STDA040 August   2026 BQ76907-Q1

 

  1.   1
  2.   Abstract
  3.   Trademarks
  4. 1Introduction
  5. 2E-latch System Architecture
  6. 3Typical Design
    1. 3.1 Motor Drivers
    2. 3.2 Network Connection
    3. 3.3 Microcontroller
    4. 3.4 Backup Power
      1. 3.4.1 Discussion of Energy Requirements
        1. 3.4.1.1 Energy Needed By A Single Motor
        2. 3.4.1.2 Requirements for Multiple Motors
        3. 3.4.1.3 Usable energy stored
        4. 3.4.1.4 Derating the Capacitance
      2. 3.4.2 Charging Circuit
      3. 3.4.3 Discharging Circuit
      4. 3.4.4 Bidirectional Charge and Discharge Circuits
      5. 3.4.5 Cell Balancing Circuit
    5. 3.5 Position Sensors
  7. 4Future Trends and Innovations
  8. 5Conclusion
  9. 6References

3.4.1.1 Energy Needed By A Single Motor

To design the backup power circuits, we need to look at the voltage and current requirements of the motors and other loads to be driven. The motors in a lock mechanism are relatively small, typically 12V-rated brushed DC motors, with maximum current of a few Amps. Operation time is short, typically significantly less than one second. The total energy necessary to actuate each motor is:

Equation 1. E t o t a l = ∫ v t ×   i t   d t

Where v(t) is the voltage supplied to the motor, and i(t) is the motor current, and the integration is over the total time of motor activation.

As an example, Figure 3-7 shows the current profile of a lock motor activation, with a 12V supply and a 500mΩ sense resistor, giving a scale factor of 2A/V. The profile can be divided into two segments, with the first (in yellow) showing the initial high current as voltage is applied to the motor, decreasing as the motor speed and back EMF increase. The second segment (in red) is when the mechanism reaches a mechanical limit, and the motor stalls after an initial bounce. Not shown is the deactivation of the motor when the control circuit recognizes the lock has reached the limit and the drive is shut off. Note that in this case, the motor is rotating for only about 25ms, and then stalls for at least 55ms. The stall current is about 2.4A.

BQ76907-Q1 Current Profile of a Lock Motor Activation Figure 3-7 Current Profile of a Lock Motor Activation

For this example, the energy dissipated during an unlock event can be approximated as the sum of the energy during motor rotation, and the energy during motor stall. The energy during the motor rotation is:

Equation 2. E r o t a t i o n =   12 V × 24 A +   0.8 A 2 ×   25 m s =   0.48 J

The energy during motor stall depends on how quickly the motor is deactivated after reaching the mechanical limit; for this example assume the motor is turned off at the end of the oscilloscope trace about 55ms after the motor stops turning. This gives a stall energy of:

Equation 3. E s t a l l =   12 V ×   24 A ×   55 m s =   1.58 J

These sum to give a total energy needed for this example of around 2 Joules. Note that the duration of the motor stall phase significantly influences the total energy needed. Thus the time for detection of end of travel and subsequent disabling the motor drive can strongly affect the design.