Boost Switching Regulator Topology.

A boost converter or step-up converter is a DC-to-DC power converter that steps up input voltage
to a higher output voltage.

DC is duty cycle, the ratio of S1 "on" time to "off" time, assuming that S1 and S2 open and close alternately. Duty cycle can take on values only between 0 and 1; therefore, the output voltage of a boost regulator is always higher than the input voltage

Average diode current is equal to I_{OUT} and average switch
current is **I _{OUT}(V_{OUT} – V_{IN})/ V_{IN}**,
both of which are significantly
less than peak current.The switch, diode and output
capacitor must be specified to handle the peak currents as
well as average currents. Discontinuous mode requires
even higher ratios of switch current to output current.

One drawback of boost regulators is that they cannot be current limited for output shorts because the current steering diode, D1, makes a direct connection between input and output.

With higher input voltages, output power levels can exceed 100W. Power loss internal to the LT1070 in a boost regulator is approximately equal to:

The first term of this equation is the power loss due to the "on" resistance of the switch (R). The second term is the loss from the switch driver.

The only other significant power loss in a boost regulator is in the diode, D1, as given by:

Total power loss in the regulator is the sum of P_{IC} + P_{D}, and
this can be used to calculate efficiency (E):

With higher input voltages, efficiencies can exceed 90%.

Maximum output voltage in the boost mode is limited by
the breakdown of the switch to 65V (standard part) or 75V
(HV part). It may also be limited by maximum duty cycle
if input voltage is low. The 90% maximum duty cycle of the
LT1070 limits output voltage to ten times the input voltage. For the simple boost mode, higher ratios of output to
input voltage require a tapped inductor.

Maximum output voltage in the boost mode is limited by
the breakdown of the switch to 65V (standard part) or 75V
(HV part). It may also be limited by maximum duty cycle
if input voltage is low. The 90% maximum duty cycle of the
LT1070 limits output voltage to ten times the input voltage. For the simple boost mode, higher ratios of output to
input voltage require a tapped inductor.

Design procedure for a boost regulator is straightforward.
R1 and R2 set the regulated output voltage. The feedback
pin voltage is internally trimmed to 1.244V, so output
voltage is equal to 1.244 (R1 + R2)/R2. R2 is normally set
to 1.24k and R1 is found from:

The 1.24k value for R2 is chosen to set divider current at 1mA, but this value can vary from 300Ω to 10k with negligible effect on regulator performance. For proper load regulation, R2 must be returned directly to the ground pin of the LT1070, while R1 is connected directly to the load.

Next, L1 is selected. The trade-offs are size, maximum
output power, transient response, input filtering, and in
some cases, loop stability. Higher inductor values provide
maximum output power and low input ripple current, but
are physically larger and degrade transient response. Low
inductor values have high magnetizing current which
reduces maximum output power and increases input
current ripple. Low inductance can also cause a
subharmonic oscillation problem if duty cycle is above
50%.

With the aforementioned considerations in mind, a simple
formula can be derived to calculate L1 based on the
maximum ripple current (∆I) to be allowed in L1.

Let ∆I = 0.5A and frequency is 40kHz

A second formula will allow a calculation of maximum power output with this size inductor:

Note that the second term in the first set of brackets is the
only one which contains "L" and that this term drops out
of the equation for large values of L. In this example, that
term is equal to ?? A, showing that maximum effective
switch current, and therefore maximum output power is
reduced by one-half the inductor ripple current in a boost
regulator. In this example, peak effective switch current is
reduced from 5A to ?? A with 0.5A ripple current, a 5%
loss. An additional 12% reduction of maximum available
power is caused by switch "on" resistance. At higher input
voltages, this switch loss is significantly reduced.

When continuous inductor current is desired, the value of
L1 cannot be decreased below a certain limit if duty cycle
of the switch exceeds 50%. Duty cycle can be calculated
from:

The reason for a lower limit on the value of L for duty cycles greater than 50% is a subharmonic oscillation which can occur in current mode switching regulators. For further details of this phenomenon, see Subharmonic Oscillation section of this application section. The minimum value of L1 to ensure no subharmonic oscillations in a boost regulator is:

Note that for VOUT ≤ 2VIN, there is no restriction on inductor size. The minimum value of ??µH obtained in this example is below the value which would yield continuous inductor current, so it is an artificial restriction. Subharmonic oscillations do not occur if inductor current is discontinuous. The critical inductor size for continuous inductor current is:

Discontinuous mode operation is sometimes chosen because it results in the smallest physical size for the inductor. The maximum power output is considerably reduced, however, and can never exceed 2.5(VIN) watts with the LT1070. The minimum inductor size required to provide a given output power in the discontinuous mode is given by:

This formula does not take into account efficiency losses, so the minimum value of L should probably be increased by at least 50% for worst-case conditions. Efficiency is degraded when using minimum inductor sizes because of higher switch and diode peak currents.

In summation, to choose a value for L1:

- Decide on continuous or discontinuous mode
- If continuous mode, calculate C1 based on ripple current and check maximum power and subharmonic limits.
- If discontinuous mode, calculate L1 based on power
output requirements and check to see that output power
does not exceed limit for discontinuous mode
(P
_{MAX}= 2.5V_{IN})

L1 must not saturate at the peak operating current. This value of current can be calculated from:

A core must be selected for L1 which does not saturate with ?? peak inductor current.

The main criteria for selecting C2 is low ESR (effective series resistance), to minimize output voltage ripple. A reasonable design procedure is to let the reactance of the output capacitor contribute no more than 1/3 of the total peak-to-peak output voltage ripple (VP-P), yielding:

This leaves 67% of the ripple attributable to ESR, giving:

After C2 has been selected, output voltage ripple may be calculated from:

If lower output ripple is required, a larger output capacitor
must be used with lower ESR. It is often necessary to use
capacitor values much higher than calculated to obtain the
required ESR. In the example shown, capacitors with
guaranteed ESR less than 0.04Ω with a working voltage of
15V generally fall in the 1000μF to 2000μF range. Higher
voltage units have lower capacitance for the same ESR.

A second option to reduce output ripple is to add a small
LC output filter. If the LC product of the filter is much
smaller than L1 * C2, it will not affect loop phase margin.
Dramatic reduction in output ripple can be achieved with
this filter, often at lower cost and less board space than
simply increasing C2. See section on Output Filters for
details.

Loop frequency compensation is performed by R3 and C1.

D1 should be a fast turn-off diode. Schottky diodes are best in this regard and offer better efficiency in the forward mode. With higher output voltages, the efficiency aspect is minimal and silicon fast turn-off diodes are a more economical choice. Turn-on time is important also with output voltages above 40V. Diodes with slow turn-on time will have a very high forward voltage for a short time after forward current starts to flow. This transient forward voltage can be anywhere from volts to tens of volts. It must be summed with output voltage to calculate worst-case switch voltage. To minimize switch transient voltage, the wiring of C2 and D1 should be short and close to the LT1070 as shown below.

Boost regulators are not short-circuit protected because the current steering diode (D1) connects the input to the output. The LT1070 will not be harmed for overloads up to 5A. Beyond that point, D1 can be permanently "on" and the LT1070 switch will be effectively shorted to the output. A fuse in series with the input voltage is the only simple means of protecting the circuit. Fuse sizing can be calculated from:

For I_{IN} = 2.4A, a 4A fast-blow fuse would be a reasonable choice in this
design.