### Abstract

A resistive feedback network is often used to set the output voltage of a power supply. A mechanical potentiometer (pot) conveniently solves the problem of adjusting a power supply. For easier automatic calibration, a mechanical pot can be replaced with a digital pot. This application note presents a calibration solution that uses a digital pot, because digipots are smaller, do not move with age or vibration, and can be recalibrated remotely. This proposed solution reduces the susceptibility of the system to the tolerance of the digital pot's end-to-end resistance, making the solution optimal for designers. This application note also explains some of the equations required to calculate the resistor chain values and to use a digital pot in this way. A spreadsheet with standard reisistor values is available for easy calculations.

*Power Management DesignLine*, February 2012.

### Introduction

### Conventional Power-Supply Feedback

**Figures 1**and

**2**.

*Figure 1. A conventional fixed-resistive feedback.*

*Figure 2. A conventional variable-resistive feedback.*

(Eq. 1) |

_{R}will have a tolerance of ±5%, this leads to a tolerance in the output voltage of 5%. If this is not acceptable in the application, the resistive-divider needs to be made variable. Replacing R1 and R2 with a mechanical potentiometer is not normally done, since it would result in a wide range of output voltages and would be very sensitive to adjustments. Over time and temperature, any drift in the pot's position would create an unacceptable amount of output voltage drift. Therefore, Figure 2 illustrates a scheme that reduces the output range, which is now easier to adjust and is more stable.

### Basic Structure of Digital Pots

*Figure 3. Typical digital potentiometer structure.*

**Figure 3**. For simplicity, the switches are shown as single MOSFETs. Typically, these switches would be two BiCMOS transistors (one "P" and one "N") to produce low on-resistances.

### Using the Digital Pot to Calibrate a Power Supply

*Figure 4. The initial resistor string.*

*Figure 5. The final resistor string.*

**Figure 4**. We will use the following example:

_{O}= 12V

V

_{R}= 2.5V ±5%

R

_{T}= R1 + R2 + R3

(Eq. 2) |

(Eq. 3) |

R1 = R_{T} - R3 - R2 |
(Eq. 4) |

_{T}, using (R1 + R2 + R3). Since this is arbitrary, we'll start with R

_{T}= 20kΩ. (We can always change it later to give more realistic values for R1, R2 and R3 if required.) From Equation 2, we find R3 = 3.598kΩ. From Equation 3, R2 = 417Ω, and from Equation 4, R1 = 15.625kΩ.

_{O}will be correct when R2 is cantered and V

_{R}is at its nominal value.

(Eq. 5) |

**Figure 5**. Thus, the parallel combination of R2

_{A}and R

_{P}makes R2.

(Eq. 6) |

_{P}, the ideal value for R2

_{A}is 690Ω. The closest 1% value is 698Ω. If we calculate the parallel combination of this and the digital pot at its tolerance extremes, we get R

_{MIN}= 642Ω and R

_{MAX}= 660Ω. This is a tolerance of only 1%, due to the 20% end-to-end tolerance of the pot. We use a 698Ω resistor for R2

_{A}, as this is the closest standard 1% value.

**Figure 6**. We obviously do not need to calculate R6.

*Figure 6. Using the star-delta transformation.*

(Eq. 7) |

(Eq. 8) |

(Eq. 9) |

(Eq. 10) |

(Eq. 11) |

_{O}= 12V (output voltage)

V

_{R}= 2.5V ±5% (reference voltage)

R1 = 15.8kΩ (upper resistor)

R2

_{A}= 698Ω (parallel resistor)

R

_{P}= 10kΩ (digital potentiometer)

R3 = 3.92kΩ (lower resistor)

_{R}= 2.375V, tap = 44, V

_{O}= 11.99842V

V

_{R}= 2.625V, tap = 210, V

_{O}= 11.99773V

### Summary