Jetijs,

OK. You need each transistor to operate in it's own little voltage divider (between the two resistors). Also, we know that if we parallel resistors, the effective resistance goes down. And finally, the little reed switch has current limitations that should not be exceeded, or it won't run very long.

So, step one is to take your motor that is running on one strand of wire and start increasing the value of the resistors until the motor starts drawing LESS current. Right now you are using 200 ohm resistors (approximately). See if the motor will run the same way if you put in two 680 ohm resistors in place of the ones you are using now. If it runs the same, try even higher values.

You may also have noticed that running your motor at 30 volts, your little 200 ohm resistors are getting quite hot. (Maybe you didn't notice.) The current in the control circuit is controlled by Ohm's Law (E=IR) So, E/R=I. 30 volts divided by 400 ohms (total of both resistors) equals .075 Amps or 75ma. Since each resistor has half of the voltage drop on it (15 volts), then each resistor is dissipating 15 volts X 75ma = 1.125 WATTS. If you are using 1/4 watt resistors, they should be getting hot.

Also, this energy is unrecoverable, so you want this control current to be minimized. So, you want these resistors to have the largest practical value that doesn't restrict the transistor from operating like a switch (transitioning from a fully ON position to a fully OFF position as quickly as possible).

So, find the highest value for your resistors that does not restrict the operation of the transistor (turning it into a resistor because it can't turn ON fully). Then use three sets of that value of resistor (or slightly less) to make the three voltage dividers to run the three transistors. Each voltage divider will begin by being connected to the ground, which is the negative of the supply and the connection to the emitter of each transistor. The mid-point of each voltage divider will connect to one of the transistor base terminals, and the top of each voltage divider will connect to the bottom of the magnetic reed switch. The top of the magnetic reed switch will connect to the positive supply voltage. This produces three branch currents to operate the three transistors from the single reed switch.

One more suggestion. Your little neon light is coming on because the radiant, longitudinal wave in the inductive collapse sees the long wires of your circuit arrangement as HIGH IMPEDANCE. Clean up your circuit and shorten the circuit paths to bare minimum, so one resistor is right on the transistor from the emitter to the base. The other resistor is right on the base going to the magnetic reed. The neon light is right across the emitter and the collector, and the diode and the bottom of the coil wire are connected directly to the collector. This will eliminate the neon triggering in the first two test set ups.

Keep up the great work!!!

Peter