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Power to the motor

5164 Views 17 Replies 7 Participants Last post by  Gone Racin
On a earlier thread on motor lead cable sizes the question was posed on how much current do slot motors actually use, and from this logically follows into what current rating of power supply is really needed.

Trying to find this out with just a meter is only effective as a snap shot of a point in time i.e. taking the highest reading, during acceleration or waiting until the reading stabilises with a steady RPM & load. The following is actual measurement of current over time taken from a digital sampling oscilloscope, which shows graphically the current peak on acceleration and how this reduces as the motor reaches max RPM simulating the slot car reaching top speed.

In the picture above is a voltage trace against time where each square going vertical is 0.2 volts (200mV) and squares going across is 0.25 seconds (250ms). This signal has been created by adding to the power lead from the motor return, a 0.2 ohm resistor across which is the test leads for the oscilloscope. The trace shown above is the volt drop across this resistor so for each 200mV square going vertical now can represent 1 amp.

(0.2 volts divided by 0. 2 ohms = 1 Amp ) Therefore as current in a series circuit is equal across all points then this is a direct representation of the current through the motor.

The motor by the way is a standard Scalex/Fly Mabuchi 130 black stripe driving a 60 grm flywheel supplied from a nominal 4 amp PSU at a stabilised 12 volts. The trace clearly shows the MAB130 is pulling 2A+ initially dropping to 0.5A after 2.25 seconds. The Mabuchi under this load was struggling simulating a heavy car or magnet fitted.

The second picture is of the same setup but with a Ninco NC 5 motor in jig. Here we can see the initial current is a lot higher at 4A peak dropping to @0.5A at 2.25 seconds.

The third picture is of a Scaleauto Yellow boxer motor and the initial current has gone off the screen i.e. greater than 8A. It can be seen that the 8A peak is of short duration and falls rapidly to @6A then declines in a similar curve as before. Reason for this is the limitation of the PSU unable to sustain the 8+ amps required by the motor the output voltage has reduced from 12 volts.

Picture 4 has the same Scaleauto motor but the PSU is now a 17A rated unit which as can be seen above will sustain the 8A plus requirement of the motor during acceleration as the voltage follows a curve similar to the lower rated motors on the 4A supply.

The final picture is the Scaleauto motor again but this time the scale on the screen in the vertical has been changed to 500mV per square to find the actual current peak which from the new setting is 2.5 amps per square = 10A.

Points to note are that the initial peak current of a motor even from what may be considered the low powered hard bodied end of slot racing can still pull large current peaks during acceleration.

Secondly a PSU which at 4A should be sufficient for these motors will not supply the power the motor needs to sustain acceleration at design limits. From re-running the NC5 & MAB130 motors on the 17A supply it can be demonstrated that even these motors will benefit from the extra power.
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Interesting measurements, thanks for sharing them.

The measurements show the highest current only occurs when starting from rest, as speed increases the current decreases.
So the maximum current is only needed when starting from rest.
In normal running this only happens at the start of a race and when a car is marshalled after a deslot.

Once the car is up and running, the maximum current needed would be when accelerating out of the slowest corner on the track.
That requires less current than the maximum current for starting from rest.
So this lower current is all that is actually needed for minimum lap times.
The full "start from rest" maximum current wouldn't make any difference to lap times, just allow maximum torque from rest.

So an interesting question is putting a figure on this "lower current for minimum lap times"?

Anybody want to discuss??
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QUOTE (hankscorpio @ 10 May 2011, 23:46) <{POST_SNAPBACK}>Any idea of what happens if you place some "drag on the flywheel" to represent what happens to a motor as you apply power and the car slides unable to turn the "push" into forward momentum placing a strain on the motor... I Guess the High amp peak would stay for considerably longer... guess the peak could last for half a second or more... gulp!
Yes I can tell you what happens in theory. It would be interesting to figure out a way of measuring what happens in a real motor and see how exactly (or not) it matches the theory.

The current taken by a motor depends on the applied voltage and the revs its doing at the time. Nothing else makes a differance (except small changes as the armature heats up.)

So for example a motor supplied with 13 volts doing 5000 rpm will take the same current whatever the drag or flywheel inertia.
(The current depends on the type of motor)
The drag or flywheel inertia will make a differance to how quickly in accelerates from 5000 rpm.
More flywheel inertia means it accelerates more slowly.
More drag means it accelerates more slowly, if there is just enough drag to balance the motor's torque it'll stay at 5000 rpm and if the drag exceeds the motor's torque it'll slow down.

Normally when cornering the controller isn't on full power, so the voltage and current is reduced.
If a motor were held on full power at low revs by drag for an extended period it would get hot. In the extreme that could be hot enough for the motor to fail.
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QUOTE (hankscorpio @ 11 May 2011, 21:20) <{POST_SNAPBACK}>A 12V 1Amp PSU faced with a 5 amp peak demand will redcuce the voltage to maybe 5 or 6 volts for a few 10ths of a second.... or overload!
Agreed, that what can happen (including overload = power supply turns itself off so all the cars stop)

Where only one car is running from the power supply, a reduction in voltage will make it slower than it should be - at least it's the same every time - a driver can deal with that.

Where two car are running from the same power supply, the reduction in voltage will caused by one car will slow the other car , then when the first car stops taking all that current, the other car suddenly speeds up - and quite frequently ends up crashing. Surges of power completely outside the driver's control is not the sort of thing a driver can deal with.
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