Designing a Boost Converter: Finding L for CCM Operation

In summary, the assignment requires designing a boost converter operating in continuous conduction mode (CCM). The input voltage is constant at 10 V and the output voltage ranges from 15 to 25 V. The output current ranges from 2 to 5 A and the switching frequency is 50 kHz. The maximum allowable output voltage ripple is 0.5 V. The main focus is on finding the correct value for the inductance to ensure CCM operation for all output current values. The design process involves solving 4 equations in 4 unknowns: T1, T2, L, and the peak inductor current I(T1). The capacitor size is determined by the ripple requirement and is calculated last. To ensure CCM
  • #1
jmaack
3
0

Homework Statement


I have some very big difficulties with this assignment, hope someone can give me some clue in the right direction

Design the boost converter represented in and operating in continuous conduction mode (CCM). The usual
hypotheses apply: the output voltage is considered to be constant and equal to Vo, the input voltage is
considered to be constant and equal to Vi, all the components are ideal. Remember to choose a correct value
of inductance in order to guarantee the operation in CCM in the whole range of output current.
Data

Input voltage: Vi = 10 V
Output voltage: Vo = [15; 25] V
Output current range: Io = [2; 5] A
Switching frequency: fs = 50 kHz
Maximum output voltage ripple admitted: Δvo,max = 0.5 V

Questions

1. Switching period: Ts
2. Output power range: Po
3. Inductance: L
4. Duty cycle range: D

Homework Equations


The main problem is that I have no idea on how to find the inductance! I know the solution for a Buck, but not for a boost. I have attached a drawing of the circuit.

Will really appreciate any kind of help or hints!

Best regards!
 

Attachments

  • Capture.JPG
    Capture.JPG
    8.7 KB · Views: 381
Physics news on Phys.org
  • #2
Regarding the load resistor, are we to assume that it can range from 15V/5A = 3 ohms to 25V/2A = 12.5 ohms?
 
  • #3
Yes, believe so - but the other way around: 15V/2A = 7.5 to 25V/5A = 5.

But to honest, I have am not at all sure! Do you have any hints then? :-)
 
  • #4
Once you double-check the above, some things to think about:

* do you understand how the circuit basically works?

* what is the worst case of Vo and Io for the worst-case load R in order for the current thru L to never = 0?

* hint 1: develop 4 equations in 4 unknowns: T1, T2, L and the peak inductor current I(T1) at the end of inductor build-up time T1. T2 is the time the inductor current is fed to the load.

hint 2: net charge addition to C over the time T1 + T2 = 0.

Once you have obtained the largest needed L, you still have to solve 4 equations in 4 unknowns even though you now know L, since the current I0 at the start of T1 is not zero for any conditions other than the worst case. So that initial current now becomes the 4th unknown.

* hint 3: assume C is arbitrarily large so the output voltage is essentially always constant. The size of C is determined solely by the ripple requirement and comes last in your computations.

When you're done you might want to set up a spread sheet where you can input any Vo an R within the permissible range and compute T1, T2, I(T1) and I0.
 
Last edited:
  • #5
jmaack said:
Yes, believe so - but the other way around: 15V/2A = 7.5 to 25V/5A = 5.

But to honest, I have am not at all sure! Do you have any hints then? :-)

You need to find out positively what the (Vo, Io) choices are. Do you believe you need design the circuit for (15V, 2A) and (25V, 5A) ONLY?

EDIT: from the problem statement I think Vo is either +15V or +25V but current can range from 2A to 5A. Notice the word "range" for current in the problem statement.
 
Last edited:
  • #6
I need to determine the inductor, and capacitor size, so I can keep it in CCM. I need to do that for all the values of Io and Vo. Meaning, if for example I use LT Spice to simulate the circuit, then it should be in CCM for the lowest values (because then it would also be the point for the other values i guess). Does that make sense?
 
  • #7
EDIT: I'm sorry, one more change:

You bring up a good point. L needs to be large enough to guarantee continuous current operation (CCM) for the four possibilities: (15V,2A), (15,5), (25,2) and (25,5). Let's assume we don't know which of the four that is.

Looking at the equations I developed it became obvious that there is a direct relation between T1, T2 and the duty cycle D = T1/(T1+T2): you can get a very simple equation relating Vout to Vin and D. Use the fact that at t=0 we throw the switch on, so IL builds up from I0 to I0 + ET1/L. Then, when the switch is switched off, IL decays until at t=T, IL(T)= IL(0) = I0 again.

So now T1 and T2 are determinable since T1+T2 = T = 1/50KHz. So do this for the two output voltages of 15V and 25V: get D(15V) and D(25V).

You also know load R for the four possible conditions of course.

By realizing that, over one cycle, the net charge into the capacitor must be zero you can now determine L for each of the four conditions. You pick the highest L of the four for your final design.

To double-check that you get CCM operation for all four conditions you now assume a finite I0 and solve again for zero net current flow into the capacitor per cycle. You should get I0 = 0 for the worst case and I0 > 0 for the other three.

You can finally solve for your C value from the output voltage ripple spec. You know charge in during T2 and charge out during T1, and delta Vout = (1/C)(charge in during T2 - charge out during T1).

BTW I don't know how much help your simulation software is for the design phase. But you can certainly check your computations with it.
 
Last edited:

1. What is a boost converter and how does it work?

A boost converter is a type of DC-DC converter that steps up the output voltage from an input DC source. It works by storing energy in an inductor during the ON time of the switching period and then releasing that energy to the output during the OFF time.

2. What is continuous conduction mode (CCM) operation?

CCM operation is when the inductor current never reaches zero during the switching period, meaning that it is always conducting. This mode of operation allows for a smoother output voltage and less output ripple.

3. How do I choose the value of the inductor in a boost converter for CCM operation?

The value of the inductor is determined by the desired output voltage, input voltage, and switching frequency. It can be calculated using the formula L = (Vout * (1 - D)) / (f * ΔIL), where Vout is the output voltage, D is the duty cycle, f is the switching frequency, and ΔIL is the desired change in inductor current during the switching period.

4. How do I choose the value of the capacitor in a boost converter for CCM operation?

The value of the capacitor is determined by the desired output voltage ripple and the switching frequency. It can be calculated using the formula C = (IL * (1 - D)) / (8 * f * ΔVout), where IL is the inductor current, D is the duty cycle, f is the switching frequency, and ΔVout is the desired output voltage ripple.

5. What are some common challenges when designing a boost converter for CCM operation?

Some common challenges include selecting the appropriate values for the inductor and capacitor, minimizing switching losses, and dealing with parasitic components. It is also important to consider the effects of component tolerances and variations in input and output voltages. Proper layout and grounding techniques are also crucial for achieving stable and efficient operation.

Similar threads

Replies
55
Views
3K
  • Electrical Engineering
Replies
23
Views
3K
  • Electrical Engineering
Replies
17
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
2K
  • Electrical Engineering
Replies
13
Views
2K
  • Electrical Engineering
Replies
14
Views
809
  • Electrical Engineering
Replies
9
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
Back
Top