I'm trying to figure out how does this DC motor controller circuit works. The circuit originally controls 2 DC motors for lifting a table. I extracted section relating to one motor only.

I added my own analysis with red text of what I think the circuit is doing.

My Questions

1- Is it a feedback loop circuit that I have identified? if yes, how does it work? any references?

2- Why is the PWM signal connected to multiple transistors and MOSFET just to drive IOUT to ground ?

enter image description here

  • 1
    \$\begingroup\$ Just a comment on schematics. Earth/ground symbols should always point downwards. The symbol represent metal plates buried in the ground. Similarly your positive supply should be at the top. That way current flows from top to bottom. Invert the section around Q202 and Q203 and you'll be looking good. Your contacts of K201 are touching each other making it difficult to see what's going on. Join the net CT00B as it simplifies reading the schematic and avoids searching for the other end. There seems to be no ground on the contacts of K201 so it's difficult to see how that could drive the motor. \$\endgroup\$
    – Transistor
    Jun 28 '20 at 18:02
  1. What you call the feedback loop is nothing more than op-amp measuring motor current over the shunt resistor and amplifying the measured voltage to a suitable level so that the MCU can read it with ADC input. Most likely a safety feature to cut power to the motor if hits the end of the movement or there is a human body part blocking the movement.

  2. The PWM output from the MCU is a 3.3V signal, and has current drive ability in the order of 10-20 mA. That is not enough voltage to turn the FET fully on, and the FET gate needs to be driven with higher current to turn it on and off faster. Therefore the PWM signal is amplified to 12V level and buffered with a push-pull stage so that the FET gate is driven with 12V signal that has fast edges.

  • \$\begingroup\$ Thank you! Makes sense. I'v read on TI datasheet something similar "push-pull output stage, which is unique for high-voltage comparators, offers the advantage of allowing the output to actively drive the load to either supply rail with a fast edge rate. This is especially valuable in applications where a MOSFET gate needs to be driven high or low quickly in order to connect or disconnect a host from an unexpected high voltage supply" \$\endgroup\$
    – Ghaith
    Jun 28 '20 at 22:00

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