# Why is buck-boost converter showing a different trend than buck and boost in efficiency peaking point?

In general I was under the impression, the efficiency is highest when input and output voltage are close to each other. This holds good for boost (text book image below) and buck (personal experience).

The below graph represents the plot of both efficiency and conversion ratio for various duty cycles. (X-axis=duty ratio)

M = Conversion ratio (with non idealities) = output voltage/input voltage But why is the trend not followed for buck-boost? I was expecting the efficiency would peak at 0.5 duty ratio ( when Vin =Vout). This link (Page 8) shows that a 4 switch buck-boost converter is having highest efficiency when Vin = Vout. Why is there is a difference in the statements between sources? Is it because 4 switch converter (during transition mode) should not be considered similar to traditional buck-boost?

Additional ref: AN-1246 Stresses in Wide Input DC-DC Converters, Table 1: Page 3 and 4

• Where did that top graph come from - please link to an online resource so it can be put in context. Feb 10, 2022 at 9:01
• scribd.com/document/386051824/SMPC-Ramnarayanan-pdf Page 116 and Page 112 Feb 10, 2022 at 9:11
• OK I'm not going to subscribe to read it unfortunately so maybe you can explain what the axes of the graph represent and what M represents (i.e. what is Vg) because the graph is definitely not clear. Feb 10, 2022 at 9:19
• @Andyaka : Added the explanation and additional TI reference Feb 10, 2022 at 10:00
• @Andy akka - Yes. Iam good with the answers. Marking it accepted. Feb 14, 2022 at 9:32

Your graph correctly shows that the maximum efficiency of a boost converter is when M (the output voltage to input voltage ratio) is lowest. This can be recognized by looking at an idealized boost converter circuit (using an ideal diode) from my no-frills website here: - And clearly, if Vout = Vin (i.e. M = 1), the MOSFET does not need to switch at all. Therefore the efficiency has to be 100% for ideal components. It's inevitable; the output is permanently coupled to the input via an inductor and diode i.e. there is nothing other than static non-idealities (inductor and diode) to lose energy.

So, when M = 1 (Vout = Vin) the booster is at its highest efficiency. The same is true for a buck converter because the MOSFET would be permanently on and switching losses would be zero.

But why is the trend not followed for buck-boost?

"Here's the rub" at the risk of badly paraphrasing Shakespeare: -

If the output voltage equals the input voltage then we cannot say whether the switching devices are operating or not. But, we could consider that the switching devices are not switching as an initial example....

Reminder: for a buck or a boost converter (as an individual module) operating with Vout = Vin, it has to mean that the MOSFET is not switching and that means ideal 100% efficiency. Not necessarily so for the buck-boost.

For the buck-boost circuit, it could mean that two of the transistors are permanently on whilst the other two are permanently off: - Picture from here.

If M1 and M4 are on and, M2 and M3 are off then, you have the same scenario as a boost converter when input and output voltages are equal. So, you can make an argument that the same trend is followed and ideal efficiency is 100% with a slight degradation due to there being two MOSFETs in the power chain.

But, on the other hand, the buck part of the buck-boost regulator might be genuinely dropping the input voltage to (say) one-third whilst the boost part might be raising that one-third by three to give an output voltage that does equal the input voltage. Under these circumstances the MOSFET losses will be significant because they are switching AND the efficiency cannot be 100%.

So, you can't make the general comparisons like what you have in your question without a bunch of details regarding exactly how the individual buck and boost parts are operating. In other words there are many ways to make a buck-boost converter produce 1:1 input/output but, the most efficient way is when two of the MOSFETs are permanently on and two are permanently off.