What causes the voltage drop across a solenoid?

[Shaybay92]

We are currently learning about solenoids in my physics class, and am very confused about where this potential drop comes from across a solenoid. I understand that an opposing electric field is created when a switch is first opened because of Faradays Law and the fact that the magnetic flux inside the solenoid is changing. Is this opposing field the reason there is a voltage? Is this voltage then just temporary until it reaches steady state? Or is there's always a voltage across the inductor?

 

[univac]

There is always a voltage of L di/dt across an inductor. This describes the behavior of the voltage across the inductor for all time. A quick change in current will cause a large voltage to develop. A constant current applied for a long time will cause a short circuit or zero voltage. This is all contained in the compact description of the voltage across the inductor Ldi/dt.
Hope that helps...

 

[Shaybay92]

Ok so there is only a voltage when there is a back emf due to self inductance? (the opposing electric field due to a changing flux) and when the flux isn't changing then it no longer has a voltage and acts like a short circuit?

 

[Phrak]

Hold on. The back emf is a voltage, due to self inductance. However, in the idealized inductors, or solenoids you are studying in your physics class, the inductor never becomes a short circuit. The impedance is always L, and L is independent of the current in the inductor.

In the non idealistic world we actually live in, the inductance L depends upon current in the inductor when there is ferromagnetic material serving as inducor's core. For an air core solenoid, L can be assumed independent of the magnitude of current.

 

[Dadface]

Hold on. The back emf is a voltage, due to self inductance. However, in the idealized inductors, or solenoids you are studying in your physics class, the inductor never becomes a short circuit. The impedance is always L, and L is independent of the current in the inductor.

In the non idealistic world we actually live in, the inductance L depends upon current in the inductor when there is ferromagnetic material serving as inducor's core. For an air core solenoid, L can be assumed independent of the magnitude of current.

It is the inductance which is always L and the reactance is wL

 

[Phrak]

It is the inductance which is always L and the reactance is wL

Yes. My mistake. The impedance is not L. The impedance is i omega L. We can never say we don't have enough sticklers around here for details.

 

[vk6kro]

Solenoids are wound with wire and this wire is often very thin.

So, it has resistance and it is this resistance which determines the ultimate current flow in a solenoid. Solenoids used with DC currents have to have this resistance or they would become short circuits very quickly after turn-on.

Solenoids have a surprising lack of inductance because they do not have complete iron cores around their windings.

Laboratory solenoids are wound with thick wire and these have to have resistors or some other current limiting in series with them when they are used with DC currents.

 

[Wetmelon]

So when I drive a solenoid with a transistor, should I have a current-limiting resistor in series, or is it fine to leave it open, assuming the solenoid won't become a short circuit? I'm driving a pneumatic solenoid valve, 300mA @ 12V

 

[dlgoff]

If the transistor can handle a collector current of 300ma (which IS the limited current @ 12v) then you are good to go.

 

[vk6kro]

Yes, the solenoid will have a resistance of about 40 ohms and, assuming it is being used on DC, this will limit its current to 300 mA maximum.

A small power transistor on a heatsink should be able to switch 300 mA OK. A power Mosfet would be an even better choice.

You need to put a diode across the solenoid, like this:


[PLAIN]http://dl.dropbox.com/u/4222062/relay%20driver.JPG