Buck - Boost Converter Buck - Boost Converter
Buck-Boost converter is a dc-to-dc converter that has the capability of stepping up or Buck-Boost converter is a dc-to-dc converter that has the capability of stepping up or stepping down the output voltage. In other words, the output voltage can be higher or stepping down the output voltage. In other words, the output voltage can be higher or lower than the input (source) voltage. It is also labeled as indirect converter because the lower than the input (source) voltage. It is also labeled as indirect converter because the source is never directly connected to the load. It is the inductor in the circuit that controls source is never directly connected to the load. It is the inductor in the circuit that controls the flow of energy from the input side to the output side. Figure-1 shows the buck-boost the flow of energy from the input side to the output side. Figure-1 shows the buck-boost converter in its simplest form. Pay attention to the directions of the currents through the converter in its simplest form. Pay attention to the directions of the currents through the capacitor and the load resistor and the polarity of the output voltage.
capacitor and the load resistor and the polarity of the output voltage.
Figure 1: Boost Converter Figure 1: Boost Converter
The switch S is usually an electronic device that operates either in the conduction The switch S is usually an electronic device that operates either in the conduction mode (when closed) or the cut-off mode (when open). The conduction and cut-off mode (when closed) or the cut-off mode (when open). The conduction and cut-off time-periods are controlled by the suitably designed gating circuits, which are usually not periods are controlled by the suitably designed gating circuits, which are usually not shown. The conduction (on) time of the switch is a fraction of its time period T such that shown. The conduction (on) time of the switch is a fraction of its time period T such that
T T D D T
TONON == , where D is the duty cycle. During the cut-off (off) time when the switch is, where D is the duty cycle. During the cut-off (off) time when the switch is
open, ,
open, TTOFFOFF == ((11−−DD))TT, the inductor current is directed toward the load via diode D. Whenthe inductor current is directed toward the load via diode D. When the switch is closed, the diode D blocks the direct current flow from the source toward the switch is closed, the diode D blocks the direct current flow from the source toward the load. During this time, the current is channeled through the inductor and the capacitor the load. During this time, the current is channeled through the inductor and the capacitor supplies the load current. Only when the switch is in its open position, the inductor
supplies the load current. Only when the switch is in its open position, the inductor current flows toward the load and the capacitor. The diode D helps maintain the current current flows toward the load and the capacitor. The diode D helps maintain the current continuity through the inductor.
continuity through the inductor.
Let us make an assumption that the circuit has been operating for a long time and Let us make an assumption that the circuit has been operating for a long time and the inductor current varies between its minimum and maximum values during each time the inductor current varies between its minimum and maximum values during each time half-time period.
We begin our analysis when the inductor current is at its minimum and the switch We begin our analysis when the inductor current is at its minimum and the switch S
S is is closed. closed. The The differential differential equation equation for for the the inductor inductor current, current, for for 00≤≤ tt ≤≤ TTONON ==DTDT , and, and its solution are
its solution are
S S L L VV dt dt )) tt (( di di L L == min min ,, L L S S L L tt II L L V V )) tt (( ii == ++ (1)(1)
According to this equation, the inductor current increases linearly and attains its According to this equation, the inductor current increases linearly and attains its maximum value
maximum value I I LL..maxmax aas s tt →→TTONON == DTDT such thatsuch that
min min ,, L L S S max max ,, L L DTDT II L L V V II == ++ (2)(2)
Defining the change in the current from its minimum to maximum value as the Defining the change in the current from its minimum to maximum value as the peak-to-peak current ripple
peak-to-peak current ripple ∆∆IILL, the above equation yields an expression for, the above equation yields an expression for ∆∆IILLasas DT DT L L V V II II
IILL == LL,,maxmax −− LL,,minmin == SS ∆
∆ (3)(3)
As soon as the inductor current reaches its maximum value, the switch is opened. As soon as the inductor current reaches its maximum value, the switch is opened. The inductor current now begins to supply the load current and charge the capacitor in The inductor current now begins to supply the load current and charge the capacitor in accordance with the following differential equation.
accordance with the following differential equation. o o L L V V dt dt )) tt (( di di L L == −−
The solution of this equation yields The solution of this equation yields
A A tt L L V V )) tt (( ii oo L L == −− ++
where A is the constant integration and is determined by applying the initial condition at where A is the constant integration and is determined by applying the initial condition at
on on T T tt == . That is. That is A A DT DT L L V V II oo max max ,, L L == −− ++
From this equation, we obtain A and the inductor current as From this equation, we obtain A and the inductor current as
DT DT L L V V II A A == LL,,maxmax ++ oo DT DT L L V V II tt L L V V )) tt (( iiLL == −− oo ++ LL,,maxmax ++ oo (4)(4)
As per this equation, the inductor current decreases linearly form its maximum As per this equation, the inductor current decreases linearly form its maximum value
value and and attains attains its its minimum minimum value value when when tt →→TT,,such thatsuch that
max max ,, L L o o min min ,, L L ((11 DD))TT II L L V V II == −− −− ++ (5)(5)
The peak-to-peak current ripple now is The peak-to-peak current ripple now is
T T )) D D 1 1 (( L L V V II II II oo min min ,, L L max max ,, L L L L == −− == −− ∆ ∆ (6)(6)
The current ripple as given by (3) must be the same as given by (6). Therefore, The current ripple as given by (3) must be the same as given by (6). Therefore, equating the two equations, we get
equating the two equations, we get
T T )) D D 1 1 (( L L V V DT DT L L V VSS oo − − = =
This equation upon simplification yields This equation upon simplification yields
D D 1 1 DV DV V V SS o o − − = = (7)(7)
Equation (7) states that the output voltage of the buck converter is directly Equation (7) states that the output voltage of the buck converter is directly
proportional to D and indirectly proportional to (1-D). When D = 0.5, the output voltage proportional to D and indirectly proportional to (1-D). When D = 0.5, the output voltage is exactly equal to the applied voltage. The output voltage is greater than the applied is exactly equal to the applied voltage. The output voltage is greater than the applied voltage
voltage as as long long as as DD >> 00..55, which corresponds to the boost operation. On the other hand,, which corresponds to the boost operation. On the other hand, the buck-boost converter behaves as a buck converter with output voltage less than the the buck-boost converter behaves as a buck converter with output voltage less than the applied
applied voltage voltage when when DD << 00..55..
When the switch, the inductor, and the capacitor are treated as ideal elements, the When the switch, the inductor, and the capacitor are treated as ideal elements, the average power dissipated by them is zero. Consequently, the average power supplied by average power dissipated by them is zero. Consequently, the average power supplied by the source must be equal to the average power delivered to the load. That is,
the source must be equal to the average power delivered to the load. That is,
o o S S o o o o S S S S II D D 1 1 V V II V V II V V − − = = = =
This equation helps us express the average source current in terms of the average This equation helps us express the average source current in terms of the average load current as load current as D D 1 1 DI DI IISS oo − − = = (8)(8)
The output current is The output current is
− − = = = = D D 1 1 D D R R V V R R V V IIoo oo SS (9)(9)Equations (8) and (9) help us express the source current in terms of the duty cycle Equations (8) and (9) help us express the source current in terms of the duty cycle as as 2 2 S S S S D D 1 1 D D R R V V II
− − = = (10)(10)Figure 2: Source current in a buck-boost converter Figure 2: Source current in a buck-boost converter
The source current as a function of time is shown in Figure-2. It is simply a plot The source current as a function of time is shown in Figure-2. It is simply a plot of the inductor current as given by (1) when the switch is in its closed position. The of the inductor current as given by (1) when the switch is in its closed position. The average source current is
average source current is
D D DT DT L L 2 2 V V II dt dt II tt L L V V T T 1 1 II LL,,minmin SS DT DT 0 0 min min ,, L L S S S S
+ + = =
+ + = =∫∫
(11)(11)Figure 3: Inductor current Figure 3: Inductor current
From Figure-3, which is a plot of inductor current, it is clear that From Figure-3, which is a plot of inductor current, it is clear that
avg avg ,, L L L L min min ,, L L S S min min ,, L L II 2 2 II II DT DT L L 2 2 V V II ++ == ++ ∆∆ == (12)(12)
Consequently, the average values of the source and the inductor currents are related as Consequently, the average values of the source and the inductor currents are related as
avg avg ,, L L S S DDII II == (13)(13)
The average inductor current from (10) and (13) is The average inductor current from (10) and (13) is
( (
))
22 S S avg avg ,, L L D D 1 1 D D R R V V II − − = = (14)(14)The expressions for the maximum and minimum currents through the inductor may now The expressions for the maximum and minimum currents through the inductor may now be written as be written as Lf Lf 2 2 DV DV )) D D 1 1 (( R R DV DV 2 2 II II
IILL,,maxmax LL,,avgavg LL SS 22 ++ SS − − = = ∆ ∆ + + = = (15)(15) Lf Lf 2 2 DV DV )) D D 1 1 (( R R DV DV 2 2 II II II SS 2 2 S S L L avg avg ,, L L min min ,, L L −− − − = = ∆ ∆ − − = = (16)(16)
The peak-to-peak current ripple can be expressed in terms of the input and output The peak-to-peak current ripple can be expressed in terms of the input and output voltages, as voltages, as )) D D 1 1 (( Lf Lf V V D D Lf Lf V V II SS oo L L == == −− ∆ ∆ (17)(17)
Figure 4: The diode current Figure 4: The diode current
The current through the diode is shown in Figure 4. Its average value is the same The current through the diode is shown in Figure 4. Its average value is the same as the average load current and can be computed as
as the average load current and can be computed as
R R V V T T T T 2 2 II II
II LLmaxmax LL,,minmin OFFOFF oo avg avg ,, D D == + + = = (18)(18)
Since the average current in the diode is equal to the average current through the Since the average current in the diode is equal to the average current through the load resistor R, the average current in the capacitor is zero.
When the switch is in its closed position, the capacitor supplies the load current. When the switch is in its closed position, the capacitor supplies the load current. Hence,
Hence, from from 00≤≤ tt ≤≤ TTONON == DTDT, the capacitor current is, the capacitor current is
R R V V II )) tt (( ii oo o o C C == −− == −− (19)(19)
When the switch is opened, the inductor current supplies both the capacitor When the switch is opened, the inductor current supplies both the capacitor ccuurrrreennt t aannd d tthhe e llooaad d ccuurrrreenntt. . TThhuuss, , dduurriinng g tthhe e ttiimme e iinntteerrvvaal l ffrroomm00 ≤≤ tt ≤≤TTOFFOFF == ((11−−DD))TT,, the capacitor current is
the capacitor current is
o o L L C C((tt)) ii ((tt)) II ii == −− (20)(20)
The maximum and minimum values of the capacitor current when the switch is in The maximum and minimum values of the capacitor current when the switch is in its open position as
its open position as
Lf Lf 2 2 DV DV D D 1 1 D D R R V V II II II SS 2 2 S S o o max max ,, L L max max ,, C C
++
− − = = − − = = (21)(21) Lf Lf 2 2 DV DV D D 1 1 D D R R V V II II II SS 2 2 S S o o min min ,, L L min min ,, C C
−−
− − = = − − = = (22)(22)It must be noted that It must be noted that
min min ,, C C max max ,, C C L L II II II == −− ∆ ∆ (23)(23)
The capacitor current waveform is shown in Figure 5. The capacitor current waveform is shown in Figure 5.
Figure 5: Current through the capacitor Figure 5: Current through the capacitor
The current waveform of Figure-5 helps us determine the change in the voltage The current waveform of Figure-5 helps us determine the change in the voltage across the capacitor. During the time the switch is closed, the charge on the capacitor is across the capacitor. During the time the switch is closed, the charge on the capacitor is decreasing because the capacitor is supplying the current to the load. The change in the decreasing because the capacitor is supplying the current to the load. The change in the charge is charge is DT DT R R V V T T II Q Q oo ON ON o o == −− − − = = ∆ ∆ (24)(24)
The decrease in the charge will result in the decrease of the capacitor voltage. Therefore, The decrease in the charge will result in the decrease of the capacitor voltage. Therefore, the magnitude of the change in the capacitor voltage is
the magnitude of the change in the capacitor voltage is DT DT RC RC V V C C Q Q V V oo o o == ∆ ∆ = = ∆ ∆ (25)(25)
Note that during the time the switch is open, the component of the inductor Note that during the time the switch is open, the component of the inductor current that flows through the capacitor will increase the capacitor voltage by the same current that flows through the capacitor will increase the capacitor voltage by the same amount. Hence, when we define the capacitor ripple as the ratio of the increase in the amount. Hence, when we define the capacitor ripple as the ratio of the increase in the capacitor voltage to its average value, it can then be expressed as
capacitor voltage to its average value, it can then be expressed as
RCf RCf D D RC RC DT DT V V V V o o o o == == ∆ ∆ (26) (26)
Note that the capacitor ripple define by (26) is not the same as the peak-to-peak Note that the capacitor ripple define by (26) is not the same as the peak-to-peak voltage ripple for the rectifiers. The peak-to-peak voltage ripple for the buck-boost voltage ripple for the rectifiers. The peak-to-peak voltage ripple for the buck-boost converter will be twice of that given by (26). Equation (26) may be viewed as one-sided converter will be twice of that given by (26). Equation (26) may be viewed as one-sided voltage ripple.
voltage ripple.
The buck-boost converter can operate either in its continuous conduction mode or The buck-boost converter can operate either in its continuous conduction mode or discontinuous conduction mode. When it operates in the continuous conduction mode, discontinuous conduction mode. When it operates in the continuous conduction mode, there is always a current in the inductor. The minimum current in the continuous
there is always a current in the inductor. The minimum current in the continuous
conduction mode can be zero at the time of switching. Consequently, there is a minimum conduction mode can be zero at the time of switching. Consequently, there is a minimum value of the inductor that ensures its continuous conduction mode. It can be obtained value of the inductor that ensures its continuous conduction mode. It can be obtained from (16) by setting
from (16) by setting IILL,,minmintto o zzeerro o aass
0 0 D D )) D D 1 1 (( f f L L 2 2 V V )) D D 1 1 (( R R V V min min o o o o −− −− == − − Hence, Hence, 2 2 min min ((11 DD)) f f 2 2 R R L L == −− (27)(27)
From the peak-to-peak current ripple, we can also obtain an expression for the percent From the peak-to-peak current ripple, we can also obtain an expression for the percent current ripple as current ripple as
= = − − = = × × ∆ ∆ = = L L L L 2 2 100 100 )) D D 1 1 (( Lf Lf R R 100 100 100 100 II II CR CR % % 22 minmin avg avg ,, L L L L (28) (28)Example: ___________________________________________________________ Example: ___________________________________________________________ A buck-boost converter operating at a frequency of 20 kHz is used to step-up a 120-V dc A buck-boost converter operating at a frequency of 20 kHz is used to step-up a 120-V dc supply to 480-V so that it can provide the rated voltage to a 480-V, 1000-W heater. To supply to 480-V so that it can provide the rated voltage to a 480-V, 1000-W heater. To ensure a reliable operation, inductor must at least be 20% greater than its minimum value ensure a reliable operation, inductor must at least be 20% greater than its minimum value and the voltage ripple should be within 1%. Design the boost converter.
and the voltage ripple should be within 1%. Design the boost converter. Solution:
Solution:
From the given data, the duty cycle, from (7), is From the given data, the duty cycle, from (7), is
8 8 .. 0 0 120 120 480 480 480 480 V V V V V V D D S S o o o o = = + + = = + + = =
The time period, the on time and off times of the switch are The time period, the on time and off times of the switch are
ss 50 50 000 000 ,, 20 20 1 1 f f 1 1 T T == == == µµ ss 5 5 .. 37 37 10 10 50 50 75 75 .. 0 0 DT DT T TONON == == ×× ×× −−66 == µµ ss 5 5 .. 12 12 )) 10 10 50 50 )( )( 75 75 .. 0 0 1 1 (( T T )) D D 1 1 (( T TOFFOFF == −− == −− ×× −−66 == µµ The equivalent resistor of the heater is
The equivalent resistor of the heater is
Ω Ω = = = = = = 230230..44 1000 1000 480 480 P P V V R R 2 2 2 2 o o
The minimum value of the inductor for the continuous conduction mode, from (22), is The minimum value of the inductor for the continuous conduction mode, from (22), is
H H 4 4 .. 230 230 )) 8 8 .. 0 0 1 1 (( 000 000 ,, 20 20 2 2 4 4 .. 230 230 )) D D 1 1 (( f f 2 2 R R L Lminmin 22 −− 22 == µµ × × = = − − = =
The minimum value of the inductor to satisfy the design requirement must be The minimum value of the inductor to satisfy the design requirement must be
H H 48 48 .. 276 276 10 10 4 4 .. 230 230 2 2 .. 1 1 L L 2 2 .. 1 1 L L == minmin == ×× ×× −−66 == µµ To allow for inductor-to-inductor variations, let us select
To allow for inductor-to-inductor variations, let us select H H 300 300 L L == µµ
The average current through the heater is The average current through the heater is
083 083 .. 2 2 4 4 .. 230 230 480 480 R R V V II oo o o == == == AA
The peak-to-peak current ripple from (12) is The peak-to-peak current ripple from (12) is
16 16 10 10 50 50 8 8 .. 0 0 10 10 300 300 120 120 DT DT L L V V IILL SS 66 ×× ×× ×× 66 == × × = = = = ∆ ∆ −− −− AA
The average source current is The average source current is
332 332 .. 8 8 120 120 083 083 .. 2 2 480 480 V V II V V II S S o o o o S S == × × = = = = AA
The average inductor current, from (13), is The average inductor current, from (13), is
415 415 .. 10 10 8 8 .. 0 0 332 332 .. 8 8 D D II II SS avg avg ,, L L == == == AA
Hence, the maximum and minimum currents through the inductor, from (8) and (9), are Hence, the maximum and minimum currents through the inductor, from (8) and (9), are
415 415 .. 18 18 2 2 16 16 415 415 .. 10 10 2 2 II II
IILL,,maxmax == LL,,avgavg ++ ∆∆ LL == ++ == AA
415 415 .. 2 2 2 2 16 16 415 415 .. 10 10 2 2 II II II LL avg avg ,, L L min min ,, L L == −− == ∆ ∆ − − = = AA
Let us now select the capacitor based upon the desired voltage ripple of 1%. Let us now select the capacitor based upon the desired voltage ripple of 1%. Using (16), we determine C as Using (16), we determine C as F F 36 36 .. 17 17 01 01 .. 0 0 000 000 ,, 20 20 4 4 .. 230 230 8 8 .. 0 0 V V V V Rf Rf D D C C o o o o µ µ = = × × × × = = ∆ ∆ = =
For this application, let us select a standard
20-For this application, let us select a standard 20-µµF capacitor.F capacitor.
The percent current ripple of the inductor current can be computed as The percent current ripple of the inductor current can be computed as
% % 6 6 .. 153 153 100 100 415 415 .. 10 10 16 16 100 100 II II CR CR % % avg avg ,, L L L L = = × × = = × × ∆ ∆ = =
In some applications, such a high percent ripple may not be acceptable. We have to In some applications, such a high percent ripple may not be acceptable. We have to increase the value of the inductor to decrease the ripple.
