Dear Sir/Madam,

Someone recommend me to have a diode for protection against back emf, may I know how I should connect the diode? Which one is correct, left or right diagram? I only know the diode can only allow current flow in one direction, could you explain how the diode can protect against back emf?

Best regards

Kelvin.

 

The left diagram is correct. When shut off, the coils in the motor generate a voltage in opposite polarity to the supply. Protection or snubber diodes are installed across the motor terminals, opposing the supply voltage. Thus, when the motor turns off, the diode presents a short circuit to the back EMF, allowing it to dissipate safely.

 

An easy way to think about inductive effects is to remember that the current in an inductor acts similar to the inertia of a mass, it wants to keep flowing, even after the voltage (force) across it it removed.
It will keep flowing until all the inductive energy (1/2 LI²) is dissipated, which can generate a very high voltage if there's not a path for that current.

So from that, it's obvious that the diode is series has no effect, since the current through it and the inductor will just keep flowing in the same direction, which will then generate a large voltage at the transistor collector when it shuts off, likely zapping the transistor.

The circuit on the left, however, provides a path for that current
As soon as the transistor collector voltage builds up to within one diode-drop of the supply voltage after it shuts off, the diode starts carrying the inductive current, preventing a high voltage spike.
It rather short-circuits the inductive current until the energy is dissipated in the diode forward voltage and the inductor/motor resistance.

Make sense?

 

And as well as preventing the high voltage spike, the current flowing via the diode as the magnetic field collapses helps to keep the motor turning so the control is smoother.

 

Thanks djsfantasi and crutschow,

I watch a video about using the diode

, is it same the reason when a switch turns off the circuit, then a high voltage will be generated oppositely polarity to the supply. In that case, does it mean every time when we use a motor, it is a good practice of adding a diode opposite to the current flow of the motor? Please advise.

Best regards,

Kelvin.

 

Thanks djsfantasi and crutschow,

I watch a video about using the diode

, is it same the reason when a switch turns off the circuit, then a high voltage will be generated oppositely polarity to the supply. In that case, does it mean every time when we use a motor, it is a good practice of adding a diode opposite to the current flow of the motor? Please advise.

Best regards,

Kelvin.

It is generally a good idea to have a suitable diode across any inductive load, unless of course you want the voltage spike, like in a car ignition.
Some H Bridge chips have diodes internally so no extras are needed. And in that case, they are just not directly across the motor but across each FET.
I always add diodes across the relay coils in any design I do. It does slow the relay operation down a bit but if you are using relays, high speed is not a design feature anyway. And the speed reduction will not really be noticeable most times.
Just make sure the diode is capable of the current pulse it will be required to carry.

 

...Is it same the reason when a switch turns off the circuit, then a high voltage will be generated oppositely polarity to the supply. In that case, does it mean every time when we use a motor, it is a good practice of adding a diode opposite to the current flow of the motor? Please advise.

Yes, it is always a good idea to include that diode. Note that this simple addition only works when the motor is run only in one direction.

 

In that case, does it mean every time when we use a motor, it is a good practice of adding a diode opposite to the current flow of the motor?

It's risky to say always. A diode won't work for AC.

A diode limits the back EMF to a diode drop, but that slows down the collapse of the field. In cases where you want a quicker decay, you'd use some other method of snubbing.

 

with regard to permanent magnet DC motors:

"Back EMF" refers to the potential produced by the motor as it turns, acting like a generator. It is "back" because it opposes the externally applied voltage that is causing the motor to turn. Back EMF will never exceed the applied voltage under steady state conditions. It can exceed applied voltage if the motor is driven to a speed greater than it would operate at with no load with the applied voltage. This can occur if the supply voltage is reduced and the power supply is able to absorb energy (an ideal voltage source has zero impedance, so it should be able to absorb energy, but with most real DC power supplies, not including batteries, the ability to absorb energy sourced into the output is very limited). In such a case, the motor has ceased to be a motor and becomes a generator. A diode across the motor that does not "short circuit" the applied voltage will not conduct due to back EMF unless the motor is forced to turn in the opposite direction. A diode in series with the motor, as shown in the right hand diagram, is what is required to prevent back EMF from being forced into the power source.

There is a very common misconception that the "inrush current" of a motor that is starting to turn is because the motor is inductive. Inductance has nothing to do with it, and in fact inductance forces the current to start at zero - you cannot instantaneously change the current through an inductor. The Inrush current is high because the motor is not producing any back EMF, so the input supply "sees" the motor winding resistance. As the motor spins up, the back EMF increases, opposing the input source, reducing the voltage across the winding resistance and reducing the current. Of course it is more complex than this because mechanical energy can be taken out once the motor is no longer stalled. Torque is highest at stall because the current through the winding is the highest it can be and the magnetic force produced is proportional to current.

The diode is there to deal with energy stored in the inductance of the winding. Such a diode is often called a "free wheeling" diode, and the term is very apt for a motor. With the diode in place, if circumstance arise such that it conducts, the energy stored in the motor's inductance contributes to driving the motor instead of being wasted - the direction of current flow in the motor is the same as it was before the diode began to conduct. This is not "back EMF." The polarity is opposite that of back EMF.

So, if there is a risk the motor will be forced to rotate at a speed that would produce a generated voltage greater than the supply voltage, you need both diodes. Note that with a BJT, if the diode weren't there the transistor would have to conduct in reverse, with current flowing into the emitter and out of the collector. BJTs do act like transistors in reverse, but very poorly (very low current gain) so without the series diode most of the voltage difference between the generated voltage and the supply voltage would likely appear across the transistor because only a small amount of current would flow. If the switch were a FET, the intrinsic body diode would allow conduction. The processor might release its magic smoke. (again - if the motor runs at 5 V, you'll only make it generate more than 5 V if you mechanically drive it at a speed higher than its no-load speed at 5 volts)

If the current path to an inductor is instantaneously opened, the voltage across the inductor will rise to an indeterminate level (it is accurately predictable if enough is known about everything in the circuit). It will rise as high as is necessary to keep the current at the instant of opening the circuit exactly the same (direction and magnitude) as it was before opening the circuit. It will "find" some place for that current to go. The capacitance of the winding will absorb some of the energy (and you'll get resonant ringing after the voltage on the capacitance reaches its peak). If the switch is mechanical, an arc across the opening contacts will be produced. If the switch is a semiconductor, it will likely be forced into breakdown and may be destroyed. A diode provides a well-defined path.
If the inductor were wound with a perfect conductor and was short-circuited, the current flowing in the coil would continue to flow in the same direction forever, assuming nothing else were removing energy (messing with the magnetic fiield). BIG "if!"

 

regarding slowing of relays by using a diode:
This actually can be detrimental to the relay because it increases the time it takes for the contacts to go from closed to fully open. With difficult loads, such as those that do have inductance and operate a high current, it can mean that an arc across the contacts will last longer. Arcs erode contacts (melts them in local areas; transfers metal from one to the other in DC circuits). If longevity of contacts is important in power circuits, letting the coil voltage go as high as driving circuitry can handle, resulting in fast opening of the contacts, is important. In low level circuits, slow opening is usually of no consequence as far as the contacts are concerned.