Design a boost converter to provide an output of 18 V from a 12-V source. The load is 20 W. The output voltage ripple must be less than 0.5 percent. Specify the duty ratio, the switching frequency, the inductor size and rms current rating, and the capacitor size and rms current rating. Design for continuous current. Assume ideal components.

 

L1 tries to keep the intensity of current constant through him.(.....)

The solution that L1 "will find" is to increase the voltage at its terminals.
See Faraday low e=-d Magnetic flux /d Time
L=d Magnetic flux /di(t)

It's a problem ... you did not draw the schematic.
https://en.wikipedia.org/wiki/Boost_converter

From
http://www.learnabout-electronics.org/PSU/psu32.php

During this time, we are talking about charging the coil through resistance.
So L charging through R series.
So if it is series then the voltage is added.
https://en.wikipedia.org/wiki/RL_circuit

Ldi/dt+iR=U
I(t)=U/R(1-e^-Rt/L)
I do not think you're interested in this calculation, but if you want we do this calculation.

The current intensity, as you can see, has the exponential shape, but it can be approximated quite well with a straight line on the on small potions.

You have online calculator if you want to
https://learn.adafruit.com/diy-boost-calc/the-calculator

After i(t) reaches its maximum it decreases.

View attachment 132795


We look at the last graph of the current intensity through inductor for continuous mode.(from wiki).
Now look at the red and blue triangle areas.

Vertical current intensity, horizontal time. I=dQ/dt => Q=area in i*t= triangle aria
=Q With which we charge the output capacitor + supply output

20W at 18V=> 1A output=>20ohm load
red triangle is on Ton
blue triangle is on Toff

The maximum inductor intensity is Imax.

On Toff Q is Imax*Toff/2 (blue triangle aria)+Is*Toff (the small rectangle under the triangle)
During Toff out 1A is Qout=1A*Toff
The rest of the charge is accumulated in the capacitor and will be used for the Ton period:

Imax*Toff/2+Is*Toff-1A*Toff=1A*Ton

Now for Ton
Q1=Imax*Ton/2(red triangle aria)+Is*Ton
but
Q2 = Imax*Toff/2+Is*Toff



We have Imax am Ton, we can calculate value of L using the above mentioned formula.

This variation in Q gives on capacitor an voltage ripple. C=Q/U

Please tell me if I am wrong somewhere.

 

For the ideal case we have

Duty cycle = 1 - Vin/Vo = 1 - 12V/18V = 0.34

The voltge across the inductor is

V = L * dI/dt = L * ΔI/Δt we solve for L

L = (Δt * V)/ΔI = (Δt * 12V)/(2 * 1.1A) and for Fswitch = 100kHz we have L_min = 18μH

C = Q/V = (I*t)/V = (1.1A * 3.3μs)/90mV > 40μF

 

This is Homework Help where we are to lead the student into finding the solution rather than just giving him the answers. That is the forum's policy. Please don't post answers in this forum until the tread starter has been given a chance to work through the problem.

It has been two weeks since the thread starter posted his question, and that apparently was the only time he was here, so probably no harm was done this time.

 

I know that, but the post is "old" and I doubt TS ould ever return here, as it is the usual case with the user with one post.

 

Understand that you took the right factors under consideration.

 

For the ideal case we have

Duty cycle = 1 - Vin/Vo = 1 - 12V/18V = 0.34

The voltge across the inductor is

V = L * dI/dt = L * ΔI/Δt we solve for L

L = (Δt * V)/ΔI = (Δt * 12V)/(2 * 1.1A) and for Fswitch = 100kHz we have L_min = 18μH

C = Q/V = (I*t)/V = (1.1A * 3.3μs)/90mV > 40μF

Jony,
How do you know that current ripple ΔI = 2 * 1.1A? The load current is 1.1A. So I think you are designing for the boost converter to work in the boundary conduction mode.

 

How do you know that current ripple ΔI = 2 * 1.1A? The load current is 1.1A.

Well, I remember that the load current is the is the average of a inductor current ( this is not exacly true for boost) also the incutor current will have "triangular shape" so, to get average of 1.1A the peak inductor current should be 2 times larger.

So I think you are designing for the boost converter to work in the boundary conduction mode.

Yes, you are right. But keep in mind that this is the minimum value for the inductor, you could always used the large one.

 

Well, I remember that the load current is the is the average of a inductor current ( this is not exacly true for boost) also the incutor current will have "triangular shape" so, to get average of 1.1A the peak inductor current should be 2 times larger.

I made a mistake. The load current is equal to average inductor current for buck converter and equal to (1-D) times average inductor current in boost converter.