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Role of diode in a circuit with coil

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insight

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I understand that diode is place in the same circuit as coil (DC motor or relay) to prevent a voltage from hitting semiconductors, like transistors really hard.

Diode is always shown in parallel with coil, but what I think is strange, what is there to prevent voltage from going through node back into transistor. I know that diode lets the current only in the one direction, but at the node current can split into two wires.
 
Think of the current going through the coil. If the current flow is interrupted by a switch, the inductance of the coil will attempt to keep the current flowing through the coil, which can generate a large voltage spike (of the opposite polarity to the voltage originally powering the coil). The parallel diode provides an alternate path for the inductive current until the stored inductor energy is dissipated. Thus the inductive spike is limited to the forward voltage drop of the diode.
If you draw a diagram of the current flow, both before and after the switch is opened, you will see how the diode prevents a large voltage spike. A common misconception is that the voltage spike is the same polarity as the original voltage applied to the coil, but it is actually the opposite polarity.
 
Current must flow in a loop. That's why it doesnt split. When the inductor spikes, becomes like a source and current from a source must flow back to the source. THat's why it doesn't branch off at the nodes and only flows in the loop formed by the inductor and diode.
 
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ok, so inductor becomes, for a very short period of time like a battery or charged capacitor. And diode simply directs this current around the inner, inductor - diode, circuit.

but I would say that this mini circuit, as a whole, is like a current source and that a little bit of current must be leaking out into the ground and thus endangering the rest of circuit.

I would like to record this spike. Is the inductive spike dangerous for the oscillator?
 
but I would say that this mini circuit, as a whole, is like a current source and that a little bit of current must be leaking out into the ground and thus endangering the rest of circuit.

I would like to record this spike. Is the inductive spike dangerous for the oscillator?
Nope. Can't leak. Current doesn't wander unless it has a path. The only spike you will see is the value of the forward diode drop.

You can view the spike as long as the diode is across the coil. Otherwise you can get some high spike voltages, as determined by the value of the inductance, the inductor current, and the stray capacitance. That could endanger your "oscillator" (I assume you mean oscilloscope).:D
 
Ground is relative and the negative end of the inductor is ground to the current flowing out of the positive end (when the inductor is behaving as a source). It will always flow back to where it came from, regardless of whether or not there is a lower voltage somewhere else.

What are you seem to be referring to is common mode currents. But in this case, nothing is floating so there aren't any common mode currents. Even though no current flows off into the branch of the node, that branch is still tying one end of the potential difference formed across the inductor to a source or ground. So when the voltage at the other end of the inductor is measured relative to system ground, it will be higher. BUT voltage doesn't flow. Current flows. Just because there's a voltage does't mean there is a current.
 
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ok, at the moment when coil is shut off, there are two grounds:

1) the other side of the coil, the side with lower potential and diode links to this one.

2) the original ground, from the initial circuit design.

What I am thinking, when that voltage spike hits the first branching node, some small current must be going into the original circuit's ground and much bigger part goes through the diode into the opposite end of the coil.

If I connect oscilloscope across the coil, in the moment when coil is shut down, will the high voltage spike damage the oscilloscope. I would like to record the event so I can understand it better.
 
Ground 2 is the ground YOU are choosing for devices to use a reference in the system. THe inductor flyback current doesn't care about what you chose though. All it knows is that it wants to get back to where it came from and that loop diode/inductor is the easiest path. THat's the whole reason you add the diode there. The inductor current WILL flows and it WILL find it's way back. It will take the easiest path and the inductor will behave like a source of whatever voltage is required for current to flow through that easiest path, even if that easiest path might not be so easy and require many many volts to force current through it. WHen the diode conducts it ideally acts like a short-circuit in parallel with the inductor, and in parallel with the rest of the circuit. The low voltage drop of the diode makes it so the inductor does't need to be such a high voltage source to force current to flow (which would damage the rest of the circuit since for things in parallel voltage is the same). And current doesn't flow through the rest of the circuit either because the diode is acting sort of like a very small resistance in parallel with a large resistance (the rest of the circuit)- all current flows through the small resistance. So I guess, yeah, nothing is ideal and i guess something does go through the rest of the circuit, but it's negligible because the diode has not only clamped the voltage down thereby reducing the ability of the inductor to force current to flow, but the diode has also provided an easier path for current to flow. HOwever, the current still is not flowing to your circuit ground. THat's not it's goal. It's goal is to make it's way back to the other end of the inductor. Current flows in loops.

Now...without the diode...that current from the inductor will indeed flow through the circuit (not to ground, but through the circuit, but maybe through ground but not to ground), but it will find the easiest way to get back to the inductor. And if the easiest path requires the inductor to act like a 1000V to force the current through, then it's going to do it and burn things along the way.

It's easy to think of ground as something that everything returns to. In reality, it's just an arbitrary point in the circuit that has been chosen to be the reference voltage for the components (often you have no choice since the IC designers have already made this choice for you). Current flows in loop and that's it- nothing says it actually needs to go your arbitrarily defined circuit ground, particularily if your the potential difference causing that current flow isn't using your system ground as a reference. Current originating from a source that is referenced to ground however will flow back to ground- but not because it's "ground". It could care less about that. What it cares is that going through ground is the easiest path that it can take to get back to the second terminal of the source that it came from. Think of a circuit with multiple batteries all over the place? What is ground? Nothing is really. It's just an arbitrary definition. What realy matters is currents coming from a battery will find their way back to the other end of the battery.

I'm not sure if you can connect your scope or not...you'll still probably be able to see the some of the spike even with the diode because the diode takes time to turn on. It's not a perfect solution after all (nothing is). The less current is flowing through the inductor the moment it is shut off, the smaller it will be as well so that would help.
 
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thanks man, that is very detailed explanation :), I'll learn a lot from it.

One thing caught my eye, are you saying that inductor can create any voltage that is necessary, even 1,000V, to find its way back to the other end of the coil.
 
Yes, I am. That's how spark plugs work. The easiest current path is through the air and so the inductor produces a voltage high enough to force current to flow through the air.

It's like this:
Inductors store energy but they only do so when current is being forced through them. If you stop forcing current through it, it will release it's energy. (Think of stuffing a kid's mouth full of grass. He's only going to keep that grass in his mouth as long as you keep stuffing more grass in). Inductors also try to keep current flowing continuously (and they use their stored energy to do this). That means if you suddenly break the current flow, the inductor will release it's stored energy and that energy will go towards trying and keep that current at that level as best as it can. If the easiest path is low resistnace path, that energy will be released in the form of a low voltage across the inductor and lots of current that takes time to collapse to zero as the stored energy runs out. If the path is a high resistance however, that energy will come in the form of very high voltage and low current that collapses to zero very quickly. Of course, the inductor only has a finite amount of energy and something has got to give eventually, so the inductor isn't going to magically keep 100A flowing forever after you've disconnected it. The current collapses eventually.
 
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must admit, I never understood inductors, because of that interplay with magnetic field.

In my mind, I use this analogy with water, which works just fine with wires, resistor, capacitors, diodes etc. Inductors are kind off, an odd man out.
 
only the example with kid's mouth is maybe not good, because kid's mouth would have a preference for the direction, once it starts spewing it back. While, coil, would throw the current in whichever direction offers the less resistance. Am I right here?
 
think of the inductor as an enclosed paddle pump with a big flywheel. water flow starts the paddle wheel turning, turning the flywheel. stop the current, and the flywheel keeps turning, building up a pressure (voltage) difference between the ends of the pump. a diode (one way valve) across the pump shunts the pressure difference, and the current flows locally between the pump and the one way valve.
 
Yes, I am. That's how spark plugs work. The easiest current path is through the air and so the inductor produces a voltage high enough to force current to flow through the air.

It's like this:
Inductors store energy but they only do so when current is being forced through them. If you stop forcing current through it, it will release it's energy. (Think of stuffing a kid's mouth full of grass. He's only going to keep that grass in his mouth as long as you keep stuffing more grass in). Inductors also try to keep current flowing continuously (and they use their stored energy to do this). That means if you suddenly break the current flow, the inductor will release it's stored energy and that energy will go towards trying and keep that current at that level as best as it can. If the easiest path is low resistnace path, that energy will be released in the form of a low voltage across the inductor and lots of current that takes time to collapse to zero as the stored energy runs out. If the path is a high resistance however, that energy will come in the form of very high voltage and low current that collapses to zero very quickly. Of course, the inductor only has a finite amount of energy and something has got to give eventually, so the inductor isn't going to magically keep 100A flowing forever after you've disconnected it. The current collapses eventually.
You dont force current through an inductor
Inductors don't try to keep current flowing
When a voltage is removed from the inductor the magnetic field collapses & in doing so generates a voltage in reverse polarity to the original voltage. The diode is to provide a short circuit to this induced voltage. The presence of the diode actually increases the induced voltage as it causes the rate of magnetic collapse to increase.
 
If you don't believe that an inductor causes a high voltage spike when its power is turned off then hold the coil terminals of a relay in one hand, apply a battery then disconnect the battery. You will feel a big zap.
 
Inductors don't try to keep current flowing
That's not true. Just place a short across an inductor which is carrying current and monitor the current flow (you can easily do this in SPICE). The current will continue to flow but goes to zero due to IR loss in the resistance of the coil, which absorbs energy from the magnetic field and causes it to collapse. If you have a superconducting coil, the current will continue forever without any power required, which is why they use superconducting magnets in the big particle accelerators.

If you interrupt that current, then you get a voltage spike as the field attempts to keep the current flowing.
 
The presence of the diode actually increases the induced voltage as it causes the rate of magnetic collapse to increase.

This also is not true. The presence of the diode clamps the voltage across the inductor to the forward voltage of the diode (if the diode is just a simple non-zener diode). This will be about .7 to maybe 1.0 volt. This low voltage will slow the rate of magnetic collapse.
 
Maybe "force current" is the wrong phrase. But you can drive current through an inductor by playing a potential difference across it. Granted, that you can't actually decide how much current you want to flow- the inductor has control of that. I didn't mean "force" current through an inductor by doing something like connecting it to a current source.
 
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