SDAA170 September   2026 TPS61383-Q1

 

  1.   1
  2.   Abstract
  3.   Trademarks
  4. 1Introduction
  5. 2Design Requirement and System Structure
    1. 2.1 Design Requirements
    2. 2.2 System Structure
      1. 2.2.1 Centralized E‑Latch System
      2. 2.2.2 Distributed E‑Latch System
  6. 3Power Train
    1. 3.1 Integrated E-Latch Power Train with TPS61383-Q1
      1. 3.1.1 Super Capacitor Charger Function Introduction
      2. 3.1.2 Boost Function Introduction
      3. 3.1.3 Super Capacitor Bleeding
    2. 3.2 Hardware Design Procedure
      1. 3.2.1 Selecting the External MOSFET
      2. 3.2.2 Inductor Selection
      3. 3.2.3 Capacitor in Super Capacitor Side
      4. 3.2.4 Boost Output Capacitor
      5. 3.2.5 Loop Compensation Design
      6. 3.2.6 Key Considerations
    3. 3.3 Software Initial Process
  7. 4Super Capacitor Energy Modeling Basis and Selection
    1. 4.1 Discharge Profile and Cut-off Voltage
    2. 4.2 Super Capacitor Discharge Energy Modeling
    3. 4.3 Calculation Example
    4. 4.4 Super Capacitor Excel Calculation Tool
  8. 5Super Capacitor Management
    1. 5.1 Voltage Monitoring with BQ76907-Q1
    2. 5.2 Cell Balancing with BQ76907-Q1
      1. 5.2.1 Balancing Strategy
      2. 5.2.2 I2C Implementation
      3. 5.2.3 Cell Balancing Current
      4. 5.2.4 Low-Side Protection Circuit Design Considerations
    3. 5.3 Super Capacitor Health Detection: Combined Solution of TPS61383-Q1 and BQ76907-Q1
  9. 6Control Unit and Communication
    1. 6.1 Control Unit Design Consideration
    2. 6.2 SBC Selection
  10. 7Execution Unit
    1. 7.1 Motor Driver Structure
    2. 7.2 Motor Driver Design Considerations
  11. 8Summary
  12. 9References

4.2 Super Capacitor Discharge Energy Modeling

After calculating cut-off voltage, it is still hard to calculate super capacitor discharge energy correctly. Super capacitor internal resistance is usually 5-10 times larger than the IC MOSFET Rds-on. So total discharge efficiency becomes much lower than boost voltage, making the calculation based on E=CV2/2 far from accurate. This section explains the basic method to calculate super capacitor energy correctly.

During the entire scap discharge, the loss on internal resistance can be given by Equation 15

Equation 15. l o s s R S C A P = ∫ 0 T I s c a p 2 R S C A P d t
Where:

  • RSCAP is the internal resistance of super capacitor
  • T is the duration of load
  • ISCAP is the discharge current on super capacitor

For ISCAP , use Equation 16:

Equation 16. I s c a p = - V o u t I o u t V I N η B S T = V o u t I o u t V S C A P - I S C A P R S C A P η B S T
Obtain ISCAP as :
Equation 17. I S C A P = V S C A P - - 4 V o u t I o u t R S C A P - V S C A P 2 η B S T η B S T 2 R S C A P
In the previous equation, Vout, Iout, RSCAP and ηBST are constant parameters, while VSCAP is a unknown function of t to be solved. Based on the equivalent circuit, build the following differential equation:
Equation 18. C S C A P d V S C A P d t = - I S C A P = - V S C A P - - 4 V o u t I o u t R S C A P - V S C A P 2 η B S T η B S T 2 R S C A P
The differential equation is separable, so the user can obtain a general solution in implicit form.
Equation 19. V S C A P 2 2 b + V S C A P V S C A P 2 - b 2 b - 1 2 l n V S C A P + V S C A P 2 - b = - t 2 R S C A P C S C A P + C 1
Where
Equation 20. b = 4 V o u t I o u t R S C A P η B S T
The now we have Vscap, and Iscap can be solved by numerical approach. But since it’s not clear enough to calculate loss by numerical methods. Do not use the implicit form to calculate ISCAP and loss. Referring back to Equation 15, substitute Equation 17 and Equation 18 into Equation 15:

Equation 21. l o s s R S C A P = ∫ 0 T I S C A P 2 R S C A P d t = ∫ 0 T - C S C A P d V S C A P d t V S C A P - - 4 V o u t I o u t R S C A P - V S C A P 2 η B S T η B S T 2 R S C A P R S C A P d t = - C S C A P 2 ∫ 0 T V S C A P ' V S C A P - V S C A P 2 - 4 V o u t I o u t R S C A P η B S T d t = - C S C A P 2 ∫ 0 T V S C A P ' V S C A P - V S C A P ' V S C A P 2 - b d t

The integral is standard integral multiplied by differentiation. This type integral can easily be solved by substitution. Let u=VSCAP:

Equation 22. l o s s R S C A P = - C S C A P 2 ∫ 0 T u - u 2 - b d u = C S C A P 4 ( - V S C A P 2 + V S C A P V S C A P 2 - b - b ⋅ ln ⁡ ( V S C A P + V S C A P 2 - b ) ) | 0 T

The VSCAP voltage at t=0 and t=T are given VCH and VCL. So the definite integral can be calculated by:

Equation 23. l o s s R S C A P = C S C A P 4 [ V C H 2 - V C L 2 + V C L V C L 2 - b - V C H V C H 2 - b + b ( ln ⁡ ( V C H + V C H 2 - b ) - ln ⁡ ( V C L + V C L 2 - b ) ) ]

The VCH, VCL and b are known parameters while the only unknown parameter left is Cscap. So the loss can be given by:

Equation 24. l o s s R S C A P = k   C S C A P

Where k is:

Equation 25. k = [ V C H 2 - V C L 2 + V C L V C L 2 - b - V C H V C H 2 - b + b ( ln ⁡ ( V C H + V C H 2 - b ) - ln ⁡ ( V C L + V C L 2 - b ) ) ] / 4

So the minimum CSCAP can be calculated as:

Equation 26. C m i n = 2 V o u t I o u t T η B S T ( V C H 2 - V C L 2 ) ( 1 - l o s s R i n V o u t I o u t T + l o s s R i n ) = 2 V o u t I o u t T η B S T ( V C H 2 - V C L 2 ) ( 1 - k C m i n V o u t I o u t T + k C m i n )

So the final minimum Cscap can be solved as:

Equation 27. C m i n = 2 V o u t I o u t T η B S T ( V C H 2 - V C L 2 ) - 2 k

Where:

Equation 28. k = [ V C H 2 - V C L 2 + V C L V C L 2 - b - V C H V C H 2 - b + b ( ln ⁡ ( V C H + V C H 2 - b ) - ln ⁡ ( V C L + V C L 2 - b ) ) ] / 4
Equation 29. b = 4 V o u t I o u t R i n η