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I am struggling to see how the "inverted buck-boost" converter works the same as a "4-switch buck-boost" converter. In my mind, a 4-switch buck-boost is simply connecting a buck and boost together, but in the inverted buck-boost, I can't visualize that. Can anyone show how the buck and boost operations actually work in the inverted buck-boost (explanations on how it converts 20v to 5v & 5v to 20v)? I've seen a lot of descriptions/videos about how it stores/releases energy from the inductor, but never how the buck and boost functionalities come in to play (explanations on how it converts 20v to 5v & 5v to 20v).

Note: this post was mentioned by someone before @ Boost operation in inverting Buck-Boost converter, but wasn't answered to my better understanding

UPDATE: I'm asking the question in reference to the 1st picture shown below.

enter image description here enter image description here

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  • \$\begingroup\$ works the same as – it doesn't. Both converters can do both buck and boost, that's what they share. \$\endgroup\$ – Janka May 21 at 2:55
  • \$\begingroup\$ @Janka I know they both do buck/boost, but I just want to visualize the workflow of the inverted as opposed to the non-inverted. Do you perhaps mind sharing a diagram of their comparisons? \$\endgroup\$ – Alex Feng May 21 at 3:07
  • \$\begingroup\$ @StefanWyss I did not post any comment on the linked question. That comment was by Nandi. I'm not Nandi. \$\endgroup\$ – Alex Feng May 21 at 4:31
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The inverting converter, your first figure, operates this way.

When S is closed, current builds up in the inductor. The node joining S and L is positive. The output node is negative, which means D isolates the input and output. Current builds up in L with no reference to the output conditions.

When switch S opens, we have a current in L, and a stored energy in L, with no reference to the input conditions. The charging phase could have been (let's say) 10us from 10v, or it could have been 2us from 50v, or 20us from 5v. All three charging sequences would have ended up with 100uVs of current in the inductor (scaled by the inductance value of course).

As the inductor discharges into the load, D connects the inductor to the output, while S isolates it from the power source. The inductor doesn't know what voltage it's going to find on the load, and doesn't care. With its 100uVs of current, if there's 10v there, discharge will take 10us, if the output voltage is 20v then discharge will take 5us.

So you see the inductor doesn't care what the input and output voltages are, and which is bigger. If input is lower than output, you'd call it a Boost. If input is bigger, you'd call it a buck. The switching sequence is exactly the same for both.

Switch 'on' time has a maximum defined by the inductor saturation current (I'm so tempted to call this parameter the inductor's 'capacity', but that would be confusing). The switch on time, and/or frequency of on pulses, has to be controlled by a servo loop that monitors the output voltage. This is essentially controlling the energy throughput of the converter.

Both the conventional buck and boost converters work in a similar way, the output voltage defines how long the inductor current takes to change. However, the input and output are not isolated. In the buck converter, if the output voltage is higher than the input, then no current can be built up in the inductor. In the boost converter, if the output voltage is lower than the input, then the charging logic has no control of the inductor current, it will simply build up until something gives. It's the isolation of input and output that allows the inverting converter to work with any input and output voltage.

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  • \$\begingroup\$ So the inverted circuit adjusts output voltage based on load (e.g. 20v load will make inductor produce 20v [just with faster current draw]), unlike a regular buck or boost converter which produces a defined output voltage regardless of load, correct? \$\endgroup\$ – Alex Feng May 21 at 5:39
  • \$\begingroup\$ No. Inductive storage converters ALWAYS have their output voltage instantaneously defined by the load. However, the driving logic often servoes the switching duty cycle to control the inductor current over many cycles so that the load voltage is correct. In a conventional buck converter, the output voltage is constrained to be less than the input. In a conventional boost, it must be more than the input. If the output voltage violates these constraints, then the switching logic doesn't have control over the inductor current during the charge period. \$\endgroup\$ – Neil_UK May 21 at 7:04
  • \$\begingroup\$ I like your explanation, Neil, but I cannot upvote because of the capital S you used for "second". The unit second uses a small s. A capital S is Siemens, 1/Ω. \$\endgroup\$ – Janka May 21 at 11:32
  • \$\begingroup\$ @Janka Hmm, I can't be the pedant that I am over the apostrophe, and leave my units sloppy. You'll be telling me to leave a non-breaking space between the figures and units next. \$\endgroup\$ – Neil_UK May 21 at 12:43
  • \$\begingroup\$ No, I won't. Using S instead of s in an electrical context creates ambiguity. Same if someone used i for the imaginary unit instead of j. And the latter wouldn't even be wrong, just very very misleading. \$\endgroup\$ – Janka May 21 at 12:47
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In first circuit, with switch on, the inductive node is pulled positive (reverse biasing the Diode). After energy is built up in the inductor, the switch is turned off and that same inductive node immediately goes below ground and pulls down on the output energy storage capacitor.

In 2nd circuit, to convert 5v to 20v (flyback mode?), you close both switches to store energy into the inductor. Once done, with both switches off, the current still flows in the same direction but the voltage polarity has switched and the current now flows thru D1 and D2. To flow thru D1, the left end of inductor goes below ground.

In 2nd circuit, to convert 20v to 5v, you leave Q1 ON ALL THE TIME.

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  • \$\begingroup\$ Perhaps it was unclear, but I was trying to understand how the buck-boost works in the 1st picture (inverted buck-boost). Could you explain how this work please? I've updated this in my question. \$\endgroup\$ – Alex Feng May 21 at 4:29
  • \$\begingroup\$ @AlexFeng He's just told you how it works. \$\endgroup\$ – Neil_UK May 21 at 4:56
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    \$\begingroup\$ There are no different operating modes in the inverted buck-boost. Duty cycle defines the output voltage directly. \$\endgroup\$ – TemeV May 21 at 5:08

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