Here I'm asking help again. This time with Boost converter.

I'm reading the same article as for Buck converter and I'm having the same issues to figure out how the equations for Imax and Imin are evaluated.

To evaluate [Equation 9] they make [Equation 7] equal to [Equation 8] but then I can't figure out the steps through until evaluate the final expression given by [Equation 9].

I tried it on paper but I get this:

\(\displaystyle{V_{s}\cdot D\cdot T=\left ( V_{s} - V_{c}\right )\cdot \left ( 1-D \right )\cdot T}\)

\(\displaystyle{V_{s}\cdot D=V_{s} - V_{s}\cdot D -V_{c} +V_{c} \cdot D}\)

\(\displaystyle{2\cdot V_{s}\cdot D - V_{s}=V_{c}\left ( D-1 \right )}\)

\(\displaystyle{V_{c}=\frac{V_{s}\left ( 2\cdot D-1 \right )}{D-1}}\)

But I know this is incorrect because if I chose for instance, Vs = 50V and D = 0.25, and plug them into article's [Equation 9] and my equation, results are different!

Article's equation result:

Vo = 66.67 V,

My equation

Vo = 33.33 V.

However, I find here a relationship which is adding both results I get ~100 V.

So, where am I going wrong?

 

You made a mistake in your starting equation.
Inductor voltage waveform:



Apply inductor volt-sec balance:



or:



Solve this and you got:

 

@anhnha this equation:

\(\displaystyle{V_{s}\cdot D\cdot T=\left ( V_{s} - V_{c}\right )\cdot \left ( 1-D \right )\cdot T}\)

 

That equation missed the minus sign. Did you see my post above?

 

That equation missed the minus sign. Did you see my post above?

I did saw your post. That equation is from AAC article! I started from it! Did you saw the article? Is it wrong then?

 

Yes, it missed the minus sign. I did correct it and showed the correct one in my post above.

 

Yes, it missed the minus sign. I did correct it and showed the correct one in my post above.

Ok, I got it.

In the Buck equations there is also an error. This should be fixed!

 

Ok, I'm now struggling again with Imin and Imax equations.

They say they used [Equation 11] and [Equation 8] to find both Imin and Imax.

[Equation 8]
\(\displaystyle{I_{min} - I_{max}= \frac{V_{s}-V_{c}}{L}\cdot \left ( 1-D \right )\cdot T}\)

[Equation 11]
\(\displaystyle{I_{max} + I_{min}= 2\cdot \frac{V_{s}}{R\cdot \left ( 1-D \right )^{2}}}\)

What steps are involved to evaluate Imax and Imin using those 2 equations?

I think they didn't made them equal to each other because if that was the case, Imin and Imax would disappear from the expression (equality).

I tried to solve [Equation 8] for Imin and then replace that result in [Equation 11] to get rid of Imin (in [Equation 11]) and solve for Imax but I got different result!

 

hi Psy,
Cleaned up and rotated your image.
E

 

Hello @ericgibbs .

Now I just need help to find out where am I going wrong!

 

Hello @ericgibbs .

Now I just need help to find out where am I going wrong!

It is just a set of equations so you can solve it.
Your approach is right but you just need to replace Vc = Vs/(1-D) to get the final result.

 

It is just a set of equations so you can solve it.
Your approach is right but you just need to replace Vc = Vs/(1-D) to get the final result.

Oh no! Why haven't I seen that detail. It made all the difference. So the article should say that Imin and Imax also uses [Equation 9] to find final result.