# boost converter inductor value

Discussion in 'Homework Help' started by notoriusjt2, Oct 31, 2010.

1. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0

where did I screw up? is the output voltage ripple % supposed to be in there somewhere?

2. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
figured it out. i was using the formula for a buck-boost converter. not just a boost converter

3. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
The output ripple is usually a function of the output filter, typically a capacitor. Sometimes useless information is added to a question, sometimes it may appear useless, but some part of it is actually required.

4. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
answer A 1.1uH was actually incorrect...

maybe that output voltage ripple % was needed?

5. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
after going over this again I found that my value for Io was incorrect

I=P/V=25/15=1.667

recalculating everything I got 4.4uH

6. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
and thats still the wrong answer, well i only have 3 choices left

7. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
are you accounting for minimum current to be 1/2 of full current? Continuous switching mode, essentially.

8. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
My bad. D is from above, Vin/Vout

$L=(V_in -V_out ) * {D * Freq \over I_ripple}$

9. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
here is another similar one

Vs=5V
Vo=15V
R=75ohms
f=100kHz
T=10uS
D=.667

the book gives me this equation for max inductor current
Imax=(Vs/(((1-D)^2)R)+((VsDT)/(2L))
solving for Imax gives me 11.9uH

is this the correct equation?

10. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
Click My Sigline.

$Imax={V_s\over {((1-D)^2 \cdot R) + \frac{V_s \cdot D \cdot T} {2 \cdot L}}$

1st, Does the equation above match your book?
2nd: How did you rearrange it to solve for L?

11. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
here is the exact equation out of my book and how I solved for L

12. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
Ok, I'll work it out here.

Vs=5V
Vo=15V
R=75ohms
f=100kHz
T=10uS
D=.667

$I_{max}={\frac{V_s}{(1-D)^2 \cdot R}+{\frac{V_s \cdot D \cdot T}{2 \cdot L}}
$

$
2={\frac{5}{(1-0.667)^2 \cdot 75}+{\frac{5 \cdot 0.667 \cdot 10\times 10^{-6}}{2 \cdot L}}
$

$
2=601.2\times 10^{-3} + {\frac{33.35\times 10^{-6}}{2 \cdot L}
$

$
1.3988=\frac{33.35 \times 10^{-6}}{2 \cdot L}
$

$
2.7976 \cdot L=33.35 \times 10^{-6}
$

$
L = 11.921 \times10^{-6}
$

L = 11.921 μH

So, we either both did it wrong in similar ways, or it isn't the correct formula they want you to use in this instance.

13. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
the professor submitted the same question with new answers and 10uH was on there. So I went with that and by golly I got it right