2
\$\begingroup\$

newbie here. I'm having some problem with my DC motor circuit.

enter image description here

As you can see from the image above, voltage does not drop off quickly when it's turned off. I understand that the sudden spike is due to sudden change of current which causes back EMF, but why does the voltage come back and drop slowly?

\$\endgroup\$
1
  • \$\begingroup\$ This is the voltage across the motor? We kinda need to see your circuit too. \$\endgroup\$
    – DKNguyen
    Mar 4 at 3:36

4 Answers 4

6
\$\begingroup\$

The fast spike is the effect of the inductance of the motor windings,

The slow decline is the effect of the spinning motor acting as a generator as it slows to a stop.

\$\endgroup\$
4
\$\begingroup\$

The spike is caused by motor winding inductance, as you noted. It happens when you try to open the motor's current path, causing that path to become very high resistance. The winding's inductance tries to maintain whatever current was flowing just prior to "switch off", which develops a huge voltage across the new, and sudden, high impedance of whatever switching element just "opened". When using relay contacts, or a single bipolar junction transistor to switch motor current, this can be very destructive (left and middle below):

schematic

simulate this circuit – Schematic created using CircuitLab

When these switching elements (relay or transistor) switch off, there will be a sudden and large voltage developed between A and B. For the relay contacts, this will cause a spark, each time degrading the contacts until the relay becomes unusable. For the bipolar transistor (middle) this might not damage the transistor immediately, but eventually it will.

On the right, using a MOSFET, there's a "body diode" present which becomes forward biased and conductive when the potential at node A falls below ground, so in that scenario (single MOSFET as a switch), the spike is somewhat constrained. It's not really recommended to rely on this body diode to protect the transistor, though.

In these single-switch scenarios, the spike is usually mitigated with a diode across the motor. This is possible because intended motor current is only ever in one direction:

schematic

simulate this circuit

If you are operating the motor using an H-bridge, or are otherwise conntrolling current in both directions through the motor, you'll have to get more creative.

The reason the voltage across the motor dies away slowly is because in the absence of current driven through it, it becomes a generator. That is, the spinning rotor has momentum, and kinetic energy, and that energy doesn't just "disappear", it has to be dissipated somewhere. Unless you take measures to electronically dissipate that energy, then there's only friction to slow it down.

An effective way to "stop" the motor spinning, is to "short-circuit" it, instead of leaving the current path open. This causes the kinetic energy of the still-spinning rotor to deliver energy to the low-resistance of that short, instead of relying on just friction to convert its kinetic energy to heat. In this way, motor speed will diminish more quickly, as will the voltage across it.

Again, when using an H-bridge to provide a low-impedance path on both sides of the motor, it becomes possible to electronically force the voltage across the motor to zero, effectively short-circuiting it with transistors. Then the transistors can absorb all that kinetic energy.

\$\endgroup\$
1
\$\begingroup\$

I'm not seeing your time scale, but there would normally be an inductive 'kick' followed by the back-EMF of the motor itself as the armature slows. It acts as a generator and the inertia maintains some voltage for a while.

\$\endgroup\$
0
\$\begingroup\$

I am answering without seeing your motor circuit but I really need to see it to be sure.

The normal inductive flyback voltage spike from interrupting current through a motor should send the voltage ABOVE the operating voltage before decaying. You can see that you do not have this since your decay starts off at a voltage BELOW your operating voltage.

If I remember, right, that very fast negative spike is due to shoot-through. So I'm guessing you have an H-bridge. I'm also guessing that you are sharing control signals between your MOSFETs. Shoot-through will suppress the inductive flyback voltage spike at the expense of having shoot-through currents, and that would explain why your inductive flyback voltage starts out below your operating voltage.

So if you correct your shoot-through by leaving more time between the when a MOSFET switches off and the other MOSFET in the same half-bridge switches on so their conduction times do not overlap and cause shoot-through this should reduce the negative going spike, but in return your inductive voltage spike will increase and unless you have measures to deal with this, could damage your MOSFETs.

NOTE: A half-bridge is a high-side MOSFET on top of a low-side MOSFET. An H-bridge is made of two half-bridges.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.