Other Parts Discussed in Thread: DRV8837, TPS62745
Hello,
We have a project which is based on reference design for smart lock TIDA-00757.
In order to get more flash memory and smaller space, we modify two blocks:
- Main processor uses MSP432+CC264R2 as SNP
- Motor driver uses DRV8837
We need help to understand more about the motor driver in relation with the supply side that uses 4 AA Alkaline batteries for our project.
Fresh batteries give us more than 6V and we drained it into 5.2V (each becomes 1.3V). When we try to turn on the motor at this 5.2V level, the board restarts itself.
Our DC motor has stall current around 2A and no-load current around 200mA on 3.5V.
Although DRV8837 is rated on 1.8A, we still can control the motor when batteries voltage is still fresh.
- We think that it is related to internal resistance of the battery. From open loop and close loop test with a resistor and voltmeter, we calculate the internal resistance of our battery at 0.8 Ohm on 1.3V while at fresh we have smaller resistance from 0.15 Ohm to 0.3 Ohm just like Alkaline Energizer E91 datasheet.
When the motor starts, it draws 2A current and the drop voltage on internal resistance becomes 4 * 0.8 Ohm * 2A = 6.4V and since the batteries itself already on 5.2V then the whole system voltage just drop below acceptable range and the microcontroller gets no power.
However at fresh batteries the drop voltage on internal resistance becomes 4 * 0.3 Ohm * 2A = 2.4V. With batteries voltage at more than 6V there is still available voltage of more than 3.6V. - On TI reference it shows the buck input accept range from 3.7V to 6V. We design our system to work on voltage range from 4.4V to 6V.
In case when the batteries drop at 5.2V, if we modify our motor to 1A stall current we would get a drop of 4 * 0.8 Ohm * 1A = 3.2 V and our available voltage from batteries 5.2V - 3.2V = 2V. This shall be the voltage that feeds in the buck.
However when batteries voltage reach 4.4V then internal resistance on each battery increments to around 1 Ohm or more from our test. It gives us drop voltage 4 * 1 Ohm * 1A = 4V. So available voltage at motor start from batteries 4.4V - 4V = 0.4V.
This might as well create same problem where microcontroller gets no power. - Above points focus on the stall current which will have effect on motor start and when motor stall. When motor already runs with a load and draws current for example 500mA.
At fresh batteries or 6V with each internal resistance 0.3 Ohm, the drop voltage would be 4 * 0.3 Ohm * 0.5A = 0.6V and actual output from batteries will be 6V - 0.6V = 5.4V.
If we drive the motor with 100% duty cycle (full on), then our motor speed would follow the torque-speed curve chart for 5.4V and not for 6V.
Questions:
Are these analysis correct? Would you please kindly advise if we have misunderstanding?
Kind regards,
Pranata