What is the purpose of a flyback diode?

Question

I'm new to electrical engineering, and I still have many stupid questions. Here's one of them.

This is what I'm reading. https://itp.nyu.edu/physcomp/labs/motors-and-transistors/using-a-transistor-to-control-high-current-loads-with-an-arduino

Next, add a diode in parallel with the collector and emitter of the transistor, pointing away from ground. The diode to protects the transistor from back voltage generated when the motor shuts off, or if the motor is turned in the reverse direction.

I thought before that polarity could be reversed in transistors, but now I'm confused. Can it not be? Why would it hurt the transistor, and how would it? Would it instantly break it? How does doing this fix it?

Thank you.

-edit to defend not being duplicate-

The proposed duplicate does not answer my question because it does not give how the transistor is affected. From searching online, I couldn't find this either; I'm probably searching the wrong thing. It also does not answer whether it is instant or how it fixes it.

Answer 1

In normal use, a small current on the base of the transistor allows a much larger current to flow between the emitter and collector.

In this case, however, we have a large voltage applied across the emitter and collector, with (presumably) little or no current flowing through the base, so the transistor is trying to stop current from flowing between the emitter and base.

In this case, it's probably useful to think of the transistor as a pair of diodes, with (depending on whether it's an NPN or PNP transistor) either their anodes or cathodes connected to each other.

One of these diodes is going to conduct current from the emitter/collector voltage. The other is going to try to block it. One question, then, is what's the breakdown voltage of that "diode" (that junction of the transistor). If the voltage being applied exceeds that value, you're liable to toast the transistor.

Now let's consider that added diode, and what it does.

We connect the diode so the transistor's power supply is trying to push current in the direction that the diode won't conduct. Therefore, when the transistor is operating normally the diode basically has no effect at all¹ (unless we choose the wrong diode, such as one with a breakdown voltage lower than our power supply voltage).

For a voltage in the opposite direction, however, the diode looks/acts pretty much like a dead short. This means all the current from that reverse voltage flows through the diode. Since it acts like (nearly) a short circuit, no voltage can be developed across the transistor, preventing any damage to the transistor.

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^{1. "basically" meaning that, for example, it will add a little extra capacitance. If we were dealing with a high enough frequency, we might need to figure that into calculations about how the circuit works--but for a typical motor driver, we're dealing with low enough frequencies we don't normally care.}

Answer 2

Inductors are funny things but can be understood fairly easily. Motors are very noisy inductors. When current flows through a piece of wire a magnetic field that is relative in strength to the current is built/formed around the wire. If the wire is a coil such as in a motor the magnetic field strength is more intense (larger). Current = magnetic field strength. When you open the switch or turn the transistor off the magnetic field collapses. The lines of magnetic flux cut through the coil of wire creating a current in the same direction as the original current. Since the coil is now the source, the polarity switches + to - and - to +. Since the circuit is now open and the magnetic field is collapsing it is going to cause the original current to flow (briefly). Let's say for example 10 ma caused the magnetic field with the switch closed and when suddenly it is opened 10 ma WILL continue to flow through the now high ohm (open) circuit. Voltage will build up until it can arc across something (your transistor....not good) and the current will flow until the magnetic field is gone. Your transistor will be toast.

Answer 3

An inductive load like a motor, when it is no longer powered, can have a sudden charge of up to thousands of volts. I believe this is called "inductive spiking" or something like that. Basically, it is caused by the collapsing electromagnetic field of the electromagnets (aka inductors) inside the motor.

A flyback diode serves to protect your circuitry from this effect. Because diodes have a forward voltage well under the thousands of volts generated by the inductive spiking, the extra power from the collapsing field instead goes through the flyback diode.

These diodes are hooked up to the circuit directly across the motor's + and - pins, but backwards from normal usage. What this means is that the positive side of the diode would go to the negative input of the motor. This way, under normal circumstances the diode would not be active.

The high voltage that the diode short circuits in a safe way would kill the transistor that you are using. It recommended to use that flyback diode.

Please see wikipedia for more info

Answer 4

A bipolar junction transistor (BJT) will fail if the appropriate polarity requirements are not observed. If you apply a reverse bias to a diode that exceeds the diodes absolute maximum rating, it will destroy the diode. A transistor can be viewed as two back-to-back diodes which also have absolute maximum ratings as well. If you exceed that value, the transistor will be destroyed. An inductor of any type such as a motor or relay coil, when power is removed, generates a reverse pulse of theoretical infinity trying to keep the current going because of the inherent characteristics of the inductor. If you have a switch connected to an inductor in series with a battery, and a current is flowing with the magnetic flux at maximum, when the power is removed, the magnetic field begins to collapse. With a physical switch, the reverse EMF, or electromotive force, would rise to the point where an arc would be generated across the contacts. Remember your basic principles that when a conductor is moved through a magnetic field, it INDUCES a current. Since the current is flowing in one direction, the collapse generates a voltage in the reverse direction. When you connect a diode across the emitter and collector of a BJT as described, that reverse current forward biases the diode which shunts the current AROUND the transistor, thereby protecting it from the excessive reverse voltage that would let the magic smoke out of the transistor. Field effect transistors CAN conduct current in either direction and would not need such a shunt diode because if they are in the on condition, they will just shunt the current in either direction, as long as their voltage ratings were not exceeded. Some kind of protection would be recommended, however, if the generated voltage was above the absolute maximum rating for the FET. The absolute maximum ratings for any device can be found in the data sheet provided by the manufacturer. I hope this answers your question.