I've made the project below, and it works. However, I'm confused about how it works, and why the diode is needed at all in this example.
As I understand it, the left pin of the Transistor receiving the off/on signal from the Arduino turns the current between the other pins of/on (thus grounding the solenoid). If I understand correctly, the solenoid at right has positive electricity flowing into it from the positive section of the breadboard, and then is able to ground the solinoid's ground wire (which is connected to the center pin) the when the transistor is turned on via the right pin which is connected to the breadboards gound. So, if that gives us the complete circuit we need there, why is the diode needed? As I see it, no current needs to flow in/out positively/negatively through it?
I understand that diodes can be used the stop the flow of current in the wrong direction. But, if that was the case here, wouldn't the diode need to be put right in line with main circuit?
In addition, I'm doubly confused by why the diode is connected to the positive power on the breadboard. Isn't the middle pin on the transistor negative? If the solenoid wishes to ground on that center transister pin, why would it "ground" to the positive breadboard?
I do notice that my push/pull solenoid heats up pretty fast, but I suspect that's another issue.
Yes put current through a coil or, as we say inductor, by applying a voltage to it and a magnetic field builds up around the coil. Then when you remove that voltage and so remove the current their is no energy to keep up the magnetic field. So the magnetic field collapses. This collapse makes the magnetic field lines cut through the windings of the coil quite quickly and induces a large voltage ( could be several hundred volts ) in the coil. This voltage is in the opposite direction to the voltage that created the magnetic field in the first place, so the coil becomes a voltage source. The diode is their to short out this voltage and thus protect the other components in the circuit.
In addition, I'm doubly confused by why the diode is connected to the positive power on the breadboard. Isn't the middle pin on the transistor negative? If the solenoid wishes to ground on that center transister pin, why would it "ground" to the positive breadboard?
The middle pin on that transistor is not negitave. When the transistor is off it is in fact at positive rail, that is +18V. This is irrelevant to your problem.
Forget talking about "grounding" it is wrong, irrelevant and just confusing you. Just think of the coil suddenly becoming a 400V battery, with its negitave terminal on your +18V rail and its positive terminal on the middle pin of your transistor. It would destroy the transistor. So the diode now conducts and shorts out the coil / battery. Once the field as collapsed fully no more voltage is generated by the coil and the diode stops conducting.
Basically to protect the circuit from reverse current from Solenoid when Solenoid is being turn off.
No, the current doesn't reverse, the voltage reverses to as large a value as needed to keep the
current flowing, the diode provides an easy path to prevent that voltage becoming destructively
large.
Inductive loads store energy in their magnetic field proportional to the square of the current - try
to rapidly reduce the current to zero and all that energy will be driven back into your circuit by force
(ie voltage). The "free-wheel" analogy is like the pedals of a bike - you stop pedalling but the bike
coasts on gently to a stop rather then slamming instantly to a stop and generating huge forces
on your feet!!
Wow, thanks, everyone! That was very helpful. It sounds like the physics of electricity are a bit more complicated than I had imagined. I've learned quite a lot.
I love the bike analogy MarkT and thanks for the thorough answer Grumpy_Mike!
