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I recently learned about buck and boost converters. Both are switched-mode power supplies. Until recently, I had the impression that the duty cycle and input voltage determine the output voltage for both types of converters. For example, wikipedia states that Vout / Vin = 1 / (1 - D), where D is the duty cycle. This formula assumes that all components are ideal (in particular, the resistance of the inductor is neglected). I assume that this is an oversimplification that causes a discrepancy with reality. Likewise, in a buck converter I would expect the output voltage to equal the duty cycle times the input voltage.

In practice, most converters use a voltage divider and feedback mechanism to ensure that the voltage stays constant when the load changes. Why is this necessary? What mechanism causes the voltage to change when the load changes?

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    \$\begingroup\$ It occurs because losses in the power supply (non ideal components) increases with load current. This loss in the power supply drops the voltage. \$\endgroup\$ – scorpdaddy Apr 24 at 12:36
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    \$\begingroup\$ Only if you would "over design" the switches and inductor to have negligible losses at a given (relatively small) load then you would be correct that the Ducy determines the output voltage. That does make the design too expensive though as "better" switches and inductor are needed. It is more cost effective to simply add feedback. \$\endgroup\$ – Bimpelrekkie Apr 24 at 12:40
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the duty cycle and input voltage determine the output voltage

This is true ONLY if the converters are operating in continuous conduction mode.

As for why a feedback mechanism is necessary — most of the time, the input voltage is unregulated, and the goal is to have a regulated output voltage.

Indeed, in applications where the input voltage has sufficient regulation, no feedback is used. You'll frequently see this in point-of-load converters, for example. There are several manufacturers of fixed-ratio converter modules for exactly that purpose.

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  • \$\begingroup\$ Good remark! Indeed I assume continuous conduction mode. So, to recap: If I have a 12v adapter (which, I assume, is regulated), I can use a fixed duty cycle to step up the voltage to some other, higher output voltage? \$\endgroup\$ – Ruben Apr 24 at 17:50
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    \$\begingroup\$ Yes, but I don't know why you would assume the input is regulated. You either know it is regulated, or you must assume that it is not. Or you don't care whether it's regulated to begin with. \$\endgroup\$ – Dave Tweed♦ Apr 24 at 18:54
  • \$\begingroup\$ I think I might be using a wrong definition of 'regulated'. I thought 'regulated' means 'with fixed voltage'. If your power supply supplies 12 volt, it must be regulated. Is this wrong, or are you merely saying my usage of the word 'assume' is awkward in this context? \$\endgroup\$ – Ruben Apr 25 at 6:36
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    \$\begingroup\$ 12 V is just a "nominal" value. For example, an automobile power system is nominally 12 V, but it actually varies from 10 V (or less, during engine cranking) to 14.4 V (when running and charging the battery). Similarly, a plug-in power supply can be nominally 12 V or a regulated 12 V. You would have to read the specifications of a specific source in order understand how tightly it holds an exact value. \$\endgroup\$ – Dave Tweed♦ Apr 25 at 10:24
  • \$\begingroup\$ Thanks, that was enlightening :) \$\endgroup\$ – Ruben Apr 25 at 10:27
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The duty cycle and input voltage merely, along with Inductor size, determine how much ENERGY is stored into the inductor. That energy is transferred onto the output capacitor.

Should the Iload change, which changes the energy demanded at a constant Vout, the regulatory action must alter the duty cycle and change the INPUT ENERGY.

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  • \$\begingroup\$ I understand, but if you have a voltage source, the voltage would stay constant while the current will increase. You don't explain why the duty cycle must be altered (and the energy can't increase due increasing current). \$\endgroup\$ – Ruben Apr 24 at 17:50
  • \$\begingroup\$ It's not instantaneous. Buck switcher. Load current increases, it comes from capacitor, output voltage decreases. Feedback recognizes voltage decreases, increases duty cycle on control element. Control element is on longer, voltage on capacitor increases. \$\endgroup\$ – StainlessSteelRat Apr 24 at 19:22
  • \$\begingroup\$ Ruben --- has StainlessSteelRat assisted your understanding? \$\endgroup\$ – analogsystemsrf Apr 25 at 2:09
  • \$\begingroup\$ Not really, it feels like I'm missing something. Your answer seems to imply that using this kind of feedback is necessary, while Dave Tweed's answer states it is not necessary. StainlessSteelRat merely describes how the feedback loop works, which was not what confused me. I'm asking why the 'explicit' feedback (with the feedback pin) is necessary. If the energy stored in the capacitor decreases, I would expect the current to it during the on-state to increase as well. Is this not the case? Or is this mechanism just too slow for some purposes? \$\endgroup\$ – Ruben Apr 25 at 8:10
  • \$\begingroup\$ No sampled-regulator system can function faster than the clock-rate. But that is not the issue here. I think the issue remains the very predictable energy we can store into the inductor. If the load current increases, then if Vout is to remain approximately constant, we must transfer more energy (done by storing energy into the inductor, then transferring that energy onto the output capacitor). Given constant Vin and constant Inductor size, our only way to store more energy into the inductor is to keep the switch closed longer, thus we alter the duty cycle. \$\endgroup\$ – analogsystemsrf Apr 26 at 4:15

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