# What exactly changes at closed loop transfer function, when circuit move from CCM to DCM - half bridge convertor (Buck derived)

I have hard time to understand what exactly changes in close loop transfer function when it goes from ccm to dcm. When we close loop, blocks which signals goes through are:

• A(s) LC post filter with load connected
• B(s) compensator
• C(s) PWM comparator stage
• D(s) Input Line with transformer

When DCM occurs, LC post filter is "disconnected" from secondary side of transformer and also duty cycle of controller becomes relevant on the load connected. So output voltage is not dutyCycle*InputVoltage anymore. At CCM, duty cycle is equal to EA_vout/rampVoltage, its "same" in DCM except EA_Vout is load dependent.

At this page, we can see derived transfed ratio dCalc for DC gain at DCM, since Half Bridge is buck derived topology i guess it will be same... BUT i just can not figure out, where should i used it... https://sciamble.com/resources/pe-drives-lab/basic-pe/buck-converter-dcm

• when current reach zero, inductor is disconnected and load is powered from capacitors, so should we change block A(s)?
• compensator B(s) will be same i guess
• PWM comparator stage C(s) - I think there will some magic occurs =). At CCM its equal to EA_vout/rampVoltage. I know it does at DCM exactly same thing, but now Vout cannot be lineary aproximated.
• D(s) Input Line with transformer - This will be also same i guess

I am really confused, sorry...

Here is block diagram in ccm:

I am interested in "how to implement DCM in the loop transfer function". Stages are multiplied together, so we have A(s)*B(s)*C(s)*D(s). But its not clear for me which equation, for what stage i should change when smps transit from ccm to dcm. I think that instead of Eaout/Vramp which is TF of comparator block there will be equation for dcalc, which can be found here https://sciamble.com/resources/pe-drives-lab/basic-pe/buck-converter-dcm

Also it will be probably necessary to derive again TF of LC post filter for DCM due high input impedance. I have made schematic in kicad. Here i will need a little bit help from you.

According to slva301 appnote from TI, i should subtitute RL1 for "r+RL1" and then subtitute r = Rload*(1-Vout/Vin).

• In EE we replace words with a schematic because it's far less ambiguous. Commented Apr 20 at 14:03
• I dont know how could i express it with schematic, Ive made block diagram for ccm. I would like to get transfer function of closed loop in DCM to have a look at root locus. But I dont know which blocks exactly should i modify and how. Commented Apr 20 at 17:52
• @andyaka I have added schematic, hope its more clear now Commented Apr 21 at 9:25
• Open-loop DCM has a different TF to open-loop CCM. Can you calculate the open-loop TFs for both. If not then trying to establish the closed-loop TFs is going too far too early. Have you tried using a simulation tool? Commented Apr 21 at 9:36
• I have not try any sim. tool. But i am able to obtain TF for CCM, or atleast i think I am =D... I have found this app note ti.com/lit/an/slva301/slva301.pdf. At the end there is appendix A, equations TF for filter in DCM. But its not "step by step", atleast not for me Commented Apr 21 at 9:56

## 1 Answer

Comparing a DCM and CCM converter with the same input and output voltage and same load:

A: You are correct. In DCM Mode, the capacitor must supply all of your load current during the part of the duty cycle when the switch is off. It will also have to receive enough charge to increase the voltage each cycle. So if you do not increase the capacitance, you will have more ripple. Similarly, the peak inductor current must be high enough so that the inductor current is equal to the load current when averaged over one period. So the peak inductor current must be higher.

B: Compensation is more dependent on the range of input voltages, since it determines how quickly your duty cycle reacts to a change in output voltage. Usually you must set your compensation to be stable at the highest input voltage you expect. This is often DCM if you have a wide input range. Your converter will operate best at this input, and your ability to react to load or input voltage changes will not be as rapid at lower input voltages.

C: No magic here really. When in DCM, you can easily calculate the peak current in the inductor, since you know the voltage and the "on" time and the starting current is zero (V = dI/dt). When the switch turns off, the energy in the inductor in joules is 1/2LI^2. Since a watt is a joule per second, and you know the frequency of how often you are creating this much energy (your switching frequency) you can easily calculate how much power gets to the secondary. It is then a simple matter extract the voltage since your load is a diode and resistor in series.

D: As mentioned in A, DCM requires more inductor current for the same load. Even if you could be 100% efficient, the input power must equal the output power. In DCM, the input is only supplying power during the part of the duty cycle when the switch is closed, so both your inductor and your line power must be capable of handling higher instantaneous currents.

Keep in mind that this discussion is a comparison of the two modes when they have the same input and output. However, if you build a dual-mode converter, you will only be in DCM mode when the load is small and/or the input voltage is high, when the design will be at the lower end of its power delivery range.

• Thanks for an answer! I am interested in "how to implement DCM in the loop transfer function". Stages are multiplied together, so we have A(s)*B(s)*C(s)*D(s). But its not clear for me which equation, for what stage i should change when smps transit from ccm to dcm. I think that instead of Eaout/Vramp which is TF of comparator block there will be equation for dcalc, which can be found here sciamble.com/resources/pe-drives-lab/basic-pe/… Commented Apr 21 at 7:44
• Also i think i need to modify TF of LC filter due high input impedance in DCM. Commented Apr 21 at 9:12