I was wondering if some one could help me understand what would happen to the motor if I suddenly change the rotation from clockwise to anticlockwise or vice verse in an H-bridge configuration, would the current flowing through double ? Would the voltage double across the motor ? enter image description here


3 Answers 3


Let's give some simple calculations based on some fictional values.

  • Motor resistance, Rm, = 1Ω.
  • Supply voltage, Vs, = 12 V.
  • With the applied load the motor runs up to the speed where the back EMF, Vb, is 11 V.

From these figures we can calculate:

  • Starting current is \$ \frac {V_s}{R} = \frac {12}{1} = 12 \ \text A \$.
  • Running current is \$ \frac {V_s - V_b}{R} = \frac {12 - 11}{1} = 1 \ \text A \$.

Now reverse the motor by reversing Vs.

  • Running current is \$ \frac {-V_s - V_b}{R} = \frac {-12 - 11}{1} = -23 \ \text A \$.

... if I suddenly change the rotation from clockwise to anticlockwise or vice verse in an H-bridge configuration, would the current flowing through double?

Worse. It would be almost double the starting current but in the opposite direction.

Would the voltage double across the motor?

No. It would just switch polarity. This assumes that the voltage supply is "stiff" and can absorb any current injected into it by the decelerating motor. Most practical systems will require some way of burning off the regenerated energy by switching in a brake resistor when the bus voltage gets too high. This will dissipate the energy as heat.

  • \$\begingroup\$ Does putting a 1000uf capacitor between the Vcc and ground help with anything ?? \$\endgroup\$ Commented Feb 2, 2019 at 22:32
  • \$\begingroup\$ Large capacitors on the supply rails will supply energy during peaks in demand and will sink some of your regenerated current. \$\endgroup\$
    – Transistor
    Commented Feb 2, 2019 at 22:39
  • \$\begingroup\$ If I’m running the motor from 36V and 60A I’m getting a back emf of 0V is that correct ?? \$\endgroup\$ Commented Feb 2, 2019 at 23:36
  • \$\begingroup\$ I think it's time for you to study some DC motor theory. If the motor is turning it will generate back EMF. If it is stalled it won't. \$\endgroup\$
    – Transistor
    Commented Feb 2, 2019 at 23:39
  • \$\begingroup\$ I just wanted to ask one more question , so if supply 60 A to the motor would that be the starting current or running current \$\endgroup\$ Commented Feb 3, 2019 at 0:19

Motors (when they are spinning) are kind of like capacitors - voltage is stored via back-emf as rotational inertia.

When you reverse the polarity on a running motor there's a surge of current (higher than normal stall current) until the motor is spinning in the right direction

if you put a capacitor next to the bridge it would increase the current available to the bridge, this would assist the power supply, but make life harder for the bridge itself and the motor (higher currents, higher mechanical stress)

Usually what is done before reversing polarity is to slow the motor by short-circuiting it (eg by turning on Q2 and Q4 at the same time) this can reduce the peak current to little more than the normal stall current.

  • \$\begingroup\$ It's really weird to see motors compared to capacitors instead of inductors, but it's true! And that makes it feel even weirder, I think. \$\endgroup\$
    – Hearth
    Commented Feb 2, 2019 at 20:05
  • \$\begingroup\$ Could you expand on this subject a little further please as my understanding of this subject is Breif \$\endgroup\$ Commented Feb 2, 2019 at 20:07
  • \$\begingroup\$ And also if I put a 1000uF capacitor between Vcc and ground would this help \$\endgroup\$ Commented Feb 2, 2019 at 20:11

DC motors are actually AC motors by self commutating DC or externally commutated as in a BLDC bridge with pole commutation.

In either case current starts with the DCR winding resistance drawing V/DCR = I current then reduces by a factor of 10(+/-20% depending on winding efficiency) up to rated speed and mechanical load with a coasting with no load voltage at some kV/RPM with a voltage available to generate with reverse mechanical load , the same for the applied voltage. However the no load current is not zero as there is often some 10% current needed for winding excitation for magnetic coupling and friction losses.

Thus reversing a motor is at full load speed of ~ 82% max no load speed is like starting from 0 RPM with 10x the current except your differential voltage is either 1.82 x Vcc from full load or 2x Vcc from no load full speed, thus Isurge is again due to the coil DCR resistance but now almost twice the voltage thus twice the current and twice the time or 4x the total surge energy to reverse to reverse full speed and thus 4x the temperature rise declining to rated power with zero acceleration for the given mechanical load.

However inertia plays a big role on energy consumption and heat losses from DCR and RdsOn. If you controlled a heavy rotating mass with a slow ramp in speed to limit the current, then suddenly reverse full voltage, you will likely see failures from energy losses and heat rise from now up to 20x the rated max current for the motor.

This explanation can be mathematically expressed if the inertia is defined with then all the motor ratings and actual mechanical load. You may compute thus by conservation of mechanical and electrical energies stored and applied against the load with some 10% losses for excitation inductive currents and a DCR about 10% of the rated V/I.


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