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Where will the current flow through?

If ON current(back emf "current") will loop through A B and C

If OFF current will(back emf "current") loop through x y and z

  • \$\begingroup\$ Already answered here \$\endgroup\$ – Biduleohm Oct 25 '14 at 20:39
  • \$\begingroup\$ can you tell me he CURRENT flow not the voltage? where CURRENT loops? \$\endgroup\$ – user16307 Oct 25 '14 at 20:40
  • \$\begingroup\$ ok thnks am gonna read it now \$\endgroup\$ – user16307 Oct 25 '14 at 20:42
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    \$\begingroup\$ Ref to your diag on imgur when Q1 is ON conventional current flow is 'downwards' thru M and Q1. D1 is reverse biased. When Q1 is OFF, the back-emf acts in such a way as to try to continue the current flow in the same direction = 'downwards'. This means the lower terminal of M is positive wrt to the top and D1 is forward biased and conducts. \$\endgroup\$ – peterG Oct 25 '14 at 21:14
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    \$\begingroup\$ but wait.. in OFF mode the motor is rotating in the same direction why would it generate an inverse voltage of ON mode? \$\endgroup\$ – user16307 Oct 25 '14 at 21:20

Where you may be confused is that there are actually two back-emfs - the voltage generated by the motor when it is spinning, and the voltage caused by inductance when the PWM switch opens.

A DC motor can be thought of as consisting of a DC generator whose voltage is proportional to rpm, a resistor which represents the resistance of the windings and brushes, and an inductor which represents the armature inductance.

Consider the equivalent circuit below. When the motor is spinning it generates a DC voltage in opposition to the supply voltage. The difference between the supply voltage and generated voltage is impressed across the motor's internal resistance, which determines how much current it draws. As the motor speeds up the voltage difference gets less so current draw reduces, until it is receiving just enough current to maintain a constant rpm. If 50% PWM is applied then the motor 'sees' 50% of the supply voltage on average, so it drops to half speed and the generated voltage is nearly half the supply voltage (slightly less due to voltage lost in the resistor).

During PWM ON time the motor receives full supply voltage, so you might expect the instantaneous current to rise dramatically. However this doesn't happen immediately because the inductance apposes any current increase, generating a back-emf voltage which adds to the generator voltage (polarity is + at the top of the inductor and - at the bottom).

During PWM OFF time the switch is open, so motor current cannot get back to the supply. However once again the inductor opposes current change, this time generating a back-emf voltage in the opposite direction (with polarity as shown by the + and - symbols in the diagram) with sufficient amplitude to maintain current flow. At this point the flyback diode becomes forward biased and motor current flows through it.

The intention is to maintain a constant current through the motor even though it is being continuously switched on and off. In practice the inductor can only slow down current change, not completely eliminate it. However if the PWM frequency is high enough then the current ripple will become a triangle wave of low amplitude.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ lets focus only on OFF mode here. in your figure the current generated by L1 passes through the diode i understand that and the theory says V=-L(di/dt), inductor polarity will change during the transition. BUT are you sure about the current generated by the Vgen in your figure will also flow through the diode? since the motor rotates in the same direction always why would its polarity change? \$\endgroup\$ – user16307 Oct 26 '14 at 10:44
  • \$\begingroup\$ besides i observed with scope the same polarity in OFF and ON modes. check out my video where i play with pwm and see voltage across the motor during ON and OFF modes:youtube.com/watch?v=1oS1zyJJTWA&feature=youtu.be what do you say?... see also Biduleohm's comments here: electronics.stackexchange.com/questions/135827/… \$\endgroup\$ – user16307 Oct 26 '14 at 10:45
  • \$\begingroup\$ I assume a motor with no inductance and Bruce's answer is right for a real motor. \$\endgroup\$ – Biduleohm Oct 26 '14 at 11:43
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    \$\begingroup\$ Vgen doesn't change polarity. The voltage generated by L1 is opposite to and greater than Vgen - just enough greater to turn on the diode (in effect, L1 is now the power supply). If the diode wasn't there then the voltage would keep increasing until it reached the transistor's breakdown voltage (~60V). This is an equivalent circuit - you cannot directly measure the voltage across Vgen or L1. However if you measure across the motor terminals you will see the total voltage reverse polarity (going from -6V to about +0.4V with the diode, or +60V without the diode) during PWM OFF time. \$\endgroup\$ – Bruce Abbott Oct 26 '14 at 11:48
  • \$\begingroup\$ is that right what i understood from your words; in OFF mode the diode is for an instant forward biased with a much higher voltage than Vgen. so the diode turns to a conductor pn junction is penetrated. and Vgen's current flows through the diode as a short, but this current is now in opposite direction of the diode's usual direction. is that true? could u tell me the direction/loop path of Vgen's current after the diode is opened by inductor's self emf for an instant? \$\endgroup\$ – user16307 Oct 26 '14 at 12:33

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