# Electronic – How to calculate the output voltage of a discontinuous inductor boost converter

dc/dc converterinductorswitch-mode-power-supply

When designing a boost converter to take a low input voltage to a high output voltage, there are many knowns that you control, such as the choice of inductor and the duty cycle of the switching that is at the heart of the boosting voltage. The below circuit is a demonstration of what I'm talking about – no values since its not a question about specifics.

If you're running the boost converter in discontinuous mode, with a known duty cycle and period (\$D\$ and \$T\$ respectively), input voltage (\$V_i\$) and inductance (\$L\$), you can calculate various important values such as peak current through the inductor. But the sticking point that I keep arriving at is how exactly do you calculate the output voltage over a theoretical load, R? Or, as I'm being led to believe, does the output voltage/current actively depend on the value of the load R?

Much of the literature I read assumes that you know \$V_o\$ and \$I_o\$ already (which when you're designing this, it does make sense to know what your desired output is), but I'm approaching this as if I don't, yet not getting anywhere, since the equation regarding the voltage gain:

\$\dfrac{V_o}{V_i} = \dfrac{V_i D^2T}{2LI_o} + 1\$

includes both output voltage and output current, which seem to depend on each other from other equations:

\$I_o = \dfrac{I_{max} \cdot \delta T}{2}\$

which if you sub in the equation for \$\delta T\$ and \$I_{max}\$ just results in the first equation. I can't find a way of isolating one variable. Trying to equate the energy across the capacitor similarly isn't leading me in the right direction.

What I don't understand is that for the given four variables at the top of my post, I'm left with the output voltage and current as free variables, which leads me to posit that these then must depend on the load? Is my thinking correct, or do I misunderstand the concept of what is going on? I know I'm going about this the 'wrong' way, it's just a curious thought experiment as I'm trying to make sense of other circuits, where the PWM/duty cycle switching and other components are already set.

The upper curve is the ideal voltage across the low-side switch while the second one is the instantaneous current in the inductor. Considering a 0-V average value across \$L\$, we can write \$<v_L(t)>=<V_{in}>-<v_{sw}(t)>\$. The average voltage across the switch is obtained calculating the area of the SW curve and stretching it across the switch period: \$<v_{sw}(t)>=D_2V_{out}+D_3V_{in}\$. We know that \$1=D_1+D_2+D_3\$ so extracting \$D_3\$ and plugging into the previous equation leads to \$\frac{V_{out}}{V_{in}}=\frac{D_1}{D_2}+1\$ if you consider \$<V_{in}>=<v_{sw}(t)>\$. If you now consider a 100%-efficient converter and determine \$D_2\$ via the average inductor current, you should find \$M=0.5\times(1+\sqrt{1+\frac{2T_{sw}RD_1^2}{L}})\$. As you can see, yes, if you change the loading resistance or the switching frequency, you will change \$V_{out}\$. The load point at which the converter operates at the border between CCM and DCM is called the critical resistance value: below it you operate in CCM, at the exact value you operate in BCM for boundary-conduction mode and above it, you are in DCM.