Cuk converter ripple current

[Baxter Burgundy]

Ok for working out the minimum inductor size I think the formula is:
L = Vin*2 x dt / 2P
Value must be maximum input voltage and minimum output power

I'm not sure how I would calculate max input voltage and mininum output power??
Is it an adjustment for the 15% ripple?

 

[Arouse1973]

The workings I gave was for 15% ripple current with 20 volts in and 12 volts out. If the input voltage changes then so will the duty cycle. The output power depends on your load so in your case it was a 10R @ 1.2A. You can put in allsorts of different figures but you will have to re-calculate.
Adam

 

[Baxter Burgundy]

I thought I must have been missing something because of the statement that it must be for maximum input voltage and minimum output power. So I'll just use the 20v input and 12v, 1.2A output.

Putting this in I get 130.2uH

 

[Arouse1973]

I must admit the question is a bit woolly. I am not sure if it's what they are after, they should explain themselves a bit better.
Adam

 

[Baxter Burgundy]

Never mind. I'll attempt the next part of the question; working out the capacitor values tomorrow.

 

[Arouse1973]

Ok have a look at C=I*dt/dv
Adam

 

[Baxter Burgundy]

Thanks Adam. In the previous question you helped me out telling me peak current is average current x 1.7 and trough current was average current x 0.3 for 15% ripple.

Can I ask how you came to 1.7 and 0.3 so that I can apply that to 10% ripple. Thank you

 

[Arouse1973]

Yep it's 2 x the percentage fraction of peak and trough. 15% is 0.15 * 2 = 0.3 and 85% is 0.85 * 2 =1.7
Adam

 

[Arouse1973]

I think the 10% ripple for the capacitor is voltage. 10% voltage ripple is it dips down to 10% i.e 1.2 Volts from 12 volts out. I guess that's what they mean? You then need to use the formula above I gave you.
Adam

 

[Baxter Burgundy]

Oh right, so which current number should I use in the formula? Average, peak or trough? Where does the 10% voltage ripple factor into the equation.

Dipping down to 1.2 Volts seems quite low, isn't it meant to be a stabilised output? I must be missing something?

 

[Arouse1973]

Oh right, so which current number should I use in the formula? Average, peak or trough? Where does the 10% voltage ripple factor into the equation.

Dipping down to 1.2 Volts seems quite low, isn't it meant to be a stabilised output? I must be missing something?

Yes I thought that was quite high. I guess they are just getting you to work something out. For this forget anything to do with the inductor as this is smoothed by the cap. Have a look at how you would calculate the capacitor for a standard PSU with bridge diodes.

Edit: For the inductor I meant the peak current, we need to look at the ripple current in L2 as this goes out to the cap and load. Let me have a look at this.
Adam

 

[Arouse1973]

Ok I came up with this. The value seems a bit low, but I guess it would be for 1.2V ripple. It's late and I am tired so I will check it again sometime tomorrow if I have time.
Adam

 

[Baxter Burgundy]

You've used 500k as frequency. I'm assuming I'll have to adjust the calculation to 40k? And sorry what is ESR?

 

[Arouse1973]

You've used 500k as frequency. I'm assuming I'll have to adjust the calculation to 40k? And sorry what is ESR?

Equivalent series resistance. It a property of a capacitor, it's the capacitors AC resistance measured at a certain frequency. For say the TAJ series of Tants it is 100 KHz.

 

[Arouse1973]

You've used 500k as frequency. I'm assuming I'll have to adjust the calculation to 40k? And sorry what is ESR?

Let me see if I can do another full example for you. Give me some time.
Adam

 

[Baxter Burgundy]

Thanks Adam, it's been a while but I remember covering equivalent series resistance during my HNC. I'd have to do some digging to figure that one out. a working example would be great.

 

[Arouse1973]

Here we go. I could simulate it for 40KHz so I made it up to prove the formula. You could put this all into Excel and make a spread sheet. Use all this first to make sure you have it right. The Vn004 graph is the voltage ripple on the cap. This is as far as I can go.
Adam

 

[Baxter Burgundy]

Thanks Adam. I'm done with that question for the moment I think. Can I ask for your help on another question. I have partially answered it so far but am not confident with my answer because, again, the notes are so vague I don't trust what I have come up with. So here is the question and what I have so far...

Design a power supply to provide two outputs from a 240Volt mains supply. The supplies must provide 5volts DC at 2Amps and 12Volts DC at 400mA. Completely specify all items and cost your design using commercially available components. Your supply must remain stable if the mains input drops to 180Volts rms.

 

[Baxter Burgundy]

In order to provide 5V DC at 2A and 12V DC at 400mA then the output of the two secondary windings must be calculated using the following formula:-


Vac = Vout + Vreg + Vrect + Vripple x Vnom x 1
Rectifier efficiency Vlowline √2


Where Vout is the output voltage required
Vreg is the voltage drop of the regulator
Vrect is the voltage drop of the diodes
Vripple is the ripple voltage
Vnom is the nominal input voltage
V lowline is the minimum voltage the circuit must still operate stably at

In this question I have made assumptions as to the values of Vreg, Vrect, Vripple and the rectifier efficiency since no figures have been given

For secondary A Vac = 5 + 3 + 1.25 + 0.5 x 240 x 1
0.9 180 √2

Vac = 10.83333 x 1.3333 x 0.707

Vac = 10.2V


For secondary B Vac = 12 + 3 + 1.25 + 0.5 x 240 x 1
0.9 180 √2

Vac = 18.61 x 1.3333 x 0.707

Vac = 17.54


Secondary A can be rounded down to 10V, secondary B can be rounded down to 17V. Therefore the ratio of the windings will be 120:5:7

 

[Arouse1973]

In order to provide 5V DC at 2A and 12V DC at 400mA then the output of the two secondary windings must be calculated using the following formula:-


Vac = Vout + Vreg + Vrect + Vripple x Vnom x 1
Rectifier efficiency Vlowline √2


Where Vout is the output voltage required
Vreg is the voltage drop of the regulator
Vrect is the voltage drop of the diodes
Vripple is the ripple voltage
Vnom is the nominal input voltage
V lowline is the minimum voltage the circuit must still operate stably at

In this question I have made assumptions as to the values of Vreg, Vrect, Vripple and the rectifier efficiency since no figures have been given

For secondary A Vac = 5 + 3 + 1.25 + 0.5 x 240 x 1
0.9 180 √2

Vac = 10.83333 x 1.3333 x 0.707

Vac = 10.2V


For secondary B Vac = 12 + 3 + 1.25 + 0.5 x 240 x 1
0.9 180 √2

Vac = 18.61 x 1.3333 x 0.707

Vac = 17.54


Secondary A can be rounded down to 10V, secondary B can be rounded down to 17V. Therefore the ratio of the windings will be 120:5:7

Sure give me time to have a look at this. Can you give me some official credit by liking the posts you liked.
Adam