A capacitive load current acquisition system and a method for calculating capacitive load current.

Abstract

This invention discloses a capacitive load current acquisition system and a method for calculating capacitive load current. The system includes a battery, an AFE module, an MCU module, a current processing module, an equivalent resistance R1, an equivalent capacitance C1, a resistor R3, a resistor R4, a first control unit, a second control unit, a third control unit, and a resistor R2. The AFE module acquires the system current through resistor R2. When the AFE module cannot acquire the system current, the MCU module calculates the system current according to a formula. Thus, even when the AFE module cannot acquire the current value, the MCU module can accurately calculate the current value, enabling the system to simultaneously detect large and small currents, effectively improving the estimation accuracy of battery SOC.

Description

Technical Field

[0001] This invention relates to the technical field of battery SOC calculation, specifically to a capacitive load current acquisition system and a method for calculating capacitive load current. Background Technology

[0002] Accurate estimation of the battery's state of charge (SOC) is crucial for the charge and discharge control and power optimization management of electric vehicle batteries. It directly affects battery lifespan and vehicle performance, and can predict the electric vehicle's driving range. Electric vehicle battery packs typically consist of several individual cells to meet the demands of higher power output and longer driving ranges.

[0003] In estimating the State of Charge (SOC) of a battery pack, it is necessary to collect the current of the equivalent sampling resistor in the battery pack. In existing battery packs, to reduce the heat generated by the sampling resistor when the current is too high, a small equivalent sampling resistor is usually selected. However, a small equivalent sampling resistor cannot accurately detect small currents in the circuit, which reduces the accuracy of the battery pack SOC estimation. Summary of the Invention

[0004] To address the shortcomings of existing technologies, a capacitive load current acquisition system and a method for calculating capacitive load current are provided.

[0005] To achieve the above objectives, the present invention provides a capacitive load current acquisition system, comprising a battery, an AFE module, an MCU module, a current processing module, an equivalent resistance R1, an equivalent capacitance C1, resistors R3 and R4, a first control unit, a second control unit, a third control unit, and a resistor R2; the negative terminal of the battery is grounded, the positive terminal of the battery is connected to one end of the equivalent resistance R1, the equivalent capacitance C1 is connected in parallel with the equivalent resistance R1, the other end of the equivalent resistance R1 is connected to the third control unit, the other end of the third control unit is connected to the first control unit, the other end of the first control unit is connected to resistor R2, and the other end of resistor R2 is connected to the negative terminal of the battery. Resistor R2 is electrically connected to the AFE module; the AFE module is electrically connected to the MCU module and the battery; one end of resistor R4 is connected to the first control unit and the third control unit, and the other end is connected to the second control unit. The other end of the second control unit is connected to resistor R3 and the input terminal of the current processing module. The other end of resistor R3 is connected to resistor R2. The output terminal of the current processing module is connected to the MCU module; the MCU module is electrically connected to the first control unit, the second control unit, and the third control unit; the resistance value of resistor R2 is less than one-tenth of the combined resistance values ​​of resistors R3 and R4; the AFE module collects the system current through resistor R2. When the AFE module cannot collect system current At that time, the MCU module follows the formula Calculate system current .

[0006] According to one embodiment of the present invention, the current processing module includes a voltage amplification unit and a voltage divider unit. The input terminal of the voltage amplification unit is connected to the connection node of the resistor R3 and the second control unit, and its output terminal is connected to the MCU module. The input terminal of the voltage divider unit is connected to the connection node of the resistor R3 and the second control unit.

[0007] According to one embodiment of the present invention, the voltage amplification unit includes an amplifier U10, an input filter component and an output filter component (412); one end of the input filter component is connected to the connection node of the resistor R3 and the second control unit, and the other end is connected to the positive input terminal of the amplifier U10, and the negative input terminal of the amplifier U10 is grounded; one end of the output filter component (412) is connected to the output terminal of the amplifier U10, and the other end is connected to the MCU module.

[0008] According to one embodiment of the present invention, the voltage divider unit includes resistor R20, resistor R21 and capacitor C18. One end of resistor R20 is connected to resistor R3 and the second control unit, and the other end is connected to the MCU module. One end of resistor R21 is connected to resistor R20 and the MCU module, and the other end is grounded. One end of capacitor C18 is connected to resistor R20, resistor R21 and the MCU module, and the other end is grounded.

[0009] This invention also provides a method for calculating capacitive load current, comprising the following steps: S01, the MCU module controls the second control unit to open and the first control unit and the third control unit to close; S02, the MCU module controls the battery output current. S03, the AFE module collects current through resistor R2. Simultaneously, the MCU module calculates the capacitance value of the equivalent capacitance C1; S4, the MCU module determines whether the AFE module has acquired the current. If yes, then execute S5a; otherwise, execute S5b. S5a: The MCU module receives the electrical signal sent by the AFE module. S5b: The MCU module executes the following according to the formula: Calculate current The value of .

[0010] According to one embodiment of the present invention, in step S5a, the MCU module determines that the AFE module has acquired the current. At this time, the MCU module controls the second control unit to remain disconnected, and the first and third control units to remain closed.

[0011] According to one embodiment of the present invention, S5b includes the following sub-steps: S5b1. The MCU module controls the second control unit to close, the first control unit to open, and the third control unit to close; S5b2. The MCU module collects the voltage across resistor R3 at time t3 through the current processing module, and records the voltage across resistor R3 at time t3 as... And (i.e., t3=τ / N, N≥10); the S5b3.MCU module is based on the formula: Calculate current The value of .

[0012] According to one embodiment of the present invention, calculating the equivalent capacitance C1 includes the following steps: S31. The current processing unit acquires the voltage across resistor R3 and amplifies or reduces the voltage across resistor R3; S32. The MCU acquires the voltage across resistor R3 at times t0, t1, and t2 respectively, and records them as follows: , and The S33.MCU module will , and Substitute into the formula: The equivalent capacitance was calculated. The value of .

[0013] According to one embodiment of the present invention, it further includes S06. The MCU module controls the first control unit to close, the second control unit to open, and the third control unit to close.

[0014] The beneficial effect of this invention is that when the system current When the value of the resistor R2 is such that a voltage drop occurs across R2, the AFE module calculates the system current by detecting the voltage across R2. and the system current Send to the MCU module. When the system current... When the value of the resistor is insufficient to cause a voltage drop across resistor R2, the AFE module cannot detect the voltage across resistor R2, and therefore the AFE module cannot calculate the system current. The MCU module cannot obtain the system current. At this point, the MCU module calculates the capacitance value of capacitor C1, limits the value range at time t3 to ensure measurement accuracy, and samples the voltage across resistor R3 at time t3. Then according to the formula Calculate the system current The value. Therefore, even when the AFE module cannot sample the current. When the value is specified, the current can also be accurately calculated through the MCU module. The value of allows the system to detect both large and small currents simultaneously, effectively improving the estimation accuracy of the battery pack's SOC. Attached Figure Description

[0015] The accompanying drawings, which are included to provide a further understanding of this application and form part of this application, illustrate exemplary embodiments of this application and are used to explain this application, but do not constitute an undue limitation of this application. In the drawings: Figure 1 This is a block diagram of the capacitive load current acquisition system in the embodiment; Figure 2 This is a schematic diagram showing the connection of the first control unit, the second control unit, and the third control unit in the embodiment. Figure 3 This is an equivalent circuit diagram showing the second control unit being open and the first and third control units being closed in the embodiment. Figure 4 This is an equivalent circuit diagram showing the closing of the second and third control units and the opening of the first control unit in the embodiment. Figure 5 This is a schematic diagram of the voltage amplification unit in the embodiment; Figure 6 This is a schematic diagram of the voltage divider unit in the embodiment; Figure 7 This is a flowchart illustrating the method for calculating capacitive load current in the embodiments.

[0016] Explanation of reference numerals in the attached figures 1. Battery; 2. AFE module; 3. MCU module; 4. Current processing module; 41. Voltage amplification unit; 411. Input filtering component; 412. Output filtering component; 42. Voltage divider unit; 5. First control unit; 6. Second control unit; 7. Third control unit. Detailed Implementation

[0017] The following drawings disclose several embodiments of the present invention. For clarity, many practical details will be described in the following description. However, it should be understood that these practical details are not intended to limit the invention. That is, in some embodiments of the invention, these practical details are not essential. Furthermore, for the sake of simplicity, some conventional structures and components will be shown in the drawings in a simple schematic manner.

[0018] Furthermore, in this invention, the use of terms such as "first" and "second" is for descriptive purposes only and does not specifically refer to any order or sequence, nor is it intended to limit the invention. They are merely used to distinguish components or operations described using the same technical terms, and should not be construed as indicating or implying relative importance or implicitly specifying the number of indicated technical features. Therefore, a feature defined with "first" or "second" may explicitly or implicitly include at least one of those features. Additionally, the technical solutions of various embodiments can be combined with each other, but only if they are feasible for those skilled in the art. If a combination of technical solutions is contradictory or impossible to implement, such a combination should be considered nonexistent and not within the scope of protection claimed by this invention.

[0019] Example 1 Please refer to Figure 1 , Figure 1 This is a block diagram of a capacitive load current acquisition system. This embodiment provides a capacitive load current acquisition system, which includes a battery 1, an AFE module 2, a current processing module 4, an MCU module 3, an equivalent resistor R1, an equivalent capacitor C1, resistors R3 and R4, a first control unit 5, a second control unit 6, a third control unit 7, and a resistor R2. The negative terminal of battery 1 is grounded, and the positive terminal of battery 1 is connected to one end of the equivalent resistor R1. The other end of the equivalent resistor R1 is connected to the third control unit 7. The equivalent capacitor C1 is connected in parallel with the equivalent resistor R1. The other end of the third control unit 7 is connected to the first control unit 5, and the other end of the first control unit 5 is connected to resistor R2. The other end of resistor R2 is grounded together with the negative terminal of battery 1. Resistor R2 is also connected to the input terminal of AFE module 2. AFE module 2 is connected to both MCU module 3 and battery 1. One end of resistor R4 is connected to both the third control unit 7 and the first control unit 5, and the other end is connected to the second control unit 6. The other end of the second control unit 6 is connected to resistor R3 and the input terminal of the current processing module 4, respectively. The other end of resistor R3 is connected to resistor R2 and the first control unit 5, respectively. The output terminal of the current processing module 4 is connected to the MCU module 3. The MCU module 3 is electrically connected to the first control unit 5, the second control unit 6, and the third control unit 7, respectively. The MCU module 3 controls the switching on and off of the first control unit 5, the second control unit 6, and the third control unit 7 by sending signals. The resistance of resistor R2 is much smaller than the sum of the resistances of resistors R3 and R4.

[0020] Please refer to Figure 2 , Figure 2This is a schematic diagram showing the connection of the first control unit, the second control unit, and the third control unit. In this embodiment, the first control unit 5, the second control unit 6, and the third control unit 7 all use MOSFETs. Specifically, the first control unit 5 is MOSFET Q1, the second control unit 6 is MOSFET Q2, and the second control unit 7 is MOSFET Q3. During connection, the source of MOSFET Q3 is connected to the equivalent resistance R1 and the equivalent capacitance C1, respectively. The gate of MOSFET Q3 is connected to the CHG_EN terminal of the MCU module 3 through a current-limiting resistor R28. A pull-down resistor R27 is connected between the gate and the source of MOSFET Q3. The drain of MOSFET Q3 is connected to the drain of MOSFET Q1 and one end of resistor R4, respectively. The gate of MOSFET Q1 is connected to the DSG_EN terminal of the MCU module through a current-limiting resistor R26, and the source of MOSFET Q1 is connected to one end of resistor R2. A pull-down resistor R19 is connected between the source and the gate of MOSFET Q1. The drain of MOSFET Q2 is connected to one end of resistor R4. The gate of MOSFET Q2 is connected to the PRE_EN terminal of MCU module 3 via current-limiting resistor R11. The source of MOSFET Q2 is connected to resistor R3. The other end of resistor R3 is connected to resistor R2 and the source of MOSFET Q1. A pull-down resistor R18 is connected between the gate and source of MOSFET Q2.

[0021] In actual use, when the capacitive load current acquisition system is activated, MCU module 3 controls the first control unit 5 and the third control unit 7 to close, while the second control unit 6 opens. Then, the MCU controls the battery 1 to output current. Current After passing through the third control unit 7 and the first control unit 5, the circuit passes through resistors R3 and R2 respectively.

[0022] Meanwhile, AFE module 2 detects the voltage across resistor R2, and current processing module 4 receives the voltage across resistor R3. If AFE module 2 can detect the voltage across resistor R2, it converts the acquired voltage into an electrical signal and then sends the signal to MCU module 3, enabling MCU module 3 to obtain the current. The value of t1. Meanwhile, MCU module 3 acquires the voltage across resistor R3 through current processing module 4, and records the voltage across resistor R3 at times t1 and t2 as t1 and t2 respectively. and And t1 < t2.

[0023] Specifically, after the current processing module 4 receives the voltage across resistor R3, it amplifies or divides the voltage across R3, and then sends the amplified or divided signal to the MCU module 3. The MCU module 3 calculates the equivalent capacitance according to formula (1). capacitance: , wherein V1 is the voltage of battery 1, and R is the sum of resistance R3 and resistance R4.

[0024] If the AFE module 2 cannot collect the voltage across resistor R2, the MCU module 3 separately controls the second control unit 6 to close, the first control unit 5 to open, and the third control unit 7 to close. Then the MCU module 3 collects the voltage of resistor R3 at time t3 through the current processing module 4 and records it as . The MCU module 3 calculates the current according to formula (2) , (2).

[0025] Further, the derivation process of formula (1) is as follows: please refer to Figure 3 , Figure 3 is the equivalent circuit diagram when the second control unit is open, and the first control unit and the third control unit are closed. At the instant when the second control unit 6 is closed, since the voltage of the battery 1 in the circuit charges the capacitor C1 through R3 and R4, the internal resistance of the circuit is R=R3+R4, therefore, the equivalent resistance of the circuit is the parallel combination of (R3+R4) and R1, denoted as Req = (R3+R4) / / R1. The charging time constant of capacitor C1 is τ = Req * . The complete response equation of capacitor C1 is listed as follows: . Wherein, R=R3+R4, is the initial state of capacitor C1, V1 is the voltage of battery 1, and V1 can be directly measured by the AFE module 2. It should be noted that when the measurement time is sufficiently short, V1 can be considered to remain unchanged throughout the measurement process.

[0026] Voltage is sampled at two time points t1 and t2, where t1 < t2 and t2 << (R*C1).

[0027] According to Taylor expansion: , then the sampling equations at time t1 and time t2 are obtained: .

[0028] Subtracting formula (4) from formula (5) and organizing the terms gives: .

[0029] Since it is inconvenient to directly measure the voltage of the equivalent capacitor C1, the capacitance voltage is obtained indirectly by measuring the voltage across resistor R3. In the circuit, resistor R3 and resistor R4 are connected in series, so the current flowing through resistor R3 and resistor R4 is the same. Let I(t) be the current flowing through R3 and R4. Then: .

[0030] Simplifying, we get: .

[0031] At the initial state of the circuit (i.e., t=0): .

[0032] Substituting formulas (7), (8), and (9) into formula (4), we get: , Simplifying, we get: .

[0033] In the above formula, Let R3 be the resistance value. Let R4 be the resistance value, and and All are known values. It is the voltage across resistor R3 at the initial moment. Let t1 be the voltage across resistor R3. The voltage across resistor R3 is measured at time t2.

[0034] The derivation of formula (2) is as follows: When AFE module 2 cannot detect current At this time, MCU module 3 controls the second control unit 6 to close, the first control unit 5 to open, and the third control unit to remain closed. The equivalent circuit diagram of the system circuit at this time is as follows. Figure 4 As shown, Figure 4 The equivalent circuit diagram is shown for the second and third control units being closed and the first control unit 5 being open.

[0035] When the first control unit 5 is disconnected and the second control unit 6 is closed, the circuit is in sleep mode. At this time, the equivalent resistance R1 is very large, and it is connected in parallel with the equivalent capacitance C1. Therefore, the charging circuit consists of resistors R3 and R4 connected in series with the equivalent capacitance C1, and then connected in parallel with the equivalent resistance R1. However, since the resistance of the equivalent resistance R1 is very large, its current shunt is negligible. Therefore, the equivalent circuit is that the battery voltage V1 charges the equivalent capacitance C1 through resistors R3 and R4. The time constant in the circuit is τ = Req * Where Req is the equivalent resistance in the circuit, then τ = ((R3 + R4) / / R1) * .

[0036] Based on the total response equation of a first-order RC circuit, the capacitor voltage equation can be obtained (initial voltage is 0): The voltage equation for resistor R3 can be obtained as follows: Among them, the current flowing through resistor R3 equal Then we have: .

[0037] Through Taylor expansion, we get: .

[0038] Simplifying, we get: .

[0039] We can obtain: .

[0040] Thus, when the system current When the value of the resistor R2 is such that a voltage drop occurs across R2, AFE module 2 calculates the system current by detecting the voltage across R2. and the system current Send to MCU module 3. When the system current... When the value of the resistor is insufficient to cause a voltage drop across resistor R2, AFE module 2 cannot detect the voltage across resistor R2, and therefore AFE module 2 cannot calculate the system current. MCU module 3 cannot obtain the system current. At this time, MCU module 3 calculates the capacitance value of capacitor C1, limits the value range at time t3 to ensure measurement accuracy, and acquires the voltage of resistor R3 at time t3. Then according to the formula Calculate the system current The value. Therefore, even when AFE module 2 cannot collect the system current. When the value is specified, the system current can also be accurately calculated through MCU module 3. The value of this value allows the system to simultaneously detect both large and small currents, effectively improving the accuracy of battery SOC estimation. Furthermore, as shown in the above formula, by measuring the R3 voltage at time t3 (much smaller than the time constant τ) and substituting it into the above formula, the steady-state current can be obtained. Furthermore, the larger N is, the smaller t3 is relative to the time constant τ, and the shorter the measurement time t3 is. Thus, the time only needs to be (R* The time is 1 / N, where N is larger and the waiting time is shorter. In this way, MCU module 3 can complete the measurement within a very short time t3, without waiting for the entire charging process to reach a steady state, effectively reducing the current detection time and greatly improving the system's response speed. Furthermore, it avoids the second control unit 6 being closed for an extended period, which could affect the system's detection of large currents.

[0041] Please refer to Figure 5 and Figure 6 , Figure 5 This is a schematic diagram of a voltage amplification unit. Figure 6This is a schematic diagram of the voltage divider unit. Further, in the actual circuit, the current processing module 4 includes a voltage amplification unit 41 and a voltage divider unit 42. The input terminal of the voltage amplification unit 41 is connected to the connection node between resistor R3 and the second control unit 6, and the output terminal of the voltage amplification unit 41 is connected to the MCU module 3. The input terminal of the voltage divider unit 42 is connected to the connection node between resistor R3 and the second control unit 6.

[0042] The voltage amplification unit 41 amplifies the input electrical signal, while the voltage divider unit 42 divides the input electrical signal. In practical applications, the MCU module 3 has preset limit values. When the voltage across resistor R3 measured by the current processing module 4 is greater than the limit value, the voltage divider unit 42 divides the measured voltage across resistor R3; when the voltage across resistor R3 measured by the current processing module is less than the limit value, the voltage amplification unit 41 amplifies the measured voltage across resistor R3. Thus, by setting the voltage amplification unit 41 and the voltage divider unit 42, different input voltage ranges can be satisfied. It should be noted that the limit values ​​are determined by the range of the voltage amplification unit 41 and the voltage divider unit 42, and are not limited here.

[0043] Furthermore, the voltage amplification unit 41 includes a voltage amplifier U10, an input filter component 411, and an output filter component 412. One end of the input filter component 411 is connected to the connection point between resistor R3 and the second control unit 6, and the other end is connected to the positive input terminal of amplifier U10. The negative input terminal of amplifier U10 is grounded. One end of the output filter component 412 is connected to the output terminal of amplifier U10, and the other end is connected to the MCU module 3.

[0044] When the voltage across resistor R3 collected by the current processing unit is less than the limit value, the voltage across resistor R3 is filtered by the input filter component 411 and then input to the positive input terminal of voltage amplifier U10. Voltage amplifier U10 amplifies the electrical signal output by input filter component 411 and outputs it. Then, output filter component 412 filters the output signal of voltage amplifier U10 and outputs it to MCU module 3.

[0045] In this example, the input filter component 411 includes a resistor R166 and a capacitor C78. Resistor R166 is connected to the positive input terminal of voltage amplifier U10 and is used for current limiting. One end of capacitor C78 is connected to both resistor R166 and the positive input terminal of amplifier U10, and the other end is grounded. Capacitor C78 is used for filtering. The output filter component 412 includes a resistor R167 and a capacitor C72. One end of resistor R167 is connected to the output terminal of amplifier U10, and the other end is connected to MCU module 3. Resistor R167 is used for current limiting. One end of capacitor C72 is connected to both resistor R167 and MCU module 3, and the other end is grounded. Capacitor C72 is used for filtering. The voltage amplification unit 41 also includes a resistor R170, which is connected between the negative input terminal and the output terminal of amplifier U10 and is used for negative feedback of amplifier U10.

[0046] Voltage divider unit 42 includes resistors R20 and R21, and capacitor C18. One end of resistor R20 is connected to the small current sensing resistor R3, and the other end is connected to the MCU module 3. Resistor R20 is used for current limiting. One end of resistor R21 is connected to both resistor R20 and the MCU module 3, and the other end is grounded. Resistor R21 is used for voltage division. One end of capacitor C18 is connected to both resistors R20 and R21, and the MCU module 3, and the other end is grounded. Capacitor C18 is used for filtering.

[0047] Example 2 Please refer to Figure 1 and Figure 7 , Figure 7 This is a flowchart illustrating a method for calculating capacitive load current. This embodiment provides a method for calculating capacitive load current, applied to a capacitive load current acquisition system, which includes the following steps: S01, MCU module 3 controls the second control unit 6 to disconnect and the first control unit 5 and the third control unit 7 to close; S02, MCU module 3 controls the output current of battery 1 ; S03, AFE module 2 collects current through acquisition resistor R2. Meanwhile, MCU module 3 calculates the capacitance value of the equivalent capacitance C1; S04, MCU module 3 determines whether AFE module 2 has acquired the current. If yes, then execute S5a; otherwise, execute S5b. S5a, MCU module 3 receives the electrical signal sent by AFE module 2; S5b, MCU module 3 according to the formula: Calculate current The value of .

[0048] When the capacitive load current acquisition system is turned on, the MCU module 3 controls the third control unit 7 and the first control unit 5 to close, and the MCU controls the second control unit 6 to open.

[0049] In step S02, MCU module 3 controls the output current of battery 1. This allows the electrical signal output from battery 1 to charge capacitor C1 through resistors R3 and R4.

[0050] In step S03, AFE module 2 detects the voltage across resistor R2 and obtains the current based on the voltage across resistor R2. Then the current The signal is sent to MCU module 3. At the same time, current processing module 4 acquires the voltage across resistor R3 and amplifies or divides the acquired voltage. Then, current processing module 4 sends the processed electrical signal to MCU module 3. MCU module 3 calculates the capacitance value of capacitor C1 based on the electrical signal sent by current processing module 4.

[0051] Specifically, calculating the equivalent capacitance C1 involves the following steps: S31. The current processing unit collects the voltage across resistor R3, amplifies or reduces the voltage across resistor R3, and then sends the processed electrical signal to MCU module 3. S32. MCU module 3 records the electrical signals sent by the current processing unit at times t0, t1, and t2, respectively, and denotes them as follows: , and , where t0 is the initial time, and t0 < t1 < t2; S33.MCU module 3 will , and Substitute the value into the following formula: .

[0052] The equivalent capacitance C1 can then be calculated using the above formula. The derivation of this formula has been explained in Example 1 and will not be repeated here.

[0053] Further, in step S04, MCU module 3 determines whether AFE module 2 has acquired current. If AFE module 2 collects the current If the AFE module 2 does not collect current, then proceed to step S5a; Then, step S5b is executed. In step S04, MCU module 3 determines whether AFE module 2 has acquired current by judging whether the electrical signal sent by AFE module 2 is greater than a preset value a. In practical circuits, the resistance of resistor R2 is usually set below 1 milliohm. When the current... When the value is less than a certain value, the current The inability to generate a voltage drop across resistor R2 prevents AFE module 2 from detecting the voltage across R2, thus preventing AFE module 2 from acquiring current. However, since noise signals may exist in the circuit, AFE module 2 may detect noise and generate an electrical signal. Therefore, a preset value 'a' is set in the MCU module. When the MCU module receives the electrical signal sent by AFE module 2, it compares the signal with the preset value 'a'. If the signal sent by AFE module 2 is greater than the preset value, it is determined that AFE module 2 has detected current. When the electrical signal sent by AF module 2 is less than or equal to a preset value, it is determined that the electrical signal detected by AFE module 2 is a noise signal of the circuit, and AFE module 2 does not detect current. ...

[0054] If MCU module 3 determines that AFE module 2 has collected the current... MCU module 3 receives electrical signals sent by AFE module 2 to determine the current. Furthermore, MCU module 3 controls the second control unit 6 and the third control unit 7 to remain closed, and MCU module 3 controls the first control unit 5 to remain open. MCU module 3 determines that AFE module 2 has not collected current. .

[0055] If MCU module 3 determines that AFE module 2 has not collected current... MCU module 3 controls the second control unit 6 to close, the first control unit 5 to open, and the third control unit 7 to close; then, MCU module 3, through current processing module 4, acquires the voltage across resistor R3 at time t3, and records the voltage across resistor R3 at time t3 as... And (i.e., t3 = τ / N, N ≥ 10). Then MCU module 3 according to Given the capacitance value of the equivalent capacitance C1, substitute it into the following formula to calculate the current. The value of is given by the following formula: .

[0056] The method for calculating the capacitive load current in this example also includes step S07, when MCU module 3 completes the current calculation. After calculation, MCU module 3 controls the first control unit 5 to close, allowing AFE module 2 to detect the current in the circuit through resistor R2. MCU module 3 then controls the second control unit 6 to open and the third control unit 7 to close.

[0057] In summary, MCU module 3 calculates the capacitance value of the equivalent capacitance C1, and thus determines that AFE module 2 cannot detect current through resistor R2. At time t3, the MCU module samples the voltage across resistor R3. Then according to the formula Calculate the system current The value. Therefore, even when AFE module 2 cannot collect the system current. When the value is specified, the system current can also be accurately calculated through MCU module 3. The value of allows the system to detect both large and small currents simultaneously, effectively improving the accuracy of battery SOC estimation.

[0058] The above description is merely an embodiment of the present invention and is not intended to limit the invention. Various modifications and variations can be made to the present invention by those skilled in the art. Any modifications, equivalent substitutions, improvements, etc., made within the spirit and principle of the present invention should be included within the scope of the claims of the present invention.

Claims

1. A capacitive load current acquisition system, characterized in that, include: Battery (1), AFE module (2), MCU module (3), current processing module (4), equivalent resistor R1, equivalent capacitor C1, resistor R3, resistor R4, first control unit (5), second control unit (6), third control unit (7), and resistor R2; the negative terminal of the battery (1) is grounded, the positive terminal of the battery (1) is connected to one end of the equivalent resistor R1, the equivalent capacitor C1 is connected in parallel with the equivalent resistor R1, the other end of the equivalent resistor R1 is connected to the third control unit (7), the other end of the third control unit (7) is connected to the first control unit (5), the other end of the first control unit (5) is connected to the resistor R2, the other end of the resistor R2 is connected to the negative terminal of the battery (1), and the resistor R2 is electrically connected to the AFE module (2); The AFE module (2) is electrically connected to the MCU module (3) and the battery (1) respectively; one end of the resistor R4 is connected to the first control unit (5) and the third control unit (7) respectively, and the other end is connected to the second control unit (6). The other end of the second control unit (6) is connected to the resistor R3 and the input terminal of the current processing module (4) respectively. The other end of the resistor R3 is connected to the resistor R2. The output terminal of the current processing module (4) is connected to the MCU module (3); the MCU module (3) is electrically connected to the first control unit (5), the second control unit (6) and the third control unit (7) respectively; the resistance value of the resistor R2 is less than one-tenth of the resistance values ​​of the resistors R3 and R4; the AFE module collects the system current through the resistor R2. When the AFE module is unable to collect system current At that time, the MCU module (3) according to the formula Calculate current ,in for The voltage across resistor R3 at any given time.

2. The capacitive load current acquisition system according to claim 1, characterized in that, The current processing module (4) includes a voltage amplification unit (41) and a voltage divider unit (42). The input terminal of the voltage amplification unit (41) is connected to the connection node of the resistor R3 and the second control unit (6), and its output terminal is connected to the MCU module (3). The input terminal of the voltage divider unit (42) is connected to the connection node of the resistor R3 and the second control unit (6).

3. The capacitive load current acquisition system according to claim 2, characterized in that, The voltage amplification unit (41) includes an amplifier U10, an input filter component (411), and an output filter component (412). One end of the input filter component (411) is connected to the connection node of the resistor R3 and the second control unit (6), and the other end is connected to the positive input terminal of the amplifier U10. The negative input terminal of the amplifier U10 is grounded. One end of the output filter component (412) is connected to the output terminal of the amplifier U10, and the other end is connected to the MCU module (3).

4. The capacitive load current acquisition system according to claim 3, characterized in that, The voltage divider unit (42) includes resistor R20, resistor R21 and capacitor C18. One end of resistor R20 is connected to resistor R3 and the second control unit (6) respectively, and the other end is connected to the MCU module (3). One end of resistor R21 is connected to resistor R20 and the MCU module (3) respectively, and the other end is grounded. One end of capacitor C18 is connected to resistor R20, resistor R21 and MCU module (3) respectively, and the other end is grounded.

5. A method for calculating capacitive load current, characterized in that, The capacitive load current acquisition system according to any one of claims 1-4 includes the following steps: S01, the MCU module (3) controls the second control unit (6) to disconnect and the first control unit (5) and the third control unit (7) to close; S02, the MCU module (3) controls the battery (1) to output current. ; S03, the AFE module (2) collects current through resistor R2. Meanwhile, the MCU module (3) calculates the equivalent capacitance. The value; S4, the MCU module (3) determines whether the AFE module (2) has acquired the current. If yes, then execute S5a; otherwise, execute S5b. S5a, the MCU module (3) receives the electrical signal sent by the AFE module (2); S5b, the MCU module (3) is based on the formula: Calculate current The value of .

6. The method for calculating capacitive load current according to claim 5, characterized in that, In step S5a, the MCU module (3) determines the current acquired by the AFE module (2). At that time, the MCU module (3) controls the second control unit (6) to remain disconnected, and the first control unit (5) and the third control unit (7) to remain closed.

7. The method for calculating capacitive load current according to claim 5, characterized in that, Step S5b includes the following sub-steps: S5b1. The MCU module (3) controls the second control unit (6) to close, the first control unit (5) to open, and the third control unit (7) to close; S5b2. The MCU module (3) processes the current through the current processing module (4) in... The voltage across resistor R3 is constantly measured and then... The voltage across resistor R3 at time is denoted as ,and =τ / N, N≥10, where τ is the charging time constant; S5b3. The MCU module (3) is based on the formula: Calculate current The value of .

8. The method for calculating capacitive load current according to claim 5, characterized in that, Calculating the equivalent capacitance C1 involves the following steps: S31. The current processing module (4) collects the voltage across the resistor R3 and amplifies or reduces the voltage across the resistor R3. S32. The MCU module (3) in time, Time and The voltage across resistor R3 is collected at specific times and recorded as follows: , and ; S33. The MCU module (3) will , and Substitute into the formula: The equivalent capacitance was calculated. The value of .

9. The method for calculating capacitive load current according to claim 5, characterized in that, It also includes S06. The MCU module (3) controls the first control unit (5) to close, the second control unit (6) to open and the third control unit (7) to close.

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