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Below is a boost converter with its inductor current waveform operating in DCM. The voltage conversion ratio of the converter is as follows:

M(D1,D2) = V/Vg= 1 + D1/D2

where V and Vg are output and input voltage respectively.

From the formula we see that there are two ways to increase the conversion ratio:

  1. Increase D1.

This makes sense because the longer the D1 the more energy is stored in the inductor in the time period.

  1. Decrease D2.

I couldn't explain it intuitively or physically how it is so. When D2 is decreased, how does it affect output voltage intuitively or physically?

Thank you.

enter image description here

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  • \$\begingroup\$ No, it is for DCM. In CCM it is 1/(1-D) where D is duty cycle. \$\endgroup\$ – anhnha Mar 18 '16 at 17:09
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When D2 is decreased, how does it affect output voltage intuitively or physically?

You have cause and effect muddled up.

In discontinuous mode the time period D2 is determined by the load resistor and output capacitor. A lower time period for D2 means that the inductor is dumping its energy into the C and R much more quickly. Because V = L di/dt, if the rate (di/dt) is bigger, then the terminal voltage produced by the inductor is bigger hence more voltage.

Think about the extreme situation where C is very small and R is open circuit. During the time period D1 energy is stored in the inductor and as soon as the switch (Q1) opens, that energy produces a back emf that charges the very small value of capacitor up to some ridiculously high voltage that will likely destroy Q1. In this scenario D2 would be extremely short.

So, D2 getting smaller is the "effect" and not the cause - the cause is increasing the load resistor and/or decreasing output capacitance.

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  • \$\begingroup\$ Thanks. Could you explain how output voltage is kept constant with a large load current? \$\endgroup\$ – anhnha Mar 18 '16 at 17:08
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    \$\begingroup\$ @user3126592 the basic boost converter shown in your question is like a car without a driver. Like a car must have a driver, a boost regulator must have a means of dynamically altering the duty cycle (D1 in your case) or the output voltage will be totally dependant on load changes. A feedback system is employed that keeps the output voltage relatively constant with varying load and input voltage conditions. \$\endgroup\$ – Andy aka Mar 18 '16 at 18:29
  • \$\begingroup\$ In fact, the basic boost circuit is more akin to a power regulator; energy is stored during D1 and that energy is released during D2. The rate of energy per transferred per second is energy x operating frequency and that equals power. In it's basic form (without a controller) it is a power regulator and will try and ramp-up to produce infinite voltage if the load resistor is removed. \$\endgroup\$ – Andy aka Mar 18 '16 at 18:33
  • \$\begingroup\$ Thanks. So to keep the constant output voltage under varying load and input voltage condition we need to change the duty cycle. How about change the switching frequency solution? For example, if increasing load current, we will reduce switching frequency so that on-time D1T will increase. \$\endgroup\$ – anhnha Mar 19 '16 at 0:39
  • \$\begingroup\$ Controlling D1 is key but altering frequency comes with some problems. At lower frequencies output ripple voltage gets worse so this has to be considered. Most controllers alter D1 and leave frequency constant. \$\endgroup\$ – Andy aka Mar 19 '16 at 10:20
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You have this relationship: $$ \frac{V}{V_g} = 1 + \frac{D1}{D2} $$ For a given \$V_g\$, with D1 held constant, reducing D2 leads to increasing V.

The thing to recognize is that you don't get to make the above choices without changing something else in the circuit. With D1 and \$V_g\$ held constant, the power input remains the same. So the only way you can decrease D2 is to increase the resistance R (i.e. reducing the load) such that the output voltage V rises which gives you a steeper falling inductor output current slope.

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