The DC Motor is question is here: DC Motor

I was wondering if you use a diode and a switch in a manner shown in the below picture if the motor would exhibit regenerative braking when SW1 is open? (Assuming that the voltage from the motor when braking exceeds the motor power) Is the use of PWM required?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Please explain the relation between your question (title) and the content. You talk about PWM, but in your detail there is no PWM - I guess that the switch is PWM controlled, but you should specify that. \$\endgroup\$ – le_top Sep 22 '19 at 15:04
  • \$\begingroup\$ Oh yeah, sorry I should add regenerative braking without using PWM since many sources say its require. \$\endgroup\$ – Jake Freeman Sep 22 '19 at 15:09
  • \$\begingroup\$ Your link to the Amazon product (code B00TE42PME ) is not working neither - I do not know if this is a limitation of the platform. \$\endgroup\$ – le_top Sep 22 '19 at 15:20
  • \$\begingroup\$ @ie_top I don't know why it works on my end \$\endgroup\$ – Jake Freeman Sep 22 '19 at 15:24
  • \$\begingroup\$ @jjake-freeman I get a certificate error - maybe it is related to the CDN network. \$\endgroup\$ – le_top Sep 22 '19 at 20:39

(Assuming that the voltage from the motor when braking exceeds the motor power)

That's the problem — it doesn't. You need a way to boost the voltage coming from the motor to a level that will actually charge the battery. You can use a separate boost converter, or you can create a more tightly integrated solution that uses the inductance of the motor itself as an element in a boost converter. Either way, it does involve some sort of PWM control in order to regulate the power flow.

  • \$\begingroup\$ So if I modify the diode branch to include a DC booster to the battery it would charge it correct? Thank you for the answer as well. \$\endgroup\$ – Jake Freeman Sep 22 '19 at 14:54
  • \$\begingroup\$ Yes. The diode by itself does nothing for you. \$\endgroup\$ – Dave Tweed Sep 22 '19 at 14:59
  • \$\begingroup\$ Ok thank you. With the Boost converter I still need the diode though, correct? \$\endgroup\$ – Jake Freeman Sep 22 '19 at 15:01
  • \$\begingroup\$ Well, in a way. A boost converter will generally contain at least one diode. \$\endgroup\$ – Dave Tweed Sep 22 '19 at 15:02

The voltage on the terminals of your motor will be a function of its speed and the load. It will not be higher than your power supply, unless some external force is trying to accellerate the motor.

Regenerative braking supposes that you get back some energy from the motor while braking. Therefore the braking must be performed by applying a load accross the motor. For example, you could "short-circuit" it using a low value/high power resistor. The resistor would force the motor to supply power and transform that power into heat. At the same time the motor slows down.

When we use a resistor to slow the motor down, the energy is lost into heat. We could use that heat and transform it into "electricity", but that is not the most effective way. It is better to change the resistor with a more complex system that would transform the power. It can for instance be an inductor. In way similar to switched power supplies we can "charge" the inductor and "discharge" it into your power source. It would be applied to your power source in a different path than the one controlling your motor through PWM.

So the diode accross the controlling switch would not do anything while slowing down your motor.

  • \$\begingroup\$ Ok thank you. Would the idea of using a booster work as mentioned by Dave Tweed \$\endgroup\$ – Jake Freeman Sep 22 '19 at 15:26
  • \$\begingroup\$ A separate boost converter used as a load is an interesting idea, but I can not guarantee it will work - I do not have the experience with that. \$\endgroup\$ – le_top Sep 22 '19 at 20:36
  • \$\begingroup\$ @Transistor Ok, I corrected it. \$\endgroup\$ – le_top Sep 23 '19 at 13:53
  • \$\begingroup\$ So it could charge something with a lower voltage e.g. power supply is 12V and it tries to charge a 3.7V battery? \$\endgroup\$ – Jake Freeman Sep 25 '19 at 23:27
  • \$\begingroup\$ If you look for MT3608 on the web, you'll easily find references to low cost implementations of a DC-DC booster or step-up. The MT3608 accepts between 2 and 24V on its input and can step it up to a regulated voltage of up to 28V. So if you have a free-running motor which generates a DC voltage on its terminals and you connect it to the input of the booster, the booster will load the motor (generator) as much as is needed to regulate its output. The output can be 12V but it can not be lower than the input voltage of the DC-DC (you could use a buck/boost for that). \$\endgroup\$ – le_top Sep 26 '19 at 12:03

Your circuit doesn't provide any braking because the motor is running free when the switch is open, and won't produce higher voltage than the battery unless it is 'over-driven' to higher speed by an external force.

To brake the motor you must put a switch across it, like this:-


simulate this circuit – Schematic created using CircuitLab

When SW2 is closed it 'shorts out' the motor. While the motor is spinning it acts as generator, producing voltage which pushes current through SW2. The current produces torque which brakes the motor. This is dynamic braking, but not regenerative.

However if PWM is applied to SW2 then each time it opens the collapsing magnetic field in the motor's winding inductance creates a 'back-emf' voltage which tries to keep the current going. The current then takes the only path available to it, through D1 into the battery. As well as charging the battery the back-emf current also produces braking torque in the motor.

If the controller uses PWM to control motor speed then this circuit can be 'free', because it uses the same switches that are used in a half-bridge configuration. The only change required is to keep the 'motor' switch open while applying PWM to the 'brake' switch. Most controllers use MOSFETs which have built-in body diodes, so an external diode is not needed either.

  • \$\begingroup\$ D1 in the original schematic was not meant to be a flyback diode, but it's purpose was to get back energy from the motor while braking. So there is no need to keep it in the schematic. \$\endgroup\$ – le_top Sep 23 '19 at 9:23
  • \$\begingroup\$ To work as described the diode is necessary. If PWM is applied to both switches the diode isn't necessary, but the motor will be forced to the PWM 'speed' and dynamic braking only occurs when it is going faster. The upper switch could be operated as an 'ideal diode', but that is still functionally a diode. \$\endgroup\$ – Bruce Abbott Sep 23 '19 at 19:13

Railway locomotives traditionally used separately-excited DC motors, in which the magnetic field is provided by a field winding rather than a permanent magnet. These were in practical use for both traction and braking long before high-power switched-mode motor control was feasible; at least one such locomotive was built in the UK by 1940 (though the railway it was intended for could not be electrified until the 1950s).

The EMF of the motor armature is proportional to the product of the magnetic field strength and its rotation speed. To brake regeneratively into the fixed-voltage overhead line, the field current had to be raised to make the armature EMF exceed the line voltage. A separate control for this was installed in the cab. For normal running in traction, the field was simply connected in series with the armature.

In diesel-electric locomotives, there is no overhead line to regenerate into. Instead, power from the main generator is used to excite the motor fields, and the armatures are connected to a high-power resistor in which the generated power is dissipated.


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