Power to the motor

5170 Views 17 Replies 7 Participants Last post by Gone Racin, May 13, 2011

JCS100

Discussion Starter · May 7, 2011

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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Gone Racin

· #12 · May 12, 2011 (Edited by Moderator)

Another point to consider with higher currents is the resistance of the track wiring.
While the tracks designed for high current motors will have perhaps a tenth or two of an ohm wiring resistance, the tracks used for lower current motors often have considerably more resistance. That adds up to a serious voltage drop at 7 amps +.

QUOTE (Screwneck @ 11 May 2011, 20:56) <{POST_SNAPBACK}>I often wondered why some folks put such high amps into their track. 10 amps for a tiny slotcar seems so high!
One reason for higher current power supplies is that the quicker 1/32 cars need the higher currents.
Really quick 1/32 cars average about 10 amps round a lap and take considerably more on acceleration from slow speed. Even more modest motors won't run at all on 1 amp.

The sort of motors JCS100 take considerably less power than this and would commonly be run on rather less than 10 amps per car.

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Gone Racin

· #15 · May 12, 2011 (Edited by Moderator)

Other things being equal, winding the armature with fewer turns produces more revs.
This frees up space in the armature which is often (but not always) used for thicker wire.
This is why higher revving motors are usually wound with fewer turns of thicker wire (going to thinner wire for a higher revving version of a motor is very unusual).

Other things being equal, torque is proportional to the number of ampere turns in the armature.
Higher revving armatures have fewer turns, so need more amps to produce the same torque.
Fewer turns have less resistance, so the current is higher. Does that balance out as more or less torque? Here's an example

Simplisticly 20% fewer turns = 10% lower armature resistance (assuming the same gauge of wire is used).
Neglecting resistances elsewhere in the circuit, 20% lower armature resistance would give 10% more current so you'd end up with the same number of ampere turns. In practice the resistances elsewhere in the circuit are important, so fewer turns of the same gauge of wire usually produces less torque, but not as much of a reduction as the 10% reduction that might at first be assumed.
It is normal practice to wind fewer turns of thicker wire for higher revving motors, as a starting point 10% fewer turn would mean around 10 % (by area) thicker wire, so the armature resistance would be reduced by around 20% (OK I knoe its npot exactly 20%). This usually adds up to higher revving motors producing more torque as long as the track is up to providing the necessary current.

Taking for example the Scaleauto range of motors, for the same size motor higher revving motors generally produce more torque.

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Gone Racin

· #18 · May 13, 2011

QUOTE (Gone Racin @ 12 May 2011, 16:08) <{POST_SNAPBACK}>Simplisticly 20% fewer turns = 10% lower armature resistance (assuming the same gauge of wire is used).
Neglecting resistances elsewhere in the circuit, 20% lower armature resistance would give 10% more current so you'd end up with the same number of ampere turns.
Sorry, I should have checked my typing more carefully
What I should have said was
Simplisticly 10% fewer turns = 10% lower armature resistance (assuming the same gauge of wire is used).
Neglecting resistances elsewhere in the circuit, 10% lower armature resistance would give 10% more current so you'd end up with the same number of ampere turns.

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