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The setup is as follows:
DC Motor: Brushed DC Motor
PNP NPN standard H Bridge with 4 numbers of 1N4448 for flyback protection.
Stall current of the Motor: 200 mA (Stalling is expected)
Typical load current is 80 mA.
Motor cannot run freely unless driven.
DC resistance of Motor in OFF condition is 13 Ohms.
Voltage applied across Motor varies from 2.2 to 3 V depending on battery voltage.

How to calculate the time taken to discharge the current through the flyback diodes and DC motor when Motor is turned OFF suddenly?
The inductance of Motor Measured using Agilent LCR meter is about 8.12 mH.

Edit: Purpose of this is to have an idea on the delay to be introduced before driving the motor in other direction, if it makes sense

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  • \$\begingroup\$ You can't, from the information given. It depends on the inductance. And back-EMF will continue as long as the motor spins (assuming a PM motor). \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented May 25, 2015 at 7:06
  • \$\begingroup\$ I got inductance as 8.12 mH \$\endgroup\$
    – User323693
    Commented May 25, 2015 at 7:58

2 Answers 2

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Assuming the motor has stalled (you seem to imply that), so there is no voltage generated by the motor spinning, and you then turn it off, it becomes just an L/R time constant. Since a charged inductor tries to keep the current flowing, the 0.6V drop across each flyback diode can be ignored, and the current decays with a time constant of L / R i.e. 8.12mH / 13 Ohms = 0.625mS.

The time constant is how long it takes to reach about 37% (e^-1) of the initial value. After a second time constant it will decay to about 37% of that, or 13% of the original (e^-2), and so on. (It will never completely discharge). (e is the base of natural logarithms, 2.71828).

So decide what fraction of the original current you would consider the discharge to have finished, and solve for e^(-t/T) = "your fraction", (where T is the time constant L / R). It would be easier to simplify this to t=-(L/R) x ln("your fraction") (ln is natural log, or log to base e)

If you say the load current at stall was 200mA when it was turned off, and you decide that when it reaches one percent of that (2mA), then it is considered "discharged", then the time will be -(0.625mS) x ln (1%) = 2.87mS

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  • \$\begingroup\$ Thank you. How that will be in case the motor was running with normal load (70 mA) and I stop driving it. It may free wheel a little (shaft will be still connected to Gear assembly and hence cannot free wheel significantly). Will the calculation still be L/R times 5 for 99% discharge? \$\endgroup\$
    – User323693
    Commented May 25, 2015 at 10:01
  • \$\begingroup\$ ln (0.01) will always be -4.605, so yes. However: how long will the motor spin for? My guess is much much more than 2.8mS, so in the case when the motor is still spinning, the inducive decay can very likely be ignored - the current will continue to flow until 2.8mS after the rotor stops sinning (ha ha) - spinning. Is that even relevant? \$\endgroup\$
    – Reversed Engineer
    Commented May 25, 2015 at 11:00
  • \$\begingroup\$ What will be the voltage across the inductor immediately after turning off the motor? will it be 13 Ohms x 0.2 A = 2.6 V? I forgot the reference where I read about this \$\endgroup\$
    – User323693
    Commented May 25, 2015 at 13:13
  • \$\begingroup\$ V = I x R (plus the voltage drops across the diodes which are in the circuit - about 0.7V each) \$\endgroup\$
    – Reversed Engineer
    Commented May 26, 2015 at 10:07
  • \$\begingroup\$ Guys, i don't think l/r time constant applies here. The equivalent circuit is absolutely not just inductor and resistor.. \$\endgroup\$
    – user76844
    Commented May 28, 2015 at 17:21
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Sorry for not reading everything.

If the motor is spinning, it generates voltage, BEMF, that is Kv * Velocity. The energy is not only stored in the iinductive winding, but also in momentum of inertia of the rotor and whatever it drives. And that thing may be discharged too, if certain requirements are met. It the goal is stopping the motor, easiest way for such small motor is just to short the winding with the two low side transistors (if they are mosfets) . It's called dynamic brake. If the energy there is more than your bridgr can stand, you can dissipate it on a power resistor. For that you need a switch to disconnect the bridge from the bus voltage and capacitors, and then turn on a resistor, that is connected to bridge's + and -.

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