Korad supply Limitations

[tjones9163]

Hello, I have the Korad 30v 5A power supply, that I bought here.

I also have a 12v 550 rpm Drill motor that I have taken apart and power it with my korad and immediately noticed that I started to pull around and over 5 amps, so I cut it off. My korad still works fine.
1.I'm just wondering if it is bad for my Korad to do that?
2. Should I have a flyback diode when I hook the motor straight to the supply and is it enough to use 2- 1N5408 3A 1000v rectifier diodes in parallel to be able to handle to 5amps?

[DrAzzy]

That bench supply will back off the voltage if the current limit is reached until the current doesn't exceed the limit. They will not be harmed - a supply that would be damaged if you accidentally let the clip-leads touch wouldn't last long. That power supply allows you to adjust the current limit, and when that limit is hit, it acts as a constant current supply. 

Supplies not meant as bench supplies usually turn off entirely, requiring power to be completely removed before they will turn on again.

Bare DC-DC converters may or may not have some sort of short circuit or overcurrent protection - those that don't can be damaged by excess current or short circuit.

[tjones9163]

That bench supply will back off the voltage if the current limit is reached until the current doesn't exceed the limit. They will not be harmed - a supply that would be damaged if you accidentally let the clip-leads touch wouldn't last long. That power supply allows you to adjust the current limit, and when that limit is hit, it acts as a constant current supply. 

Supplies not meant as bench supplies usually turn off entirely, requiring power to be completely removed before they will turn on again.

Bare DC-DC converters may or may not have some sort of short circuit or overcurrent protection - those that don't can be damaged by excess current or short circuit.

Thank you for the reply, so is a flywheel diode needed in my situation where i am hooking up bare leads of the motor and power supply?

[raschemmel]

Thank you for the reply, so is a flywheel diode needed in my situation where i am hooking up bare leads of the motor and power supply?
 

The flywheel diode is necessary across any motor regardless of anything else you are doing.

[tjones9163]

The flywheel diode is necessary across any motor regardless of anything else you are doing.

Thank you for the reply, is my pic of the 1N5408 3A 1000v hooked up to the motor correct?
and is this one diode safe for the motor and current draw that I am using?
Thanks

[DrAzzy]

Connections are correct, but you say it pulls up to 5A, but you're using a single 3A diode. Flyback diodes should be rated at least for the maximum current that the motor can draw, so you should parallel two of them (or use a bigger diode in the first place).

[raschemmel]

When the motor is pulling 5A and you switch it off the magnetic field generated by the 5A will collapse , resulting in a polarity reversal of the 5A current. The end of the diode that was previously POSITIVE, (the banded cathode end), then becomes negative. The end that was previously NEGATIVE (the unbanded end) then becomes positive. The forward bias direction for a diode is with the unbanded end more positive than the banded end. With a voltage greater than the operating voltage the diode will conduct, shunting the voltage (and current) of the collapsing magnetic field. Are you sure you want to use a 1N5408 3A 1000v diode ?

[Paul__B]

So where did the OP mention anything about switching?

Getting ahead of yourselves fellas!  

[raschemmel]

So where did the OP mention anything about switching?

Ok, then :

When the motor turns off...

Just an example to explain fly back diode

[tjones9163]

So where did the OP mention anything about switching?

Getting ahead of yourselves fellas! 

Hello, what do you mean?

[Paul__B]

Hello, what do you mean?

Your original question was:

Should I have a flyback diode when I hook the motor straight to the supply

If indeed, you are simply connecting a motor to a power supply and not attempting to control it with a semiconductor switch, there is no reason to fit a "kickback" diode because you will not be interrupting the circuit.

If you use a common toggle switch or similar, or a relay to turn the motor on and off while the supply remains powered, there may be some minor arcing at the switch contacts, but it is uncommon to consider that the "kickback" diode may be needed.

Undue emphasis is placed in discussions here about the "kickback" potential of a motor, with a tendency to compare it to the behaviour of an inductor when switched on and off.  In fact, the only reason for a DC motor of the type you illustrate (permanent field magnet) to exhibit "kickback" is due to engineering imperfections.  The only motor whose design does necessarily have a "kickback" is a series wound field or "universal" motor.

A permanent field magnet motor exhibits "Back EMF" due to its equivalent action as a generator.  The only difference between the Back EMF and the applied voltage is due to the load on the motor and the internal resistance of the windings.  If you suddenly disconnect - switch off - the motor, the voltage across the motor promptly drops back to the back EMF.  If the motor is unloaded, this will be almost the same as the voltage previously applied, in a "perfect" motor with no drag (including the bearings), the voltage would not change at all and the motor would keep spinning at the same speed.

This situation is totally different to an inductor.  While the motor windings are essentially inductors, the voltages across them are defined by the movement of the armature.  The better the quality of the motor and the more windings and segments on the commutator, the more effectively any "kickback" effect of stray inductances will be cancelled by other windings.

[SteveMann]

Yes, a single 5408 diode is quite sufficient.  It is rated at 3A continuous, and with a 150A Peak Forward Surge Current for 8.3ms.  Depending on the load, the motor will likely stop in that 8ms.  Even if it has enough inertia to keep spinning, it will have slowed to a point that the current through the diode will likely be way less than 3A.

It is a good practice to have a flyback (or kickback by any other name) diode on any DC inductive load regardless of how you power it.  

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.

If you are powering the motor with, let's say, a MOSFET or bipoplar transistor, the reverse voltage could easily exceed the drain (or collector) current limits because reverse-biased, the device would look like a short.

[tjones9163]

Yes, a single 5408 diode is quite sufficient.  It is rated at 3A continuous, and with a 150A Peak Forward Surge Current for 8.3ms.  Depending on the load, the motor will likely stop in that 8ms.  Even if it has enough inertia to keep spinning, it will have slowed to a point that the current through the diode will likely be way less than 3A.

Thank you, everyone, for the great and informative replies, and thank you SteveMann, you are correct. 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.
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.
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?

[Paul__B]

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.  

My concern is that you now have thoroughly muddled the OP.  

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.

[SteveMann]

You are right that a motor turned generator does not reverse polarity.  My error there.  My Electrical Engineering degree predates microcomputers, so I'm rusty on the math.  Motor theory was not my favorite class.

The formulas for an inductor and a DC motor are almost identical, except the motor adds torque and angular velocity factors.  A collapsing field in a motor generates a reverse EMF just like an inductor.  The flyback diode gives thar reverse EMF a path to dissipate as the field collapses.

But my answer to the OP's first question is correct- one 5408 diode is sufficient as a flyback diode in his circuit.