Flyback diode & motor confusion

Hey everyone

I'm just starting out with Arduino and have been working my way through the Vilros starter kit.

I don't quite get circuit 10 (attached)

I understand the purpose of the diode is to handle the voltage spike when the transistor is switched off and the motor's EMF collapses.

But why doesn't the current flow into 5V instead of going in a loop, since the spike's voltage will be much higher than 5?
And wouldn't it be easier to just have the motor's negative terminal connected to GND and have the transistor control current flow between the positive terminal and 5V? Wouldn't that let the current drain into GND without a diode?

One last question: if I connected the motor the other way around, so it spins counterclockwise, would it change the resulting voltage spike in any way?

I'm sensing my understanding of electricity is way off.. thanks in advance for your answers

You need a complete path for current to flow.
When the transistor turns off, this path is opened.
The motor spike, being large, can blow through the off transistor damaging it.

The diode acts as a complete path when the transistor is off, thus saving it from damage.
This diode circuit absorbs the spike energy.

Turning the motor around, has little effect on the generated voltage spike.
.

Shazool,

Electricity will always look for the easiest way to flow, this is the way with the least resistant, being either ohm wise or another voltage. The EMF spike is created by the motors coil, or any other coil you use in a circuit. The easiest way, with the least resistance is ti just shortcut it with a diode. This also reduces the chance of the EMF to do damage to other components. Just always keep one thing in mind, "Electricity always takes the easiest, least resistant way to flow" If you want we can calculate a few currents with resistors in parallel or in series, if you think that will help you understand what larryd and myself are trying to explain.

shazool:
Hey everyone

I'm just starting out with Arduino and have been working my way through the Vilros starter kit.

I don't quite get circuit 10 (attached)

I understand the purpose of the diode is to handle the voltage spike when the transistor is switched off and the motor's EMF collapses.

No, the purpose is to completely prevent any spike from happening.

But why doesn't the current flow into 5V instead of going in a loop, since the spike's voltage will be much higher than 5?

The current keeps going in the same direction through the inductor or motor, gradually dying
down as the small diode voltage opposes it weakly.

And wouldn't it be easier to just have the motor's negative terminal connected to GND and have the transistor control current flow between the positive terminal and 5V? Wouldn't that let the current drain into GND without a diode?

No, the problem is quite fundamental to any switched inductive load - try to stop the current dead and the energy stored in
the magnetic field will burst out with whatever voltage is necessary to force the current through in the short
term - you have to arrange that the current is reduced slowly, which limits the voltage. With a diode that
limit is 1V or less. Without it there might be a 100V, 1000V, or even more for large inductors.

One last question: if I connected the motor the other way around, so it spins counterclockwise, would it change the resulting voltage spike in any way?

the motor is an inductive load, whichever way you look at it

I'm sensing my understanding of electricity is way off.. thanks in advance for your answers

Inductors store energy in a magnetic field when current is flowing, the energy goes back to the circuit when
the current stops. The rate the energy is transfered depends on the rate the current changes and the amount of inductance.

You can think of the difference between stopping a car using the brakes (slow reduction in speed) or a
concrete barrier (sudden reduction in speed). The energy of the moving car has to go somewhere...

Electricity will always look for the easiest way to flow, this is the way with the least resistant, being either ohm wise or another voltage

That is wrong. It implies some sort of choice or intelligence that electricity is making. It is rubbish anyway because electricity will flow through all possible paths regardless of the resistance. It is just that the resistance of each path determines the current in that path.

shazool:
But why doesn't the current flow into 5V instead of going in a loop, since the spike's voltage will be much higher than 5?

If we have DC conditions to begin with, then the current through the motor (coil) will be flowing away from (ie. out of) the 5volt terminal.

When the transistor is switched to 'off', there is some circuit theory that teaches current through a coil (eg. inductor) is time-continuous, or won't change instantly. So the current in the coil maintains the same value of instantaneous current at some moment just before transistor switching. The current just keeps flowing in the same direction for this case.

Also, since there's no L and C combination (ie. no C), we don't expect oscillation behaviour.

The current is expected to reduce at an exponential rate.

Probably need to analyse this circuit ..... like figure out the current through the diode as a function of time.

The situation will be something like ...... one side of the diode will be kept at the forward diode voltage (eg. Vf) until the current through the diode reaches zero, which is the case when the coil voltage drops to the same value as the diode forward voltage ---- where both sides of the model coil resistance 'r' has the same voltage 'Vf' (so no current flows).