SLUAAX7 August 2024 BQ25758

2 Derivation

Composed of four additional components, an analog feedback circuit, as highlighted in blue in Figure 2-1, can account for the voltage drop across the cable as a function of system load.

Figure 2-1 BQ25758 with Cable Compensation Circuit Block Diagram

The host software programs the BQ25758 VOUT_REG register to the desired regulation voltage. R_IOUT is sized to be least 10% higher than the maximum expected system load current. The converter then regulates the VO_SNS pin to VOUT_REG voltage, regardless of load current. But, we want the voltage at the other side of the cable, VSYS_LOAD to be VOUT_REG, which means:

Equation 1.

V (S R N) = V_{O U T_R E G} + I_{S Y S} \times R_{C A B L E} {= V}_{S Y S_L O A D} + I_{S Y S} \times R_{C A B L E}

The IOUT pin voltage, V(IOUT), is proportional to the system load current as sensed across resistor R_SNS by SRN and SRP pins. Specifically,

Equation 2.

V (I O U T) = \frac{{I_{S Y S} \times V}_{R E F_I O U T} \times R_{I O U T}}{K_{I O U T (A \times k Ω)}}

Where V_{REF_IOUT} is 2V ± 0.5% and K_IOUT is 50A x kΩ ± 4%. The LMV321A amplifier negative feedback loop regulates the voltage across compensation resistor RC1 to be V(IOUT) so

Equation 3.

V_{O U T_R E G} + \frac{V (I O U T)}{R_{C 1}} \times R_{C 2} = V (S R N)

Combining these equations gives

Equation 4.

V_{S Y S_L O A D} + I_{S Y S} \times R_{C A B L E} = V_{O U T_R E G} + \frac{{I_{S Y S} \times V}_{R E F_I O U T} \times R_{I O U T}}{K_{I O U T (A \times k Ω)}} \times \frac{R_{C 2}}{R_{C 1}}

Where as,

Equation 5.

V_{S Y S_L O A D} + I_{S Y S} \times R_{C A B L E} = V_{O U T_R E G} + \frac{{I_{S Y S} \times V}_{R E F_I O U T} \times R_{I O U T}}{K_{I O U T (A \times k Ω)}} \times \frac{R_{C 2}}{R_{C 1}}

Which reduces to

Equation 6.

\frac{R_{C 2}}{R_{C 1}} = R_{C A B L E} \times \frac{K_{I O U T (A \times k Ω)}}{V_{R E F_I O U T} \times R_{I O U T}}

With R_SNS<< R_CABLE and knowing that K_IOUT/V_{REF_IOUT} is approximately 125Ω / R_SNS, the equation below can be used to find the value of one compensation resistors after selecting the other to be in the 10kΩ range.

Equation 7.

\frac{R_{C 2}}{R_{C 1}} = \frac{125 Ω \times R_{C A B L E}}{R_{S N S} \times R_{I O U T}}