In a dc-dc switch mode buck converter, why the output voltage is different from the desired value when the inductor current is discontinuous? How can the voltage be adjusted to the required value? What would be the effect of open-circuiting the load terminals on the output voltage? Will the capacitor keep on charging until it blows out if the load is removed?

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2 Answers 2


The average value of the output of a buck converter in discontinuous mode is still the same as continuous mode. However, the ripple voltage increases.

This happens on light loads when the switcher cannot produce pulses of low enough duty cycle and it transfers a little too much energy per switching cycle for the light load.

This means the average output voltage starts to rise too high and the control system shuts down the switcher for several cycles resulting in a slightly lower than normal output voltage. Until the load starts to take more power this situation persists.

Discontinuous mode also happens when the input voltage to the buck rises too high.

EDIT - more information:-

If the buck converter is operating at 100kHz and the inductor is 10uH. Let's also say that the supply is 12V and duty cycle is 50%. The "on" pulse of the mosfet will charge the inductor with current and this current is determined by input voltage, output voltage and inductance. Let's say the output voltage is 5V - this means the voltage across the inductor is 7V when the mosfet is "on".

V = L\$\dfrac{di}{dt} \therefore 7 = 10\times 10^{-6} \times\dfrac{di}{dt} \therefore\dfrac{di}{dt} =\space \$700,000

Because dt is 50% of 10us we can calculate I, which is \$700000\times 5\times 10^{-6}\$ = 3.5A.

The energy transferred per cycle is therefore \$\dfrac{L\times 3.5^2}{2}\$ = 57.8uJ.

This transfers 100k times per second \$\therefore\$ the power to the load is \$ 57.8\mu J \times 100000\$ = 5.78W.

If the load resistor is too high to take that power at 5V then the simple fact is that the output voltage will rise and the buck convertor will enter discontinuous mode unless the control loop reduces the duty cycle.

This duty cycle reduction will of course happen because it would be a poor buck converter that couldn't produce less than 50% duty but, it will have a minimum value and at this point, if the energy-load equation isn't in equilibrium the converter will enter discontinuous mode.

  • \$\begingroup\$ What if the output voltage becomes higher? \$\endgroup\$
    – Anusha
    Oct 13, 2013 at 10:55
  • 1
    \$\begingroup\$ The energy into the inductor may become too much for the load and the output rises because of minimum duty cycle of switcher. \$\endgroup\$
    – Andy aka
    Oct 13, 2013 at 11:11
  • \$\begingroup\$ I've added a question above. Can you please answer it? \$\endgroup\$
    – Anusha
    Oct 13, 2013 at 14:53
  • 1
    \$\begingroup\$ If load is removed, energy in the inductor will directly charge the capacitor, voltage will rise and continu to rise unless the control system forces discontinuous operation. Then voltage will remain steady because switcher is inhibited. Cap voltage will slowly self- discharge due to leakage and eventually, switcher will restart for one or two cycles and process repeats. \$\endgroup\$
    – Andy aka
    Oct 13, 2013 at 15:16

The average voltage of a buck converter in DCM is more than CCM, because the inductor current will become zero before switching period, say 100 \$\mu\$sec switching time and duty_on(d)=0.5, duty_off(d2)=0.3(because of DCM). So inductor current becomes zero after 80 \$\mu\$sec, then diode will turn off (current flowing through the diode is zero). After diode is turned off the capacitor voltage will appear across the load. By this way the average voltage of buck converter in DCM is mode (\$V_o/V_g = d/(d+d2)\$).


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