500mW 13v zener. So using ohms law V/R=I so with R8 being 200R we get 14/200R = 0.07A or 70mA
calculating the power then P=VI so 14v x 0.07 = 0.98W or nearly 1W. As you can see from the result This will be too much for the 13v zener and will surely blow. However a 1w zener may not cope either as it is worked to its limits. If the zener burst open circuit then the full voltage from the auxiliary which is 27v will also blow the zener within the IR2153. You will also need to use a 1A fast diode in the auxiliary as 1N4148 is not strong enough also. You can use this guide to help you calculate a decent series resistor with the auxiliary winding so that the zener will not blow.
I must stress that Genuine IC must be used here as fakes do not work in this circuit.

Good luck
 
If you look at the datasheet of the IR2153 you will find that the rising voltage lockout has a minimum voltage of 8.1 a typical voltage of 9v and a maximum of 10v. This means that the IC starts oscillation at this voltage hence the 10v Zener. However as the auxiliary voltage kicks in then the voltage to pin 1 will rise to 13v which is clamped by the 13v zener. You can change R6 to say 39K to enhance a little more the collector current of TR2 thus pushing a little more current on startup. You can exchange R7 with a small diode (1n4148) You need also to change ZD2 from 10 to 12v to compensate for the voltage drop in the diode and the base emitter junction of TR2. This will get the start up voltage around 11v or so. It is important that the voltage at pin 1 will be higher than the startup voltage so that the startup circuit around TR1 and TR2 will be switched off during operation.

Some calculations
Primary 28 turns, so 155v / 28 turns = 5.5v per turn. So having 5 turns you get a peak auxiliary voltage of 27.5v ( I hope you are using a fast diode here).
So now for the current in the 13v zener. Let us say the rectified voltage drops to 27v due to the drop in the diode.
Then 27v-13v = 14v so assuming a 500mW 13v zener. So using ohms law V/R=I so with R8 being 200R we get 14/200R = 0.07A or 70mA
calculating the power then P=VI so 14v x 0.07 = 0.98W or nearly 1W. As you can see from the result This will be too much for the 13v zener and will surely blow. However a 1w zener may not cope either as it is worked to its limits. If the zener burst open circuit then the full voltage from the auxiliary which is 27v will also blow the zener within the IR2153. You will also need to use a 1A fast diode in the auxiliary as 1N4148 is not strong enough also. You can use this guide to help you calculate a decent series resistor with the auxiliary winding so that the zener will not blow.
I must stress that Genuine IC must be used here as fakes do not work in this circuit.

Good luck
Silvio, thank you for your reply.

Reviewing what I currently have:
ZD2: 15V
ZD3: 13V (actually I put Z12V because I ran out of so much burning)
FR at the output of the auxiliary winding
R6: 33K
R7: 100 ohms / 0.25W
R8: 100 ohms / 1W

With this conformation I have 19Vdc at the output of the fast diode of the auxiliary devando. Recalculating everything as you did taking into account these values, gives me:
19V-13V = 6V
6V / 100 ohms = 60mA (0.06A)
6V * 0.06A = 0.36W so you should use a 13V 0.5W zener.
Following your comments and advice, I will replace R7 with 1n4148.
ZD2 must be less than ZD3? if so I will replace 10V zener for ZD2 and leave 12V for ZD3.
R6 change for 39K but did R8 leave 100 or do I put 200 ohms?
 

Silvio

Active member
I think you better change R8 to 120-150R to compensate a little for the lesser voltage or you can put another series diode to the FR diode to drop the auxiliary voltage a little more. This will not stress the12v zener.
Have you tried to measure the voltage across the transformer winding? I always measure around 155v here at 220vac input. It is not making any sense to me that your auxiliary with 5 turns is getting a lower voltage of 19v rather not around 27v. Is your DMM reading correctly? I am saying this because I once had a DMM which when close to smps reads erratically.
One other tip for you. You can add a series diode with a zener to add around 0.6 volts to it. Do not forget to put the series diode in the opposite direction to the zener otherwise there will be no current flow. (put it on the ground side)
 
I think you better change R8 to 120-150R to compensate a little for the lesser voltage or you can put another series diode to the FR diode to drop the auxiliary voltage a little more. This will not stress the12v zener.
Have you tried to measure the voltage across the transformer winding? I always measure around 155v here at 220vac input. It is not making any sense to me that your auxiliary with 5 turns is getting a lower voltage of 19v rather not around 27v. Is your DMM reading correctly? I am saying this because I once had a DMM which when close to smps reads erratically.
One other tip for you. You can add a series diode with a zener to add around 0.6 volts to it. Do not forget to put the series diode in the opposite direction to the zener otherwise there will be no current flow. (put it on the ground side)
I put 1n4148 instead of R7 (anode in emitter), then having ZD2 = 10V and ZD3 = 12V
I changed the R6 (which was 56K on board, 33K in your schematic) for a 39K one.
Still the source does not start, I still have 8.47Vdc in ZD3 and 9.65Vdc in ZD2.
R8=100 ohms
 

Silvio

Active member
Your IC is not genuine and is drawing more current than the specs of the ir2153, why not try removing the IC and measure the voltage at pin 1?


IR2153 datasheet.PNG
 
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Your IC is not genuine and is drawing more current than the specs of the ir2153, why not try removing the IC and measure the voltage at pin 1?


View attachment 7388
I have been making changes and calculations regarding the current consumed by the zeners and IC.
The changes I have made:
R7: 1N4148
R8: 100 (1W)
R6: 39K (in schematic it says 33K and the silkscreen of my PCB says 56K)
ZD2: 10V
ZD3: 12V
ZD1: 2 * 47V in series I put an R between the anode of the ZD3 and GND of 1K.
Now it was good, having 230Vac I have 19.33V after the FR of the auxiliary winding. After R8 13.48V (which is the same as I have in ZD3). It was with their help and that of other colleagues that I was able to finish this SMPS.
I was able to get 800W out of it, I got 140V in no-load and 126V with the 800W load for more than 2hs with EE42 / 21/15 core, the turns that I already mentioned in other comments. In the datasheet I saw that the IC in question consumes 25mA and the minimum operating Vcc is 10V, then the Vclamp is 15.6V. I didn't find what you told me about it should range between 8.1V and 9V. Thank you very much for the explanations in each message, all this information is enriching.

Greetings!

IvanElectric
 

Silvio

Active member
Here is the startup current for a genuine IC .
Below R6 value on my schematic. I never made a silk screen for this pcb

IR2153-1.PNGInkedIR2153-2_LI.jpg
 

Silvio

Active member
I have adjusted the overcurrent protection and connecting a powerful load protects and disables the output, but removing the load does not reset the protection. I have also disconnected it from the NETWORK and it still does not reestablish, it turns on protected.
For the protection to reset you need to disconnect from mains and wait for the bulk capacitor to discharge.
 
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