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 I²C functionality, as a compatible power rail should already exist to power the I²C host microcontroller. In a standalone system not using I²C, 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 I²C 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.

1. 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 = V²/R = (V_IN - V_IO)²/R_TOP = (12V - 2V)²/14.846KO = 6.728mW
P_BOT = V²/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 = V²/R = (V_IN - V_IO)²/R_TOP = (12V - 1.708V)²/14.864KO = 7.126mW
P_BOT = V²/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 = V²/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 = V²/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 = V²/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 = V²/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.

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 I²C 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