Hi,

When reading about the Keppe motor working principles you might be tempted to believe such motor doesn't follow the classical laws of electrical circuits. To check that, I used 2 approaches to try to calculate the current in the circuit:
(1) Calculate the circuit using the classical electrical theory
(2) Simulate the Keppe circuit with a simulation package.

The results of approach (1) were presented in my posts of 4 and 6 May (= solve a differential equation). To simplify the calculations, I neglected the circuit capacitance.

For approach (2), I used Simetrix freeware simulation software ( http://www.simetrix.co.uk/ ) and the results seem quite OK too. Below you can see the results of a typical run as well as the simulation circuit set-up (detailed figures omitted for clarity).

Compared to measurements with an osciloscope, the general trends are OK in both cases, though the simulation software provides more accurate results and is also able to predict quite well the transient phenomena that occur when the switch re-opened. The calculated current peak values are not equal to the measured ones but well in line (say +/- 10%).

The main problem is the correct value of the motor (and entire circuit) capacitance.
* In the simulation attached below I used a value of 2 pF and that gives a realistic result.
* When I use the value of 456 nF (= 456000 pF) as advised by the Administrator in his post of May 6, the voltage curves during the transient period are not at all in line with my measurements.
* When I use a value of 200 to 400 pF (the values I though would be correct), the results for the transient are very bad too.

Anyhow, with the expected values in this case, the capacitance value only has a significant impact on the curves during the transition period (= re-opening of switch).

For me, the main conclusion of this exercise is that classical electrical theories quite adequately predict the current and voltage curves. This would also mean that inside a Keppe motor, no exotic, unbelievable or unexplainable things happen at least not when the switch is closed.
What happens during the transient periods may be more difficult to calculate and explain but I think a clever electrical engineer should be able to do so (correct measurements are also very important to validate calculation results).

Please comment and try yourself to calculate.

Regards,

J Greef



The periodic opening and closing of the reed switch is simulated by a solenoid actuated switch that is controlled by a pulse generator.

Hey Greef,

Could you send us your phone number
then I can call you and understand
better what you need? I wish I can
help you.

Att.

Ângela Stein

Dear Keppe Motor Team,

I refer to you answer below (posted June 4, 2009): have you already been able to to those measurements ? As indicated earlier, in order to allow the forum members and visitors to try to understand the curves you posted, the coil or switch voltage curves are also needed.

Best regards,

J Greef

PS:
Of course the Keppe motor V3 curves would also be quite welcome ...

Mr. Greef,

This questions needs measurements that take time to do.
If you can wait for the answer, we apreciate.

Best Reagards,

Keppe Motor Team

Dear Keppe motor team,

I have tried to repeat the test you suggest by putting the measuring probe on a lot of different points of the circuit but I haven’t found any curves that look like the ones you posted. Could you explain more in detail how you detected those curves ?

To facilitate discussions about the exact measuring points, I add a circuit sketch. The shunt to measure the current could be mounted at other positions. If I understand correctly your post, you measured the current with a 0.1 Ohm shunt between positions 1 and 2 of the sketch below ?



Unless classical electrical theory contains some flaws, I don’t see how the position of the measuring probe can have a major influence on the current that is flowing through a closed circuit. That current should be the same in all points.
To better understand what is happening, I still suggest that you also post the coil or the switch voltage curves.

For the time being, the only explanation I can imagine for those current peaks is the presence of some capacitance in the circuit (when the switch opens or closes, there is a sudden change in voltage over it and theory tells us that for a capacitor, C * dU/dt = i (current). However, with the 456 nF capacitance you detected and a voltage drop of say 4 to 6 volts on the switch, you would need a switching time ("dt") of +/- 4 * 10 exp-9 seconds to obtain a current of around 1 Amp (as in your post). That looks quite unrealistic for me.


Regards,

j greef

Those peaks were taken at the end of the reed switch, not the coil.
Also, we cannot forget that the rotor spinning inside the coil works
as a generator and the composition of these two overlaping waves need
to be better studied.

Best Regards,
Keppe Motor Team

There are at least 3 quite strange things on your curves:

1) The duration of the current peaks at beginning and end of that switching = +/- 2msec ??
2) After the first peak you have a negative current (approximately -200 mA if I interpret correct your data = -20 mV / 0.1 Ohm). That would mean that the motor would be charging the battery but that is contradicted by the voltage drop on your battery.
3) Current peak when closing the switch. Have seen none myself nor on any post on this forum before.

Are you sure this is the standard Keppemotor V1.0 ??

The only explanation I can imagine is that there is a sampling problem with your scope but even that doesn't explain my second observation. Do you also get such curves when using an analogue oscilloscope ? I am also missing the dampened oscillations of the current after the opening of the switch alike on fig. A3 of your manual V1.0.

To better understand what is happening, the curves of the voltage over the coil could help a lot.

Curves for keppemotor 1.0

colors:
red battery voltage
blue: current measure with a 0,1 ohms shunt
green: power (volts x amps)

The curves represent the period where the reed switch is closed(closing and opening )

Question: Why does the current peak only during the opening andclosing of the switch?





Best Regards,
Keppe Motor Team

We are going to post soon the osciloscope plots for the 1.0 Keppe Motor.

Keppe Motor Team

Dear admin,

Thanks for the values. However, I have a question about that capacitance. Is it the capacitance of the coil or is it of another component of the circuit ?
Based upon your data and assuming a resistance of 2 Ohm for the battery internal resistance and resistances of other circuit components I would guess that the oscillation frequency of the voltage spikes after the opening of the switch is about 32 kHz (f = 1/ 2*Pi*sqrt(L*C)).
If the capacitance can be considered as in series with the coil, in my opinion the equation would become:

Us = Rt*i + L di/dt + Uac * sin(w*t) + S idt / C (" S " is meant to be the integration symbol)

or, after derivation:

Rt*di/dt + L*d²i/dt² + Uac*w*cos(w*t) + i/C = 0

My knowledge about mathematics is insufficient to solve that equation.

When I use the equation presented earlier (without capacitance), as described the general trend of the currents’ evolution are quite correct but the measured curves are mostly steeper. One could assume that by including a capacitance in the equation, curves would indeed become steeper.

Hello Greef,
If this helps, here you are some values taken from the Keppe Motor 1.0 (Kit):

Capacitance = 456 nF
Resistance = 48 Ohms
Indutance = 55 mH

For simulation we have checked the coil resistance (Rc) and it is indeed in series with the coil inductance (L).
Nevertheless we have been unable to determine if the capacitance (C) of the coil can be mathematically simulated in series or parallel with it. There is some weird factor still unknown to us here.
Tests don't fit any alternative so far.

Very good work of yours!!!
Admin

Hi,

Here I would like to check with you how the Keppe motor behaviour can be predicted / simulated. To model the motor, I used the following equivalent schematic:



My main doubt is if a capacitance is present in the motor. In a first approximation, I have assumed the circuit capacitance to be zero (0).

The above circuit can be simplified to a DC source, a resistance (Rt) a coil L and an AC source Uac. If I am not mistaken, Kirchoff’s law applied to this circuit would lead to:

Us = Rt*i + L di/dt + Uac * sin(w*t)

With
* Uac = amplitude of induced AC voltage
* w = angular velocity (Greek letter Omega)

My mathematics are not that good but as far as I can work it out, the solution of the above equation will look like:

I = A*exp(-t*R/L) - Uac*sin(w*t - delta)/(Z) + Us/Rt

With
* tan(delta) = w*L/R
* Z = total circuit impedance (= sqrt((w*L)^2+R^2) )
* A = constant

Angle phi is defined as the moment (expressed as an angle) the switch closes. The value of constant A can be determined taking into account that when the switch closes, current is equal to zero (i.e. solve the equation at moment t = phi / omega).



Definition of angle phi (assume sine pattern for induced AC voltage). Picture shows simplified oscilloscope plot. Example calculation of angle shown.

The formula quite well follows the general trends of experimental data but I am still quite far from real curve fitting. In my case, main uncertainty is about the coil inductance L and the DC source internal resistance. If I play a bit with those parameters, I can obtain pretty good curve fitting.

Mind that the above formula predicts that the final value of the current is not equal to = battery voltage / total resistance (Us/Rt) but can lead to values that can both higher or lower than that value.

Please comment and try out with your own test data.