Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

H-Bridge transformer design help

Status
Not open for further replies.

usama14

Member level 5
Joined
Oct 31, 2015
Messages
90
Helped
0
Reputation
0
Reaction score
0
Trophy points
6
Activity points
756
Hello. I want to design a transformer that steps up the Vp=48VDC to Vs=330VDC. I am using ETD59 3C90 ferrite core and the switching frequency is 100KHz. I want to build the h bridge configuration for the dc-dc converter. Hence, the maximum duty cycle that I can have is 45%. So, this nominal voltage needs to be at 30% duty cycle so that after the feedback, if the output voltage exceeds 330VDC I will decrease the duty cycle from a range of 0 to 30% and if the output voltage falls short of 330VDC, I will increase the duty cycle from a range of 30% to 45% accordingly. I want to calculate the accurate values of turns ratio relating to this max duty cycle accordingly.
 

Your design requires a certain Ampere flow in the primary. Calculate the Henry value that will produce this, within your duty cycle. The goal is to wind sufficient turns that will create the Henry value to cause your desired Ampere flow.

Too few turns, too much Amps.

Too many turns, too few Amps.

I have slight experience winding inductors. The tutorials I have seen speak of putting one turn on your core. Measure its inductance. Scale that up to the number of turns to create your target inductance.

Make sure it is possible to add or remove turns later, as necessary.
 

Usually the starting point is the expected minimum dc input voltage at full load, which might correspond to the end point of battery discharge.

If that were to be (say) 40v for a 48v lead acid battery, the turns ratio would be calculated for 40v/330v allowing just a little a bit extra for the expected voltage drops at full load plus dead time at maximum full duty cycle.

So your turns ratio might end up being close to 8.5:1.

At anything more than 40v the duty cycle reduces accordingly.

The low input voltage maximum full duty cycle condition is also used to calculate the required number of primary turns at the desired flux density and frequency.

*edit*

3C90 material is fairly old and not optimum for 100Khz, but anyhow the recommended flux density at that frequency for 3C90 is 100mT

At 40v input and 5uS on time:
6 primary turns works out to 90.5mT
5 primary turns 108.6mT
Either should work.
 
Last edited:

Usually the starting point is the expected minimum dc input voltage at full load, which might correspond to the end point of battery discharge.

If that were to be (say) 40v for a 48v lead acid battery, the turns ratio would be calculated for 40v/330v allowing just a little a bit extra for the expected voltage drops at full load plus dead time at maximum full duty cycle.

So your turns ratio might end up being close to 8.5:1.

At anything more than 40v the duty cycle reduces accordingly.

The low input voltage maximum full duty cycle condition is also used to calculate the required number of primary turns at the desired flux density and frequency.

*edit*

3C90 material is fairly old and not optimum for 100Khz, but anyhow the recommended flux density at that frequency for 3C90 is 100mT

At 40v input and 5uS on time:
6 primary turns works out to 90.5mT
5 primary turns 108.6mT
Either should work.

Actually I am using solar panels to supply 48V DC to the primary side. I did my calculations using two different approaches as described below:
1) I used the following equations given in the datasheets

Np=E*10^8/(4*F*A*B)=3.26
where
B is the flux density in Gauss=0.1T,
E is the square wave peak voltage in volts=48V,,
F is the frequency in Hertz=100KHz,
A is the cross sectional area of the core in square centimeters=3.68,
Np is the number of primary turns.
And for the secondary as:
Ns=E*10^8/(4*F*A*B)=22.4
where
B is the flux density in Gauss=0.1T,
E is the square wave peak voltage in volts=330V,,
F is the frequency in Hertz=100KHz,
A is the cross sectional area of the core in square centimeters=3.68,
Ns is the number of primary turns.

By doing these calcultions, I got the turns ratio of n=Ns/Np=6.87

2) The second method I followed was based on the duty cuycle value. I took Minimum voltage =42V and the max duty cycle for it to be=0.7. I took the absolute value of the duty cycle to be =0.81 to compensate for the transformer.
I used the equation (Vin*D=(Vout+Diode drop)*n).=(42*0.7=(330+0.4)*n)
By doing these calcultions, I got the turns ratio of n=Ns/Np=11 And according to its further calculations the turns become as Np=5 and Ns=56.

Now I am confused as what calculations to take as the final ones. :sad:
PS: My system is of 600W(4 solar panels of 300W each) but to extend it in the future, I'm making the transformer to handle upto 1000W. So, my primary current becomes Ip=1000/48=21Amperes.
And that of secondary becomes Is=1000/330=3.03Amperes.

- - - Updated - - -

Can you kindly confirm as how to calculate the necessary henry? Also tell me whether the duty cycle value I have used is according to the full-Bridge principles or not as I am confused whether to use D/2 or D for duty cycle as 2 of the 4 MOSFETs would be switching in each half cycle.
Here is the further explanation to my design. :)
Actually I am using solar panels to supply 48V DC to the primary side. I did my calculations using two different approaches as described below:
1) I used the following equations given in the datasheets

Np=E*10^8/(4*F*A*B)=3.26
where
B is the flux density in Gauss=0.1T,
E is the square wave peak voltage in volts=48V,,
F is the frequency in Hertz=100KHz,
A is the cross sectional area of the core in square centimeters=3.68,
Np is the number of primary turns.
And for the secondary as:
Ns=E*10^8/(4*F*A*B)=22.4
where
B is the flux density in Gauss=0.1T,
E is the square wave peak voltage in volts=330V,,
F is the frequency in Hertz=100KHz,
A is the cross sectional area of the core in square centimeters=3.68,
Ns is the number of primary turns.

By doing these calcultions, I got the turns ratio of n=Ns/Np=6.87

2) The second method I followed was based on the duty cuycle value. I took Minimum voltage =42V and the max duty cycle for it to be=0.7. I took the absolute value of the duty cycle to be =0.81 to compensate for the transformer.
I used the equation (Vin*D=(Vout+Diode drop)*n).=(42*0.7=(330+0.4)*n)
By doing these calcultions, I got the turns ratio of n=Ns/Np=11 And according to its further calculations the turns become as Np=5 and Ns=56.

Now I am confused as what calculations to take as the final ones. :sad:
PS: My system is of 600W(4 solar panels of 300W each) but to extend it in the future, I'm making the transformer to handle upto 1000W. So, my primary current becomes Ip=1000/48=21Amperes.
And that of secondary becomes Is=1000/330=3.03Amperes.
 

The conventional method is to run the duty cycle right up to maximum at the minimum design input voltage, whatever you decide the minimum input will be.

I assumed a typical lead acid battery in my example, but obviously that need not be the case.
 

So it means that the second approach I am taking is the correct one?
 

Hi,
are u using batteries for storage ?
I made up a small solar system myself (~40W) using SLA batteries for storage. I made up the system in June when i had more than 6 hours of sun, so the batteries seemed to charge up completely. Now i have sunlight for ~4hours so i had to add a charger connected to the power grid to complete the charging cycle .
 

I am using batteries only to provide surge current to the submersible pump connected at the load.
 

The conventional method is to run the duty cycle right up to maximum at the minimum design input voltage, whatever you decide the minimum input will be.

I assumed a typical lead acid battery in my example, but obviously that need not be the case.

You mean to say that my second approach is the correct one? :thinker:
 

You mean to say that my second approach is the correct one?

2) The second method I followed was based on the duty cycle value. I took Minimum voltage =42V and the max duty cycle for it to be=0.7. I took the absolute value of the duty cycle to be =0.81 to compensate for the transformer.
I used the equation (Vin*D=(Vout+Diode drop)*n).=(42*0.7=(330+0.4)*n)
By doing these calcultions, I got the turns ratio of n=Ns/Np=11 And according to its further calculations the turns become as Np=5 and Ns=56.

First issue is that you will never be able to run a high power inverter direct off solar panels without using a battery.
The problem is that solar panels are current sources, the voltage will vary hugely with the current drawn, and the maximum usable power from solar input is only available over a fairly narrow voltage range.
Another worry is that inverters are negative resistance loads. As the input voltage falls, duty cycle increases and so does input current.
The result is a highly unstable system that can never work.

A battery is a necessity to stabilise operating conditions, and maintain a somewhat constant load for the panels, and a reasonably stiff voltage source for the inverter.

Try to imagine how smoothly a single cylinder gasoline engine might run with zero flywheel mass, and you get the general idea.

Once you have "something" to stabilise the operating voltage (a battery is suggested) you design your inverter operating voltage around the battery type and number of cells.

The only duty cycle you are interested in is the absolute maximum at the minimum design input voltage.
At voltages higher than minimum, the duty cycle will obviously self adjust via feedback to be lower, but the actual figure is of no interest.
 
Last edited:

First issue is that you will never be able to run a high power inverter direct off solar panels without using a battery.
The problem is that solar panels are current sources, the voltage will vary hugely with the current drawn, and the maximum usable power from solar input is only available over a fairly narrow voltage range.
Another worry is that inverters are negative resistance loads. As the input voltage falls, duty cycle increases and so does input current.
The result is a highly unstable system that can never work.

A battery is a necessity to stabilise operating conditions, and maintain a somewhat constant load for the panels, and a reasonably stiff voltage source for the inverter.

Try to imagine how smoothly a single cylinder gasoline engine might run with zero flywheel mass, and you get the general idea.

Once you have "something" to stabilise the operating voltage (a battery is suggested) you design your inverter operating voltage around the battery type and number of cells.

The only duty cycle you are interested in is the absolute maximum at the minimum design input voltage.
At voltages higher than minimum, the duty cycle will obviously self adjust via feedback to be lower, but the actual figure is of no interest.

Hmm...Actually I am using a battery in parallel to the solar panels to provide the surge current and maintain the voltage.
So no worries about that. :thumbsup:
But the problem is with the turns calculation. According to your aseesment, you said a turns ratio of 8.5:1 should be used. But mine comes out to be 11:1. I am just confused whether to use which one? :roll:
And the last question I may ask is which material would be more accurate for 100kHz beside 3C90? Can you plz specify.
Thankyou!
 

Ratio 8.5:1 has a step up ratio of 39v to 330v
That would be about right for a 48 volt lead acid battery that has an end of discharge voltage of typically 40v.

Ratio 11:1 has a step up ratio of 30v to 330v
That would be about right for a minimum expected battery voltage of maybe 31v or 32v.
That is rather low for a nominal 48 volt battery ?

3C90 is old, 100mT at 100 Khz is about all it can do.
3C94 and 3C95 are far more recent and capable, 200mT at 100 Khz is possible.
That means in theory only half as many turns might be required, a big advantage.

https://www.farnell.com/datasheets/1468410.pdf
 

This simulation illustrates how different primary values affect response, in terms of secondary output V and A.

The layouts mimic an H-bridge, 50 percent duty cycle, AC square wave.



Some amount of parasitic resistance must be factored in. I installed 1/2 ohm, as a guess.

A step-up ratio of 1:8 appears satisfactory.

It looks as though you'll get satisfactory results with a primary value between 10 and 30 uH.

With real hardware, you'll wind the transformer according to your best ability, then adjust your switching frequency to get your desired output.
 

The Al value for un gapped 3C94 is 6uH/turn, so three turns (200mT approx) = 54uH

Al value for 3C95 is 7.34uH/turn, so three turns of that = 66uH

So plenty of inductance with either material.
 

My 2 cents worth.

Np=E*10^8/(4*F*A*B)=3.26

This is correct as far as you went. But you have to cover the full voltage range from 39vdc to max charge rate 55vdc? To be safe use 4 primary turns.

Your low voltage is 39vdc your secondary voltage is 330vdc, a rule of thumb is to add 5% for duty cycle loss. 330vdc x 5% = 347vdc

347vdc/39vdc = 8.9 ratio

4 primary turns x 8.9 ratio = 35.6 turns = 36 turns Minimum.

You may need one more turn for misc losses, diodes, copper losses, ect.
 

I agree, 3 turns is pushing the limit.
Four turns is much more sensible.

You are going to drop some voltage across your switching mosfets, and the resistance of all the internal wiring, as well as the transformer.
These might be around a volt (2.5% conduction loss) in a good design.
Or might be as high as two volts (5% conduction loss)

So something like 330/39 8.46:1 ratio
Or 330/38 8.68:1 ratio

You need to round up to full turns, and if you split the secondary into two halves it will be an even number.

So 4 Turns to 36 turns sounds ideal.

First wind 18 turns, then the foil 4 turn primary, then another 18 turns on top.
 

I agree, 3 turns is pushing the limit.
Four turns is much more sensible.

You are going to drop some voltage across your switching mosfets, and the resistance of all the internal wiring, as well as the transformer.
These might be around a volt (2.5% conduction loss) in a good design.
Or might be as high as two volts (5% conduction loss)

So something like 330/39 8.46:1 ratio
Or 330/38 8.68:1 ratio

You need to round up to full turns, and if you split the secondary into two halves it will be an even number.

So 4 Turns to 36 turns sounds ideal.

First wind 18 turns, then the foil 4 turn primary, then another 18 turns on top.

Ok. Got it.
But kindly tell me that at what value of duty cycle will I get 330V? How can I calculate that?
 

My 2 cents worth.

Np=E*10^8/(4*F*A*B)=3.26

This is correct as far as you went. But you have to cover the full voltage range from 39vdc to max charge rate 55vdc? To be safe use 4 primary turns.

Your low voltage is 39vdc your secondary voltage is 330vdc, a rule of thumb is to add 5% for duty cycle loss. 330vdc x 5% = 347vdc

347vdc/39vdc = 8.9 ratio

4 primary turns x 8.9 ratio = 35.6 turns = 36 turns Minimum.

You may need one more turn for misc losses, diodes, copper losses, ect.

But what about the duty cycle calculations as I did in second point? I want to have the nominal value of 330V at some lower value of duty cycle to adjust the feedback. :|
 

Ok. Got it.
But kindly tell me that at what value of duty cycle will I get 330V? How can I calculate that?
you will get 330v at full maximum duty cycle at the minimum design input voltage.
If that is 40v then it will work right down to 40v.

At voltages higher than 40v, the feedback reduces the duty cycle to hold the output constant at 330vdc.

At 48v the input has increased by 48/40 = 20%
The duty cycle will decrease by 20% from maximum.

At 52v the input has increased by 30%
The duty cycle will decrease by 30% from maximum.
 

you will get 330v at full maximum duty cycle at the minimum design input voltage.
If that is 40v then it will work right down to 40v.

At voltages higher than 40v, the feedback reduces the duty cycle to hold the output constant at 330vdc.

At 48v the input has increased by 48/40 = 20%
The duty cycle will decrease by 20% from maximum.

At 52v the input has increased by 30%
The duty cycle will decrease by 30% from maximum.

OKAYYYY...It means you are following the reverse logic??? I mean the transformer would be designed for lowest voltage at the primary(40V) with 90%(max)Duty cycle.??? And then if the voltage will increase towards the nrmal voltage of 48V, then the duty cycle has to be decreased accordingly??
Am I right? 8-O
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top