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

 

figured it out. i was using the formula for a buck-boost converter. not just a boost converter

 

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.

 

answer A 1.1uH was actually incorrect...

maybe that output voltage ripple % was needed?

 

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

 

and thats still the wrong answer, well i only have 3 choices left

 

are you accounting for minimum current to be 1/2 of full current? Continuous switching mode, essentially.

 

My bad. D is from above, Vin/Vout

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

 

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?

 

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?

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?

 

here is the exact equation out of my book and how I solved for L

 

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.

 

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