In the case of a motor, when the supply power is removed, the motor becomes a generator with the polarity reversed. The flyback diode shorts out this reverse voltage generated by the motor acting as a generator. As a side benefit, this shorting out of the reverse voltage dynamically slows the motor turned generator more quickly than free spinning down.
Unfortunately what you have just described is utter nonsense. May I recommend you do some study on the subject and then come back with an apology?
What you have done, is to confuse a motor with a simple inductor. I did attempt to explain this in my posting #10 but do not have the time to go into more detail on the behaviour of the motor to you. Please go and research it but you may find clues from the following. :grinning:
My concern is that you now have thoroughly muddled the OP. :astonished:
What i noticed when I soldered and reconnected the motor to the diodes in parallel and hooked it up to the power supply( After setting the max current/constant current to 4.8 amps), that my power supply only spikes above 4 amps when i first hook it up, then levels off to 2.5 amps after a second.
What you have described is the "stall current". The "back EMF" of a motor when spinning appears as a voltage of the same polarity as that used to make it spin. Consider it like a "battery" inside the motor and like connecting two batteries of the same voltage in parallel with the same polarity, when there is no difference between them, no current will flow in either direction. Only when one is slightly less will some current flow into that one.
So when the motor is stopped, it has no "back EMF" so the current is limited only by the winding resistance - nothing else. As the motor starts, the current reduces in proportion to the speed of the motor as that determines the back EMF. As I explained above, if the motor were "perfect" and unloaded, it would speed up until the back EMF was exactly the same as the applied voltage and no current would flow.
In practice, it is unable to do so due to loading, including friction in the bearings and the commutator (which is quite significant in small motors) and hysteritic losses in the armature iron and windings, so it draws a minimum current unloaded, while putting a load on it causes the current to rise to some proportionate figure between the unloaded and the stall current.
1) Is the reason there is a current spike on startup because the inductor is resisting current flow? I get that inductors try to keep the current from changing by applying its own force, im just not sure why there is a sudden spike on startup? ( On turning the motor off it makes sense to me why the would be a large current spike because there is built up energy in the form of a magnetic field.
Well there is where Steve has managed to confuse you further. The inductor does not resist current flow, it resists a change in current flow. Connecting an inductor to a power supply causes a slow rise in the current until the maximum limited by the circuit resistance, including the winding resistance in the inductor itself.
2) I am trying to learn more about flywheel diodes and what makes the diode work harder(more heat given off), Is it when power is being supplied and it is blocking current in the reverse direction? Or is it when the power is removed and the large current spike travels forward bias through the coil and diode and around and around again?
The converse is when you attempt to disconnect the inductor. Again there is no spike of current whatsoever. If (with a diode,) you provide a path for the current to continue, then the current slowly decreases to nothing. The law of induction is that how fast it decreases depends on the voltage across it; if you cause the current to decrease very rapidly by completely opening the circuit, then it will generate a correspondingly high voltage, in fact whatever voltage is necessary to maintain the current momentarily flowing either by causing breakthrough in any component connected across it or else by breaking down the insulation, even the insulation of the air across the terminals where it is switched off.
If however, you have a diode across it which conducts at less than a volt, then it does not need to generate any more than that voltage and the current reduces quite slowly. In fact, the diode as such does not "work hard" at all - it only conducts that current between the time the supply is cut off and the current falls away to zero, the current is never greater that what was flowing in the inductor just before the disconnection. So in either case, the current in no way "surges"; it only reduces from the previous level. What does "surge" is the voltage.
And a reminder - this refers to an inductor, not a motor. The two are quite different.