Talk:Buck–boost converter

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Equations I_L_off in continuous mode

Why are the equations for \Delta I_L_off diffrent in these two articles (http://en.wikipedia.org/wiki/Boost_converter#Continuous_mode and http://en.wikipedia.org/wiki/Buck%E2%80%93boost_converter#Continuous_mode)?

\Delta I_{L_{Off}}=\int _{DT}^{T}{\frac {\left(V_{i}-V_{o}\right)dt}{L}}={\frac {\left(V_{i}-V_{o}\right)\left(1-D\right)T}{L}}

and

\Delta I_{{\text{L}}_{\text{Off}}}=\int _{0}^{\left(1-D\right)T}\operatorname {d} I_{\text{L}}=\int _{0}^{\left(1-D\right)T}{\frac {V_{o}\,\operatorname {d} t}{L}}={\frac {V_{o}\left(1-D\right)T}{L}}

— Preceding unsigned comment added by 91.179.98.45 (talk) 21:59, 4 December 2014 (UTC)[reply]

Source for schematic

I added a source per anon's skepticism as to the accuracy of the schematic. Mak17f 19:23, 25 February 2006 (UTC)[reply]

Hyphen? What about en dash?

Using Wikipedia's dash guidelines, shouldn't this article be Buck–boost and not Buck-boost? —TedPavlic | (talk) 18:21, 23 December 2008 (UTC)[reply]

Equation of contention

This equation has been repeatedly changed. I do not know whether the changes are well-intentioned or a subtle and malicious form of vandalism. The original text reads as follows:

This can be written as:

{\frac {V_{o}}{V_{i}}}=\left({\frac {-D}{1-D}}\right)

This in return yields that:

D={\frac {V_{o}}{V_{o}-V_{i}}}

I calculate that the second equation, which has been repeatedly subjected to change, is correct as it stands, using the following algebra:

Multiply both sides of the first equation by (1 - D):

{\frac {V_{o}-DV_{o}}{V_{i}}}=-D

Multiply both sides by $V_{i}:\,\!$

{V_{o}-DV_{o}}={-DV_{i}}\,\!

Subtract $V_{o}\,\!$ from both sides and add $DV_{i}\,\!$ to both sides:

{DV_{i}-DV_{o}}=-V_{o}\,\!

Factor out $D:\,\!$

D(V_{i}-V_{o})=-V_{o}\,\!

Divide both sides by $(V_{i}-V_{o}):\,\!$

D={\frac {-V_{o}}{V_{i}-V_{o}}}

Multiply the right side by ${\frac {-1}{-1}}:$

D={\frac {V_{o}}{V_{o}-V_{i}}}

and we're back with the original equation. Did I make any errors? Wildbear (talk) 06:22, 2 September 2010 (UTC)[reply]

I did not look closely at your working, but I believe the end result to be correct for the definitions given in this article. I suspect the problem is many people are just looking at the bottom equation for D without reading the definition of V_o earlier on, which is a negative value if V_i is positive. What’s the best solution? Change the definition to a positive value? Clarify the definition again at the bottom? Add an invisible comment into the wiki markup code?

Perhaps a quick concrete example would help people wise up?

V_i = 5 V, V_o = −5 V, D = ½ = 50%

Vadmium (talk, contribs) 01:56, 18 December 2012 (UTC)[reply]

I think a simple definition next to the equation regarding the sign of Vo would be the best. — Preceding unsigned comment added by 129.27.123.14 (talk) 08:26, 22 February 2013 (UTC)[reply]

Why do the images look so blurry?

The equations & schematics on this page (and all wiki pages?) are very blurry. Is it because the image size is 2 pixels larger than the actual image? That makes things look terrible.

For example, the style="width:202px" is applied to a 200-pixel-wide image:

<div class="thumbinner" style="width:202px;"><a href="/wiki/File:Buckboost_operating.svg" class="image"><img alt="" src="http://upload.wikimedia.org/wikipedia/commons/thumb/8/82/Buckboost_operating.svg/200px-Buckboost_operating.svg.png" width="200" height="200" class="thumbimage" /></a>

Is there a way to fix this?

Inverting behaviour is not shown in diagram

This is supposed to be an inverting buck-boost converter, yet in the schematic, the polarity on the input and output sides is the same. On the off-cycle, the inductor current flows in the same direction as on the on-cycle, so the polarity has to be opposite. This is very confusing for anybody trying to learn this stuff. --Rhombus (talk) 18:43, 20 October 2011 (UTC)[reply]

I assume you are referring to “Fig. 1”, where V_o is measured relative to the same bottom rail as V_i is measured. The circuit does actually invert the voltage, but that will just affect the sign of V_o. However this seems to be a common misunderstanding; see #Equation of contention above. How to do suggest to make it less confusing? Vadmium (talk, contribs) 01:44, 18 December 2012 (UTC).[reply]

It should be said that the buck-boost is an inverting topology. "inverting buck-boost" adds to the confusion since it implies that there is a non-inverting "buck-boost" topology. The article's use of "inverting" should be changed with a little more consistency using the term.

Perhaps this is too advanced for the article but there are two configurations if you take ground as the common connection between the input and output. You have a positive input which creates a negative output or a negative input which creates a positive output. The subtle difference on how the input and output are "stacked" could be addressed given that an N-CH device is typically used as the switch. In the first case with a positive input, the drain is on the switching node like a buck. This is diagramed already as the "buck-boost". In the second case with a negative input, the drain is on a DC node like the boost. What do you call this? "boost-buck". If the classic "buck-boost" generates a negative output then an "inverting buck-boost" would generate a positive output while still inverting the input (second case example) but potentially still confusing. I would instead say the buck-boost is an inverting topology and a "negative buck-boost" (classic buck-boost) creates a negative output and a "positive buck-boost" creates a positive output. Both inverting input to output using the common reference. Discuss... — Preceding unsigned comment added by 135.23.95.150 (talk) 08:47, 12 September 2020 (UTC)[reply]

Principle of operation

Figure 2 has clipped off the "On-State" and "Off-State" figure descriptors. Can that be fixed? Dan Watts (talk) 19:05, 6 March 2012 (UTC)[reply]

Fixed by simplifying the SVG code. Looks like it was a limitation of the SVG renderer being used. Vadmium (talk, contribs) 23:25, 6 March 2012 (UTC).[reply]

Solving equation for D is algebraically incorrect

The second last equation in the section "Continuous Mode" states: V_o / V_i = - D / (1 - D). This is correct. When I solved this equation for D, purely algebraic, I get: D = V_o / (V_o - V_i). So in the denominator we have a minus instead of a plus sign. Or am I wrong? I am learning this topic for school. TheSmallSwiss (talk) 12:02, 18 September 2022 (UTC)[reply]