Fundamentals of Power Electronics Chapter 5: Discontinuous conduction mode1Chapter 5. The Discontinuous Conduction Mode5.1. Origin of the discontinuous conduction mode, andmode boundary5.2. Analysis of the conversion ratio M(D,K)5.3. Boost converter example5.4. Summary of results and key pointsFundamentals of Power Electronics Chapter 5: Discontinuous conduction mode2Introduction toDiscontinuous Conduction Mode (DCM)G Occurs because switching ripple in inductor current or capacitor voltagecauses polarity of applied switch current or voltage to reverse, suchthat the current- or voltage-unidirectional assumptions made in realizingthe switch are violated.G Commonly occurs in dc-dc converters and rectifiers, having single-quadrant switches. May also occur in converters having two-quadrantswitches.G Typical example: dc-dc converter operating at light load (small loadcurrent). Sometimes, dc-dc converters and rectifiers are purposelydesigned to operate in DCM at all loads.G Properties of converters change radically when DCM is entered:M becomes load-dependentOutput impedance is increasedDynamics are alteredControl of output voltage may be lost when load is removedFundamentals of Power Electronics Chapter 5: Discontinuous conduction mode35.1. Origin of the discontinuous conductionmode, and mode boundaryBuck converter example, with single-quadrant switches+–Q1LCR+V–D1VgiL(t)iD(t)iL(t)t∆iLI0DTsTsconductingdevices:Q1D1Q1iD(t)t0DTsTs∆iLI∆iL=(Vg– V)2LDTs=VgDD'Ts2Lcontinuous conduction mode (CCM)Minimum diode current is (I – ∆iL)Dc component I = V/RCurrent ripple isNote that I depends on load, but ∆iL does not.Fundamentals of Power Electronics Chapter 5: Discontinuous conduction mode4Reduction of load currentIncrease R, until I = ∆iL+–Q1LCR+V–D1VgiL(t)iD(t)∆iL=(Vg– V)2LDTs=VgDD'Ts2LCCM-DCM boundaryMinimum diode current is (I – ∆iL)Dc component I = V/RCurrent ripple isNote that I depends on load, but ∆iL does not.iL(t)t0DTsTsconductingdevices:Q1D1Q1∆iLIiD(t)t0DTsTsI∆iLFundamentals of Power Electronics Chapter 5: Discontinuous conduction mode5Further reduce load currentIncrease R some more, such that I < ∆iL+–Q1LCR+V–D1VgiL(t)iD(t)∆iL=(Vg– V)2LDTs=VgDD'Ts2LDiscontinuous conduction modeMinimum diode current is (I – ∆iL)Dc component I = V/RCurrent ripple isNote that I depends on load, but ∆iLdoes not.The load current continues to bepositive and non-zero.iL(t)t0DTsTsconductingdevices:Q1D1Q1IXD1TsD2TsD3TsiD(t)t0DTsTsD2TsFundamentals of Power Electronics Chapter 5: Discontinuous conduction mode6Mode boundaryI >∆iLfor CCMI <∆iLfor DCMInsert buck converter expressions for I and ∆iL :DVgR<DD'TsVg2L2LRTs< D'Simplify:This expression is of the formK < Kcrit(D) for DCMwhere K =2LRTsand Kcrit(D)=D'Fundamentals of Power Electronics Chapter 5: Discontinuous conduction mode7K and Kcrit vs. DKcrit(D) = 1 – D0D1012K = 2L/RTsK < Kcrit:DCMK > Kcrit:CCMKcrit(D) = 1 – D0D1012K = 2L/RTsK > Kcrit:CCMfor K < 1:for K > 1:Fundamentals of Power Electronics Chapter 5: Discontinuous conduction mode8Critical load resistance RcritR < Rcrit(D) for CCMR > Rcrit(D) for DCMwhere Rcrit(D)=2LD'TsSolve Kcrit equation for load resistance R:Fundamentals of Power Electronics Chapter 5: Discontinuous conduction mode9Summary: mode boundaryK > Kcrit(D) or R < Rcrit(D) for CCMK < Kcrit(D) or R > Rcrit(D) for DCMTable 5.1. CCM -DCM mode boundaries for the buck, boost, and buck-boost convertersConverterKcrit(D) max0 ≤ D ≤ 1( Kcrit)Rcrit(D) min0 ≤ D ≤ 1( Rcrit)Buck (1 – D) 1 2L(1 – D)Ts 2LTsBoost D (1 – D)2 427 2LD (1 – D)2Ts 272LTsBuck-boost (1 – D)21 2L(1 – D)2Ts 2LTsFundamentals of Power Electronics Chapter 5: Discontinuous conduction mode105.2. Analysis of the conversion ratio M(D,K)Inductor volt-second balanceAnalysis techniques for the discontinuous conduction mode:vL=1TsvL(t) dt0Ts=0Capacitor charge balanceiC=1TsiC(t) dt0Ts=0Small ripple approximation sometimes applies:v(t) ≈ V because ∆v << Vi(t) ≈ I is a poor approximation when ∆i >IConverter steady-state equations obtained via charge balance oneach capacitor and volt-second balance on each inductor. Use care inapplying small ripple approximation.Fundamentals of Power Electronics Chapter 5: Discontinuous conduction mode11Example: Analysis ofDCM buck converter M(D,K)+–Q1LCR+V–D1VgiL(t)iD(t)+–VgLCR+v(t)–iC(t)+ vL(t) –iL(t)+–VgLCR+v(t)–iC(t)+ vL(t) –iL(t)+–VgLCR+v(t)–iC(t)+ vL(t) –iL(t)subinterval 1subinterval 2subinterval 3Fundamentals of Power Electronics Chapter 5: Discontinuous conduction mode12Subinterval 1+–VgLCR+v(t)–iC(t)+ vL(t) –iL(t)vL(t)=Vg– v(t)iC(t)=iL(t)–v(t)/RvL(t) ≈ Vg– ViC(t) ≈ iL(t)–V / RSmall ripple approximationfor v(t) (but not for i(t)!):Fundamentals of Power Electronics Chapter 5: Discontinuous conduction mode13Subinterval 2Small ripple approximationfor v(t) but not for i(t):+–VgLCR+v(t)–iC(t)+ vL(t) –iL(t)vL(t)=–v(t)iC(t)=iL(t)–v(t)/RvL(t) ≈ – ViC(t) ≈ iL(t)–V / RFundamentals of Power Electronics Chapter 5: Discontinuous conduction mode14Subinterval 3+–VgLCR+v(t)–iC(t)+ vL(t) –iL(t)vL=0, iL=0iC(t)=iL(t)–v(t)/RSmall ripple approximation:vL(t)=0iC(t)=–V / RFundamentals of Power Electronics Chapter 5: Discontinuous conduction mode15Inductor volt-second balancevL(t)0TstD1TsD2TsD3TsVg – V– VVolt-second balance:Solve for V:vL(t) = D1(Vg– V)+D2(–V)+D3(0) = 0V = VgD1D1+ D2note that D2 is unknownFundamentals of Power Electronics Chapter 5: Discontinuous conduction mode16Capacitor charge balanceiL(t)t0DTsTsD1TsD2TsD3Ts<iL> = IipkVg–VL–VLLCR+v(t)–iC(t)iL(t) v(t)/Rnode equation:iL(t)=iC(t)+V /Rcapacitor charge balance:iC=0henceiL= V / Rmust compute dccomponent of inductorcurrent and equate to loadcurrent (for this buckconverter example)Fundamentals of Power Electronics Chapter 5: Discontinuous conduction mode17Inductor current waveformiL(t)t0DTsTsD1TsD2TsD3Ts<iL> = IipkVg–VL–VLpeak current:iL(D1Ts)=ipk=Vg–VLD1Tsaverage current:iL=1TsiL(t) dt0Tstriangle area formula:iL(t) dt0Ts=12ipk(D1+ D2)TsiL=(Vg– V)D1Ts2L(D1+ D2)equate dc component to dc load current:VR=D1Ts2L(D1+ D2)(Vg– V)Fundamentals of Power Electronics Chapter 5: Discontinuous conduction mode18Solution for VTwo equations and two unknowns (V and D2):V = VgD1D1+ D2VR=D1Ts2L(D1+ D2)(Vg– V)(from inductor volt-second balance)(from capacitor charge balance)Eliminate D2 , solve for V :VVg=21+ 1+4K / D12where K