Boost Converter: DCM vs CCM. Or why not be afraid to take it yourself

Recently, the popularity of various calculators for calculating electrical circuits has increased. On the one hand, this leads to a decrease in the entry threshold for beginners, which is obviously good, since it leads to the development of the industry, but on the other hand, the level of understanding decreases, which leads to a decrease in the service life of devices and their cost. Should you trust such sources? Let's try to figure out an example.

An example is the boost converter. At first glance, the thing is simple, but if you look in more detail, it turns out it's not so simple.

We will compare the online calculator, hands-on manual training and calculation, taking into account the theory of the converter. No need to be afraid, we won’t go deep into physics.

First of all, as always, we start with the requirements for our converter:

  • Input voltage - 9V;
  • Output voltage - 200V;
  • Output current - 60mA;

Of course, there are several ways to achieve these requirements: the use of special microcircuits, a transformer or a charge pump. But we will look at the classic boost converter (eng. Boost converter ), since comparing voltage boosting methods is beyond the scope of this article.

Just in case, let me remind you the principle of operation of the boost converter.


The converter consists of only 5 components: inductance, diode, key in the form of a field effect transistor and two capacitors. C in capacity is optional.

When the key is turned on, current passes through the inductance and energy is stored in the magnetic field of the inductance L. The diode is closed.

As soon as the key is turned off, the current through the coil changes abruptly and an increased reverse polarity voltage appears on the inductance terminals, while opening the diode, which provides the path for the current to flow.


Since the key works very quickly, the EMF of self-induction increases significantly. This voltage passes through the diode and charges the capacitance, which, in turn, smooths out the ripples that occur when switching the key, leaving only a constant current. Quickly turning the key on and off, we can increase the voltage on the load.

The final output voltage of the circuit will depend on the input, inductance and the ratio of time when the key is in the “open” position to the “closed” position, that is, duty cycle D (duty cycle is the ratio of the time during which the load or circuit is in on, by the time they are off.).


The output voltage will tend to infinity at an infinitely close to unity duty cycle. In practice, the output voltage is the ratio of the parasitic resistance of the coil R L to the load R. The losses in the magnetic core (if any), losses on the diode and losses on the capacitor, etc. are slightly less. [1. 44-45 p.]. Well, of course, with a duty cycle = 1, the inductance will always be shorted to ground and nothing will work.


Let's estimate our converter on the fingers. Let me remind you the requirements: 200V output, 60mA current.
Fill factor:D=1Vin/Vout=19/200=0.955=95.5%
Load:R=200/60=3.3K,
Dependence of R to R L :


Substitute, we get R L = -0.833. So you need an inductance with an internal resistance of less than 0.8 Ohms. That sounds good. It remains to calculate the inductance itself and its currents.

Let's count in the old way, from the directory of rubber balls TI [2].

Approximate Inductance:


where ΔI L - average ripple current through inductance:


There is some constant K. The

directory suggests choosing it in the range from 0.2 to 0.4. I will take 0.2, at a frequency of 30 kHz, so I get ΔI L = 0.26A. We substitute in the formula above and obtain the inductance L = 1074 μH.

We clarify the current through the inductance:


We get 0.27A, we check the peak current through the converter:


We get 1.33A.
It seems easy. Framed, got the value. Let's check using another source - an online calculator [3]. We substitute the values ​​in the plate, set the switching frequency to the same - 30 kHz:


Note the magic constant 2 in the minimum induction formula.

Total we get:

ParameterManual calculationCalculation online calculator
Inductance1074 μH107.4 μH
Current through inductance ΔI L (D = 0.955)0.267A2.668A
Current through the converter1.33A2.66A

As you can see, the difference is many times. The current is two times lower, in the case of calculating with pens, the inductance is ten times more.

One could stop at this, declaring one of the results heresy. But which one is wrong?

Obviously, the calculations differ due to the coefficient K.

The coefficient expresses the ratio of current ripples in inductance to the input current of the entire converter. It can be expressed in terms of the coefficient K rf .


And this ratio affects the operation mode of the entire converter.

What differences does this coefficient cause besides currents and inductance sizes?

To answer these questions, you will have to understand the details of the operation of these modes.

There are two main modes of operation of such converters: DCM and CCM.

CCM - Continuous Conduction Mode. The operation mode of the converter, in which the current in the inductance does not drop to zero.


DCM - Discontinuous Conduction Mode. In each cycle, the current through the inductance drops to zero.


CCM is used in high-power converters in order to reduce currents through components. DCM, in turn, offers less inductance and eliminates the loss of polarity reversal on the diode. Read more about the pros and cons of the modes here .

Thus, DCM is possible only for K rf > 2. If K = 2, then the converter is in the BCM - Boundary Conduction Mode, that is, the switch turns on at the same moment when the current in the inductance drops to zero.

When the load R decreases, the inverter switches to DCM mode. The load at which the inverter is in BCM mode is called the critical load I CRIT . The inductance value when operating in BCM mode is called the critical inductance LCRIT and is calculated based on the maximum load.

It is known that for CCM boost converters, the maximum ripple of the current through the inductance is 50% of the key duty ratio.

Evidence
= 100%.





:


:





In order to select the inductance for the CCM converter, it is necessary to determine the maximum value of K rf .

Usually it is chosen in the range from 0.2 to 0.4, but, obviously, it can reach 2. We determined that the maximum ΔI L occurs at D = 50%, now we calculate the duty factor for the maximum value of K rf .
More formulas!




We ignore D = 1, since with such a duty cycle, the operation of the converter is physically impossible and we obtain a maximum of K rf with a duty factor of 33%.


For operation in CCM mode, the minimum inductance value is best calculated relative to the input voltage closest to the point 2/3 V out (V in (CCM) ).


We take the coefficient K rf = 0.2 and we get L min = 1074 μgH.
For critical inductance, K = 2, L = 107.4 μH. Everything here coincides with the calculations above.

Critical load, just in case:


I CRIT = 0.006A
This was a calculation for the CCM mode.

Thus, the DCM mode will be stable when the inductance is less than L CRIT , with operating V in and current I out . For DCM converters, the minimum idle time t idle is selected in such a way as to provide from 3 to 5% of the switching time, as the idle time, but may be longer to ensure a stable voltage, up to skipping cycles. The maximum value of the inductance L max will be calculated based on this time t idle . L max must be less than L CRIT , otherwise DCM mode will not be possible.


To calculate L max , with the selected t idle , we find the maximum allowable time for switching on the key. In our case, we take t idle as 2%, the frequency is 30 kHz, therefore the period = 0.000033 (3) s.
t idle = 0.000033 (3) -98% = 6.66 * 10 ^ -7c.

I said: more formulas!

tdis – .


, , ton :





Thus we get:


Substitute, we get 103.187mkGn. Pretty close to previous calculations. The result is different, because the calculation of the calculator used to take downtime as 0%.

L max repeats the graph of L crit and also has a peak at V in = 2 / 3V out . To ensure minimum downtime, L max is calculated at the rated voltage V in .

When the output current I out of the converter is less than the maximum I crit (for a certain V in ), the converter will operate in DCM mode.


Do not forget about I crit for this inductance:


We equate to zero and look for the boundaries of the input voltage:


InductanceVinVinΔIL (D=0.955)
1074 (CCM)30.17195.970.0060.2671.33
107.4 (DCM)8.99199.610.062.662.66

The table shows that the CCM mode will be stable at previously set input parameters. But the calculated DCM mode is close enough to critical points, which causes some uncertainty in further stable operation.

So which mode will be optimal in our case?

Obviously, the lower the current, the lower the requirements for the converter components, but the inductance becomes larger. B of lshaya inductance is more expensive and takes up more space, which is critical for mobile devices and mass production. On the other hand smaller inductance requires more of the other components, leading to relatively b of lshim losses and reduced efficiency.

Thus, it is necessary to find a compromise for a specific application, choosing the coefficient K and the switching frequency.

In my case, this is a desktop converter assembled in a single copy, so I will choose the CCM mode of operation, since the dimensions of the converter are not critical, and the smaller the current through the components, the lower the requirements for them. True, the switching frequency in my case will be slightly higher, but this is the topic of another article.

Conclusion


Do directories and online calculators give the right results? Definitely yes. Are these results optimal? Probably not.

Thus, without understanding the operating principles of a particular scheme and thoughtlessly using directories and calculators, it is quite possible to collect more or less working schemes. But if the task is to be done economically and cheaply, fundamental knowledge is indispensable. Now you have this knowledge. The calculations presented in the article are quite enough, and with modern means of solving equations, for example, WolframAlpha, it is very easy to calculate the necessary parameters.

Good luck with your inventions!

PS


I express my gratitude for the support and invaluable help in writing the article: Radchenko to Eugene, Bobrov Vladislav, Karpenko Stanislav.


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