Diode protection with inductors - Page 1
Forum section: electronics / beginners

author: Legion
author_rank: Frequent Contributor
timestamp: April 26, 2014, 05:18:41 pm
content: I was reading about using diodes for protection from large voltages when working with inductors and switches in AOE. I'm not sure if it's an error in the text or I'm misunderstanding something. The text states "When the switch is opened, the inductor 'tries' to keep current flowing from A to B, as it had been. That means that terminal B goes positive relative to terminal A." I know the book uses conventional flow, so wouldn't that be backwards? Wouldn't A go positive relative to B?The quote and attached diagrams are found on pages 52 and 53 of AOE 2nd edition.Second question. From the protected circuit diagram you can see that the diode routes the inductive kick away from the switch and back towards the positive supply. Does this re-routed voltage spike pose a threat to the positive supply source?

author: SeanB
author_rank: Super Contributor
timestamp: April 26, 2014, 05:44:44 pm
content: True, the voltage at the switch end does go higher than the supply. The diode merely then acts to short out the coil, dissipating the energy in the coil resistance and with some dissipated in the diode as well. As the circuit is only then completed by the coil and the diode, with only a single other connection to the supply no current can flow back into the supply so it will not be damaged. If the diode is not there the voltage across the switch rises to a very high voltage briefly, and this can cause either the switch to arc over the small gap as it opens, or for the coil insulation to break down. This then will show up as a voltage spike on the power supply.

author: TerminalJack505
author_rank: Super Contributor
timestamp: April 26, 2014, 06:07:11 pm
content: The best way that I've found to think about your first question is to realize that terminal A is held at a constant voltage--20V in this case--so it can't technically go anywhere, voltage-wise. It will always be 20V. So the other end of the inductor will see the voltage change.

author: David Hess
author_rank: Super Contributor
timestamp: April 26, 2014, 06:10:04 pm
content: Quote from: Legion on April 26, 2014, 05:18:41 pmThe book gets it right. When the switch opens, the current continues to flow as the magnetic field in the inductor collapses raising the voltage at point B until either high voltage breakdown occurs somewhere or the energy is exhausted charging the capacitance to point B.QuoteNo. Since point B is disconnected by the switch and left floating, the voltage rises until the diode is forward biased and all of the energy is dissipated in the diode. However . . .Since the voltage across the inductor is low, about 0.8 volts from the diode's forward voltage drop, the magnetic field collapses slowly which may be unacceptable. The alternative is to either place the diode across the switch and direct the energy into the positive supply or use a snubber circuit across the inductor which controls the peak voltage at a higher level allowing the magnetic field to collapse more quickly.

author: Kremmen
author_rank: Super Contributor
timestamp: April 26, 2014, 06:10:24 pm
content: While SeanB's explanation is correct the initial word "true" is ambiguous. At least if it is understood as a reply to OP's question regarding polarity.So - no, A wouldn't go positive relative to B. It happens just as the original text says.Faraday's induction law states that the voltage between the terminals of an inductor are directly proportional both to the inductance and rate of change of current through the inductor winding, i.e:V = L*di/dtIf you drive a current from a source through the inductor then energy is transferred from the source, and stored into the magnetic field of the inductor. The amount of energy is proportional to the instantaneous current passing through the inductor. If you now suddenly disconnect the source (by opening a switch), the inductor turns from a load into a source and starts to dissipate the stored magnetic energy (via the current flowing through it), into whatever load is available in the circuit. The point to note is that the current does not suddenly change direction - indeed it cannot do so because by the above formula step changes in inductor current would imply infinite voltages across the terminals. Which brings us to the second point. If there is no load for the energy to dissipate into (i.e. open circuit thus zero current) _in theory_ the voltage across the inductor would suddenly rise to infinity. Again, note that since the inductor in this case is the source and the direction of the current is known, then so is the polarity. It is B positive in relation to A because only that alternative preserves the direction of current.In a practical inductor the open circuit voltage will of course not rise to infinity, but a good quality coil will give quite a kick - several kilovolts if nothing breaks down before that. You can imagine what will happen to a poor transistor or a fet in such a case.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 27, 2014, 04:24:16 am
content: In other words, while the switch is on, dI/dt (in the direction of positive I) is positive, and when it opens, it's negative, so B becomes excessively positive because A is "anchored" by the voltage source.While the diode (if present) is conducting, the source sees no current whatsoever -- the switch is allowed to do its job. That is, reducing current to zero, more-or-less instantaneously.Note that, although the voltage *can* be very high,it only goes as high as you let it.With the diode in there, it's only allowed to get to Vf at most (if we're talking something like a relay coil and 1N914, all of 0.6V), andthe peak current is no more than the current flowing through the switch at the instant of turn-off.Suppose further, you have a little 12V relay driven by a 2N3904. It draws 100mA when on. How big of a diode do you need? If you were thinking 1N4007, you're wrong for three reasons (voltage, amperage and speed!) -- a 1N914 is more than enough, as it can handle 300mA easily (more than the ~200mA a 2N3904 can handle).Another question for you...How fast does the relay open?You'll have to know an awful lot about the relay to even begin to answer that question, but you can at least guess:it will be slower than normal. Why? Well, if V = L * dI/dt, and once the switch opens, V is limited to 0.6V (by the diode), then dI/dt can only be so much. An inductor takes a whole lot longer to discharge into a short, than a higher voltage load, or an open circuit. Typically, a relay will take on the order of 10ms to open, and might take double or triple the time if the coil is bypassed with a diode.Indeed, an ideal short circuit, to an inductor with no resistance, will maintain its charge forever: exactly the complementary case of an open-circuit capacitor, which maintains its charge with a load of infinite resistance (an open circuit)!So, naturally, if you want to run a relay as fast as possible (as fast as the spring and moving mass allows), you need a clamped -- higher, but finite -- voltage for the coil to discharge into. A transient voltage suppressor is ideal for this.Tim

author: johansen
author_rank: Super Contributor
timestamp: April 27, 2014, 04:33:47 am
content: relays can deliver pretty incredible volts per nanosecond if you ask them to.it really depends on the parasitic capacitance of that inductor, and how much current, relative to the capacitance of the diode.with a fast oscope, most any relay, and a cheap diode, you should be able to make out a voltage spike while the current builds up in the inductance of the diode.

author: Legion
author_rank: Frequent Contributor
timestamp: April 27, 2014, 11:48:57 am
content: How can B go more positive than A and have the current go through the diode? Current flows from positive to ground, if B is more positive the current will just go back up the inductor.Second, if the transient voltage spike can reach ~1000V, how can a diode with a 0.6V drop dissipate that energy? You should end up with 999.4V heading back up to the supply.

author: David Hess
author_rank: Super Contributor
timestamp: April 27, 2014, 01:20:12 pm
content: Point B goes more positive than A because the inductor current continues in the same direction when the switch opens. The voltage *has* to go up.The diode restricts the voltage to its forward voltage drop. It can handle the energy because the current through the inductor does not change immediately. If the diode was not there, *then* the voltage would rise uncontrolled. If the diode were a reverse biased zener diode, then the instantaneous power dissipated would be much higher because the inductor flyback voltage would be much higher at the same current level although for a much briefer time. That is what would damage the transistor if the diode were not present although some transistors are specified to operate under those conditions.

author: lewis
author_rank: Frequent Contributor
timestamp: April 27, 2014, 02:17:12 pm
content: As Kremmen says, when the switch is closed, the inductor is aload. Current is flowing through it which generates a magnetic field in the core and windings of the inductor which is great if we want to move some metal to make a relay contact close. But when you open the switch the inductor suddenly becomes asourcebecause the magnetic field stored in the inductor core needs to go somewhere. When the field collapses itgenerates(induces) current in the winding in the same direction as the current was first applied.Since (conventional) current was flowing from A to B with the switch closed, when the switch is opened the inductor will try tokeepthe same current flowing from A to B. But the inductor is now a source - it wants to get rid of the magnetic energy stored in its core. Think of it now as a battery: replace the inductor on the schematic with the battery symbol. But which terminal is positive? If current was flowing through abatteryin the direction marked by the arrow, terminal B would be the positive terminal. So B is more positive than A.With the diode in circuit, B is still more positive than A, but only by a diode drop. (B will be about 20.7V and A is 20V, relative to ground). The current that was flowing through the switch when it was closed will now be flowing through the diode with the switch open (remember, inductors want to keep the current the same at all times, they hate changing currents). The diode effectively short-circuits the inductor 'battery' and dissipates the energy stored in the inductor's magnetic field as heat in the diode junction.Without the diode, the inductor will try to keep the current flowing when the switch is opened. But there's nowhere for the current to go! There's 'infinite' impedance. So by Ohm's law the voltage at point B will 'instantaneously' rise and rise 'forever' until something breaks down and the current can go somewhere. In the case of a switch, the breakdown occurs as an arc across the switch contacts. In the case of a transistor, the breakdown occurs when the voltage exceeds the Vce rating of the transistor which causes it to go 'boom'. It won't go back in the winding because the winding itself is thesourceof the current flow.

author: Legion
author_rank: Frequent Contributor
timestamp: April 27, 2014, 02:25:28 pm
content: Quote from: David Hess on April 27, 2014, 01:20:12 pmCurrent flows from high to low potential. If point B reaches a higher potential than point A and they are connected, by definition current will flow from B to A. How is it ok that this fundamental principle is violated in this instance?Quote from: David Hess on April 27, 2014, 01:20:12 pmHow is the diode able to restrict the voltage to it's forward drop? Why is it able to absorb all this energy in this instance but if I have a circuit like the below it would be bad news?

author: Legion
author_rank: Frequent Contributor
timestamp: April 27, 2014, 02:42:06 pm
content: Quote from: lewis on April 27, 2014, 02:17:12 pmThe battery model is pretty good since it allows for point B to be at a higher potential than A and still maintain current flow in the expected direction. The only problem with the model is that in a battery there is some form of electron barrier that prevents (for the most part) the battery from discharging internally, but there is no such barrier here.I understand that the magnetic field is collapsing which provides kinetic energy to the electrons, inducing a current. I'm just stuck on this issue of current flowing from low to high potential.

author: Legion
author_rank: Frequent Contributor
timestamp: April 27, 2014, 02:47:38 pm
content: To clarify, I'm ok with V = L*di/dt. If AOE's explanation had stopped at "as the magnetic field collapses it attempts to maintain current flow..." I wouldn't have given it a second thought. But the sentence regarding B reaching a higher potential than A is what throws me.The above, and how a diode is able to limit the voltage of the inductive kick doesn't make sense to me.

author: AG6QR
author_rank: Frequent Contributor
timestamp: April 27, 2014, 03:28:34 pm
content: Quote from: Legion on April 27, 2014, 02:42:06 pmIf you don't like the battery model, consider the model of a transformer or generator. Those actually ARE inductors where the changing magnetic field causes current to flow.If you're stuck with the issue of current flowing from low to high potential, remember that it does this in every circuit you've ever seen. Current always flows in a closed loop. If it goes from a high to low potential somewhere, then it has to go back from a low to high potential somewhere else. You can't have a closed loop that goes "downhill all the way around".The inductor acts as a current source. The energy that was stored in the inductor before the switch was opened is what provides the power source.

author: lewis
author_rank: Frequent Contributor
timestamp: April 27, 2014, 03:48:11 pm
content: Quote from: AG6QR on April 27, 2014, 03:28:34 pmExcellent explanation.Regarding this circuit:That would be absolutely fine if the 1000V supply was current limited to, say, 1A. The diode would just cause the power supply to reduce its output voltage to about 0.7V maintaining 1A through the diode. The inductor works the same way. It's current is limited by whatever the applied current was when the switch was closed, no more can flow through the diode when the switch opens.

author: Legion
author_rank: Frequent Contributor
timestamp: April 27, 2014, 04:19:58 pm
content: Quote from: lewis on April 27, 2014, 03:48:11 pmOK. That makes a lot more sense to me, to view the inductor as a current source instead of a voltage source.