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Suppose we are controlling a dc motor via pwm. he motor has a flyback diode across it.


Question : During pwm off period, the motor current freewheels through diode. Won't the motor experience a braking torque then. Is it desirable for efficiently controlling the motor?enter image description here

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  • \$\begingroup\$ Unless you reverse the motor, the back EMF keeps the diode reverse biased. \$\endgroup\$ Apr 20 '20 at 18:52
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    \$\begingroup\$ @BrianDrummond - no - when the transistor turns off the inductance results in current flowing forward through the diode. \$\endgroup\$ Apr 20 '20 at 22:44
  • \$\begingroup\$ @KevinWhite Not quite. You are correct there will be a brief conduction spike but that's only from the motor's inductance. The braking in the question is from the back EMF which has the same polarity as the applied EMF keeping the diode reverse biased : therefore no braking as per your answer. Two separate phenomena, but often confused with each other. \$\endgroup\$ Apr 21 '20 at 11:46
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    \$\begingroup\$ @BrianDrummond - In a normal PWM motor control system the frequency of the PWM is such that the diode conduction spike is the full length of the off-cycle so the back-emf never keeps the diode reverse biased as the next on-cycle will drive forward current into the motor and reverse bias the diode again before that happens. Although the effect you're describing would happen at a low-frequency PWM, that is not the normal scenario. \$\endgroup\$ Apr 21 '20 at 14:25
  • \$\begingroup\$ @KevinWhite Fair point. There is no mention of PWM freq in the question so ... \$\endgroup\$ Apr 21 '20 at 14:49
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During pwm off period, the motor current freewheels through diode. Won't the motor experience a braking torque then. Is it desirable for efficiently controlling the motor?

Firstly, Any PWM signal frequency should have a time period that is many times shorter than the physical response of the motor due to it's mechanical inertia.

And, as Brian Drummond reminded me, the diode will only conduct due to the initial back emf from the inductance of the motor. After this has settled-down, if the motor is continuing to free-wheel in the same direction, the diode won't remain forward biased. However, the motor free-wheeling will generate a voltage and so, it might be advisable to put a diode across the BJT - anode to ground - to prevent any excessive motor free-wheeling from reverse biasing the BJT to any great extent.

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  • \$\begingroup\$ The above question was derived from this question, could you take a look at it? electronics.stackexchange.com/questions/494442/… \$\endgroup\$
    – ASWIN VENU
    Apr 20 '20 at 15:47
  • \$\begingroup\$ I don't use simulink or matlab. \$\endgroup\$
    – Andy aka
    Apr 20 '20 at 15:50
  • \$\begingroup\$ I could use your insight as to why any simulation model might behave like that \$\endgroup\$
    – ASWIN VENU
    Apr 20 '20 at 15:52
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    \$\begingroup\$ @Andyaka I think this explanation is off here : the back EMF does not put the diode into conduction (though the inductive kick does, briefly). As a sanity check, consider that a short circuit would provide the hardest braking force : a series resistor would only reduce it (and take dissipation out of the motor) \$\endgroup\$ Apr 21 '20 at 11:52
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    \$\begingroup\$ For PWM to be efficient the on-time must be much shorter than the L/R time constant or most of the power will be dissipated in the motor resistance. The off time will be the same as the on-time @50% duty cycle so it will also be << L/R. The inductive spike will be much longer than the off-time. Low-speed PWM is no different than using a resistor to drop the voltage. \$\endgroup\$ Apr 23 '20 at 15:55
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The current through the motor is in the same direction during the PWM off period as during the on period - the voltage reverses but not the current.

So there is no braking torque.

It is a very efficient way of driving a motor - virtually all PWM motor driver use this arrangement or versions of it.

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