"flyback diodes" / "back emf diodes" how big ? - Page 1
Forum section: electronics / projects

author: Simon
author_rank: Unknown
timestamp: May 25, 2010, 05:27:24 pm
content: having ascertained that where i work we definitely need to start making use of fly back diodes on electric fans they have turned to me to ask: how big a diode do we need ? and that's something I've always just eyed up and said: "yea that one will do it", but is there some sort of rule of thumb for example based on the amperage the motor or other inductive load uses ?

author: Unknown
author_rank: Unknown
timestamp: May 25, 2010, 05:38:49 pm
content: The basic principle of an inductor is that it wants to keep a constant current flowing, even if it's switched off. So I would expect the current not to exceed the regular operating current, but I admit that I've always eyeballed these things, too (and my stuff is low-power, so almost anything is overkill). Depending on the sensitivity of other things that may be damaged, speed may be a larger issue. The diode should be faster than anything it's protecting (so an 1N400x might be too slow).

author: Simon
author_rank: Unknown
timestamp: May 25, 2010, 06:02:40 pm
content: I suppose an amperage rating equal to the load is a good rule of thumb ? after all the power has just been removed and a high voltage is being generated therefore the amperage is not going to be any higher than the working amperage.Now as for the actual occurrence of back emf am I correct in thinking (as I am the one that will have to lecture my superiors on this): the removal of power leave the coil with no power flowing through it but a magnetic field that is 90 degrees out of phase that induces in the coil now longer powered a voltage that is another 90 degrees out of phase so creating a negative (180 degree out of phase) spike. or is it that the magnetic field is already 180 degrees out of phase ?

author: jahonen
author_rank: Super Contributor
timestamp: May 25, 2010, 06:40:19 pm
content: Short answer: diode current rating must be equal to current flowing in the load.Long answer: The "back EMF spike" (I somewhat hate the term, since it really is just a characteristic of the inductance) results from the basic equation of the inductor, voltage is equal to inductance times the rate of change of the current (U=L*di/dt). Now, if you abruptly just stop the current, then the resulting voltage is large. Resulting voltage polarity becomes so that it resists the change in the current. Current tries to flow same direction after the disconnection of the supply. Thus the diode direction must be so that current can flow through it. As it happens, it settles reverse biased across the load to satisfy this condition.BTW, this "back EMF" is perfectly analogous with what will happen if you short-circuit a fully charged capacitor (I=C*du/dt). Just current and voltage change roles (big current spike when voltage changes fast). It is kinda strange that shorted capacitor case is easier for people to understand.But if you can arrange alternative path for the current so that it decays slowly to zero, then there is no spike. By connecting diode across the inductive load, a controlled path is provided for the current, current decay is slower and resulting "back EMF spike" is small. This causes same current going through the diode, what was going from outside supply before the turn-off. Thus diode must take initially same current what load takes, decaying slowly to zero.Regards,Janne

author: hans
author_rank: Super Contributor
timestamp: May 25, 2010, 07:07:10 pm
content: Please note 2 things:The diode must be to handle the load of the motor (so as much as the motor pulls is indeed a good rule of thumb).Make sure the diodes are fast. You could those really cheap heavy duty ones, but they need to switch on/off fast. Well, what is fast? A 1N5401 or something isn't in most cases I think . I would consider something like a BYV 27-200 if you don't need to switch that much current around. I don't know exact figures on how many nanosecond you should look for, maybe someone can replenish me on that

author: tecman
author_rank: Frequent Contributor
timestamp: May 25, 2010, 07:11:02 pm
content: As stated, the diode needs to have at least as high a current rating as the load current. Voltage needs to be at least as high as the operating voltage. For automotive I would not go below 100 volts due to the various anomalies that can spike the supply to sometimes outrageous levels. The other factor is power rating. If the on-off duty cycle is high, the thermal conditions may require a large diode and heat sinking. If it is a low rate, 10s or more seconds between switching, power dissipation should not be an issue. Remember when the circuit is switched off, the diode will initially handle the load current due to the inductance. With a motor load, the inertia of the rotating load will add to the time that the diode is conducting.paul

author: jahonen
author_rank: Super Contributor
timestamp: May 25, 2010, 07:12:18 pm
content: I don't think that fast diode is necessarily better for this kind of application, since there is no reverse recovery situation, unless the load is re-powered while the current is decaying through the diode. But it won't hurt if fast diode is used.Regards,Janne

author: Unknown
author_rank: Unknown
timestamp: May 25, 2010, 08:04:29 pm
content: Quote from: Simon on May 25, 2010, 06:02:40 pmThe point of the diodes is that no high voltage will be generated. The high voltage is generated in an effort to keep the current flowing, if it can keep the current flowing with a low voltage, the voltage won't rise (except in the short time until the diode start conducting). The magnetic field will quickly disappear (depending on the inductance), and the current will stop flowing.These links might be helpful:Surge suppression and diodes installation tipsUsing FREE Inductive Kick Back current in a motorInductorsQuote from: Simon on May 25, 2010, 06:02:40 pmI'm no expert in magnetics, but as far as I know, back emf is generated when the coil is charged, and has a voltage opposite of the charging voltage (which is why it reduces efficiency in motors). If the voltage is removed, the magnetic field collapses, and the current will keep flowing in the same direction. It's hard to talk about out of phase for DC current. For AC current, the current is 90 degrees behind the voltage, and the magnetic field is in phase with the voltage. Each time the voltage rises, back emf is generated. Each time the voltage falls, the emf opposes this motion. I'm still puzzled why the voltage would turn negative when the power is removed (i.e. in the same direction as the change in magnetic field), maybe the positive pulse is exciting something that's resonating (ringing)? Just went back and looked at your measurements, it's definitely ringing, but it starts with a negative pulse.Quote from: jahonen on May 25, 2010, 07:12:18 pmOK, the time from reverse biased to fully conducting may technically be forward recovery, but aren't most slow diodes also slow in this regard? The voltage will increase until the diode starts conducting, although I'm not sure how fast. A schottky diode might work, but if you really need 100V reverse voltage, that might be hard to find.

author: Pyr0Beast
author_rank: Contributor
timestamp: May 26, 2010, 01:12:17 am
content: You need a fast diode to clamp the load before voltage becomes excessive.No shottky diodes are needed. Diode's voltage drop isn't that criticall, perhaps you could go with 4 V drop or even more.Diode rating may be lower since power stored in the motor isn't that high and not continous.

author: Simon
author_rank: Unknown
timestamp: May 26, 2010, 06:49:53 am
content: well surely if the diode is reversed during normal operation it will have a time lag before reversing again and conducting. Is there a chance that ringing will still occur ? should we also consider something to kill positive spikes or will all of the energy be absorbed on the first negative pulse ?I'll recommend a Schottky diode where practicable, we are dealing with up to 200 volts here, which hopefully will never happen if the diodes work and maybe a positive spike killer will have a double benefit of protecting the diode as we don't know what else in the customers vehicle is creating problems

author: jahonen
author_rank: Super Contributor
timestamp: May 26, 2010, 08:09:09 am
content: There is always some unclamped inductance in the system (like interconnecting wires), so you may get some minor spikes even after diodes, but I think that should not be a problem,as the energy stored is usually negligible.Ringing comes from the fact that stray capacitances form a resonant circuit with circuit inductances, which is triggered by the energy stored in the inductance. Spike from pure inductance is always unipolar, for ringing behaviour, capacitance is required.I'd first see how the flyback diodes improve the situation before making any further conclusions. If you want to clamp both polarities, then "Transzorb" diode would be one to use. They come in various clamp voltages and transient power ratings.I came across ameasurements of forward recovery time. It seems that for this kind of application the forward recovery time is fast enough, even for slow diodes (even if the reverse recovery is miserably slow, in µs-regimen).But like I said, it doesn't hurt to use fast/schottky diodes, as they might be more convenient for other reasons.Regards,Janne

author: tecman
author_rank: Frequent Contributor
timestamp: May 26, 2010, 01:53:22 pm
content: Based on may years experience with similar equipment, you absolutely do not need schottky diodes. The rise times of the spikes are limited by the distributed capacitance in the system and will be slow enough for a standard diode to catch. While line inductances can contribute to some fast rise voltages, the inductive energy contained by even 10's of feet of wire is easily swamped by the distributed capacitances in typical industrial wiring.Schottky diodes are expensive, have limits on reverse voltage and are static/ESD sensitive. Not needed so I would stay away.Paul

author: Unknown
author_rank: Unknown
timestamp: May 26, 2010, 04:41:21 pm
content: The scope measurements showed a (negative???) pulse about 30ns wide, if that's the case (does sound fast to me in a system with lots of capacitance and inductance), you would need pretty fast diodes. But maybe there's something wrong with those measurements?I always assumed that forward recovery would be worse than reverse recovery, since it's often not specified, but I never looked into it either, thanks for the link.