Need help understanding importance of Inductance

[bonedoc]

I have been trying to learn about the importance of inductance in circuits. I believe that a person could create a solenoid and when current is applied, it produces a back EMF. To me, it seems as though it is acting like a resistor, or maybe even a way to store energy in a field.

But, in the event that one wants to create a solenoid for a door bell, lock, or whatever....what does inductance mean? In these situations, it seems as though a larger magnetic field would produce a stronger solenoid, and a larger inductance. Is this ok? Does this mean the circuitry that is driving the current to the solenoid needs protection from a larger back EMF? Should a person designing such things try to minimize inductance by altering the dimensions of the solenoid?

Seems pretty confusing, but the ratio of the diameter to the length seems to produce a predictable inductance number....but what this means in the practice of design, I have no clue.

 

[Resqueline]

In short for all questions; yes, except one doesn't usually try to minimize inductance in a solenoid (it would be counter-productive) - one just protects the drive circuits.

 

[bonedoc]

Interesting. I was wondering why nearly everything I read just said how to calculate it, but not manage it. In the case of a solenoid, doesnt the back EMF produce massive voltages?

How does one protect the circuitry from the reverse emf? Zener? Shotsky?

In the case of line voltage applied to the solenoid, does one just reverse bias a diode from source (ground) to drain (solenoid input).

In the case of a capacitor powering a solenoid, is it possible to reverse bias the diode from the positive side of the capacitor to the drain (solenoid input). This way, when the back EMF happens, it will slightly recharge the cap?

Finally, how does one choose a good protection value? Say the switch is rated at 500V, does one choose a diode whos reverse voltage threshold is just below this?

Thanks for the help!

 

[Rleo6965]

In the case of a capacitor powering a solenoid, is it possible to reverse bias the diode from the positive side of the capacitor to the drain (solenoid input). This way, when the back EMF happens, it will slightly recharge the cap?

It won't recharge filter capacitor because emf current was smaller than charge current of capacitor.

 

[bonedoc]

I guess I am confused. Why would it not charge the capacitor, yet cause significant damage to switches?

 

[Laplace]

The following article, A TRANSIENT SUPPRESSION GUIDE, might help your understanding:
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1338123/pdf/jeabehav00172-0065.pdf

 

[(*steve*)]

A capacitor and an inductor are essentially the opposite of each other. You really can think of an inductor as an anti-capacitor.

A capacitor "stores" voltage, and inductor "stores" current.

When you remove the source that is charging a capacitor, the capacitor attempts to keep the voltage across itself constant. Current will flow, if required to try to maintain it. In the process of the attempt, energy stored in the capacitor will change. If your load becomes a short circuit, the current rises to a very high value (to try to maintain the voltage), and this can cause damage.

When you remove the source that is "charging" an inductor, the inductor attempts to keep the current through itself constant. The voltage across it will change, if required to try to maintain it. In the process of the attempt, energy stored in the inductor will change. If your load becomes a open circuit, the voltage rises to a very high value (to try to maintain the current), and this can cause damage.

In both cases, the device can only dump the amount of energy stored. So a capacitor is likely to generate a brief, high current pulse which will damage stuff due to over-current. An inductor is likely to generate a short high voltage pulse that will damage stuff due to over-voltage.

If you connect a capacitor in parallel with an inductor, each can discharge into the other. Because of their nature, discharging one charges the other, and the process is like a pendulum that will swing back and forth (with energy going from one to the other and back again) until the energy is dissipated in losses (such as wiring resistance).

Breaking a circuit containing an inductor will cause the voltage to increase to the point at which the same current can flow. This may be enough to cause an arc in a switch (and high instantaneous power leading to damaged or melted contacts) or a destroyed transistor as the voltage gets high enough to break it down.

Providing an alternate path for this energy is the role of diodes, and snubber networks.

Protection from capacitive loads is generally done with resistors to leak away the stored charge slowly.

 

[bonedoc]

Thanks guys! that makes a lot of sense. I am reading that paper now. There is a formula for the calculation of a RC protection. It needs the "Amperes of load current immediately
before the opening of the contacts." How does one calculate this value, say for a 10v, 10uF cap?

 

[jackorocko]

Steve that was awesome, but I see a mistake. I don't believe you meant to say capacitor.

When you remove the source that is "charging" an inductor, the inductor attempts to keep the current through itself constant. The voltage across it will change, if required to try to maintain it. In the process of the attempt, energy stored in the capacitor ( <-- inductor ) will change. If your load becomes a open circuit, the voltage rises to a very high value (to try to maintain the current), and this can cause damage.

 

[(*steve*)]

I see a mistake

Well spotted that man!

 

[BobK]

The voltage across an inductor is proportional to the rate of change in current in the inductor. Think about what that means when an inductor has some current flowing in it and you suddenly break the circuit. In an ideal world the current has to go instantly to zero and thus the voltage becomes infinite. In the real world this does not happen: the current finds some path to continue the while it collapses, which can cause large enough voltages to overcome even an air gap, causing a spark or to exceed the breakdown voltage of a semiconductor. Note that since the current is desceasing, the voltage across the coil is in the opposite direction of the voltage that cause the original current. So a reverse biased diode across the inductor allows the current to continue flowing (since the voltage has now reversed and the diode is forward biased) and therefore the voltage spike is kept manageble (i.e. to a diode drop). Meanwhile, the resistance in the coil dissapates the enegy until the current stops altogether.

Bob

 

[(*steve*)]

The voltage across an inductor is proportional to the rate of change in current in the inductor.

A far more technically correct answer than mine