answer's to question's
Luka
In question number four is that 65 Amps or 6.5 Amps, if 65 Amps you surely want to use a Bridge for 74 Volt @65 Amps is aprox 5kW however if the value 6.5 Amps then 74 Volts @ 7 amps is aprox 500 watts but the turn's ratio is not the limiting factor for output current, topology,core size along with frequency and wire size are the determining factors for power :
Example an ETD-49 core used in a bridge has the following Power capabilties.
Freq: 48 kHz @ 1200 watts , 72 Khz @ 1800 watts, 96 kHz@ 2400 watts.
However you must be able to wind the number of turns with the necessary wire size on the bobbin.
For the current value of 65 Amps requires copper foil of the proper dimension.
Refering to the datasheet for MC33025 or MC34025
The 1k resistor feeds the current ramp input (pin 7) and connected to pin 9 which is current limit input.
For the other info refer to page 127 & 128 of Marty Brown's book
all is detailed there
One question, what are the specifications of your design?
chas1
Luka
In question number four is that 65 Amps or 6.5 Amps, if 65 Amps you surely want to use a Bridge for 74 Volt @65 Amps is aprox 5kW however if the value 6.5 Amps then 74 Volts @ 7 amps is aprox 500 watts but the turn's ratio is not the limiting factor for output current, topology,core size along with frequency and wire size are the determining factors for power :
Example an ETD-49 core used in a bridge has the following Power capabilties.
Freq: 48 kHz @ 1200 watts , 72 Khz @ 1800 watts, 96 kHz@ 2400 watts.
However you must be able to wind the number of turns with the necessary wire size on the bobbin.
For the current value of 65 Amps requires copper foil of the proper dimension.
Refering to the datasheet for MC33025 or MC34025
The 1k resistor feeds the current ramp input (pin 7) and connected to pin 9 which is current limit input.
For the other info refer to page 127 & 128 of Marty Brown's book
all is detailed there
One question, what are the specifications of your design?
chas1
chas
You did not understand me.For the moment I am talking about BUCK converter,where I need to limit secondary current to 65 Amps @ only 13v output, and input being 45v running at ~50kHz.By turn ratio I mean ratio for current trafo(1 pri, sec=?), to limit output to 65 Amps.THIS IS NOT FOR HALF,FULL BRIDGE.
You did not understand me.For the moment I am talking about BUCK converter,where I need to limit secondary current to 65 Amps @ only 13v output, and input being 45v running at ~50kHz.By turn ratio I mean ratio for current trafo(1 pri, sec=?), to limit output to 65 Amps.THIS IS NOT FOR HALF,FULL BRIDGE.
Current transfomer design
Luka
It matters not the topology, a current transformer design is the same. In your previous post you made no mention that this was for a current transformer.
I would think that you might want to visit Magnetics website and download this software CTD-1.3 , it will help you with the design part.
The link below will explain and help you calculate the proper values for the burden resistor.
http://www.coilwinder.com/current_transformer_design_and_theory.htm
Most current transformers use one turn for the primary and up to 200 for the secondary. The burden resistor calculation is the meat of current transformer design along with core selection. If you are
winding your own and you don't know the core properties use a ATX yellow core wind about 200 turns for secondary put a 1k 1% resistor across secondary follow this with a diode with a 1k 1% resistor for a load,filter load with about .2ufd, set the supply for a 50% duty cycle and measure the voltage across the 1k 1 % load. Note: Don't forget the 1 turn thru center for primary, proper wire size for the current you expect.
Note: the secodary can be wound with #36 AWG or smaller
The voltage you measure should represent your normal output current. If this is 65 amps then the voltage you measured represents 65 amps. Divide that by the 1k and that will give you the volts per amp.
The trick is on the secondary side you want a voltage that represents a magnitude of current thru your transformer ie 1 volt per amp or in your case maybe 1 volt per 6.5 amps.
Since you have the book "High Frequency Switching Power Supplies theory and design " take a look at pages 161 & 162 all the info you need is also there.
If this does not help let me know and I will post a circuit along with calculations.
chas1
Luka
It matters not the topology, a current transformer design is the same. In your previous post you made no mention that this was for a current transformer.
I would think that you might want to visit Magnetics website and download this software CTD-1.3 , it will help you with the design part.
The link below will explain and help you calculate the proper values for the burden resistor.
http://www.coilwinder.com/current_transformer_design_and_theory.htm
Most current transformers use one turn for the primary and up to 200 for the secondary. The burden resistor calculation is the meat of current transformer design along with core selection. If you are
winding your own and you don't know the core properties use a ATX yellow core wind about 200 turns for secondary put a 1k 1% resistor across secondary follow this with a diode with a 1k 1% resistor for a load,filter load with about .2ufd, set the supply for a 50% duty cycle and measure the voltage across the 1k 1 % load. Note: Don't forget the 1 turn thru center for primary, proper wire size for the current you expect.
Note: the secodary can be wound with #36 AWG or smaller
The voltage you measure should represent your normal output current. If this is 65 amps then the voltage you measured represents 65 amps. Divide that by the 1k and that will give you the volts per amp.
The trick is on the secondary side you want a voltage that represents a magnitude of current thru your transformer ie 1 volt per amp or in your case maybe 1 volt per 6.5 amps.
Since you have the book "High Frequency Switching Power Supplies theory and design " take a look at pages 161 & 162 all the info you need is also there.
If this does not help let me know and I will post a circuit along with calculations.
chas1
I don't know next thing. I have connected my SG so that both outputs drive the fet.So if freq of SG is set to 50kHz, is then my output freq. 100kHz?
If output is 50 then my inductor will have value 17.8uH, I'll use 19 or 20uH.How does that sound?
If output is 50 then my inductor will have value 17.8uH, I'll use 19 or 20uH.How does that sound?
luka,
Your output frequency will be twice the switching frequency. This is true for all double-ended converters: push-pull center-tap, half-bridge, full-bridge, and single-ended Buck PWMs where both outputs drive the main transistor. Simply put, the output frequency is the same as the oscillator frequency, 100kHz, which is double the switching freq. Substituting 20uH for 17.8uH should be fine.
Steve
Your output frequency will be twice the switching frequency. This is true for all double-ended converters: push-pull center-tap, half-bridge, full-bridge, and single-ended Buck PWMs where both outputs drive the main transistor. Simply put, the output frequency is the same as the oscillator frequency, 100kHz, which is double the switching freq. Substituting 20uH for 17.8uH should be fine.
Steve
Since didn't think ahead to much I realised that my SG conf. as it was couldn't function properly.When the output D.C. should be biggest, was oposite of that therefor I would need to use SG1527.So I have decided to go with Uc3842.And it is current mode🙂 .Freq. will be 50kHz so I will put 20uH for inductor.
Since this UC was made for flyback, I have a problem fully open mosfet, because Uc drives it to gnd, but here it would have to drive it to mosfets drain.
Would it be OK if I would make floating gnd for uc, so that it would drive mosfet fully
Would it be OK if I would make floating gnd for uc, so that it would drive mosfet fully
1. or the 2.
The value of the inductor depends on how much ripple current you want to allow in the inductor. The page you referenced allows .4*Iin. Often the inductor ripple is chosen so that it does not go to 0 at the minumum value of load current expected.
The value of the inductor depends on how much ripple current you want to allow in the inductor. The page you referenced allows .4*Iin. Often the inductor ripple is chosen so that it does not go to 0 at the minumum value of load current expected.
Iductor size
Luka
The problem could be the method use to calculate the inductor, take a look at the ripple current in the calculations of the site you reference. When you pump in your numbers it comes back with about 16 uH @ 12 Amps of ripple current and this is calculated at full load current. If you can live with this high ripple current then 16 uH could be used.
The calculation in Pressman is based on keeping the inductor from
running dry or a percentage of load cuurent (in this case .2*iL) which will always give a larger value for the inductor.
You might want to read the help file for inductor calculations on the site you made reference to , this will help explain the difference in your values.
Example
V(out) = 14
V(in) = 42
I(out) = 30 Amps
L = 5*(42-14)*14*20usec / (42*30) = 30uH
This is the calculation at 50 % duty cycle
L = 5*(42-14)*14*10usec / (42*30) = 15.5 uH
Pressman pg 20- 21
The value .1*I(out) is normal for calculations for output inductors.
chas1
Luka
The problem could be the method use to calculate the inductor, take a look at the ripple current in the calculations of the site you reference. When you pump in your numbers it comes back with about 16 uH @ 12 Amps of ripple current and this is calculated at full load current. If you can live with this high ripple current then 16 uH could be used.
The calculation in Pressman is based on keeping the inductor from
running dry or a percentage of load cuurent (in this case .2*iL) which will always give a larger value for the inductor.
You might want to read the help file for inductor calculations on the site you made reference to , this will help explain the difference in your values.
Example
V(out) = 14
V(in) = 42
I(out) = 30 Amps
L = 5*(42-14)*14*20usec / (42*30) = 30uH
This is the calculation at 50 % duty cycle
L = 5*(42-14)*14*10usec / (42*30) = 15.5 uH
Pressman pg 20- 21
The value .1*I(out) is normal for calculations for output inductors.
chas1
OK so for D.C. of let say 97%@ 50kHz the time in us would be almost 20.So I have to calculate value for this time, since then will be max output power, right?
And I also don't want ripple current bo be to high, so that I won't have overheating problems.
I guess I'll go with 30uH
And I also don't want ripple current bo be to high, so that I won't have overheating problems.
I guess I'll go with 30uH
Ton
Luka
The period for 50kHz is 20usec, then the power supply should be stable at 1/2 this time minus the dead time. If you allow 1usec for dwell time (dead time) then that gives 9 usec at 50% duty cycle
more or less, so when the supply is stable then the inductor has to supply the energy for the load for 9 usec and is charged during the next 1/2 period. note: these number's are not exact, just a rough aproximation and there are methods in Pressman , Browns book to calculate them and should be used in all your calculations.
chas1
Luka
The period for 50kHz is 20usec, then the power supply should be stable at 1/2 this time minus the dead time. If you allow 1usec for dwell time (dead time) then that gives 9 usec at 50% duty cycle
more or less, so when the supply is stable then the inductor has to supply the energy for the load for 9 usec and is charged during the next 1/2 period. note: these number's are not exact, just a rough aproximation and there are methods in Pressman , Browns book to calculate them and should be used in all your calculations.
chas1
Hello
I wasnt here for few monthst and I see now lots of new posts
Are there maybe some finished and working schematics for HALF bridge 220V(line voltage) power supply
using sg3525
I wasnt here for few monthst and I see now lots of new posts
Are there maybe some finished and working schematics for HALF bridge 220V(line voltage) power supply
using sg3525
My was working on 230v at aaa 230w or so.But output voltage was prety low for use with amp, but hey it worked.
I can not opet that PNG file!!!
I downloaded that soft , and still can not open it
do you have maybe PCB file?
I downloaded that soft , and still can not open it
do you have maybe PCB file?
You have to install the circad98 from post #96, then download the amp smps1 from post #95,and open the file inside (.sch)