Electronic – DC choppers output voltage in ccm and dcm mode with no load

Here is an idealized buck converter without a capacitor: -

Forget about transistors for the minute and concentrate on two switches, SW1 and SW2. These take turns at switching and never switch on at the same time. If they take equal turns the junction of the two switches will be a square wave; it has a positive peak of +Vs (when SW1 closes) and then returns rapidly to 0V when SW2 closes.

The L and the R (load) form a low pass filter and the voltage seen at the output is a dc level with what closely resembles a triangle wave superimposed. The DC level seen at the output is dependant on the time ratio between SW1 and SW2. If both are on for equal amounts of time, the DC level on the load resistor is 50% of Vs.

So, if Vs is 12V and both switches were active for equal lengths of time, the output would be 6V. If SW1 were on all the time and SW2 was off all the time, the output would be 12V and if SW2 was on for twice as long as SW1, the output would be one-third of 12V = 4V.

So, altering the ratio of the timings between SW1 and SW2 can produce different DC levels on the load. This type of regulator is called a synchronous buck converter and these are the flavour of the month; they'll have two FETs in place of SW1 and SW2 and can be made to alternate at well over 2MHz.

Why the high frequency? the triangle shape superimposed on the output will get smaller as you increase the frequency and, you might as well capitalize on this by picking a smaller and smaller inductor and living with a little bit of ripple (application dependant).

Adding a capacitor across the load has a benefit of course because it helps reduce ripple to an even lower level.

This is the easier of the two common buck topologies to explain because the output inductor (and capacitor) and resistor form a simple low pass filter that turns the square wave into a near-ripple-free DC level. Non synchronous buck regulators are a bit trickier to understand because they don't use a FET at SW2, instead they use what is known as a flyback diode.

Hope this helps.

With a boost converter you need to transfer energy to the load by charging an inductor with energy and then releasing that stored energy. In CCM, inductor current never falls to zero so the energy transferred is dependant on the peak current and the min current.

What the current peaks at during charging represents the maximum inductor stored energy and, what the current drops to (during load replenishment) represents the energy left in the inductor just as the cycle repeats.

The difference in energy is what is "given" to the load.

For a fixed value inductor and input power supply voltage, the rate at which current linearly rises (di/dt) is constant and depends entirely on V = L di/dt, that well-known formula. I'm assuming perfect lossless components of course!

So, Imax, for a given operating frequency will always end-up at some fixed-value above Iaverage and, Imin will be the same fixed-value below Iaverage. We could call that fixed value Ipeak.

So, energy given to the load, W is: -

\$\dfrac{L}{2}[I_{MAX}^2-I_{MIN}^2]\$

Re-arranging using \$I_A\$ and \$I_P\$

W = \$\dfrac{L}{2}[(I_A+I_P)^2-(I_A-I_P)^2]\$

where \$I_A\$ is average current and \$I_P\$ is the peak above (or below) \$I_A\$. You also might be able to get to this formula (hopefully): -

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

This means that both the average current and the peak current determines the energy transfered to the load. But, for a given input supply voltage, operating frequency, duty cycle and inductor value you cannot control \$I_P\$.

So, if the load resistance increases in value BUT you wanted to keep the average output voltage the same, the only option (other than D) is to increase operating frequency to reduce the peak current attained by the inductor during charging. This of course reduces average current gradually over a few cycles and what you find, in some controllers, is that the average current becomes lower and the frequency slews back to the original value.

More load current means lower frequency, less load current means higher frequency.

At the end of the day, I believe you still need a "clever" controller (to alter frequency) so I'm not sure that this question and the optional answers have much bearing on the practical world.

That's the way I see it anyway. Good (but flawed) question!!