Discussed
here are the discontinuous conduction mode, mode boundary, and conversion ratio
of simple converters.
During continuous conduction
mode, the inductor current in the energy transfer never reaches zero value. In
the case of the discontinuous conduction mode, the inductor current falls to
zero level which is very common in DC-to-DC converters.
If the peak of the inductor
current ripples is less than the DC component of the inductor current, the
diode current is always positive and the diode is forced to turn on when the
switch S (either a transistor or thyristor) is
off. On the other hand, if the peak of the inductor current ripples becomes
more than the DC component of the inductor current, the total current falls to
zero value while the diode is conducting. Thus, the diode will stop conducting
and the inductor current will remain at zero value until the switch S will be
gated again due to the polarity reversal across the switch. This gives rise to
the discontinuous conduction mode in the chopper or the DC-to-DC converter.
In the discontinuous
conduction mode, inductor current is not persistent throughout the complete
cycle and reaches zero level earlier even before the end of the period.
Discontinuous conduction mode inductance is less than the minimum value of the
inductance for the continuous conduction mode,
LDCM < LCCM.
Thus, this condition generally
arises for the light-load condition.
Let the value of inductance
in the case of the discontinuous conduction mode be,
LDCM=ξ LCCM where 0<ξ<1
for the discontinuous conduction.
The discontinuous conduction
mode usually occurs in converters which consist of single-quadrant switches and
may also occur in converters with two-quadrant switches. Two-level DC buck, and
boost and buck-boost converters will be discussed further in this article.
There are two levels indicated here towards the two-voltage level for the
inductor voltage.
The energy stored in the
inductor is proportional to the square of the current flowing through it.
Having the same power through the converter, the requirement of the inductor
current is higher in the case of the discontinuous conduction as compared to
the continuous conduction mode. This causes more losses in the circuit of the
discontinuous conduction. As the energy stored is not yet released to the output
in the discontinuous conduction, the output gets affected by the ringing. This
may also cause a noise in the discontinuous conduction mode.
Moreover, the value of the
inductance required for discontinuous conduction mode is lesser as compared to
the continuous conduction mode since it allows the
fall of the inductor current to zero level. This causes higher values for the
root-mean-square and the peak current. Thus, the size of the transformer required in isolated
converters is bigger as compared to the continuous-conduction
transformer size to suit the larger flux linkage and the losses.
Conversion ratio is
independent of the load during the continuous conduction mode but when it
enters in discontinuous conduction mode, it becomes dependent to the load. This
complicates the DC-circuit analysis because the first-order equations become
second order.
In most of the applications,
continuous conduction mode is employed. Yet, discontinuous conduction mode can
also be used for certain applications such as for the low-current and
loop-compensation applications.
Consider the simple buck
converter circuit shown in Fig. 1. The current in the converter is controlled
here by two switches labeled as S (MOSFET)
and D (Diode).
Figure 1.
Circuit for Buck Converter
This is a single-quadrant
converter with the following waveforms for the continuous conduction mode shown
in Fig. 2.
Figure 2. Supply
Current IS, Diode Current ID, Inductor Current I, and Inductor
Voltage VL Waveforms respectively (Buck
Converter)
The buck converter in
discontinuous and continuous conduction modes are in the second-order and
first-order systems respectively.
For continuous conduction
mode,
IL+=1L∫t0(VS−VO)dt+IminIL+=1L∫0t(VS−VO)dt+Imin
For 0≤t≤DT.
⇒IL+=VS−VOLt+Imin⇒IL+=VS−VOLt+Imin
At t=DT, inductor current is
at maximum value,
Imax=VS−VOLDT+IminImax=VS−VOLDT+Imin
[Equation
1]
IL−=(1L)∫tDT−VOdt+ImaxIL−=(1L)∫DTt−VOdt+Imax
For DT≤t≤T.
⇒IL−=VOL(DT−t)+Imin⇒IL−=VOL(DT−t)+Imin
The average value of the
inductor current for the buck converter is
Iavg=VOR.Iavg=VOR.
Because the inductor is
always connected to the load whether the switch is on or off.
The average value of the
current through the capacitor is nil due to the capacitor charge balance
condition.
From Fig. 2, the area under
the inductor current waveform is,
(Area)L=TImin+12T(Imax−Imin)(Area)L=TImin+12T(Imax−Imin)
Average value of the inductor
current is,
Iavg=VOR=Imin+12(Imax−Imin)Iavg=VOR=Imin+12(Imax−Imin)
[Equation
2]
From Equations 1 and 2 we can
get,
Iavg=VS−VO2LDT+IminIavg=VS−VO2LDT+Imin
⇒Iavg=D(VS−VO)2Lf+Imin=VOR⇒Iavg=D(VS−VO)2Lf+Imin=VOR
The value of inductance is,
L=D(VS−VO)R2f(VO−IminR)L=D(VS−VO)R2f(VO−IminR)
The boundary of continuous
condition is when Imin=0. If the value of Imin<0, the converter enters in the discontinuous conduction
mode.
Thus,
L=LCCM for Imin=0
Hence,
LCCM=D(VS−VO)R2fVOLCCM=D(VS−VO)R2fVO
The value of inductance for
the discontinuous conduction is given by
L=LDCM=ξLCCM=ξD(VS−VO)R2fVO,L=LDCM=ξLCCM=ξD(VS−VO)R2fVO,
where 0<ξ<1.
For discontinuous conduction
mode, when L< LCCM, the waveforms
for the inductor current and inductor voltage are shown in Fig.3.
Figure 3.
Inductor Current and Voltage for the Discontinuous Conduction Mode of Buck
Converter
It is clear from the Fig.3
that the value of the minimum inductor current is zero i.e. Imin=0.
As the current across the
inductor current is reduced to zero, the value of the voltage across the
inductor is also reduced to zero value while VC =VO during
the entire cycle.
For
the time duration 0 ≤ t ≤ TON
IL+(t)=VS−VOLtIL+(t)=VS−VOLt
[Equation
3]
As the value of the peak
inductor current occurs at t = TON,
⇒Imax=VS−VOLTON=VS−VOLDT=VS−VOLfD.⇒Imax=VS−VOLTON=VS−VOLDT=VS−VOLfD.
For the time
duration TON ≤ t ≤ TX,
IL−(t)=∫tTON−VCLdt+ImaxIL−(t)=∫TONt−VCLdt+Imax
⇒IL−(t)=VCL(TON−t)+VS−VOLf⇒IL−(t)=VCL(TON−t)+VS−VOLf
[Equation
4]
At t = TX, current reduces to zero value,
0=VCL(TON−TX)+VS−VOLf0=VCL(TON−TX)+VS−VOLf
⇒TX=DVSfVO⇒TX=DVSfVO
Compared to the continuous
condition, the amount of energy needed by the load is lesser in the
discontinuous condition.
It is considered that the
converter is operated in the steady state. Thus, the energy in the inductor
remains the same at the start and at the end of the cycle. The volt-time
balance condition can also be applied here.
The above equation can also
be derived using the inductor volt-second balance condition as,
(VS−VC)TON+(−VC)(TX−TON)=0(VS−VC)TON+(−VC)(TX−TON)=0
⇒(VS−VC)DT+(−VC)(TX−DT)=0⇒(VS−VC)DT+(−VC)(TX−DT)=0
⇒TX=VSDfVO⇒TX=VSDfVO
For the time
duration TX ≤ t ≤ T
IL0(t)=0IL0(t)=0
From the Fig. 3, it is clear
that the average value of the inductor current is equal to the area under the
load current curve divided by T.
Iavg=12TXImaxTIavg=12TXImaxT
For the DC supply,
Iavg=VORIavg=VOR
Hence,
VOR=VS(VS−VO)D22LVOfVOR=VS(VS−VO)D22LVOf
The duty cycle ratio for the
discontinuous conduction mode in the case of the buck converter is,
D=VO2LfRVS(VS−VO)−−−−−−−−−−−−√D=VO2LfRVS(VS−VO)
[Equation
5]
The duty cycle ratio of the
buck converter in its continuous conduction mode is
D=VOVi.D=VOVi.
The duty cycle ratio for the
buck converter is also dependent on the inductance L, load resistance R, and
the switching frequency f.
For discontinuous conduction
mode,
L=LDCM=ξLCCM=ξD(VS−VO)R2fVOL=LDCM=ξLCCM=ξD(VS−VO)R2fVO
[Equation
6]
Substitution of the Equation
5 into Equation 6 gives,
D=VOVSξ√D=VOVSξ
[Equation
7]
Since 0 < ξ < 1,
duty cycle ratio of the buck converter in the discontinuous conduction mode is
less than its value in the continuous conduction mode. Thus, less amount of
energy is transferred through the converter which is not enough to maintain the
inductor current throughout the entire period. This is the reason the
discontinuous current flows through the inductor.
The conversion ratio of buck
DC-to-DC converter is,
VOVS=Dξ√VOVS=Dξ
where 0<ξ<1
If the value of ξ is
greater than 1, the converter enters in the continuous conduction mode. We can
easily know the conduction state of the buck converter, which is either
continuous or discontinuous, if we know the value of input and output voltages
of the converter by simply measuring the value of ξ.
Instantaneous value of the
capacitor current is given by subtracting the value of the inductor current to
the load current. When the inductor current value is reduced to zero value, the
load current is supplied by the capacitor.
From Equations 3 and 4 we can
get:
For the time duration 0 < t
IC+(t)=VS−VOLt−IOIC+(t)=VS−VOLt−IO
[Equation
8]
For the time duration DT <
t < TX,
IC−(t)=VOL(DT−t)+D(VS−VO)Lf−IOIC−(t)=VOL(DT−t)+D(VS−VO)Lf−IO
[Equation
9]
And for the time duration TX < t < T,
ICO=−IOICO=−IO
[Equation
10]
If the capacitance is assumed
to be ideal, the capacitor current will not decay even after the inductor
current value is reduced to zero value. For that case, the waveforms for the
capacitor and inductor current are shown in Fig. 4.
From Fig.4, it is clear that
the value of the capacitor current is zero at time t=Ta and at t=Tb.
Equation 8 at time t =Ta gives,
0=VS−VOLTa−IO0=VS−VOLTa−IO
⇒Ta=LIOVS−VO⇒Ta=LIOVS−VO
[Equation
11]
And Equation 9 at time t=Tb gives,
0=VO(DT−Tb)L+D(VS−VO)Lf−IO0=VO(DT−Tb)L+D(VS−VO)Lf−IO
⇒Tb=DT−LIOVO+D(VS−VO)fVO⇒Tb=DT−LIOVO+D(VS−VO)fVO
[Equation
12]
Figure 4.
Inductor Current and Capacitor Current respectively for the Discontinuous
Conduction Mode
of the Buck Converter
The positive time interval
for the charge accumulation i.e. Tb-Ta from Equations 11 and
12 is given by:
Tb−Ta=DVSfVO−LIOVSVO(VS−VO)Tb−Ta=DVSfVO−LIOVSVO(VS−VO)
[Equation
13]
From Equation 6 and Equation
7 we can get,
Tb−Ta=2ξ√−ξ2fTb−Ta=2ξ−ξ2f
[Equation
14]
From Fig.4, it is also clear
that the maximum value of the capacitor current occurs at the time t=DT.
At t = DT,
IC(DT) = Ihp
From Equation 8 we can get,
Ihp=(2ξ√−1)IOIhp=(2ξ−1)IO
[Equation
15]
The charge accumulated is the
integration of the capacitor current (area under the capacitor current
from Ta to Tb) which is also
given by the expression:
∆Q=C∆V [Equation
16]
Thus,
C∆V=12(2ξ√−ξ)(2ξ√−1)IO(12f)=VO(2−ξ√)24RfC∆V=12(2ξ−ξ)(2ξ−1)IO(12f)=VO(2−ξ)24Rf
The ripples in the load due
to the ripples in the capacitor are given by the following expression:
r=∆VVO=(2−ξ√)24Rfr=∆VVO=(2−ξ)24Rf
Circuit for the boost
converter is shown in Fig. 5.
Figure 5.
Circuit for Boost Converter
The waveform for the
continuous conduction mode is shown in Fig. 6. When it is in the discontinuous
conduction mode, the waveform is shown in Fig. 7.
We can assume that the
inductor is connected to the load for the time Ty such that
IO =Y Iavg [Equation 17]
where
Y
= Ty/T
Figure 6. Supply
Current, Diode Current, Inductor Current and Inductor Voltage respectively
(Boost Converter)
Figure 7.
Inductor Current and Voltage for the Discontinuous Conduction Mode of Boost
Converter
When the converter operates
in the steady-state condition, the energy at the start and at the end of the
cycle is the same. Thus, volt-time balance condition can be applied here too.
From the figure and the
volt-time balance condition it is clear that,
TON VS+(TX-TON).(VS-VC)=0
⇒DTVS+(TX−DT).(VS−VO)=0⇒DTVS+(TX−DT).(VS−VO)=0
⇒TX=DVO(VS−VO)f⇒TX=DVO(VS−VO)f
From Fig. 6, it is also
evident that the value of the minimum and maximum currents are as follows:
Imin=0;
and
Imax=VSLTON=VSLfDImax=VSLTON=VSLfD
Thus, the average value of
the inductor current is,
Iavg=11−DVOR=12TXImaxTIavg=11−DVOR=12TXImaxT
⇒VOR=(12)(DVO(VS−VO)f)(VSDLf)(1−D)T⇒VOR=(12)(DVO(VS−VO)f)(VSDLf)(1−D)T
From Equation 17 we can get,
D=2(VO−VS)LfRYVS−−−−−−−−−−−−√D=2(VO−VS)LfRYVS
[Equation
18]
The duty cycle ratio of the
buck converter for the continuous conduction mode is equal to
VO−VSVS.VO−VSVS.
In the discontinuous
conduction mode, the duty cycle ratio of the boost converter is not only
dependent on the input and output voltages but it also depends on the
inductance L, load resistance R, and the switching frequency f.
The discontinuous inductance
for the boost converter is,
LDCM=ξYRVS(VO−VS)2fVOLDCM=ξYRVS(VO−VS)2fVO
Substituting this value of
inductance in Equation 18 we can get,
D=VO−VSVOξ√D=VO−VSVOξ
⇒VOVS=VS1−Dξ√⇒VOVS=VS1−Dξ
Hence, the complete
conversion ratio for the boost converter in the discontinuous conduction mode
is given by the above expression.
The circuit for the
buck-boost converter is shown in Fig. 8 and the related waveforms of the
buck-boost converter in the case of continuous conduction mode are shown in
Fig. 9.
Figure 8.
Circuit for the Buck-Boost Converter
Inductor is connected to the
load during the switch-off period; where Y= (1-D).
Thus,
Iavg=YIO=YVOR=(1−D)VORIavg=YIO=YVOR=(1−D)VOR
Figure 9. Supply
Current, Diode Current, Inductor Current and Inductor Voltage respectively
(Buck-Boost Converter) in Continuous Conduction Mode
Figure 10.
Inductor Current and Inductor Voltage for the Discontinuous Conduction Mode
of the Buck-Boost Converter
Assume that the converter is
operating in steady state; therefore, energy at the start up to the end of the
cycle must be equal. Thus, volt-time balance condition is applied here.
Applying the volt-sec balance
across the inductor using the Fig. 9,
VS TON + (TX -
TON) (-VO) = 0
⇒VSDT−(TX−DT)VO=0⇒VSDT−(TX−DT)VO=0
⇒TX=D(VS+VO)VOf⇒TX=D(VS+VO)VOf
From the Fig. 9, it is also
noticed that the value of the minimum and maximum currents are as follows:
Imin = 0
Imax=VSLTON=VSLfDImax=VSLTON=VSLfD
Thus, the average value of
the inductor current is,
Iavg=YVOR=12ImaxTXTIavg=YVOR=12ImaxTXT
⇒VOR=12VSDLfD(VS+VO)YVOff⇒VOR=12VSDLfD(VS+VO)YVOff
In the discontinuous
conduction mode of the buck-boost converter, the value of the duty cycle ratio
is given by
D=VO2LfRYVS(VS+VO)−−−−−−−−−−−−−√D=VO2LfRYVS(VS+VO)
The duty cycle ratio of the
buck-boost converter for the continuous conduction mode is equal to
VOVO+VS.VOVO+VS.
In the case of the
discontinuous conduction mode, the duty cycle ratio of the buck-boost converter
is also dependent on the inductance L, load resistance R, and the switching
frequency f.
The conversion ratio for the
buck-boost converter is,
VOVS=Dξ√−D