Buck Boost Converter
Design
Buck Boost Converter Design
Before reading this section, please read the
Introduction to DC to DC Converter Design.
All of the circuits in this tutorial can be
simulated in LTspice^{®}. If you are new to
LTspice,
please have a look at my
LTspice Tutorial.
__
Introduction__
A buck boost power supply generates a constant
output voltage when the input is either above or
below the output voltage.
The SEPIC Converter (Single Ended Primary Inductance
Converter) and the 4 Switch Buck-Boost Converter are
the two main buck-boost architectures each with its
benefits and drawbacks. The SEPIC converter will be
discussed on this page. The 4 Switch Buck Boost
architecture is discussed here:
4
Switch Buck Boost Architecture
__
The SEPIC Converter__
The SEPIC Converter is the most common and long
standing buck boost architecture and is illustrated
in FIG 1.
**
FIG 1**
The LTspice circuit can be downloaded here:
SEPIC Converter.
The architecture is very similar to a boost
converter. However with a boost converter there is a
dc path that flows through the inductor and
rectifier diode into the output capacitor meaning
the output voltage can never be lower than the input
voltage. With a SEPIC converter, capacitor C5 in FIG
1 breaks this dc path enabling the output of a SEPIC
converter to start from 0V.
Another benefit of the SEPIC converter is that it
can also act as a low noise buck converter. The
conventional buck converter has sharp edges on the
input current that can interfere with circuitry
connected to the input voltage. With the SEPIC
converter, the input inductor, L1, slows down the
input current edges thus providing a lower noise
buck regulator solution.
To analyse FIG 1 we will assume the input voltage is
fixed at 10V and the output is in regulation at 5V.
MOSFET Q1 turns on, shorting the right hand side of
L1 to 0V (ignoring the small sense resistor R5) thus
a fixed voltage is applied across L1. According to
if the voltage across the inductor is constant the
inductor current ramps up linearly with time (di/dt
is constant).
When the MOSFET switches off, the inductor tries to
maintain its current flow. It does this by
generating a voltage across its terminals very
similar to a battery, where the current flows from
the negative terminal, through the battery, to the
positive terminal.
In the circuit of FIG 1, we can see that to maintain
current flow, the right hand side of the inductor
has to increase in voltage with respect to the left
hand side. The left hand side is connected to the
input voltage (so cannot change), thus the right
hand side voltage increases above the input voltage.
The inductor current flows into capacitor C5, thus
charging C5.
To save a lot of maths, it is worthwhile considering
the operation of the circuit from a common sense
viewpoint. The dc voltage across an inductor, when
averaged over time, must always be 0V. Thus, the
average voltage at the drain of Q1 (labelled DRAIN)
must be equal to the input voltage. Likewise the
average voltage at the top of L2 (labelled ANODE)
must be equal to 0V. Thus the average voltage across
C5 is equal to the input voltage, Vin.
When Q1 switches ON, it puts a negative going edge
on DRAIN. If the voltage across C5 is constant, this
means a negative going edge of equal magnitude
appears at ANODE. When Q1 switches OFF, the voltage
at DRAIN rises until something conducts (to maintain
the current flow through L1). If voltage across C5
is constant, the voltage at ANODE rises until D1
conducts, thus the voltage at DRAIN rises to (Vin +
Vout).
Therefore the voltage at DRAIN node is a square wave
oscillating from 0V to (Vin + Vout) and the voltage
at ANODE is a square wave of the same amplitude, but
offset by the voltage across C5 (Vin), thus
oscillating from –Vin to Vout.
From the equation
when Q1 switches ON, the current in L1 ramps up
linearly with time and flows from left to right.
Since the voltage across the inductor is equal to
Vin, the change in current with time is represented
by
When Q1 is ON, the voltage at ANODE is equal to
–Vin, so current flows UP L2 increasing in
amplitude.
Since D1 is reverse biased, no current flows to the
output, so all of the current flowing UP L2 flows
into Q1. This has important consequences when
choosing the current sense resistor, R7, as well as
the characteristics of the MOSFET, Q1.
When Q1 switches off, DRAIN rises to (Vin + Vout)
meaning there is a voltage of Vout across L1 so the
current in L1 ramps down linearly according to
Thus we can see that the current in L1 increases
proportional to Vin and decreases proportional to
Vout.
Now we will consider the current in L2.
When Q1 switches ON, the voltage at ANODE goes to
–Vin, so the current in L2 flows towards ANODE,
increasing linearly with time and is given by
When Q1 switches off, the voltage at ANODE goes to
Vout, so the current in L2 decreases linearly with
time and is given by
Thus we can see, like L1, the current in L2
increases proportional to Vin and decreases
proportional to Vout.
In FIG 2, we can clearly see the charge and
discharge characteristics of L1 and L2 are
identical. The input voltage here is 10V and the
output voltage is 5V/1A and we can see a steeper
charging slope than discharging slope. (LTspice
probes the current flowing *down* L2. To invert
this waveform, right click over the plot icon I[L2]
and insert a ‘-‘ sign in front).
**
FIG 2**
We can also see DRAIN oscillating between 15V (Vin +
Vout) and ground and ANODE oscillating between –Vin
and Vout.
It is useful to calculate the duty cycle of the
SEPIC converter. Assuming the current flow in both
inductors is continuous we can say that the change
in inductor current during charging is the same as
the change in inductor current during discharging.
Hence, for L1,
Where *dt*_{1} is the charge time and
*dt*_{2} is the discharge time.
Hence
If Duty Cycle (DC) is represented by
then
so
so
so
therefore
This is an approximation (but is good enough for
most designs) since we have ignored the voltage drop
across the diode. The higher the output voltage, the
more the diode drop can be ignored and the more
accurate the above statement becomes. This equation
only holds true if the inductor currents never fall
to zero. This is known as Continuous Conduction Mode
(CCM) operation.
The same calculation can be done for L2 and since L1
and L2 charge and discharge at the same rate, the
equation works out the same.
Now, we stated above that the average voltage across
an inductor is zero. If this were not the case, the
current in the inductor would rise over time.
Likewise with a capacitor, the average *current*
in a capacitor is zero, otherwise the capacitor
voltage would rise over time.
We can use this observation to determine the
currents in the input inductor and output inductor.
If the average current flowing in C5 is zero, we can
assume that C5 represents an open circuit when
averaged over one switching cycle. Thus the average
output current is equal to the average current
flowing in L2. Likewise the average input current is
equal to the average current in L1. If this were not
the case then the voltage across C5 would increase
over time.
__
SEPIC Converter Design Procedure__
We are going to use the LT3757 to design a SEPIC
converter that converts and input voltage of 4.5V-8V
to an output voltage of 5V supplying a load of 1A
with a switching frequency of 500kHz. The LT3757
datasheet can be downloaded here:
LT3757 Datasheet.
The outline schematic is shown in FIG 3
**
FIG 3**
We need to design for the lowest input voltage,
since this is when the input current is at its
highest.
The duty cycle of the converter when operating in
continuous conduction mode is given by
The datasheet says we need a 24.3kOhm resistor to
set the switching frequency to 500kHz. With a
switching frequency of 500kHz, this equates to a FET
ON time of
At this point it is worth checking that we are not
violating the minimum ON time of the controller
(220ns).
__
Inductor Choice__
The output current is 1A, so the average current
flowing in L2 is 1A. If the output power is 5W and
the circuit operates with an efficiency of 85%, this
means the input power is 5.88W. At 4.5V this results
in an average input current of 1.31A.
When the MOSFET switches ON we have seen that the
currents in both L1 and L2 increase and both
currents flow through Q1. Thus the average current
in Q1 is 2.31A. It is generally good design practice
to keep the ripple current in the switch equal to
about 40% of the total average current. Thus the
ripple current should be approximately 924mA.
To divide the ripple current equally between L1 and
L2 we should design to make the ripple current in
each inductor equal to 462mA. Therefore from
we get
Implying the inductors need to be 10uH.
The 462mA ripple current is superimposed on top of
the average inductor current.
The average current in L2 is 1A so L2’s current
should ramp from 769mA to 1.231A. The average
current in L1 is 1.31A, so L1’s current should ramp
from 1.08A to 1.54A. Naturally the average current
in L1 is dependent on the efficiency of the
converter.
Obviously if the input voltage is higher, the input
current will be proportionately lower.
Therefore for L1, we need to choose an inductor with
a saturation current of at least 1.54A and for L2 a
saturation current of at least 1.231A. If too much
current flows in the inductor, the ferrite that the
inductor is wound on saturates and the inductor
loses its inductive properties making the inductance
value fall. From the equation
if the inductor value falls, the current ramp
increases causing the ferrite to further saturate
causing more current to flow……
Therefore must make sure that the inductor never
saturates. To keep things simple, it is wise to
choose both inductors to have a saturation current
of somewhat greater than 1.54A.
Looking at the Wurth Website:
http://www.we-online.com
or
using the
Wurth Electronics Component Simulation
Software
we can see that the 744778610 is 10uH with a
saturation current of 1.8A
744778610 Datasheet
__
Using a Transformer__
Since both inductor currents are in phase there is
no reason why a coupled inductor cannot be used. A
coupled inductor is simply 2 inductors wound on the
same ferrite. However, when 2 inductors are wound on
the same ferrite the mutual inductance between the
inductors effectively doubles their inductance.
Wurth have an application note (Transformer
windings in series and parallel) that explains this effect
and it is illustrated in FIG 4. If each inductor is
10uH and wound on the same ferrite, the net
inductance of each inductor is 20uH.
**
FIG 4**
Since we have calculated a desired (individual)
inductance of 10uH, if we use a coupled inductor we
need the inductance of each coil to be 5uH. A
coupled inductor often works out smaller in size
than 2 separate inductors. Also, note the winding
phase of the transformer as shown in FIG 7. A useful
way to remember the phasing of the SEPIC is to
consider removing the coupling cap - the SEPIC then
becomes a flyback converter and indeed the circuit
operation of a coupled inductor based SEPIC and a
flyback are very similar.
A suitable
coupled inductor from Wurth is the
74489430056. This is a 5.6uH, 2.2A coupled inductor
74489430056 Datasheet
On a note of simulation, LTspice needs a coupling
coefficient of <1 to model coupled inductors. If we were modelling
the ideal transformer (with a coupling coefficient
of 1) we would insert an LTspice statement to the
effect of:
K1 L1 L2 1
However, with a coupled inductor, we use a coupling
coefficient of 0.9
K1 L1 L2 0.9
This can be seen in FIG7 below.
As an
exercise, it is interesting to evaluate FIG 7 with
varying values of coupling coefficient. FIG 7 shows
the coupling coefficient as '0.9' so not all of the
energy from the primary is coupled to the secondary.
This means some of the primary current flows through
the coupling capacitor during its discharge phase.
This allows the current in the primary to smoothly
ramp downwards which means a transformer with an
imperfect coupling coefficient can give less
upstream interference than a circuit with a perfect
transformer. If the coupling coefficient was '1' the
current in the primary suddenly collapses and
transfers its energy to the secondary, but the fast
collapsing current can cause interference to
upstream circuitry. If the coupling coefficient is
'1', no current flows in the coupling capacitor and
the coupling capacitor can be removed. The circuit
is now identical to a flyback converter.
For a non
unity coupling coefficient, current from both
windings flows in the FET during the charge cycle.
This is easy to picture if we consider the discharge
cycle first. When the FET switches OFF, some of the
primary current flows through the coupling
capacitor. Since the average current in the
capacitor has to be zero, this implies that when the
FET switches ON, an equal and opposite current flows
in the capacitor. Since the output diode is reverse
biased, this current can only come from the
secondary. So an imperfect transformer means lower
upstream interference, but it reduces the efficiency
since we now have primary and secondary currents
flowing in the FET, thus increasing the FET losses.
The current flowing from the secondary also has to
flow through the ESR of the coupling capacitor, thus
further increasing the losses.
__
Rsense Calculation__
The minimum current sense threshold voltage for the
LT3757 is 100mV. Since the currents in both L1 and
L2 flow through the FET and sense resistor at the
same time, the peak current can reach (1.231A +
1.54A = 2.78A). We have chosen inductors each with a
saturation rating of 1.8A, so we can allow a peak
current flowing through the sense resistor of up to
3.6A before encroaching on the saturation rating of
the inductors. This implies a sense resistor value
of 27mOhms. Choosing Rsense to be 30mOhms should
allow enough margin for correct circuit operation
without saturating the inductors.
__
MOSFET Choice__
The MOSFET needs to be able to handle the peak
current from both inductors so in this design a
drain source current rating (Id) of 10A is more than
sufficient. The Drain–Source voltage rating (Vds)
needs to be in excess of the (Vin + Vout + Vdiode).
Since our design specification has a maximum input
voltage of 8V the drain voltage should never exceed
13V, so a MOSFET with a VDS rating in excess of 20V
is suitable.
The Gate-Source turn on voltage of the MOSFET (Vgs)
needs to be less than the input voltage, to ensure
that the Gate drive voltage can actually activate
the MOSFET. Logic level MOSFETs have a low turn on
voltage, are widely available and usually perfect
for low voltage dc/dc converters.
The above parameters represent the bare minimum
characteristics of the MOSFET. However, to get a
good design, we must ensure that the losses in the
MOSFET are as low as possible. The MOSFET switch
presents 2 losses in the circuit: switching losses
and conduction losses.
The switching losses result from current flowing
through the MOSFET at the same time that a voltage
is across the MOSFET (so power is generated in the
MOSFET), during the turn on and turn off times of
the MOSFET. For a given gate drive from the
controller, the lower the Gate-Source capacitance of
the MOSFET, the quicker the MOSFET will turn on.
Thus the Qg specification of the MOSFET is important
and should be as low as possible. The Qg of the
MOSFET also has an impact on the heat dissipation of
the chip, especially if the input voltage to the
chip is high.
Charge is dictated by the equation:
Charge (Q) = Current (I) x Time (s)
Since Frequency is the inverse of Time, we can write
So we can calculate the current needed to flow into
the chip, just to charge the gate capacitance of the
FET. Since heat is the product of voltage and
current, if the gate charge is high and/or the
switching frequency is high, the heat dissipation in
the chip will be high if the input voltage is high.
Once the MOSFET has switched on, the MOSFET presents
a small dc resistance between its Drain and Source
terminals. This is the MOSFETs ‘Drain Source on
resistance’ or Rdson. Again, this needs to be as low
as possible, especially since both inductor currents
flow at the same time though the MOSFET.
Now, MOSFET manufacturers reduce the ON resistance
of the MOSFET by constructing many parallel
conduction paths between the Drain and Source. Thus,
like connecting resistors in parallel, the ON
resistance comes down with more parallel paths.
However, in connecting Drain Source paths in
parallel, a negative effect is that the Gate Source
capacitance (Qg) is also connected in parallel, so a
low ON resistance (and hence low conduction loss)
sometimes implies a high gate source capacitance
(hence high switching loss). Thus the MOSFET that is
chosen should be a compromise between these two
characteristics. In addition, high current MOSFETs
tend to come in much larger packages, so meeting the
ideals of low ON resistance and low Qg might violate
a space requirement spec, so the selection process
has to start over. Engineering, as ever, is a
compromise.
Indeed looking at the selection tables of the MOSFET
manufacturers, it is better to select a MOSFET with
a low ON resistance (less than 10mOhms), then filter
this selection to remove MOSFETs with a Qg of
greater than 10nC, then select a MOSFET from this
list, as long as the Gate turn on voltage, Vds and
Id can be met. Starting by selecting MOSFETs with a
Vds of between 20V and 30V might rule out some
higher voltage FETs that are better suited to lower
voltage designs.
Failing that, download all the results to a
spreadsheet and sort from there. I have never had
much luck with the parametric searches on MOSFET
websites.
Alternatively, download all the MOSFET
characteristics into a spreadsheet, remove the ones
that don't meet the VDS and ID requirements, then
add a column called FOM (Figure of Merit). This
column should contain the value RDSON x QG. Then
sort by this column and pick the FET with the lowest
FOM. This part will be the best compromise between
RDSON and QG and ideal for the top MOSFET.
The standard LTspice circuit for the LT3757 uses a
Fairchild FDS6680A (QG = 27nC, RDSON = 15mOhms).
Changing this to an Infineon BSC018NE12LS (QG =
19nC, RDSON = 1.8mOhms) gives a 5% efficiency
improvement in LTspice simulations.
BSC018NE12LS
Datasheet
FDS6680A Datasheet
__
Rectifier Diode Choice__
When the MOSFET switches off, the inductor voltage
ramps rapidly in order to maintain current flow.
Many diodes are not fast enough to react to this
voltage change, resulting in a large spike on the
Drain of the MOSFET. This can (and does) destroy the
MOSFET.
Therefore Schottky diodes should be used in all
dc/dc converter designs where the inductor voltage
has to be rectified. Ultra fast diode have a
response time of 10’s of nanoseconds, standard
rectifier diodes have a response time of several
microseconds, whereas a Schottky has a response time
in the order of a few nanoseconds. Schottky diodes
also have a much lower forward voltage drop (0.3V)
compared with standard rectifiers (0.6V) so half the
power is wasted as a result of VxI losses.
When choosing a Schottky diode, the key parameters
are: forward voltage drop (should be as low as
possible), forward current (this should be greater
than the sum of the peak currents in L1 and L2) and
reverse voltage rating. When the FET is charging the
inductor, the anode of the Schottky diode will be at
-Vin and the cathode will be at Vout, so the reverse
voltage rating of the Schottky should be greater
than (Vin + Vout).
In this design example, the MBRS340 is a good choice
with a reverse voltage rating of 40V and a forward
voltage of 0.53V at 3A peak current.
MBRS340 Datasheet
__
Output Capacitor Choice__
Unlike the buck converter that has a continuous
current flowing from the inductor into the output
capacitor, the buck-boost converter output capacitor
has to keep the output voltage alive when the
inductors are being charged (and is hence
disconnected from the output). Therefore there will
be a component of the output ripple due to the
discharge of the output capacitor. In addition, when
the inductors are discharging, the output capacitor
will experience an inrush of current and any ESR
(effective series resistance) in the capacitor will
also result in ripple.
Therefore the output ripple is made up of 2
components: the ripple caused by the output
capacitor discharging when the inductors are being
charged and the ripple caused by the inrush current
from the inductors into the ESR of the output
capacitor.
The ripple caused by the discharge of the output
capacitor while the inductor is charging is dictated
by
where *i* is the load current in Amps, *C*
is the output capacitance in Farads and *dv/dt*
is the change in output voltage during the ON time
of the MOSFET.
Earlier we calculated that the MOSFET switches on
for a period of 1.06us. If we require a discharge
ripple of 0.5% (25mV) with a load current of 1A,
this implies we need a capacitance of
or 42uF.
Note that when the inductors are charging, there is
zero current flowing in the rectifier diode. When
the MOSFET switches off, the diode current jumps
from 0A to the sum of the peak inductor currents, so
it is the peak inductor current, not the ripple
current amplitude that determines this component of
the output ripple (compare this with the ripple in a
buck converter that is determined by the ripple
current amplitude, not the peak inductor current).
The ripple caused by the ESR is a product of the
peak inductor current and the ESR. In our example
the peak current is 2.78A and the ESR is of a
typical ceramic capacitor is 10m Ohms, giving a
ripple of 28mV. Two capacitors in parallel yields an
effective ESR of 5m Ohms and halves the ESR ripple
to 14mV.
Therefore two 47uF ceramic capacitors should ensure
our output voltage ripple is within spec.
FIG 5 shows the output voltage ripple and diode
current.
**
FIG 5**
The peak diode current is 2.8A and it can be seen
that this generates a sharp rising edge on Vout of
14mV as it flows into the effective 5mOhm ESR. The
output capacitor voltage continues to increase, but
at a reducing rate as the diode current ramps down.
__
Other points to note__
The feedback resistor values can be calculated using
the Feedback Resistor Calculator:
Feedback Resistor Calculator
The feedback resistors were picked as 20k and 42.2k
so that a 5V output keeps the feedback point at
1.6V. Some engineers make these resistor values way
too big in the hope of conserving wasted current in
the feedback loop. However, this can have a negative
effect in that too high resistor values (over 500k
Ohms) can cause a phase shift created by the
internal capacitance of the feedback pin and the
large external resistor values which will lead to
poor stability. In lower power designs (where
feedback current is important), bypassing the top
feedback resistor with 100pF overcomes this issue by
providing a phase lead that counteracts the phase
lag created by the input capacitance.
The final LTspice circuit using individual inductors
is shown in FIG 6. It can be downloaded here:
SEPIC
Converter using Single Inductors
**
FIG 6**
The final LTspice circuit using coupled inductors is
shown in FIG 7. It can be downloaded here:
SEPIC
Converter using Coupled Inductors
**
FIG 7**
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