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- A capacitor integrates current
- Capacitor i-v equation in action
- Inductor equations
- Inductor kickback (1 of 2)
- Inductor kickback (2 of 2)
- Inductor i-v equation in action
- RC natural response - intuition
- RC natural response - derivation
- RC natural response - example
- RC natural response
- RC step response - intuition
- RC step response setup (1 of 3)
- RC step response solve (2 of 3)
- RC step response example (3 of 3)
- RC step response
- RL natural response
- Sketching exponentials
- Sketching exponentials - examples
- LC natural response intuition 1
- LC natural response intuition 2
- LC natural response derivation 1
- LC natural response derivation 2
- LC natural response derivation 3
- LC natural response derivation 4
- LC natural response example
- LC natural response
- LC natural response - derivation
- RLC natural response - intuition
- RLC natural response - derivation
- RLC natural response - variations
Inductor kickback (2 of 2)
We give the inductor's kickback current a path to flow by adding a diode. Created by Willy McAllister.
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- Under ideal conditions would the current in the diode, inductor circuit remain unchanged forever? Or at least until the button is pressed again?(5 votes)
- Hello Hstrathman,
An ideal diode has zero voltage drop when forward biased and zero leakage current when reversed biased. By these definitions it has zero resistance.
An ideal inductor has zero resistance.
If we have current flowing in this ideal inductor and then disconnect it the current will continue to flow through the fly-back diode. Since neither part has resistance the current would continue on forever.
In the real world this is not possible and the inductor's energy will be "burned" in the resistor.
There are some interesting application related to this concept. Suppose you have a solenoid such as the injector for a diesel engine. Ref: https://www.youtube.com/watch?v=NUvWnOd5lFw
Here it is important to turn off the solenoid as quickly as possible to stop the flow of fuel. Know that the solenoid will remain energized as long as current flows. Once again the trick is to dissipate the inductors energy as fast as possible. This can be done with resistors or other more complex methods.
- Is there somewhere where there is effectively an endless amount of circuit problems and videos that walks through the process of how to solve certain circuits? (RC, LC, RLC, Op-amp, etc.? )(2 votes)
- Hey Willy, then can we replace the diode with a capacitor for the same result? If yes, does the connected polarity of the capacitor matters? Thank you in advance.(2 votes)
- No. A capacitor can't be used in place of the diode. The action of the circuit needs to have the forward conducting path through the diode in order to protect the switch. A capacitor does not have a conducting path.(2 votes)
- Is there a similar effect with capacitors...?(1 vote)
- Capacitors are the "dual" of inductors. Exchange terms: C for L, v for i, and short for open...
If you build up a current in an inductor and then suddenly open the circuit, you get a sudden jump in voltage.
If you build up a voltage in a capacitor (storing charge) and then suddenly short the leads of the capacitor, you get a sudden jump in the current (all that stored charge comes out in a rush).
Both events can damage nearby components.
The capacitor effect does not have a special name like 'kick back'. Usually it's just called 'Dang it! Stop doing that!".(3 votes)
- sir,I am confused a little bit on the issue that when we close the circuit inductor should oppose current and voltage should be less than 3?please help me if i am missing something.(2 votes)
- why is the stored energy around the conductor preventing or opposing the current that create this stored energy? is the reason behind that the principle of energy conservation?(2 votes)
- Hey Willy, thanks for tutorial. I learnt alot. However, I have some questions and hope you or anyone could help me with it.
1) At4:58, you mentioned that the current is headed up. At5:44, why is there a current flowing down through the inductor when the circuit is still open?
2) As there is a new current path to flow, would the 3.6V keep going up to 100,000V as calculated before?
3) Would the inductor be activated again(even though the circuit is open) as the current is going in a loop as described at5:55, flowing through the inductor in a forward biased manner?
Thank you in advance!(1 vote)
- 1. There is a current flowing in the inductor when the circuit is open because the inductor says so. You cannot force the current in an inductor to instantly go to zero. It's like trying to stop a spinning bicycle wheel with your hand. The momentum of the wheel will force your hand to move, even if you don't want to.
2. No. The voltage stays "clamped" at 3.6V by the diode. The inductor current is happy because it has a new path to flow, so there is a much smaller change of current (di/dt) and therefore a much smaller voltage = L di/dt.
3. The current flowing up through the diode comes around and flows down through the inductor. This loop current continues until the it eventually falls to zero. When we talk about inductors we don't need terms like activated or forward biased. All you need is current, slope of current, and voltage.(1 vote)
- Hey Willy,
I have a feeling that I may have been confused with "Back EMF" and "Inductive Kickback" generated by inductor(s) in the circuit.
Further search and reading on Google didn't help to distinctively differentiate them, except with common suggestions to "always use capacitor to help regulate voltage ripple in the circuit caused by inductor(s)".
1) But how are those voltage ripples different from the inductive kickback mentioned here?
2) And if they are different, how was it created(differently) compared to the inductive kickback taught here?
Hope you could shed some light on the matter pls.(1 vote)
- Will the low voltage generated from a diode enough to create a spark to actuate the switch that is supposed to be actuated by a 100,000V?(1 vote)
- The switch is only actuated by pushing it closed. The spark is created when there's both a small gap while the switch is being opened, and a high voltage from the inductor kickback (ie the 100,000V). Because the diode stops the high voltage from being made, there won't be enough voltage to make a spark across the air gap.(1 vote)
- Hi !
At3:22, when the button is pushed down, what stops the current to flow into the diode after it leaves the inductor ? Is it because there is no voltage difference over the diode ?
Thank you.(1 vote)
- When the switch is closed there is a voltage difference across the diode. 3V on top, 0v on the bottom. But, if you look carefully at the diode direction, its symbol is pointing up, so it is reverse biased. Pretty much zero current flows in the diode. All the current is in the inductor.(1 vote)
- [Voiceover] So, the problem with allowing this spark to happen across here is if this is not a mechanical switch, we can build switches out of electronic devices, as well. But this is what we use transistors for. And a transistor is a rather small, delicate device. And so, if I go in here and I somehow cause a huge spike in voltage to occur, very, very likely I'm gonna destroy that transistor that's used in this position right here as an electronic switch. So, the way to deal with this is to provide a place for this current to go. Here's this current that's flowing through this conductor, and the magnetic energy that's stored around this inductor is gonna force that current to flow, even if we open this switch. And in the case here we saw when we opened the mechanical switch, it sparks right across there, so the inductor wins, the inductor current wins in that case. And it can be very damaging to whatever is in this position down here. The best way to solve this is with a device called a diode. And this is another semiconductor device, it's not a switch, but a diode has, a diode has an IV curve. Here's V, here's I. And it looks like this. The current, I'm just gonna sketch this, the current is zero when the voltage on the diode is negative. And it's somewhere around here that the current goes up like that. So, when we have a diode, this is the symbol for a diode, like that, and that's plus and minus V. And this is the current through the diode. So, positive current and positive voltage is in this part of the graph over here, and when I have a positive voltage, this is about .6 volts, something like that, could be .5, could be .7, when you have a positive voltage on this, then the current starts to flow freely through this. If this voltage goes negative, so that means that this voltage is higher than this voltage, then the current actually goes pretty much to zero, very, very small value. So, I'm gonna take advantage of this diode device to help me with this problem I have over here with the current, and the way we do that is we do this. We hook up a diode pointing in this direction here. Pretty distinctive. But anytime you have a circuit that has an inductor in it, so it's a solenoid, or a motor, or a relay, this is a way you can protect the devices driving your coil. So, let's look at what happens here when we push this push button down and let it go again. When we started out, we had no current flowing, the push button was open, so there's no current going through here, and there's no voltage difference across here, both these points are at three volts. And so, there's no voltage across the diode. And that's that point right there, no current, no voltage. So, all the currents are zero, and the voltage across the diode is zero, so there's no current. Nothing's happening. Now we push the button down. And a current starts to flow through the inductor, and there's three volts across the inductor. So, let's look at that, there's plus three volts here, and there's zero volts down here. So, there's three volts on this side of the diode, this is the side where it doesn't conduct. That three volts represents negative three volts across the diode. Let me mark the diode voltage on here. Here's minus, plus, V diode. And we'll call this V diode. So, this is a V diode of negative three, so we're operating over here out at negative three volts on the diode, and the current through the diode is zero, there's no current flowing over here. All of this current is going down and going through the inductor, and doing whatever this inductor is supposed to do. Okay, now we open the switch. Now we open the switch. And as you recall a minute ago, what happened was this voltage right here, this voltage went big. This went plus big, and we had like, 100,000 volts or something. That's where it was headed. Well, we're not gonna let it get there, so I'm gonna take that away. So, this voltage here, as soon as this switch opens, this voltage right here, this is headed up. This is going up, hard. But we're not gonna let it get very high, and that's the job of this diode. So, as soon as this voltage here gets to about 6/10ths of a volt higher than right here. So, when this voltage gets to 3.6 volts, what that means is that the V diode... Now equals about .6 volts, right? This is at three volts. This is at 3.6 volts, so V diode is .6 volts. And that puts us about here on the curve. And what's gonna happen is this current that we had here, this current that was flowing down through this inductor is gonna do what? It's gonna go this way. It's gonna flow this way through the diode, and that's this portion of the curve right here. And it'll keep going like that. Look at that, we've provided a current path for this inductor that's not gonna destroy what's down here. The highest voltage this will get to is whatever voltage it climbs to on this diode curve here, which is gonna be between .6 and point, say .8 volts. So, the highest this is gonna get is point, 3.6 to 3.8 volts. And typically, and typically a transistor down here will easily withstand this kind of a voltage on its terminals. So, this is called, the effect of having this voltage go way north is called kickback. That's kickback from an inductor. This diode is a protection diode, and basically what it does is it supplies a pathway for that inductor current when this voltage goes above the top end of the inductor. So, whenever you build something that has an inductor in it, and it could be, you could build a circuit that drives a motor, or an actuator, or a solenoid, or something like that, or part of a robot, this is one of the circuits you wanna keep in mind. Whenever you have an inductor and you're gonna switch it on and off, you basically, you wanna design in a diode like this to protect your circuit from those unstoppable inductor currents.