Hello everyone,

I just got a question on a project which I'm working on in my university. I am designing a switching power supply using flyback topology. I have calculated all the parameters which are necessary but somehow my professor tells me that the duty cycle is wrong. I assumed a duty cycle of 0.5 and he keeps telling me that the duty cycle changes. I understand that but I can not regulate that, I think an IC can do that. He wants me to calculate using a variable duty cycle. Does anybody here have experience in designing and implementing power supplies and can help me? Following are the requirements:
input voltage: 12V-24V
output voltage: 4.95V-5.05V
Output power: 20W

Assumptions:
duty cycle: 0.5
Efficiency: 80%
switching frequency: 100kHz
Voltage drop Schottkey diode: 0.3V

Calculations:
Prim. winding: L1= 8.5 uH
Sec. winding: L2= 1.9 uH
Transformer core: EF20
N1= 13 windings
N2= 6 windings

I will appreciate so much if someone helps me since I am working on this problem for more than a month.

 

As always a schematic would be helpful. There are several ways to approach this depending on what kind of control you are comfortable with.

1. Why is your professor not providing guidance in this matter?
2. How do you currently set the duty cycle?
3. What alternatives for having a variable duty cycle are there?

 

I don't know why he does not help. I assume my duty cycle as 0.5.

 

I assume my duty cycle as 0.5.

And I assume that assumption is arbitrary.
Is it just that 0.5 sounded like a good value?

The duty-cycle depend upon the input and output voltages, the transformer turns ratio, and the load currnt.

You apparently don't know the details of how flyback converters really work, so I suggest you read up on that first.
Then you will be able to ask more pointed and less general questions.

An important parameter not given, is the saturation current of the transformer core.
If you exceed that, you will likely zap the driver transistor.

 

I understand that the duty cycle depends on the turns ratio and input as well as output voltages. I used this formula to calculate the Turns ratio otherwise I will be in a loop because N1/N2 is unknown too. the formula is: N1/N2 = (Vin*D)/((Uo+Udiode)*(1-D)). Is this formula wrong? I don't see any current here!

 

Hello everyone,

I just got a question on a project which I'm working on in my university. I am designing a switching power supply using flyback topology. I have calculated all the parameters which are necessary but somehow my professor tells me that the duty cycle is wrong. I assumed a duty cycle of 0.5 and he keeps telling me that the duty cycle changes. I understand that but I can not regulate that, I think an IC can do that. He wants me to calculate using a variable duty cycle. Does anybody here have experience in designing and implementing power supplies and can help me? Following are the requirements:
input voltage: 12V-24V
output voltage: 4.95V-5.05V
Output power: 20W

Assumptions:
duty cycle: 0.5
Efficiency: 80%
switching frequency: 100kHz
Voltage drop Schottkey diode: 0.3V

Calculations:
Prim. winding: L1= 8.5 uH
Sec. winding: L2= 1.9 uH
Transformer core: EF20
N1= 13 windings
N2= 6 windings

I will appreciate so much if someone helps me since I am working on this problem for more than a month.

Hi,

Did you take into account the efficiency?

I did not go over your calculation(s) yet but basically you consider the base transfer function which depends on the duty cycle, the transformer ratio which depends on the turns, the load power, and the efficiency which tells you how much loss there is that also has to be accounted for because if the output is 20 watts then the required power for the converter is 20 watts plus whatever the efficiency kicks in too.
Do you know how to calculate the effective power output knowing the load power and the efficiency?

If you do that with ONE input voltage level it will provide you with ONE duty cycle calculation. However, if you vary the input voltage by say 1v per step that will give you several other duty cycles because it changes with input voltage even with constant load.

 

Hi, thank you for your contribution. Yes I considered the efficiency. I will also attach all my calculations in details here. I am really sorry but it is in German I hope that you can understand it.

 

Hi, thank you for your contribution. Yes I considered the efficiency. I will also attach all my calculations in details here. I am really sorry but it is in German I hope that you can understand it.

Who wrote that paper?

Yeah it is a little hard to read, but what did you get for 5v output and 12v input, what duty cycle?

 

I wrote this paper. I got 0.3 duty cycle for the maximum input voltage and and max output voltage but didn't include it in the paper. you think my professor meant I should make every calculation for each parameter twice? once with the Dmax and once with Dmin?

 

I wrote this paper. I got 0.3 duty cycle for the maximum input voltage and and max output voltage but didn't include it in the paper. you think my professor meant I should make every calculation for each parameter twice? once with the Dmax and once with Dmin?

I'll check that one, but in the mean time what did you get for 12v input and 5.000v output?
This particular example might be more important than the others that's why i am mentioning this one.
Also, you should specify the duty cycle with a minimum of two digits like 0.33 or 0.42, etc.

 

I'll check that one, but in the mean time what did you get for 12v input and 5.000v output?
This particular example might be more important than the others that's why i am mentioning this one.
Also, you should specify the duty cycle with a minimum of two digits like 0.33 or 0.42, etc.

To be honest I didn't check this one. Ok I understand now I will try it. Thank you very much

 

To be honest I didn't check this one. Ok I understand now I will try it. Thank you very much

You're welcome.

You should also spec your 0.3 duty cycle with more digits because i cant compare yours to mine without at least one more digit like 0.31, 0.32, 0.33, etc.
I may have done it slightly different than you did so we can then compare.

 

For 12V input and 5V output I have gotten a duty cycle of 0.47. For the max. input voltage 24V and max. output voltage 5.05V I got 0.308.

 

For 12V input and 5V output I have gotten a duty cycle of 0.47. For the max. input voltage 24V and max. output voltage 5.05V I got 0.308.

Is there some reason why you insist on doing output voltages other than 5v?
5 volts is the nominal output voltage.

How are you calculating the 12v,5v duty cycle?

 

yes I consider the tolerance which is 5% from the nominal output voltage (5V). this is the formula I use to calculate all the duty cycles . The index max is just for the case when I use maximum input and output.

 

Hi,

OK, so what is Ua, Ub and Ue and what values did you use for them?
I assume then you want to use 12v in and 5.05v out.
Also, how did you account for the efficiency?

 

hi, sorry I didn't explain what are Ua, Ue and Ud. Ua= output voltage, Ue= input voltage and Ud= voltage drop on the diode (indexes in German). Ud= 0.3V (Schottky Diode). The values for the input voltage varies in the range from 12V and 24V and the output range from 4.95V to 5.05V. This is the problem that the values are changeable so I can not calculate the duty cycle for only one input and one output. And that is my problem with y professor! I designed the flyback for 12V input and 4.95V output but he said the duty cycle must be variable.

 

hi, sorry I didn't explain what are Ua, Ue and Ud. Ua= output voltage, Ue= input voltage and Ud= voltage drop on the diode (indexes in German). Ud= 0.3V (Schottky Diode). The values for the input voltage varies in the range from 12V and 24V and the output range from 4.95V to 5.05V. This is the problem that the values are changeable so I can not calculate the duty cycle for only one input and one output. And that is my problem with y professor! I designed the flyback for 12V input and 4.95V output but he said the duty cycle must be variable.

Well then what he could have meant was that you have to come up with an expression that includes Vin and D *only*. In other words, a function like this:
D=f(Vin)

But we should digress here. Your result for 12v in and either of your outputs does not look right. That is because you still did not account for the total efficiency (all the losses).
The output power for 5v output is 20 watts. The factional efficiency is 0.80. The efficiency is:
Eff=Pout/Pin
and since Eff=0.8 we must end up with:
0.8=Pout/Pin
and since Pout=20 watts we have:
0.8=20/Pin
and so we can calculate Pin as:
Pin=25 watts.

Now, if we have 5v output and 20 watts that means the output current is 4 amps.
Since the diode drop is 0.3v, that adds 0.3v to 5v output and that is the output just before the diode. The output current remains the same, which is 4 amps.
So the total effective output power with this configuration is:
PoutEffective=(5+0.3)*4=5.3*4=21.2 watts
Since that is also the total input power, and the actual useful output power is 20 watts, the efficiency is:
Eff=20/21.2=0.9434 rounded to 4 significant digits.
So that is the overall efficiency of the converter with this configuration. Since this is not the expected efficiency of 0.80 there is still something wrong.
That something is the efficiency is lost not just in the diode alone, there are other losses that you have to account for.

Do you understand now?

Also, PLEASE stick to 12v input and 5v output until you learn how to do that right. Thanks in advance. This makes it easier to compare notes. Later, we can look at other variations.

 

No schematic?

 

Well then what he could have meant was that you have to come up with an expression that includes Vin and D *only*. In other words, a function like this:
D=f(Vin)

But we should digress here. Your result for 12v in and either of your outputs does not look right. That is because you still did not account for the total efficiency (all the losses).
The output power for 5v output is 20 watts. The factional efficiency is 0.80. The efficiency is:
Eff=Pout/Pin
and since Eff=0.8 we must end up with:
0.8=Pout/Pin
and since Pout=20 watts we have:
0.8=20/Pin
and so we can calculate Pin as:
Pin=25 watts.

Now, if we have 5v output and 20 watts that means the output current is 4 amps.
Since the diode drop is 0.3v, that adds 0.3v to 5v output and that is the output just before the diode. The output current remains the same, which is 4 amps.
So the total effective output power with this configuration is:
PoutEffective=(5+0.3)*4=5.3*4=21.2 watts
Since that is also the total input power, and the actual useful output power is 20 watts, the efficiency is:
Eff=20/21.2=0.9434 rounded to 4 significant digits.
So that is the overall efficiency of the converter with this configuration. Since this is not the expected efficiency of 0.80 there is still something wrong.
That something is the efficiency is lost not just in the diode alone, there are other losses that you have to account for.

Do you understand now?

Also, PLEASE stick to 12v input and 5v output until you learn how to do that right. Thanks in advance. This makes it easier to compare notes. Later, we can look at other variations.

Yes I did consider the efficiency of course. In the paper which I attached you can find it but unfortunately it is in German.
but anyway I am going to do what you mentioned with 12V and 5V and recalculate all the parameters again. Thank you for your help and your time.