I am having a problem finding the conducting period for the diode in a boost converter operated in DCM (theoretically). It is part of a multiple-choice assignment with three questions related to the boost converter operated in DCM:

a) Find the duty cycle

b) How long time is diode conducting?

c) How long time is the inductor without current?

Given parameters:



\$V_{in}=325 V\$

\$V_{out}=1500 V\$

\$L=22\mu H\$

The converter is ideal (lossless). Capacitor and inductor are big enough so no ripple is conceived. I have deduced the following:

Deduced parameters:

\$T_s=33.3\mu s\$

\$t_{on}= 26.11 \mu s\$

\$t_{off}= 7.22 \mu s\$

\$D= 0.7833\$

\$ I_{in} = 15.38 A \$

\$ I_{out} = 3.33 A \$

And it is known that the slope of the inductor current is:

\$ \frac{dI_L}{dt} = \begin{cases} \frac{V_{in}}{L}, & 0 < t < t_{on} \\ \frac{V_{in}-V_{out}}{L}, & t_{on}< t < T_s \end{cases}\$

My approach:

I have tried to take the solve the equation for the inductor currents at each state:

\$ solve\left(\int_{0}^{D\cdot T_s}\frac{Vin}{L} dt = \int_{D\cdot T_s}^{t}\frac{Vin - Vout}{L} dt \right) = 18.88 \mu s\$

But it does not seem right (it is a multiple-choice test and I am worried that I have missed a key point).

Boost converter:

Boost converter

On state:

enter image description here

Off state:

enter image description here

  • L = 22 μH
  • \$t_{on}\$ = 26.11 μs
  • I have tried to take the solve the equation for the inductor currents at each state
  • But it does not seem right

The problem might be your inductor value of 22 μH

It should be much higher for a \$t_{on}\$ of 26.11 μs (D = 0.7833 at 30 kHz). For instance, with an applied DC voltage of 325 volts, after 26.11 μs, the inductor current would be nearly 386 amps.

Or, your problem might be the stated \$t_{on}\$ time of 26.11 μs.....

Back to (some) basics

For an output power (P) of 5 kW at 1500 volts, the load current is 3.3333 amps. The inductor has to raise 325 volts to 5000 volts at 3.3333 amps so, the power needed to be supplied by the inductor is 3.3333 x 1175 = 3916.67 watts. That's an energy (W) of 130.556 mJ per switching cycle.

enter image description here


$$\text{Energy (W)} = \dfrac{\text{Power (P)}}{\text{Frequency(F)}} = \dfrac{L\cdot I^2}{2} \therefore \text{ }\Longrightarrow I = \sqrt{\dfrac{2\cdot P}{F\cdot L}}$$

But we also know this: -

$$V_{in} = L\cdot \dfrac{dI}{dt}$$

  • Where \$dI\$ is the peak current in the inductor (used for calculating inductor energy) and
  • \$dt\$ is the ON time duration of the switching cycle (\$t_{on}\$).

So, given that we know that the current ramps up linearly we can say: -

$$V = L\dfrac{I}{t_{on}} \Longrightarrow I = \dfrac{V_{in}\cdot t_{on}}{L}$$

Combining the two formulas to remove \$I\$, we can find \$L\$: -

$$L = \dfrac{V_{in}^2\cdot t_{on}^2\cdot F}{2\cdot P}$$

This results in an inductor value of 275.78 μH.

Sanity check

In 26.11 μs, with 325 volts applied, the current will rise to 30.77 amps. As a sanity check, this is an energy in the 275.78 μH inductor of 130.555 mJ (as previously calculated above).

If you did use 22 μH, the problem would be your \$t_{on}\$ time

  • \$t_{on}\$ = 26.11 μs

Re-arranging the inductance formula above gives us: -

$$t_{on} = \sqrt{\dfrac{2\cdot P\cdot L}{V_{in}^2 \cdot F}}$$

Then, \$t_{on}\$ = 7.375 μs and D = 0.2213.

And the peak current would become 108.9 amps. This peak current is also the current that falls to zero during the energy transfer into the output capacitor.

To calculate the time taken for this current to fall to zero (during the 2nd period of the switching cycle aka \$t_{off}\$) we use \$V_{L}\$ = (1500 - 325) volts, L = 22 μH and this formula: -

$$V_{L} = L\dfrac{I_P}{t_{off}} = L\dfrac{108.9}{t_{off}} \Longrightarrow t_{off} = 22 μH\cdot\dfrac{108.9A}{1175V}$$

Hence, \$t_{off}\$ = 2.039 μs.

Finding conducting period for diode in DCM (theoretically)?

This is the time that the diode is conducting for, 2.039 μs.

Charge time = 7.375 μs, transfer in 2.039 μs and hold for 23.91 μs before the next cycle begins.

  • \$\begingroup\$ I found the duty cycle by \$ \frac{V_{out}}{V_{in}}=\frac{1}{1-D}\$. The inductor is defined as 22 \$\mu\$H and can't be the problem. I will look into your formula tomorrow. Thank you so much for your elaborate answer. \$\endgroup\$ Mar 1 '20 at 20:21
  • \$\begingroup\$ That formula is perfectly correct only if the circuit is operating in CCM as far as I remember. Your circuit isn’t and, the math becomes more elaborate. \$\endgroup\$
    – Andy aka
    Mar 1 '20 at 21:01
  • \$\begingroup\$ If it were that formula then the conducting period for the diode would be 1-D converted to time. \$\endgroup\$
    – Andy aka
    Mar 1 '20 at 21:29
  • \$\begingroup\$ @DisabledWhale did you get anywhere understanding that the formula used is inappropriate? Have we finished with this question now? \$\endgroup\$
    – Andy aka
    Mar 2 '20 at 20:36
  • \$\begingroup\$ yes - I did understand why the formula is inappropriate. The thing is that the calculation is for a multiple-choice assignment and the discharge time does not add up. I believe I got an approvable approach (since qualitative descriptions for alternative results are also evaluated) but I am just curious why it is not the same now. 1) The energy transfer formula was new to me. Thanks - it elaborates a lot. 2) Alternatively i found: \$V_{in} \cdot D \cdot T_{s}+\left(V_{in}-V_{out}\right) \cdot \Delta_{1} \cdot T_{s}=0 \implies t_{off,1} = \Delta_1 \cdot T_s = 2.3 \mu s\$ \$\endgroup\$ Mar 3 '20 at 3:16

You wrote that no ripple is needed to take into the account. Then you simply calculate how long it takes until the loaded current of your inductor is diminished to zero when the voltage over the inductor is Vout-Vin.

If your loaded current in the inductor is Ix then the needed time is Ix/((Vout-Vin)/L). You have actually tried this but got stucked with your solver machine. At least Vin-Vout is different thing than Vout-Vin. Windows's calculator can be a better choice but try to swap the terms at first.

Things become complex if you take into the account the following factors:

  • Vout rises because the output capacitor gets charged
  • the load consumes part of the current, so all current from the L doesn't rise Vout

This is still an ideal case, but needs the differential equation or numerical simulation.


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