How does this motor drive circuit work


The Drive terminals are connected to a PWM signal generated by an arduino micro controller. The purpose of this circuit is to isolate the arduino from the motors to avoid too much current being drawn.

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This is my best guess, but I'm completely unsure: When the NPN BJT is in forward active mode (i.e when the LED is turned on), current will flow from the +6V rail to the -6V rail. There will be a potential difference across the 1K resistor and therefore a potential difference between the gate and source of the n channel FET. The FET is now in conducting mode, which will then reverse bias the diode and turn on the motor.

When the NPN BJT is in cut-off (i.e when the LED is turned off), the FET will act like an open circuit, so no power goes through the motor. Under no circumstance is the diode ever forward biased.

Is this correct?

And what is the capacitor and Rx for?

Best Answer

The FET controls the motor. The diode is not part of the control as such. The diode is there to kill off any voltage spikes that occur when the FET opens and the motor is turned off. The diode will only be forward biased when the motor shuts off (the FET opens.) The coils in the motor can generate a pulse higher than the supply voltage. The diode would short out any such pulse and prevent it killing other parts of the circuit.

The capacitor is there to filter out the RF garbage that the motor puts out - the sparking you can see on a DC motor causes RF interference. The capacitor absorbs that and makes the motor much quieter (RF wise, at least.)

I would assume that Rx is planned for use when a different opto-isolator is used - an isolator in which (for what ever reason) the base of the BJT floats up and could cause spurious operation of the motor.