### 要約

Recommendations on how to create a compatible power rail for MAX77831's V_{IO} pin, with options of using a resistor voltage divider, Zener diode, or TLV431 voltage reference IC. Comparing each solution's cost, size, power loss, other benefits, and limitations.

### Introduction

The MAX77831 is a high-efficiency, buck-boost regulator with wide input and output ranges, and a lot of useful features. Besides the main input voltage, it also requires a 1.08V to 2V supply for the V_{IO} pin, and this voltage sets the logic level for the digital pins (EN, POKB/INTB, SCL, and SDA). This is usually not an issue in a system utilizing MAX77831's I^{2}C functionality, as a compatible power rail should already exist to power the I^{2}C host microcontroller. In a standalone system not using I^{2}C, however, this is an extra voltage rail required to use MAX77831. It is not possible to connect the V_{IO} pin directly to the main input IN pin, because the voltage on the IN pin is higher than the V_{IO} pin's allowed maximum voltage V_{IO(MAX)} 2V. Although the internal regulator's voltage on the VL pin is compatible, it is also not possible to use the VL pin to power the V_{IO} pin because the presence of a valid V_{IO} voltage is required to turn on the internal regulator. Therefore, an external circuity is needed to power the V_{IO} pin. This application note offers some suggestions on how to create a compatible voltage rail for MAX77831's V_{IO} pin from a separate power rail (if there is one in the system) or from the main input.

### V_{IO} Load Current

Before starting, it is necessary to know how much load current the V_{IO} rail can supply to the MAX77831. When the I^{2}C interface is not being utilized, the MAX77831 is expected to draw a maximum 2µA from the V_{IO} pin. It is also expected that both the EN pin and POKB/INTB pin are connected to the V_{IO} rail. So, current consumption from those two pins are also accounted for. For the EN pin, there is an internal 800kO pulldown resistor. So, there is about 1.35µA to 2.5µA drawn from the EN pin when the MAX77831 is enabled, depending on the V_{IO} rail voltage (1.08V to 2V). For the POKB/INTB pin, the recommended pullup resistor connected to the V_{IO} rail is 15kO. So, there is about 72µA to 133µA drawn from the POKB/INTB pin when this functionality is in use, again depending on the V_{IO} rail voltage. Therefore, the maximum current drawn from the V_{IO} rail is about 75.35µA to 137.83µA depending on the V_{IO} rail voltage. If the application uses a different pullup resistor, the current consumption of the V_{IO} rail changes. The bulk of the current goes to the POKB/INTB pin. So, if the POKB/INTB pin is not in use (connected to AGND), then the V_{IO} supplies about 5µA max. The rest of this application note assumes that POKB/INTB is in use with the default 15kO pullup resistor.

### Design Considerations

The requirement for this V_{IO} supply is to maintain a voltage within MAX77831's V_{IO} voltage range (1.08V to 2V) regardless how much load MAX77831 draws from it. This application note introduces three common circuit options (resistor voltage divider, Zener diode, and TLV431 voltage reference IC) to achieve such a goal, with detailed design procedures. Each option has its own advantages and limitations. Note that the following options result in a V_{IO} voltage varying inside the valid V_{IO} voltage range depending on the operating condition. So, the POKB/INTB logic HIGH level also varies based on the V_{IO} voltage.

### Option 1: Resistor Voltage Divider

If there is a second power rail in the system, then the simplest and cheapest option with the smallest footprint is to use a resistor voltage divider as in Figure 2. Simply calculate the ratio of the resistors to bring the second power rail's voltage down to a V_{IO}-compatible voltage (1.08V to 2V). Note that the resistor voltage divider has high output impedance, and the output voltage starts to drop as soon as it is being loaded. Therefore, when selecting resistor values, make sure to satisfy the following: when the resistor voltage divider is supplying full load current (see the previous section for the expected load current from the V_{IO} rail), the output voltage must still maintain above the minimum V_{IO} voltage V_{IO(MIN)} 1.08V. In general, the smaller the resistor values, the less the voltage drop as load increases, but at the same time the power loss on the resistor increases. With this specific application, when the total resistance R_{TOP} + R_{BOT} is around 20kO, the power loss is minimized while having an acceptable voltage drop at full load. The expected voltage drop can be determined by a simple circuit analysis (KVL, KCL, Ohm's Law) or through circuit simulation.

If there is no separate power rail available in the system and the V_{IO} rail needs to be created off the main input, the resistor voltage divider might still work if the main input's voltage variation is not too large. With some applications that take advantage of MAX77831's wide input range (2.5V to 16V), this is probably not the case. As an example, some applications using the MAX77831 have a wide input range from 2.7V to 12V, and the resistor voltage divider definitely does not work in this case, because it is not possible for a fixed ratio resistor voltage divider to output a V_{IO}-compatible voltage (1.08V to 2V) from such a wide input voltage range. In general, the narrower the input voltage range, the higher chance the resistor voltage divider works.

To check if the resistor voltage divider is suitable for a particular input voltage range, perform the following calculation. An example is also shown for an application with maximum input voltage V_{IN(MAX)} = 12V and minimum input voltage V_{IN(MIN)} = 7.6V.

- Divide the application's V
_{IN(MAX)}by MAX77831 V_{IO(MAX)}2V to get the resistor divider ratio.

E.g., Resistor Divider Ratio = V_{IN(MAX)}/V_{IO(MAX)} = 12V/2V = 6

Divide the application's V_{IN(MIN)} by the calculated ratio. If the result is greater than V_{IO(MIN)} 1.08V, then the resistor voltage divider might work. Otherwise, the resistor voltage divider is not a valid option. See other options recommended in the following sections.

E.g., No Load V_{OUT} with V_{IN(MIN)} = V_{IN(MIN)}/Ratio = 7.6V/6 = 1.267V

Since this is greater than V_{IO(MIN)} 1.08V, it passes the check.

If the main input's voltage range passes the above check, the resistor voltage divider likely works. The remaining uncertainty comes again from that the resistor voltage divider's output voltage starts to drop as soon as the load current increases. The goal is to select the resistor values for the resistor voltage divider so that with V_{IN(MAX)} and at no load from the V_{IO} rail, the V_{IO} supply voltage does not exceed V_{IO(MAX)}, and with V_{IN(MIN)} and at full load from the V_{IO} rail, the V_{IO} supply voltage does not drop below V_{IO(MIN)}. Compared to the previous case in which the input voltage is fixed, there is now additional output voltage variation for the resistor voltage divider due to the variation of the input voltage. Therefore, the allowable output voltage drop due to increase in loading is now smaller, and to achieve that, smaller resistor values are required, which means the power loss on the resistors increase. In general, wider input voltage variation ultimately results in higher power loss on resistor voltage divider.

Let us continue with the example of V_{IN(MAX)} = 12V and V_{IN(MIN)} = 7.6V. Let us look for a resistor pair R_{TOP} and R_{BOT} such that with 12V input at no load from V_{IO} rail, the output voltage is V_{IO(MAX)} 2V, and with 7.6V input at full load from V_{IO} rail, the output voltage is V_{IO(MIN)} 1.08V. This resistor pair represents the maximum suitable resistor values, as any larger resistor values result in an output voltage below V_{IO(MIN)} with input being V_{IN(MIN)} and at full load on the V_{IO} rail. From the previous calculation, it is determined that the resistor voltage divider should have a ratio of 6. Plugging this into the resistor voltage divider equation, the first equation is:

V_{IN}/V_{OUT} = (R_{TOP} + R_{BOT})/R_{BOT}

6 = (R_{TOP} + R_{BOT})/R_{BOT}(Equation 1)

With V_{IN(MIN)} 7.6V at full load, the target output voltage is V_{IO(MIN)} 1.08V. As explained in the previous section, the MAX77831 draws about 76µA maximum from V_{IO} rail at V_{IO(MIN)} 1.08V. Therefore, the load resistance in this condition is:

R_{LOAD} = V_{IO(MIN)}/I_{LOAD} = 1.08V/75.35µA = 14.333kO

R_{BOT} and R_{LOAD} are in parallel. Combine these two resistances and get an effective bottom resistor R_{EFF} (Figure 3). Using the same R_{TOP} and the new effective bottom resistor R_{EFF}, view the whole circuit as a different resistor voltage divider, with the same input voltage 7.6V and V_{IO(MIN)} 1.08V as the output voltage. Plugging these into the resistor voltage divider equation, the second and third equations are:

V_{IN}/V_{OUT NEW} = (R_{TOP} + R_{EFF})/R_{EFF}

7.6V/1.08V = (R_{TOP} + R_{EFF})/R_{EFF}(Equation 2)

R_{EFF} = (R_{BOT} × R_{LOAD})/(R_{BOT} + R_{LOAD})(Equation 3)

There are three equations here with three unknowns. Solving the equations, the following results occur:

R_{TOP} = 14.864kO, R_{BOT} = 2.973kO

With these resistor values, calculate the power loss from the resistor voltage divider. Power loss is highest with input at V_{IN(MAX)}. No load number represents the condition when the MAX77831 is disabled and virtually draws no current from the V_{IO} rail, whereas the full load number represents the condition when the MAX77831 is enabled, and POKB/INTB pin is outputting logic LOW (signaling normal operation).

At V_{IN(MAX)} 12V and no load:

From the calculation above, V_{IO} = 2V at this condition. The power loss is:

P_{TOP} = V^{2}/R = (V_{IN} - V_{IO})^{2}/R_{TOP} = (12V - 2V)^{2}/14.846KO = 6.728mW

P_{BOT} = V^{2}/R = (V_{IO})^{2}/R_{BOT} = (2V)2/2.973KO = 1.345mW

P_{TOTAL} = P_{TOP} + P_{BOT} = 6.728mW + 1.345mW = 8.073mW

At V_{IN(MAX)} 12V and full load:

In this case, calculate V_{IO} by solving the following equation:

(V_{IN} - V_{IO})/R_{TOP} = V_{IO}/R_{BOT} + I_{LOAD}

I_{LOAD} = 2µA + V_{IO}/800kO + V_{IO}/15kO

Plugging in V_{IN} = 12V, R_{TOP} = 14.864kO, R_{BOT} = 2.973kO, and get V_{IO} = 1.708V. Next, calculate the power loss:

P_{TOP} = V^{2}/R = (V_{IN} - V_{IO})^{2}/R_{TOP} = (12V - 1.708V)^{2}/14.864KO = 7.126mW

P_{BOT} = V^{2}/R = (V_{IO})^{2}/R_{BOT} = (1.708V)^{2}/2.973KO = 0.981mW

P_{TOTAL} = P_{TOP} + P_{BOT} = 7.126mW + 0.981mW = 8.107mW

Again, this resistor pair represents the theoretical maximum resistances suitable for the resistor voltage divider. In practice, select resistors with nominal values smaller than these for tolerance and to achieve a narrower V_{IO} voltage range. But remember, the smaller the resistors, the higher the power loss on the resistor voltage divider. So, the power loss number calculated above is the lowest possible number for a V_{IO} supply solution using a resistor voltage divider to support the 7.6V to 12V input voltage range. When selecting specific resistor parts, pay attention to the power rating of the resistor package.

### Option 2: Shunt Regulator with Zener Diode

If the application has a wide input voltage range and resistor voltage divider does not work, the shunt regulator is a viable option. The simplest shunt regulators consist of a resistor and a Zener diode (Figure 4). The Zener diode shunt regulator operates using the characteristic that when a Zener diode is in reverse breakdown, it can maintain a relatively stable voltage across its terminals (within a certain range of reverse current, and the Zener voltage varies based on the amount of reverse current flowing through the Zener diode).

Zener diodes are available with a lot of different breakdown voltages, with the lowest one being within the 1V to 2V range, which is the perfect candidate for the V_{IO} supply voltage. However, comparing to Zener diodes with higher breakdown voltages, the one with breakdown voltages less than 5V tend to have large voltage variation within its operating range, and this variation can be more than 100% of the nominal Zener voltage. Figure 5 shows an example breakdown characteristic curves from a Zener diode family. The Zener voltage of the 2V variant, the one we are interested in, varies from less than 1V at 10µA to about 2.5V at 20mA. Therefore, make sure the Zener diode is biased at an operating point within MAX77831's V_{IO} voltage range. The bias current is set by the series resistor R_{S}. There is a theoretical resistance range suitable for a particular input voltage range. Let us explore it using the same input range 7.6V to 12V as an example.

First, select a Zener diode with a nominal Zener voltage between 1.08V and 2V. From the data sheet, obtain its Zener current I_{Z} vs. Zener voltage V_{Z} curve. Pay attention to the Zener voltage variation. The larger the variation, the smaller the acceptable series resistance range. So, choose a Zener diode with relatively smaller voltage variation. The one chosen for this example is the 2V variant in Figure 5. From its I-V curve, estimate the equation of the curve. This equation is used later to calculate the operating point of the Zener diode. Microsoft Excel is a great tool for this task. Simply pick some points spread evenly on the original I-V curve, enter them into MS Excel, create a scatterplot, and then use the trendline function to generate a best-fit curve and get its equation. Figure 6 shows the best-fit curve for the I-V curve of the 2V Zener diode selected, generated using the method described above, and with an equation V_{Z} = 1.6121 × (1000 × I_{Z})^{0.1497}.

Let us calculate the theoretical maximum value for R_{S}. The theoretical maximum R_{S} is limited by the requirement that the Zener voltage cannot drop below V_{IO(MIN)}. Besides determining the Zener bias current, this series resistor is also related to the maximum load current the shunt regulator can supply. The higher the R_{S}, the less the load capability. Moreover, with a certain value or R_{S}, the higher the input voltage, the more the load capability. Therefore, the maximum R_{S} should be calculated with input voltage at V_{IN(MIN)} and output voltage at V_{IO(MIN)} supplying full load. The amount of current flowing through R_{S} is the combination of the Zener current and load current. From the Zener diode I-V characteristic equation, determine the Zener current when V_{Z} is V_{IO(MIN)} 1.08V. Plug V_{Z} = 1.08V into V_{Z} = 1.6121 × (1000 × I_{Z})^{0.1497} and get I_{Z} = 68.85µA.

And, the amount of resistor current is:

I_{R} = I_{Z} @ [V_{Z} = V_{IO(MIN)}] + [I_{LOAD(MAX)} @ [V_{Z} = V_{IO(MIN)}] = 68.85µA + 75.35µA = 144.20µA

With this, maximum R_{S} can be determined using Ohm's Law.

R_{MAX} = (V_{IN(MIN)} - V_{IO(MIN)})/I_{R} = (7.6V - 1.08V)/144.20µA = 45.215kO

This represents the theoretical maximum value suitable for R_{S}. Any higher value of R_{S} results in V_{IO} voltage dropping below V_{IO(MIN)} 1.08V when providing full load with input voltage at V_{IN(MIN)} 7.6V.

Next, calculate the theoretical minimum value for R_{S}. Similarly, the minimum R_{S} is limited by the requirement that Zener voltage cannot exceed V_{IO(MAX)}. Because a higher I_{Z} results in a higher V_{Z}, perform the calculation at condition where I_{Z} is maximized. Therefore, as opposed to the previous calculation, minimum value for R_{S} is calculated with input voltage at V_{IN(MAX)} and output voltage at V_{IO(MAX)} supplying no load. Again, to determine I_{Z} when V_{Z} is V_{IO(MAX)} 2V, plug V_{Z} = 2V into V_{Z} = 1.6121 × (1000 × I_{Z})^{0.1497} and get I_{Z} = 4.222mA. Since there is no load current, the resistor current is the same as the V_{Z}.

I_{R} = I_{Z} @ [V_{Z} = V_{IO(MAX)}] = 4.222mA

With this, minimum R_{S} can be determined using Ohm's Law.

R_{MIN} = (V_{IN(MAX)} - V_{IO(MAX)})/I_{R} = (12V - 2V)/4.222mA = 2.369kO

So far, it is determined the series resistor should be between 2.369kO and 45.215kO, to make sure the output voltage of this shunt regulator stays within MAX77831's V_{IO} voltage range. Next, calculate the power loss from this solution. Like the resistor voltage divider solution, the higher the resistor value chosen, the smaller the power loss on both the series resistor and Zener diode. Let us calculate the power loss with R_{S} = 45.215kO, and this represents the minimal power loss from this solution for input range 7.6V to 12V, with the example Zener diode. With a specific series resistor, the power loss is higher with higher input voltage. So, calculate at V_{IN(MAX)} 12V for the example. Again, no load number represents the condition when MAX77831 is disabled and virtually draws no current from the V_{IO} rail, whereas full load number represents the condition when MAX77831 is enabled, and POKB/INTB pin is outputting logic LOW (signaling normal operation).

At V_{IN(MAX)} 12V and no load:

First, determine Zener diode's operating point. To do so, solve the equations involving the series resistor and I-V characteristics of the specific Zener diode:

(V_{IN} - V_{Z})/R_{S} = I_{Z}

V_{Z} = 1.6121 × (1000 × I_{Z})^{0.1497}

Plugging in V_{IN} = 12V and R_{S} = 45.215kO, get V_{Z} = 1.299V and I_{Z} = 236.663µA. Now, calculate the power loss.

P_{R} = V^{2}/R = (V_{IN} - V_{IO})^{2}/R_{S} = (12V - 1.299V)^{2}/45.215kO = 2.533mW

P_{Z} = V_{Z} × I_{Z} = V_{IO} × I_{Z} = 1.299V × 236.663µA = 307.425µW

P_{TOTAL} = P_{R} + P_{Z} = 2.533mW + 307.425µW = 2.840mW

At V_{IN(MAX)} 12V and full load:

Again, first, determine the Zener diode's operating point by solving the following equations:

(V_{IN} - V_{Z})/R_{S} = I_{Z} + I_{LOAD}

I_{LOAD} = 2µA + V_{Z}/800kµ + V_{Z}/15kµ

V_{Z} = 1.6121 × (1000 × I_{Z})^{0.1497}

Plugging in V_{IN} = 12V and R_{S} = 45.215kO, get V_{Z} = 1.218V and I_{Z} = 153.737µA. Now, calculate the power loss.

P_{R} = V^{2}/R = (V_{IN} - V_{IO})^{2}/R_{S} = (12V - 1.218V)^{2}/45.215kO = 2.571mW

P_{Z} = V_{Z} × I_{Z} = V_{IO} × I_{Z} = 1.218V × 153.737µA = 187.252µW

P_{TOTAL} = P_{R} + P_{Z} = 2.571mW + 187.252µW = 2.758mW

The power loss number calculated above is the lowest possible number for a V_{IO} supply solution to support the 7.6V to 12V input voltage range using the shunt regulator with the example Zener diode. These numbers only give an estimation and the calculation should be repeated for a different Zener diode. In practice, resistors with nominal values smaller than 45.215kO and within the calculated allowable range are selected for this example, to allow for tolerance. But remember, the smaller the resistors, the higher the power loss on the shunt regulator. When selecting specific resistor and Zener diode parts, pay attention to the power rating of component package.

### Option 3: Shunt Regulator with TLV431 Voltage Reference IC

Another flavor of the shunt regulator is to replace the Zener diode and use the TLV431 voltage reference IC instead (Figure 7). Comparing to the one with a Zener diode, the TLV431 IC offers a much more stable output voltage (1.5% or less variation) regardless of the bias current (after some small amount of minimum cathode current typically around 50µA), as illustrated in Figure 9. Therefore, unlike the previous two options introduced in this application note, where the V_{IO} voltage varies based on the operating condition, using TLV431 ensures a constant V_{IO} voltage. TLV431 has a reference voltage of 1.24V. Select this as the V_{IO} voltage if minimizing component count is desired (Figure 7). If a different V_{IO} voltage is desired, a pair of resistor divider can be added to the circuit to achieve a different V_{IO} voltage (Figure 8).

Let us do another example again using the same input range 7.6V to 12V. Remember from the previous example with Zener diode that there is a theoretical upper and lower limit for the series resistor R_{S}, limited by MAX77831's V_{IO} voltage range 1.08V to 2V. In the case of TLV431, the output voltage is constant. So, there is no theoretical lower limit for R_{S} (except for power loss in practice). There is an upper limit, however, because the TLV431 has a minimum cathode current requirement to maintain the reference voltage 1.24V. For the specific TLV431 IC chosen, this value is listed as 80µA maximum in the data sheet. The maximum R_{S} should ensure at least 80µA flowing through the TLV431 when the MAX77831 is drawing maximum current from the V_{IO} rail. The maximum load current when V_{IO} = 1.24V is:

I_{LOAD} = 2 µA + V_{IO}/800kO + V_{IO}/15kO = 2µA + 1.24V/800kO + 1.24V/15kO = 86.22µA

So, the total R_{S} current at this condition is:

I_{R} = I_{LOAD} + I_{K(MIN)} = 86.22µA + 166.22µA

With this, the maximum value for R_{S} is:

R_{S} = (V_{IN(MIN)} - V_{IO})/I_{R} = (7.6V - 1.24V)/166.22µA = 38.263KO

Next, calculate the power loss at V_{IN(MAX)} 12V, because power loss is higher at higher input voltage.

At V_{IN(MAX)} 12V and no load:

P_{R} = V^{2}/R = (V_{IN} - V_{IO})^{2}/R_{S} = (12V - 1.24V)^{2}/38.263KO = 3.026mW

Because there is no load, the current flowing through the TLV431 is the same as the resistor current.

I_{K} = I_{R} = (V_{IN} - V_{IO})/R_{S} = 12V - 1.24V)/38.263kO = 281.212µA

P_{TLV431} = V_{KA} × I_{K} = V_{IO} × I_{K} = 1.24 × 281.212µA = 348.702µW

P_{TOTAL} = P_{R} + P_{TLV431} = 3.026mW + 348.702µW = 3.375mW

At V_{IN(MAX)} 12V and full load:

P_{R} = V^{2}/R = (V_{IN} - V_{IO})^{2}/R_{S} = (12V - 1.24V)^{2}/38.263kO = 3.026mW

I_{K} = I_{R} - I_{LOAD} = (V_{IN} - V_{IO})/R_{S} - I_{LOAD} = (12V - 1.24V)/38.263KO - 86.22µA = 194.992µA

P_{TLV431} = V_{KA} × I_{K} = V_{IO} × I_{K} = 1.24V × 194.992µA = 241.790µW

P_{TOTAL} = P_{R} + P_{TLV431} = 3.026mW + 241.790µW = 3.268mW

Again, the power loss number calculated above is for maximum R_{S}, and therefore, it is the lowest possible power loss number for a V_{IO} supply solution to support the 7.6V to 12V input voltage range using the shunt regulator with TLV431 IC. In practice, resistors with nominal values smaller than 38.263kO are selected for this example, to allow for tolerance. But remember, the smaller the resistors, the higher the power loss on the shunt regulator. When selecting specific resistor and TLV431 parts, pay attention to the power rating of the component package.

### Choosing the Right Option

So far, we have walked through some simple options available for V_{IO} supply, but how to decide which one to go with? Let us look at the following comparison table.

Option | Resistor Voltage Divider | Shunt Regulator with Zener Diode | Shunt Regulator with TLV431 IC |

Component Count | 2 resistors | 1 resistor + 1 Zener diode | 1 resistor + 1 TLV431 (1.24V V_{IO})3 resistors + 1 TLV431 (other V _{IO} voltage) |

Cost | Cheapest: cost of 2 resistors | Not as cheap: cost of 1 resistor + about 4 cents | More expensive: cost of resistors + about 8 cents |

Component Size | Smallest: size of 2 SMD resistors | Not as small: size of 1 SMD resistor + SOD523 (1.2mm x 0.8mm) | Bigger: size of SMD resistors + SOT323 (2mm x 1.25mm) |

Minimum Power loss (for V_{IN} 7.6V to 12V example) |
About 8.1mW* | About 2.8mW* | About 3.3mW* |

Other Notes | 1. Only works if input voltage is constant or has a narrow range. 2. V _{IO} voltage is not constant and varies depending on conditions. |
1. Works with all input voltage ranges. 2. V _{IO} voltage is not constant and varies depending on conditions. |
1. Works with all input voltage ranges. 2. V _{IO} voltage is constant. |

* Power loss number heavily depends on the input voltage range and resistor value selected. Perform the calculation to get the actual power loss number for the specific application. |

If the application has a constant input voltage or the input voltage is relatively narrow, the cheapest option and with smallest footprint is a resistor voltage divider. If the application wants a constant V_{IO} voltage, then the best option is a shunt regulator with TLV431 IC. Otherwise, go with the shunt regulator with Zener diode.

### Conclusion

This application note explored the options of using a resistor voltage divider, Zener diode, and TLV431 IC to create V_{IO} voltage supply for the MAX77831 when I^{2}C is not present in the system. Each option has its benefits and limitations. Besides the ones discussed, there are other available options, such as a linear regulator. However, considering the cost, size, and other aspects, unless there are other components in the system that also plan to use this power rail, having a dedicated linear regulator just for MAX77831's V_{IO} supply seems to be overkilled and thus not recommended.

The MAX77831 is a high-efficiency (97% peak), buck-boost regulator with wide input voltage range (2.5V to 16V) and wide output voltage range (4.5V to 15V with internal feedback, and 3V to 15V with external feedback). It can provide 18W of continuous output power. It packs lots of useful features including DVS, programmable current limit, overcurrent protection (OCP), overvoltage protection (OVP), Power-OK, output active discharge, and much more. Learn more: MAX77831