Table 1. Design Parameters

8.2.2.1 Selecting the Switching Frequency

The first step is to decide on a switching frequency for the regulator. Typically, the user will want to choose the highest switching frequency possible since this will produce the smallest solution size. The high switching frequency allows for lower valued inductors and smaller output capacitors compared to a power supply that switches at a lower frequency. The switching frequency that can be selected is limited by the minimum on-time of the internal power switch, the input voltage and the output voltage and the frequency shift limitation.

Equation 8 and Equation 9 must be used to find the maximum switching frequency for the regulator, choose the lower value of the two equations. Switching frequencies higher than these values will result in pulse skipping or the lack of overcurrent protection during a short circuit.

The typical minimum on time, t_onmin, is 130 ns for the TPS54140A. For this example, the output voltage is 3.3 V and the maximum input voltage is 18 V, which allows for a maximum switch frequency up to 1600 kHz when including the inductor resistance, on resistance and diode voltage in Equation 8. To ensure overcurrent runaway is not a concern during short circuits in your design use Equation 9 or the solid curve in Figure 34 to determine the maximum switching frequency. With an maximum input voltage of 20 V, assuming a diode voltage of 0.5V, inductor resistance of 100 mΩ, switch resistance of 200 mΩ, an output current of 2.8 A, the maximum switching frequency is approximately 1600kHz.

Choosing the lower of the two values and adding some margin a switching frequency of 1200kHz is used. To determine the timing resistance for a given switching frequency, use Equation 7 or the curve in Figure 32.

The switching frequency is set by resistor R_t shown in Figure 39.

8.2.2.2 Output Inductor Selection (L_O)

To calculate the minimum value of the output inductor, use Equation 10.

K_IND is a coefficient that represents the amount of inductor ripple current relative to the maximum output current.

The inductor ripple current will be filtered by the output capacitor. Therefore, choosing high inductor ripple currents will impact the selection of the output capacitor since the output capacitor must have a ripple current rating equal to or greater than the inductor ripple current. In general, the inductor ripple value is at the discretion of the designer; however, the following guidelines may be used.

For designs using low ESR output capacitors such as ceramics, a value as high as K_IND = 0.3 may be used. When using higher ESR output capacitors, K_IND = 0.2 yields better results. Since the inductor ripple current is part of the PWM control system, the inductor ripple current should always be greater than 100 mA for dependable operation. In a wide input voltage regulator, it is best to choose an inductor ripple current on the larger side. This allows the inductor to still have a measurable ripple current with the input voltage at its minimum.

For this design example, use K_IND = 0.2 and the minimum inductor value is calculated to be 7.6 μH. For this design, a nearest standard value was chosen: 10 μH. For the output filter inductor, it is important that the RMS current and saturation current ratings not be exceeded. The RMS and peak inductor current can be found from Equation 12 and Equation 13.

For this design, the RMS inductor current is 1.506 A and the peak inductor current is 1.62 A. The chosen inductor is a MSS6132-103. It has a saturation current rating of 1.64 A and an RMS current rating of 1.9 A.

As the equation set demonstrates, lower ripple currents will reduce the output voltage ripple of the regulator but will require a larger value of inductance. Selecting higher ripple currents will increase the output voltage ripple of the regulator but allow for a lower inductance value.

The current flowing through the inductor is the inductor ripple current plus the output current. During power up, faults or transient load conditions, the inductor current can increase above the calculated peak inductor current level calculated above. In transient conditions, the inductor current can increase up to the switch current limit of the device. For this reason, the most conservative approach is to specify an inductor with a saturation current rating equal to or greater than the switch current limit rather than the peak inductor current.

Equation 10.

Equation 11.

Equation 12.

Equation 13.

8.2.2.3 Output Capacitor

There are three primary considerations for selecting the value of the output capacitor. The output capacitor will determine the modulator pole, the output voltage ripple, and how the regulators responds to a large change in load current. The output capacitance needs to be selected based on the more stringent of these three criteria.

The desired response to a large change in the load current is the first criteria. The output capacitor needs to supply the load with current when the regulator can not. This situation would occur if there are desired hold-up times for the regulator where the output capacitor must hold the output voltage above a certain level for a specified amount of time after the input power is removed. The regulator also will temporarily not be able to supply sufficient output current if there is a large, fast increase in the current needs of the load such as transitioning from no load to a full load. The regulator usually needs two or more clock cycles for the control loop to see the change in load current and output voltage and adjust the duty cycle to react to the change. The output capacitor must be sized to supply the extra current to the load until the control loop responds to the load change. The output capacitance must be large enough to supply the difference in current for 2 clock cycles while only allowing a tolerable amount of droop in the output voltage. Equation 14 shows the minimum output capacitance necessary to accomplish this.

Where ΔIout is the change in output current, ƒsw is the regulators switching frequency and ΔVout is the allowable change in the output voltage. For this example, the transient load response is specified as a 4% change in Vout for a load step from 0A (no load) to 1.5 A (full load). For this example, ΔIout = 1.5-0 = 1.5 A and ΔVout = 0.04 × 3.3 = 0.132 V. Using these numbers gives a minimum capacitance of 18.9 μF. This value does not take the ESR of the output capacitor into account in the output voltage change. For ceramic capacitors, the ESR is usually small enough to ignore in this calculation. Aluminum electrolytic and tantalum capacitors have higher ESR that should be taken into account.

The catch diode of the regulator can not sink current so any stored energy in the inductor will produce an output voltage overshoot when the load current rapidly decreases, see Figure 40. The output capacitor must also be sized to absorb energy stored in the inductor when transitioning from a high load current to a lower load current. The excess energy that gets stored in the output capacitor will increase the voltage on the capacitor. The capacitor must be sized to maintain the desired output voltage during these transient periods. Equation 15 is used to calculate the minimum capacitance to keep the output voltage overshoot to a desired value. Where L is the value of the inductor, I_OH is the output current under heavy load, I_OL is the output under light load, Vƒ is the final peak output voltage, and Vi is the initial capacitor voltage. For this example, the worst case load step will be from 1.5 A to 0 A. The output voltage will increase during this load transition and the stated maximum in our specification is 4% of the output voltage. This will make Vƒ = 1.04 × 3.3 = 3.432. Vi is the initial capacitor voltage which is the nominal output voltage of 3.3 V. Using these numbers in Equation 15 yields a minimum capacitance of 25.3 μF.

Equation 16 calculates the minimum output capacitance needed to meet the output voltage ripple specification. Where fsw is the switching frequency, V_oripple is the maximum allowable output voltage ripple, and I_ripple is the inductor ripple current. Equation 17 yields 0.7 μF.

Equation 17 calculates the maximum ESR an output capacitor can have to meet the output voltage ripple specification. Equation 17 indicates the ESR should be less than 144mΩ.

The most stringent criteria for the output capacitor is 25.3 μF of capacitance to keep the output voltage in regulation during an unload transient.

Additional capacitance de-ratings for aging, temperature and dc bias should be factored in which will increase this minimum value. For this example, a 47 μF 6.3V X7R ceramic capacitor with 5 mΩ of ESR will be used.

Capacitors generally have limits to the amount of ripple current they can handle without failing or producing excess heat. An output capacitor that can support the inductor ripple current must be specified. Some capacitor data sheets specify the Root Mean Square (RMS) value of the maximum ripple current. Equation 18 can be used to calculate the RMS ripple current the output capacitor needs to support. For this application, Equation 18 yields 66mA.

Equation 14.

Equation 15.

Equation 16.

Equation 17.

Equation 18.

8.2.2.4 Catch Diode

The TPS54140A requires an external catch diode between the PH pin and GND. The selected diode must have a reverse voltage rating equal to or greater than Vinmax. The peak current rating of the diode must be greater than the maximum inductor current. The diode should also have a low forward voltage. Schottky diodes are typically a good choice for the catch diode due to their low forward voltage. The lower the forward voltage of the diode, the higher the efficiency of the regulator will be.

Typically, the higher the voltage and current ratings the diode has, the higher the forward voltage will be. Since the design example has an input voltage up to 18 V, a diode with a minimum of 20 V reverse voltage will be selected.

For the example design, the B220A Schottky diode is selected for its lower forward voltage and it comes in a larger package size which has good thermal characteristics over small devices. The typical forward voltage of the B220A is 0.50 volts.

The diode must also be selected with an appropriate power rating. The diode conducts the output current during the off-time of the internal power switch. The off-time of the internal switch is a function of the maximum input voltage, the output voltage, and the switching frequency. The output current during the off-time is multiplied by the forward voltage of the diode which equals the conduction losses of the diode. At higher switch frequencies, the ac losses of the diode need to be taken into account. The ac losses of the diode are due to the charging and discharging of the junction capacitance and reverse recovery. Equation 19 is used to calculate the total power dissipation, conduction losses plus ac losses, of the diode.

The B220A has a junction capacitance of 120 pF. Using Equation 19, the selected diode will dissipate 0.632 Watts. This power dissipation, depending on mounting techniques, should produce a 16°C temperature rise in the diode when the input voltage is 18V and the load current is 1.5 A.

If the power supply spends a significant amount of time at light load currents or in sleep mode consider using a diode which has a low leakage current and slightly higher forward voltage drop.

Equation 19.

8.2.2.5 Input Capacitor

The TPS54140A requires a high quality ceramic, type X5R or X7R, input decoupling capacitor of at least 3 μF of effective capacitance and in some applications a bulk capacitance. The effective capacitance includes any dc bias effects. The voltage rating of the input capacitor must be greater than the maximum input voltage. The capacitor must also have a ripple current rating greater than the maximum input current ripple of the TPS54140A. The input ripple current can be calculated using Equation 20.

The value of a ceramic capacitor varies significantly over temperature and the amount of dc bias applied to the capacitor. The capacitance variations due to temperature can be minimized by selecting a dielectric material that is stable over temperature. X5R and X7R ceramic dielectrics are usually selected for power regulator capacitors because they have a high capacitance to volume ratio and are fairly stable over temperature. The output capacitor must also be selected with the dc bias taken into account. The capacitance value of a capacitor decreases as the dc bias across a capacitor increases.

For this example design, a ceramic capacitor with at least a 20 V voltage rating is required to support the maximum input voltage. Common standard ceramic capacitor voltage ratings include 4 V, 6.3 V, 10 V, 16V, 25 V, 50 V or 100V so a 25 V capacitor should be selected. For this example, two 2.2 μF, 25 V capacitors in parallel have been selected. Table 2 shows a selection of high voltage capacitors. The input capacitance value determines the input ripple voltage of the regulator. The input voltage ripple can be calculated using Equation 21. Using the design example values, Ioutmax = 1.5 A, Cin = 4.4 μF, ƒsw = 1200 kHz, yields an input voltage ripple of 71 mV and a rms input ripple current of 0.701 A.

Equation 20.

Equation 21.

8.2.2.6 Slow Start Capacitor

The slow start capacitor determines the minimum amount of time it will take for the output voltage to reach its nominal programmed value during power up. This is useful if a load requires a controlled voltage slew rate. This is also used if the output capacitance is very large and would require large amounts of current to charge the capacitor to the output voltage level. The large currents necessary to charge the capacitor may make the TPS54140A reach the current limit or excessive current draw from the input power supply may cause the input voltage rail to sag. Limiting the output voltage slew rate solves both of these problems.

The slow start time must be long enough to allow the regulator to charge the output capacitor up to the output voltage without drawing excessive current. Equation 22 can be used to find the minimum slow start time, tss, necessary to charge the output capacitor, Cout, from 10% to 90% of the output voltage, Vout, with an average slow start current of Issavg. In the example, to charge the 47 μF output capacitor up to 3.3 V while only allowing the average input current to be 0.125 A would require a 1 ms slow start time.

Once the slow start time is known, the slow start capacitor value can be calculated using Equation 6. For the example circuit, the slow start time is not too critical since the output capacitor value is 47 μF which does not require much current to charge to 3.3 V. The example circuit has the slow start time set to an arbitrary value of 1ms which requires a 3.3 nF capacitor.

Equation 22.

8.2.2.7 Bootstrap Capacitor Selection

A 0.1-μF ceramic capacitor must be connected between the BOOT and PH pins for proper operation. It is recommended to use a ceramic capacitor with X5R or better grade dielectric. The capacitor should have a 10-V or higher voltage rating.

8.2.2.8 Undervoltage Lock Out Set Point

The Under Voltage Lock Out (UVLO) can be adjusted using an external voltage divider on the EN pin of the TPS54140A. The UVLO has two thresholds, one for power up when the input voltage is rising and one for power down or brown outs when the input voltage is falling. For the example design, the supply should turn on and start switching once the input voltage increases above 7.7 V (enabled). After the regulator starts switching, it should continue to do so until the input voltage falls below 6.7 V (UVLO stop).

The programmable UVLO and enable voltages are set using a resistor divider between Vin and ground to the EN pin. Equation 2 through Equation 3 can be used to calculate the resistance values necessary. For the example application, a 332 kΩ between Vin and EN and a 61.9 kΩ between EN and ground are required to produce the 7.7 and 6.7 volt start and stop voltages.

8.2.2.9 Output Voltage and Feedback Resistors Selection

For the example design, 10.0 kΩ was selected for R2. Using Equation 1, R1 is calculated as 31.25 kΩ. The nearest standard 1% resistor is 31.6 kΩ. Due to current leakage of the VSENSE pin, the current flowing through the feedback network should be greater than 1 μA in order to maintain the output voltage accuracy. This requirement makes the maximum value of R2 equal to 800 kΩ. Choosing higher resistor values will decrease quiescent current and improve efficiency at low output currents but may introduce noise immunity problems.

8.2.2.10 Compensation

There are several industry techniques used to compensate DC/DC regulators. The method presented here yields high phase margins. For most conditions, the regulator will have a phase margin between 60 and 90 degrees. The method presented here ignores the effects of the slope compensation that is internal to the TPS54140A. Since the slope compensation is ignored, the actual cross over frequency is usually lower than the cross over frequency used in the calculations.

Use SwitcherPro software for a more accurate design.

The uncompensated regulator will have a dominant pole, typically located between 300 Hz and 3 kHz, due to the output capacitor and load resistance and a pole due to the error amplifier. One zero exists due to the output capacitor and the ESR. The zero frequency is higher than either of the two poles.

If left uncompensated, the double pole created by the error amplifier and the modulator would lead to an unstable regulator. To stabilize the regulator, one pole must be canceled out. One design approach is to locate a compensating zero at the modulator pole. Then select a cross over frequency that is higher than the modulator pole. The gain of the error amplifier can be calculated to achieve the desired cross over frequency. The capacitor used to create the compensation zero along with the output impedance of the error amplifier form a low frequency pole to provide a minus one slope through the cross over frequency. Then a compensating pole is added to cancel the zero due to the output capacitors ESR. If the ESR zero resides at a frequency higher than the switching frequency then it can be ignored.

To compensate the TPS54140A using this method, first calculate the modulator pole and zero using the following equations:

Equation 23.

where

• I_OUT(max) is the maximum output current
• C_OUT is the output capacitance
• V_OUT is the nominal output voltage

Equation 24.

For the example design, the modulator pole is located at 1.5 kHz and the ESR zero is located at 338 kHz.

Next, the designer needs to select a crossover frequency which will determine the bandwidth of the control loop. The cross over frequency must be located at a frequency at least five times higher than the modulator pole. The cross over frequency must also be selected so that the available gain of the error amplifier at the cross over frequency is high enough to allow for proper compensation.

Equation 29 is used to calculate the maximum cross over frequency when the ESR zero is located at a frequency that is higher than the desired cross over frequency. This will usually be the case for ceramic or low ESR tantalum capacitors. Aluminum Electrolytic and Tantalum capacitors will typically produce a modulator zero at a low frequency due to their high ESR.

The example application is using a low ESR ceramic capacitor with 10 mΩ of ESR making the zero at 338 kHz.

This value is much higher than typical crossover frequencies so the maximum crossover frequency is calculated using both Equation 25 and Equation 28.

Using Equation 28 gives a minimum crossover frequency of 7.6 kHz and Equation 25 gives a maximum crossover frequency of 45.3 kHz.

A crossover frequency of 45 kHz is arbitrarily selected from this range.

For ceramic capacitors use Equation 25:

Equation 25.

For tantalum or aluminum capacitors use Equation 26:

Equation 26.

For all cases use Equation 27 and Equation 28:

Equation 27.

Equation 28.

Once a cross over frequency, ƒc, has been selected, the gain of the modulator at the cross over frequency is calculated. The gain of the modulator at the cross over frequency is calculated using Equation 29 .

Equation 29.

For the example problem, the gain of the modulator at the cross over frequency is 0.542. Next, the compensation components are calculated. A resistor in series with a capacitor is used to create a compensating zero. A capacitor in parallel to these two components forms the compensating pole. However, calculating the values of these components varies depending on if the ESR zero is located above or below the cross over frequency. For ceramic or low ESR tantalum output capacitors, the zero will usually be located above the cross over frequency. For aluminum electrolytic and tantalum capacitors, the modulator zero is usually located lower in frequency than the cross over frequency. For cases where the modulator zero is higher than the cross over frequency (ceramic capacitors).

Equation 30.

Equation 31.

Equation 32.

For cases where the modulator zero is less than the cross over frequency (Aluminum or Tantalum capacitors), the equations are:

Equation 33.

Equation 34.

Equation 35.

For the example problem, the ESR zero is located at a higher frequency compared to the cross over frequency so Equation 32 through Equation 35 are used to calculate the compensation components. For the example problem, the components are calculated to be: R_C = 76.2 kΩ, C_C = 2710 pF, and Cƒ = 6.17 pF.

The calculated value of the Cƒ capacitor is not a standard value so a value of 2700 pF will be used. 6.8 pF is used for C_C. Rc resistor sets the gain of the error amplifier which determines the cross over frequency. The calculated R_C resistor is not a standard value, so 76.8kΩ will be used.