Cuk converter ripple current

[Baxter Burgundy]

I've been given the following homework assignment and the study notes I have are far from clear. Any guidance would be much appreciated.

A C'uk converter is required to produce a stabilised 12V output voltage at 1.2A from an input supply of 20V. Assume a frequency of 40kHz.
Calculate:

The inductor size for a 15% ripple converter.
The minimum inductor size for continuous current operation.
The capacitor value for a 10% voltage ripple.

 

[Arouse1973]

Do you have a circuit diagram we can see. It's just the Cuk converter has two inductors and offers a very clean D.C output. So 10% ripple seems quite high. Also should that be 15% ripple current? and not ripple converter. This seems quite low?
Adam

 

[Baxter Burgundy]

I think the question means a converter with 15% ripple, the phrasing is a little confusing. I've written it word for word to the question I have been given.

I don't have a circuit diagram to show. I don't think it matters in this case that it may be unrealistic, it's a hypothetical circuit purely for the purpose of this homework exercise.

 

[Arouse1973]

Oh ok so I think you need to be looking at the rate of change of current over time. And this can be used for both the capacitor and the inductor. As this is homework I can't just give you the answer. But you need to look at the voltage across the inductor and the on time of the MOSFET and the % of current required, in this case it's 15% of 1.2A.
Adam

 

[Baxter Burgundy]

Thanks for your help Adam. I'm still a little confused however, it's a distance learning course I'm doing and the study notes provided are not clear at all, and I've nobody to ask at the institution.

Can you maybe give me some pointers back to basics as to how a C'uk works, and how then that will lead me to answering those questions.

 

[Arouse1973]

Ok as it's a distance learning course I'll give you a little more help. I'll see if I can put something together tonight. But because it's quite unclear and you don't have a circuit I will have to generate one so the answers may not be exactly what you want but close enough.
Adam

 

[Baxter Burgundy]

Thanks Adam, I've written the question as posed to me, so if there's any assumptions that have to be made I'll state that I've made them in my answer.

 

[Arouse1973]

Ok here is what I have so far. See if you can follow it? You will have to do a bit of work yourself.

I am going to be a bit mean here and get you started and see if you can fill in the blanks. Ok first thing I would do is work out the duty cycle (DC) needed.

This is DC Vout / (Vin+???) x100 =??%

The MOSFET on time is the period of the running frequency multiplied by the??

Which = 1 /?? x Vout / (Vin+???) = ??us

So the average input current = (Pout x 0.15) / ?? = ??.??A (Assume 85% efficiency)

Peak Current = Average current x 1.7 (15% Ripple)

Trough current = Average x 0.3 (15% ripple)

The difference is therefore ???A

So using we can work out the inductor value dt is the MOSFET on time and di is the difference in current above.

So using Vin / L = di / dt
L = ????

The output inductor is calculated the same. I know there is different currents in and out which would result in a different ripple percentage but for most applications it’s minimal.

I'll work the cap out later, then we can simulate it all and see if we are close. I am tired now.
Adam

 

[Baxter Burgundy]

Great thanks Adam I'll have a look through that tomorrow night after work and see how I get on.

 

[Arouse1973]

Great thanks Adam I'll have a look through that tomorrow night after work and see how I get on.

Ok

 

[Baxter Burgundy]

Ok Adam, I had a look at it last night and I've got my head around it I think. Have a look at what I've worked out and tell me if I'm correct.

Duty cycle is DC Vout / (Vin + Vout) x 100
12 / (20+12) x 100 = 37.5

MOSFET turn on time is running frequency x duty cycle
1/40khz x 35.5 = 937.5 microseconds

Average input current = (Pout x 0.15) / Vin
(12 x 1.2 x 0.15) / 20 = 108mA

Peak current = 108mA x 1.7 = 183.6mA

Trough current = 108mA x 0.3 = 32.4mA

Difference is 151.2mA

So using Vin / L= di / dt
12 / L = 0.1512/0.0009375
12 / L = 161.28
L = 12 / 161.28
L = 74.4mH

I've just realized as I'm typing this that I never made an 85% efficiency calculation as you suggested, so do you agree with my answer assuming 100% efficiency?
Thanks again for your help

 

[Arouse1973]

Good try but nope.mH dont you think thats a bit big?
Adam.

 

[Arouse1973]

Very good try. Look at the power calc.
Adam

 

[Arouse1973]

Clue your looking for about 164uH. In your defence I will have a look and make sure I have not confused you.
Adam

 

[Arouse1973]

It is strange that your course has asked you about such an unusual terminology. Most Cuk designs are for a negative inverter not a converter. For a positive version of thisyou would have to have a transformer of some kind.
Adam

 

[Baxter Burgundy]

Thanks Adam, looking at the power calculation I would intuitively think it would be Pout/Vin, if that's not the case I'll need another hint as to why it's not.
I imagine they have chosen the questions because it's all hypothetical and they are not too concerned about it being unrealistic in practical terms.

 

[Arouse1973]

Ok here is my attempt. I'll simulate it later and see if we are close.
Adam

 

[Arouse1973]

Right here is my design. I had to change the frequency because I couldn't go that low with the device I chose. But goes to show how the formula works in a simulation.


Fig 1. Circuit Diagram



Fig 2. Inductor Current in L1





Fig 3. Formula and Calculations

So from this you can see we are not far off.
Adam

 

[Baxter Burgundy]

Thanks Adam you're a legend. I'll have a look later on tonight at working out the minimum inductor size and capacitor value, hopefully your guidance will have given me enough pointers that I can answer those questions now.

 

[Arouse1973]

Good luck.
Adam