How to use LTspice to select peripheral components? This article tells you ~

What are the input voltage and output voltage? This is the first question to ask when selecting a buck or boost DC-DC converter. The second question is, what is the output current required to meet the expected load? Although the input and output questions are the same for buck and boost, the process of selecting the right IC to meet the input and output requirements is very different for both.

The first clue that the boost selection process is different from the buck selection process is apparent if you compare a buck IC product selection table to a boost IC product selection table. Figure 1 shows a selection table for some internal power switch buck products. As can be seen, output current is one of the main selection parameters.

Let’s compare Figure 1 (Internal Power Switch Buck Product Selection Table) to Figure 2 (Internal Power Switch Boost Product Selection Table). In the boost selection table, output current is not even shown as a selection parameter, but is replaced by switch current.

Another reminder that boosts follow different rules is that there is a subtle but important current statement in the boost data sheet title. For example, Figure 3 shows the front page of the LTC3621 monolithic buck regulator data sheet, which clearly states a 17 V maximum V _IN and 1 A continuous load capability.

In contrast, the LT8330 monolithic boost regulator data sheet header specifies the maximum switch (internal power MOSFET) voltage (60 V) and current (1 A), rather than the typical maximum values ​​for load current and input voltage.

Why is there such a difference? In a buck regulator, the average inductor current is approximately equal to the output (load) current, but in a boost topology, this is not the case. Let’s compare the boost and buck topologies to understand why.

Figure 5 shows a simplified schematic of the asynchronous boost topology, and Figure 6 shows a simplified schematic of the asynchronous buck topology. The D block in both is a PWM signal that drives the power MOSFET, and the duty cycle of the switching cycle is determined by the input and output voltage ratio. In this article, for simplicity, I use the lossless continuous conduction mode (CCM) equation because the result is close enough.

Using LTspice, we can clearly see the difference between the input and output currents of these two different topologies. Figure 7 shows the basic open-loop design of a buck regulator to convert a 12 V input voltage to a 3.3 V output voltage, delivering 1 A (3.3 W) to a resistive load R1. The PWM D block is implemented with the V2 floating supply, as we need V _GATE > V _SOURCE to establish conduction for the N-channel MOSFET M1. V2 is used as the PULSE voltage source to achieve a 0 V to 5 V pulse, which starts at time 0 of the simulation, transitions from 0 V to 5 V in 5 ns, and back in 5 ns, with a T _ON of 550 ns and a T _P (full switching cycle) equal to 2 µs.

After running a simulation of the circuit in Figure 7, you can probe the currents in L1 and R1. The current in L1 follows a triangular pattern when charging and discharging because M1 switches according to the timing of T _ON (the time M1 is on) and T _OFF (the time M1 is off).

The L1 current switches at a 500 kHz switching frequency. As can be seen, the inductor current is an AC + DC waveform. It switches from a minimum value of 0.866 A (at the end of T _{OFF ) to a maximum value of 1.144 A (at the end of T ON} ). As the AC signal seeks the path of least impedance, the AC portion of the current flows through the ESR of the output capacitor C2. This AC current and the charging and discharging of C2 result in the output voltage ripple, while the DC current flows through R2.

By comparing the triangles formed by the inductor current above and below the load current, we can see that they are equal. Simple algebra shows that:

The average inductor current is equal to the load current.

When searching for a buck regulator IC, one can assume that the data sheet shows the maximum allowed output current because I _IN ≈ I _OUT , but this is not the case with the boost topology.

Let’s look at Figure 9, which shows an open-loop boost design from 3.3 V to 12 V output at 0.275 A, or about 3.3 W. What is the average inductor current at this point?

In Figure 10, the output current is 291 mA, the DC locus of I(R2) – close to the calculated value. Although the simulated load current is 291 mA, the simulation shows that the average inductor current is 945 mA, with a peak value of over 1 A. This is more than 3.6 times the output current. During T _ON (the time when M2 is on and V3 is on L2), the inductor charges from its minimum value to its maximum value. During T _ON , D2 is off and the load current is provided by the output capacitor.

Figure 10. LTspice simulation results for an open-loop boost from 3.3 V to 12 V at 0.275 A.

During T _ON , the inductor is in series with the MOSFET, so any current flowing through the input inductor will flow through the switch. Because of this, the data sheet specifies a maximum current, I _{SW ,} that can flow through the switch . When selecting a boost IC for a new design, you should understand the maximum expected current through the switch.

For example, selecting a boost regulator for the following application:

• V _IN = 12 V

• V _OUT = 48 V

• I _OUT = 0.15 A

To select the correct boost regulator, you need to find the average input current, which is the current flowing through the inductor and MOSFET during T _ON . To find this current, work backwards from the output to the input based on the output power and efficiency:

• P _OUT = V _OUT × I _OUT = 48 V × 0.15 A = 7.2 W

• Assume an efficiency of 0.85 (use the value from the data sheet if there is an efficiency curve with input and output parameters similar to the desired design).

• P _IN = P _OUT / Efficiency = 7.2 W / 0.85 = 8.47 W

• I _IN _AV = Average input current. This is the average current flowing in the inductor and switch during the on-time and is calculated as P _IN /V _{IN = 8.47 W/12 V = 0.7 A.}

• Again, I _IN is the average inductor current, and the maximum peak current will be 1.15 to 1.20 higher than I _IN , providing a 30% to 40% ripple current. Therefore, I _PEAK = I _IN × 1.2 = 0.7 A × 1.2 = 0.847 A.

The data sheet usually specifies the V _IN range of an IC—both the recommended range and the absolute maximum. In the data sheet, the highest output voltage that a boost regulator with an internal power switch can produce is stated as its maximum V _{SW rating. If you use a boost controller with an external MOSFET as the power switch, the V DS} rating specified in the MOSFET data sheet is the value that limits the maximum output voltage.

For example, the LT8330 boost regulator has an input voltage range of 3 V to 40 V, an absolute maximum switch voltage of 60 V, and a fixed switching frequency of 2 MHz. Although the 60 V absolute maximum switch voltage rating enables the part to produce a 60 V boosted output, best practice is to stay at least 2 V below this value.

The output voltage is also limited by the duty cycle. The maximum and minimum duty cycles can probably be found in the data sheet or calculated. By using the LT8330 to convert from 12 V to 48 V, CCM ignores the diode drop to achieve a high conversion ratio, and the duty cycle can be calculated from the input and output voltages:

• D = (V _O – V _IN )/V _O = (48 V – 12 V)/48 V = 0.75 or 75%

• Check whether the IC can operate at the required duty cycle.

• The IC minimum duty cycle calculation formula is as follows:

• D _MIN = 最小 T _ON(MAX) × f _SW(MAX)

• The IC maximum duty cycle calculation formula is as follows:

• D _MAX = 1 – (最小 T _OFF(MAX) × f _SW(MAX) )

The minimum T _ON and minimum T _OFF can be found in the Electrical Characteristics table of the data sheet. Use the maximum value from the "Minimum", "Type", and "Maximum" columns of the table. Using the published values ​​for the LT8330 and the equations for D _MIN and D _MAX , we get D _MIN = 0.225 and D _MAX = 0.86. From the results, we can see that the LT8330 should be able to convert from 12 V to 48 V, since the design calls for a duty cycle of 0.75.

The schematic shown in Figure 12 implements the design concepts introduced previously, using the LT8330 in a 12 V input to 48 V output converter supporting a 150 mA load.

From the LTspice simulation, we can plot and measure various parameters that can help you choose the parameters of the IC, as shown in Figure 13.

After running the simulation, you can view the SW node behavior as a waveform to understand what voltage is present on the power switch during switching. To do this, hover your mouse over the SW node so that the crosshairs change to a red voltage probe. Click to plot the switch node behavior on the waveform viewer. The resulting graph corresponds to the drain of the internal power MOSFET.

As expected, when the MOSFET is on, the voltage potential is close to ground, but more importantly, during TOFF _, the MOSFET is off and the drain voltage is affected by the output voltage and the diode drop. Now we know what the stress on the MOSFET's VDS _is . If we had chosen a controller design that used an external MOSFET as the power switch, the selected MOSFET would have a VDS rating of 60 _V.

In the LTspice waveform viewer, cursors can be used to make horizontal and vertical measurements, similar to the cursors on an oscilloscope. To invoke the cursors, click on the V(sw) label in the LTspice waveform viewer. This will attach the first cursor to the trace, and a second cursor can be attached to the same trace by clicking again. Alternatively, right-click on this label and select the desired cursor for a given probed trace. Use these cursors to measure T _ON and calculate the duty cycle by dividing T _ON by the period.

T _PERIOD = T _ON + T _OFF = 1/f _SW . Earlier, we calculated this value to be 75% or 0.75. Using LTspice, this yields a value of approximately 373 ns. The LT8330 uses a fixed switching frequency of 2 MHz, so T _P = 1/2e6 = 500 ns, giving a duty cycle of 373 ns/500 ns = 0.746.

Peak current and voltage on the inductor

To select an inductor for a boost application, you need to know if the inductor can handle the current and voltage you are dealing with—that is, the peak inductor current and the T _ON and T _OFF voltages. This can also be estimated in LTspice using a differential probe. To differentially probe the inductor, hover your mouse over the IN node and the crosshairs will turn into a red probe. Click and drag your mouse to the SW node. The cursor color will turn black. Release the mouse when it is on the second node.

In Figure 15, the voltage between nodes IN and SW is probed differentially across the inductor. During T _ON , the MOSFET is on and the right side of the inductor is at ground while the left side is at V _IN , causing the voltage across the inductor to be 12 V during T _{ON . During T OFF} , the MOSFET is off and the right side of the inductor is placed at 48 V while the left side is at V _IN during T _ON . Since the differential probe subtracts V _{SW from V IN} , –36 V is obtained, but the sign is not important now. What is important is that the inductor varies between 12 V and 36 V.

During T _ON , the voltage across the inductor draws a positive di/dt, which is the slope of the blue I(L1) plot. The maximum point of this trace is I _PEAK , which is calculated to be 0.847 A. Using LTspice, we can see that the peak current is approximately 866 mA.

Understanding this peak current is important to properly select an inductor with adequate current rating (IR) and saturation current (I _{SAT ). IR is more about how much heat is generated at the specified current, while I SAT} applies to the event that short-circuit protection is invoked. If a regulator with an internal MOSFET is used, (I _SAT > regulator current limit), and a controller is used with an external MOSFET, when the current limit is triggered, (I _SAT > peak inductance value).

It is important to note that the boost topology described here has no current limiting in the inductor or diode. If the switch is not used, or the IC is disconnected, there is a direct path between the input and output. Some ICs offer additional protection features such as output disconnect during shutdown, inrush current limiting, and other features to address this direct input to output connection issue—for example, the LTC3122 and LTC3539.

To improve efficiency, an inductor with low DCR (DC resistance) and low core losses should be used. The DCR at a specific temperature is specified in the inductor data sheet—it rises with temperature and has a tolerance. The DC losses can be easily calculated by P _{INDUCTOR_LOSS} = I _{IN_AV} ² × DCR, while the AC losses and core losses can be found in the manufacturer's simulation or other documentation. LTspice can integrate the power to calculate the associated power dissipation. Providing LTspice with the recorded DCR and other known parasitic parameters of the inductor improves the accuracy of the LTspice simulation.

Figure 17 shows the simulated differential voltage across the diode V _{SW,OUT , the diode forward current I(D1) and the inductor current I(L1) . When the switch is on (during T ON} ), the anode is close to ground and the cathode is at the output voltage, so the diode will be reverse biased and exposed to its maximum voltage, which is V _OUT . The first criterion is to select a diode with a V _RRM (maximum peak repetitive reverse voltage) higher than V _OUT .

The inductor's peak current flows through the diode at the beginning of the T _OFF period after the MOSFET turns off , so the diode peak current is the same as the inductor peak current. Diode data sheets include a parameter called I _FRM (repetitive peak forward current), which is specified in terms of duration and duty cycle. This parameter is usually higher than the average current that the diode can provide.

Once the simulation is complete, LTspice can integrate all the waveforms in the waveform viewer to get the rms and average values ​​and use the same calculations to calculate the average current that the diode will handle. First, zoom in on the portion of the waveform you want to integrate—zooming effectively sets the boundaries of integration. In this case, you zoom to cover a large number of steady-state cycles (not startup or shutdown). To set the boundaries of integration, drag-select a steady-state time period and hover the mouse over the graph name. For example, the integrated results shown in Figure 18 cover 0.75 ms, or over 1000 cycles. The cursor changes to a hand icon. Press the CTRL key and click to invoke the integration window of the waveform viewer.

The integration dialog box shown in Figure 18 shows that the average current through the diode is 150 mA. This value should be less than the maximum average forward current IF(AV), which is specified in the diode data sheet at a specific temperature.

The power dissipation in a diode can also be calculated through simulation. The total power dissipation, PTOT (total power), and the thermal resistance from junction to ambient, RTH, at 25°C are specified in the diode data sheet _. In _LTspice , hover the cursor over a diode to display the power dissipation in the waveform viewer. When hovering over a discrete component or voltage source, the cursor changes to a current probe. Pressing the ALT key changes the cursor to a thermometer, and clicking displays the simulated power dissipation in the diode. Zooming in on steady-state operation, integrate the waveform using the same procedure described previously for integrating the diode current. The diode power capability includes both the voltage across the diode and the current flowing through it.

Some of the diode's capacitance is charged during its conduction period. When the diode is no longer conducting, the accumulated charge must be discharged. This damping of charge movement results in power losses, so choosing a low capacitance value is recommended. This capacitance value varies with the reverse voltage of the diode, and the diode data sheet should include a graph showing this effect. This internal capacitance is usually shown as C _{d in the diode data sheet and as C jo} in the LTspice database .

Using a low capacitance diode relaxes the requirement for maximum reverse recovery current, thus improving efficiency. Figure 20 shows what is happening with the recovery current. The power dissipation inherent in reverse recovery is left as an exercise for the reader.

When selecting a boost IC, start with the output. Work backward from the desired output voltage and load current to find the input power, taking efficiency into account. From there, determine the average and peak input current values. In a boost converter, the average current flowing in the inductor is higher than the load current, making the IC selection process different from that of a buck converter. Selecting the properly rated components for a boost converter requires knowledge of the regulator peak and average voltages and currents, which can be determined using LTspice.

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