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| "flyback diodes" / "back emf diodes" how big ? |
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| Simon:
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 ? |
| alm:
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). |
| Simon:
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 ? |
| jahonen:
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 |
| hans:
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 ;) |
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