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I've never taken a formal power electronic course and am trying to learn a little bit about boost converters.

I assume there are different boost converter designs based on what you're trying to do, and I believe I understand the basic principle for a simple step up DC-DC converter, however I'm struggling to understand how different variations operate. Specifically, I'm trying to understand how a boost converter that takes varying DC voltage levels and is able to maintain a constant output voltage.

Referring to this Texas Instruments DC-DC boost converter, you will see that it can take an input ranging anywhere from 0.9 - 6V. There are several different converters apart of the TPS6102x family, and I'm specifically interested in the TPS61025 which accepts a 0.9 - 6V input voltage, but will output a constant 3.3V. Other chips in the family allow for variable voltage output in which you can use a resistor divider to 'program' the output voltage.

In my case of using the TPS61025, I will not be using the 'FB' pin since the chip can't be programmed to output a certain voltage.

So getting to my question, when a DC-DC boost converter accepts varying input voltages, can it be assumed that there will always be some kind of internal feedback to maintain that output voltage? If a chip were designed to step up a precise voltage of 1.5V to 3.3V, I would assume that it could be implemented either open loop or closed loop since the chip can make the assumption that it will ONLY have one voltage (1.5V) to ever worry about. However in the case of accepting a wider range of voltage inputs, the chip can make no asssumption.

TLDR: Do DC-DC boost converters that accept a wide voltage range input always rely on internal feedback to maintain a constant voltage output?

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Essentially, yes. Even if the input voltage is fixed, feedback is still required due to component tolerance variations and changing output current load - regulation would be poor without some form of feedback.

Isolated DC-DC (or AC-DC) converters such as the flyback topology can make it inconvenient to implement feedback, since traditionally the feedback signal would have to cross the isolation barrier. This can be done with optoisolators, but primary-side sensing is also an option, to cut down on part counts at the expense of a little load regulation (which can be made up by an output linear regulator)

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At its most basic, a dc boost converter stores input energy in one half of the switching cycle and releases that energy to the output in the 2nd half of the cycle. That energy, multiplied by the number of times per second it switches, equates to a power transfer.

This makes an unrefined boost converter into a power regulator and not a voltage regulator. This means that if it stores 1 uJ each cycle and releases that to the output at 100,000 times a second, the power it delivers to the load is 0.1 watts. If the load is a resistor of 1 kohm then the output voltage becomes 10 volts. If the load resistor is 10 kohm then the output voltage is 31.6 volts. For no-load situations, the output rises (charging the output capacitor) until either the capacitor fails or the diode fails or the switching transistor fails.

This is why on low loads you need a sophisticated feedback system to control the duty cycle of the switching and usually, there is a minimum load value needed to guarantee regulation. That sophisticated feedback system turns a power regulator into a voltage regulator.

Additional information about discontinuous mode

What is described above is called discontinuous mode of operation and it represents the most problematic control scenario for a boost converter - changes to both the input supply voltage and the load have to cause modifications to the duty cycle or the output will be unregulated and potentially capable of damaging the load due to over-voltages.

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  • \$\begingroup\$ A DC boost converter acts as constant-power source only while working in discontinous mode (where all energy from the inductor is dumped before the cycle ends). In continous mode of an ideal converter, the output voltage solely depends on the duty cycle and the input voltage. In continous mode, the energy transmitted per cycle will be bigger than in discontinous mode (assuming the same frequency and duty cycle), and thus the power will be bigger, too. If the load might need less power than at the crossover point between continous and discontinous mode, you do need regulation. \$\endgroup\$ – Michael Karcher Oct 29 '16 at 17:56
  • \$\begingroup\$ @MichaelKarcher thanks for painting the fuller picture. Sometimes when you write answers it's easy to get fixed into one little problematic area and forget other areas. \$\endgroup\$ – Andy aka Oct 29 '16 at 18:51

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