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I'm looking into the design of the control circuit for the mosfet. If I know the output voltage and input voltage then I can find the duty cycle required. I found this schematic that uses feedback

enter image description here

Source

https://www.maximintegrated.com/en/app-notes/index.mvp/id/2031

Why would I need feedback that changes the duty cycle like above ?, if the output voltage is dependent on the duty cycle and Vin and if those are fixed values then Vout should be constant regardless of load ?, also I'm a bit confused on the working of the above schematic, is it trying to maintain Verror of 0V regardless of changes in Vout ?

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  • \$\begingroup\$ Search for "Discontinuous mode". There is a region of operation, when the load current is very low, where the duty cycle needs to be much lower than a naïve calculation would lead you to believe. \$\endgroup\$ – Lawrence NK1G Feb 11 '17 at 21:40
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    \$\begingroup\$ You're completely ignoring the fact that the output voltage also depends on the load at the output. You're also confusing the buck converter with this design where the unloaded output voltage is indeed determined by the duty cycle. But this is a boost converter. Without feedback and load, the output voltage can theoretically reach an infinitely high voltage (Why is that ?? I know why, I want to hear the explanation form you). \$\endgroup\$ – Bimpelrekkie Feb 11 '17 at 21:41
  • \$\begingroup\$ The boost converter is a power regulator and not a voltage regulator. \$\endgroup\$ – Andy aka Feb 11 '17 at 22:11
  • \$\begingroup\$ I think the transient response will be very undesirable without feedback. I simulated this for a buck converter just the other night because I was curious about it. In a buck, Vout = D * Vin. There is overshoot at startup and any time the load changes. The damping factor depends on parasitic resistance in the filter capacitor and inductor. In the buck converter, the voltage does settle back to the correct voltage, though, after some ringing. For some applications, it might be good enough. \$\endgroup\$ – mkeith Feb 12 '17 at 6:08
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A boost converter follows the relationship \$ V_o = \frac{V_i}{1-D} \$ (ignoring non-ideal effects) when in continuous conduction mode (CCM). But with the diode as one of the switch, the converter would drop out of CCM at light load when the inductor current wants to reverse. If you replace the output diode with another active synchronous switch that would allow current to flow both ways, then the proportional duty cycle relationship can hold down to zero load because that would force the converter to stay in CCM. And not surprisingly, this is sometimes called forced CCM. (By the way, many boost converter does not operate in CCM at all in the first place.)

But the more important reason for feedback is probably this: Without feedback, the output voltage would depends directly on the input voltage (as seen in the equation) and the impedance of all the elements (the non-ideal effects of the switches, inductor, wiring) in between. That is usually not acceptable in most cases. For example, if the input voltage regulation is 5%, without feedback, the output regulation would always be worse than 5% and could be much worse for significant and changing load.

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First, yes, the circuit maintains zero error. This is a control circuit, and to understand it you need control theory course.

Second, you could for some cases calculate D and work without the control circuit- in open loop mode. But then any change of input or some changes of load would affect the output voltage. By closing the loop (working "controlled" way) you ensure that in wide range of situations your output is stable. Actually, almost doesn't depend on input or load.

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