I had a quiz on power electronics and the following questions were asked: Buck Chopper

Boost Chopper

Question 1: Consider the buck chopper in Figure 1, operating at 100 kHz frequency, with the switch duty ratio of 0.5 and input voltage of 100 V(DC). The load is resistive with R = 10 ohms, put across the filter capacitor. Load voltage is found to be nearly ripple free and inductor current is continuous. What should be the smallest value of inductor to keep peak value of the inductor current within 10% of its mean value?

Question 2: Consider the Boost chopper of Figure 2. With input voltage 10Volts and a fixed duty ratio of 0.8. The converter feeds a resistive load of 10 ohms and operates with continuous inductor current. Due to some fault, the load gets disconnected. What will be the resulting output voltage under the steady state? Why?

So, in the test, I was able to think and write only for the 2nd question: My answer was: The output voltage across capacitor keeps on rising and rising since the diode will never be reverse biased thus the capacitor gets a continuous charge current from the inductor, which ultimately will lead to burning of the components of the whole setup.

As for the 1st question, I was just able to just find that the mean Inductor current will be 0.8 Amperes.

So, I was hoping if someone could help me with what might be the right answer for Question 1.

Also, did I write my 2nd answer right? What will be the correct answer?

Please, review and help.

  • 1
    \$\begingroup\$ Welcome to EE.SE. There is a CircuitLab schematic button on the editor toolbar. With this you can easily create decent schematics of your circuit and name and number the components, if you wish to. They're weird transistors in your schematics. \$\endgroup\$ – Transistor Sep 21 '18 at 19:47
  • \$\begingroup\$ I think, that does not matter if they are weird. They are just to be considered switches as long as it is clear from the figure that they are transistor switches, which I think is quite clear in both figures \$\endgroup\$ – user58802 Sep 21 '18 at 19:56
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    \$\begingroup\$ @user58802 Transistor's comment about the odd choice in symbol was just incidental. Everyone is asked to use the schematic editor as much as possible to improve overall clarity and consistency. You could probably shrink them down and smooth them out a bit if you really like the images, they're just a little bit of an eyesore. Functional though. \$\endgroup\$ – K H Sep 22 '18 at 3:30

I will try to answer question 1. In "continuous operation mode" (current in coil never falls to zero), the output voltage of a buck converter is Vout=D*Vin, where D=50% is the duty cycle and Vin=100V. So you know that Vout=50V and thus Iout=Vout/R=5A.

Next, remember the inductor equation V=L* dI/dt. V is Vin-Vout=50V. dt is the time the switch is on or off. dI is 0.2*Iout.

This equation can be formed to L=V*dt/dI, where V=50V, dt=1/(2*f)=5us and dI=0.2*Iout=1A.

So Lmin=250uH


For giggles another approach for continuous mode is using the frequency domain.

a) if ZL(f) fundamental is 9x the load then 1/(1+9) = 0.1 load current variation so From inductor impedance at 100kHz find X=90 Ohms or L=90 /2pi100kHz =1.4mH , close but not quite.

so Fourier harmonics must be considered in X(f) to get 1mH. Like fundamental is 24% higher than Vp dc and so on,

so time domain analysis of LdI/dt is the way to go.


If the converter is continuous, then the duty cycle defines the output voltage hence, with Vin = 100 volts and D = 50%, the output voltage is 50 volts and this is a power of 250 watts in the 10 ohm load.

With a buck converter, output energy is naturally transferred during the inductor charge cycle and this means that the energy storage requirement of the inductor only needs to service the second half of the switching cycle.

Given that 250 watts requires 2.5 mJ to be transferred at a switching frequency of 100 kHz every cycle, the energy storage in the inductor need only be 1.25 mJ when D = 50%.

The formula for energy transfer via an inductor operating in CCM is: -

$$W = 2\cdot L\cdot I_A\cdot I_P$$

Where: -

  • \$I_A\$ is the inductor average current (5 amps) and,
  • \$I_P\$ is the peak above or below that average current (0.5 amps)

Rearranging to solve for L, gives a value of 250 uH


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