There could be some small back EMF effect when the slug moves. However, I seriously doubt that is what is causing the high voltage on closing. There may be other effects:
- Relay contacts bounce. That means the solenoid will be disconnected multiple times even during a overall "on" operation. These short disconnects that happen after some current has built up could cause high voltage for a short time.
- Ringing. There is inevitable capacitance in the system accross the coil. When the coil is switched on, it is like energizing a tank circuit. In ideal conditions, this could ring up to twice the input voltage especially with contact bounce. In practise, the DC resistance of a solenoid is usually substantial enough to damp the system well enough, and the R and L of the solenoid dominate.
- It's not really there. The scope may be showing you things that aren't really happening at transients and especially with poor probe grounding.
I don't know what exactly is happening, except that I'm quite skeptical the EMF is really going to 200 V. I also don't like the PTC fuse being in series for testing these things. Try shorting it out and see what that does. Also try putting a reverse diode immediately accross the solenoid, not at the other end of a wire or on the other side of the PTC fuse. This should be a fast diode.
Interesting. The back-emf (modeled as a voltage source proportional to speed) is not equivalent to an inductance that depends on speed. Furthermore, there is no possible L(w) you can come up with that will make that assertion true.
I will describe a simple experiment, but in essence I'll be saying that they can't be equivalent because upon a motor load change, an inductor dependent on speed L(w) will not affect the stationary state current (torque after all transients have died down, becoming a contradiction), while a voltage source dependent on speed v(w) will (which makes sense).
Assuming a DC motor, a simple proof is to imagine that the load on the motor gets reduced. Because there is less load, the motor speeds up. Also imagine we wait for some time so that all transients go away (t=inf.). Now let's see what happens with both models:
With the back-emf modeled as a voltage source, its voltage increases because speed increased. This means that the current decreases, because the difference between the power voltage source and the back-emf voltage got smaller. This means torque decreased, which makes sense because we reduced the load on the motor.
On the other hand, no matter what inductance value you give to the "back-emf inductor", the current on the motor would remain the same, because inductors are short-circuits in dc. But this does not make sense, because torque is proportional to current and if the current remains the same, torque remains the same, but we started this analysis saying that we reduced the load on the motor.
Best Answer
simulate this circuit – Schematic created using CircuitLab
Figure 1. OP's schematic redrawn with switch replacing the transistor and normal current flow from top to bottom (as is standard in schematics).
Now it should be clear that when SW1 opens no current can flow into the battery positive as no "return" current can flow from the battery negative.
Instead, think of the inductance as trying to maintain the current in the direction it was flowing. With the diode present it will circulate from the bottom of the motor back to the top until it decays to zero due to the power loss in the diode and the motor winding resistance.
Your concern is valid if regenerative braking is required. In this case the generated power is passed back to the battery (or power supply). The usual way of dealing with this in variable speed drives or servo drives is to monitor the DC bus voltage and if it exceeds a certain voltage to switch in a load resistor to burn off some power. But that's a different question ...