Inductor kickback 1 of 2

I want to talk about a new example of an inductor circuit, and we have one shown here where this inductor is now controlled by a switch. This is a push button switch that we move in and out, and this metal plate here will touch these two contacts and complete this circuit here. This is a lot like the circuit we looked at before without the switch, where the voltage source had a pulse going up and down.

This circuit has a real interesting side effect when this switch opens up. So, let me give you a couple of examples, first some practical examples of where this kind of an inductor will appear connected to a voltage source and controlled by a switch. There are a couple of examples. One of them is called a solenoid. A solenoid is a coil of wire that looks like this and has a metal rod going through it. When you put a pulsive current into this, what happens is the rod moves. The rod will move back and forth, so this is a way, for instance, a doorbell. A doorbell is this kind of actuator.

Another kind of device that has an inductor in it, or a coil, is called a relay. That one again has a coil like this, and these are all forms of an electromagnet. This one actually has a switch, some sort of a piece of metal that will move back and forth. When the coil is energized, this piece of metal will actually tip over here, and when the current stops, it'll move back to its original state.

So, for example, if there's a connection point here and a connection point here, this piece of metal will go from making a short across there and move away and open and close this bigger switch. So, a relay is some sort of a switch, and you might find this in a car where this relay is controlling the windshield wipers of a car. The current for the windshield wiper goes through this switch, and there's a smaller current over here that pulls the switch open and closed to start and stop the windshield wipers or the motor.

That's just two examples of where inductors or windings of wire can show up in real applications, and there's some fairly large currents that flow through these windings. So, we want to look at what happens when we switch this current on and off, and this is the example circuit that we'll use. We'll assume we have an inductor of 10 mH, and what we're going to do is we're going to push this push button switch down and connect it up, and then we're going to let it go.

We're going to look at a couple of different voltages here. This is going to be, we'll call this voltage V, and that's going to be measured with respect to ground. Then there's another voltage here that's interesting too, which is VL, and that's the voltage across the inductor. So, V plus VL is equal to 3 volts all the time.

Now, we look at what happens when we push the button down, and then after that, we'll look at what happens when we let go of the button. Something interesting is going to happen when we let go of this button. As we look at our circuit here, we see there's an open circuit, so there's no current flowing in here, and that means I equals zero. We'll call that I zero because it's the initial current when we close our switch.

So now let's close our switch. We'll do that like this, and we just closed our switch at time T equals 0. Let's look at what happens. All of a sudden now, our inductor has a voltage across it. It's a voltage of this node is at 3 volts, and this node is at 0 volts. So, all of a sudden we have three volts across our inductor.

Let's use the integral form of the inductor equation to solve for the current that's going to happen here. I = (1/L) * ∫(from 0 to T) V dt + I0. Let's fill in what we know. I equals 1 over 10 mH, and V is a constant. V is a constant 3 volts multiplied by 3, and it's the integral of dt from 0 to T, and I0 is zero.

Then we get the final form which is I equals (3 over 10 mH) * ∫(from 0 to T) dt, which is just T. That's the answer. So what we have here, just like we did before when we had the switching power supply, is we're going to get a current that has a ramp that looks like that. The slope of that is 3 over 10 mH, which equals 300 amps per second.

Oh my goodness, this current is going up really fast! That's the slope of that current right there. This is going up really fast, and that's what it does. Now, in a real circuit, there'll be real resistances in here, and so there will be a limit to the amount of current that will be determined by the resistor. But for the purposes of showing you just how this inductor equation works, that's the kind of slope you would see at the initial closing of the switch.

Okay, I'm going to clean off the screen here so I can keep my same circuit, and now we're going to look at what happens here when we open this switch. Let me open the switch, and now we've opened our switch back up. It went that way. All right, so we have an initial current, and it's going to be some value depending on how long we held the switch down. So it's going to be some current, and that's flowing in the inductor.

Now let's look at what happens when we open the switch, and all of a sudden I goes directly and sharply to zero. That's what the switch does when we open this switch. Contact one moment it's touching, and the other moment it's not touching. So if we look at what's ΔI, or what's dI, it's the ending current minus the starting current minus I. If we look at what's the time involved, what's the change in time involved in opening a switch? Well, it's something like zero. The switch was closed, then it was open, and that's how much that took. Let's say that took zero time to happen.

Now here's something that happens with inductors that is kind of strange. Let's calculate the voltage on the inductor right when this happens, and we know that V = L * (dI/dt). Let's plug in some numbers here. So we have V = L times what dI is, which is minus I over what? Over zero, and that equals what? That equals negative infinity. What the heck is going on here? Is that possible? No, it's not.

Well, let's take a second. Let's say the switch didn't open in zero time. Let's say the switch, let's say dt was, um, let's say it was one nanosecond. Let's give it some time to open up. All right, maybe that'll save us; maybe that'll make more sense. Okay, let's do that. Let's go V = L * (dI/dt), and dI we decided was minus I over it took one nanosecond to open the switch. One nanosecond is 1 * 10^-9 seconds.

Okay, so what does that equal? That equals L times I times 10^9. Okay, let's plug in some real values and see what we get here. Let's say V equals, let's say L was 10 mH, and let's say I was 10 mA flowing down through. So ΔI is minus 10 mA, and the time involved is 10^-9 seconds. What does that work out to?

10 mH, that's minus 3 and minus 3, so it's going to be 100 minus (-3) -3 is -6 over -9 * 10^3. Okay, that is, and there's a minus sign here. This says that V, this point right here, or the voltage across the inductor, our calculation just said it's going to be 100 minus 100,000 volts.

Now, minus 100,000 volts means that the negative terminal is 100,000 volts above the positive terminal. So this voltage V is actually at 110,000 volts. How can this be? This is a puzzle about inductors that we actually have to solve by looking right up close at what's going on here, right in this switch area here.

So this is where reality comes in and saves us from the crazy results we're getting from our ideal equations here. So, let's pretend here's the terminal of the switch right here. Let's do a blowup, and here's that switch plate. As soon as this switch moves away, there's an air gap created here. In our ideal world, this air gap is a perfect insulator, but what's going to happen because of these extreme numbers here, what's happening is this air gap is actually not an ideal conductor.

What's going to happen is we're going to get a bright spark that goes right across here. This really happens in real life. There is actually, when you open a switch, there can be a little spark that happens in between, and that gives this equation here, this dI/dt, enough time to release that energy, and that current from the inductor continues to flow.

That current will flow, and it does it by breaking down air molecules. When you do that, that voltage there for air, if this gap here, if this gap is 1 millimeter for air, that's 3,000 volts. 3,000 volts will cause that spark to happen there. So, this is what actually happens, and you can build switches that will take this spark and work for many years. But it’s not always a good idea to let this happen.