I think this term is often used to describe the application where a diode allows a peak voltage (usually created by an inductor that has had its current source switched off) to be limited by conducting current into a capacitor that is precharged to a voltage just below the desired limit. Often that precharge is just what is left after a resistor has bled the effect of the previous pulse from the capacitor.
-- John Popelish
used to absorb the reverse (flyback) voltage release from a coil when energized source is removed quickly. the flyback voltage is in reverse of what went in and can get very high in level which will short out things. the trick is to place a diode across the coil connections. the Cathode is connected to the + side and Anode to the - side when energized, the diode does not conduct. when source is removed quickly, the release will generate high voltage in reverse polarity. at this point the diode will conduct and anything above the cut off voltage of the diode will get absorb in the diode and protect other voltage sensitive components.
Anyone know?
tnx,
steve
--
Fat, sugar, salt, beer: the four essentials for a healthy diet.
Jamie wrote
The voltage isn't "absorbed" it is prevented from developing.
(What do you mean when you use the word "flyback"? Do you know where the term came from?)
I'll give you 3 points out of 10 for English, and 3 out of 10 for the clarity of your explanation.
The same point is dealt with in this item from a forum:
If the coil were made of superconductor, this might well be the case, but in the more practical copper wound coils, the wire resistance consumes most of the energy.
-- John Popelish
That makes sense. But the diverted engery has to be dissipated somewhere. If the diodes simply in antiparallel with the source, itll act as a short circuit on the back emf. Why doesn't that (the energy of that reverse pulse) destroy the diode? Someonne else said a cap and bleed resitor can be used to store and discharge the pulses harmlessly, but no one esle has verfified this. Can we have some clarification, please?
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There seems to be some confusion here, a flywheel diode is a diode connected across an inductor. It provides a path for the inductor current when the drive voltage is removed (switch off). If say there was 1 amp flowing and the drive was removed then the current (1 amp) would circulate through the coil and the diode, as there is resistance as well as the diodes forward voltage drop, the current will decay to zero.
Sometimes diodes are used with snubber networks, typicaly accross transistors, igbt or thyristors. They are in series with the cap and allow the switch off pulse to pass into the cap,as normal, but prevent the enrgy flowing back into the transistor when it switches on again. A small resistor accross the cap dissipates the charge from the cap during the "off" time.
If you short circuit a charged up inductor then absolutely nothing happens!. On the other hand, a charged capacitor would destroy the shorting wire. .... Use a mechanical switch and switch off a relay coil without a diode across the coil. The relay discharges and drops out near instantly and you'll probably get a spark across the switch contacts as thousands of volts come back off the coil. Now put a ('flywheel')diode across the coil and again switch off the relay. The relay takes ages to discharge and drop out and there is no spark as only
0.7V can come off the coil. Now put a 33V Zener diode (+ a reverse diode!) across the relay and switch it off. It drops out quite fast with a peak of 33V across the coil. Similar effect occurs if you put a diode in series with (say) a 100 ohm resistor. This time the discharging coil voltage can be quite high and depends on the inductance value. Put a capacitor across the relay coil and you've made a nice resonant circuit that takes many cycles to decay before the relay drops out. (it's the relay wire resistance that finally dissipates all the stored energy). Capacitor + resistor across the coil gives a very lossy 'damped' tuned circuit ' but a number of measurements and calcs then needed to allow a balancing act on the numbers.Idea is that with an inductor the stored energy can be removed (discharged) by allowing it to develop a voltage across some kind of load hence lose it's stored energy as heat. Bigger the discharge load resistance, then bigger the voltage, then bigger the power loss, then quicker the discharge. Put a short circuit across the inductor and a big current would try to flow but that same current will also be charging up the inductor so nothing actually happens. Put an open circuit across the inductor and the voltage screams upwards until something gives. Tiny current flowing but high voltage and power dissipation hence fast discharge.
The only inductor formula worth noting are ...
Stored Watt seconds(Joules)=1/2 x Inductance Value x [current through it ^2].
Amps per second though inductor = V across inductor / Inductor value.
[snip 8" of tedious relay coil example calcs]These simple calcs are useful for test purposes. Generally it's easier and much more accurate, just to use Spice.
regards john
Tnx, john (and others). Thats a pretty comprehensive answer, i guess. I'll ponder on it for a while. I must say your first bold pronouncement about coils discharging harmlessly and caps desroying had me confused, but your explanation of these pheonomena is of considerable help!
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Nope. Faraday's Law states that a time varying flux (which is caused by the "field") induces an EMF.
d(Phi) EMF = - ------- dt
That flux can't drop to zero instantaneously, but you can get a pretty respectable voltage, anyway.
it is in a cap because it takes work to get that charge in there.
Yeah, like C and V.
-- Best Regards, Mike
Somebody check me on this.
AIUI a cap stores voltage and an inductor stores current, so the result is that when you discharge a cap you can't see a voltage higher than what you had put into it, and when you discharge an inductor you can't see a current higher than what it carried before discharging.
If the cap sees zero resistance then it produces a large current. If the inductor sees infinite resistance then it produces a large voltage. Both of these conditions must be avoided.
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IIrc,. they both store *charge*.
-- Fat, sugar, salt, beer: the four essentials for a healthy diet.
Actually the inductor stores a magnetic field, which is converted back to a current. People say inductors store current because there is a direct relationship, but there is no direct relationship to charge. Charge is a function of the number of electrons (6 * 10^18 electrons per coloumb I think), but that parameter is not characteristic to the stored energy in either device, and measurement of columbs only takes you farther away from the parameters that matter.
One big and one small cap/inductor in parallel/series both have the same voltage/current but not the same charge within them.
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to a
Mike, can you reconcile that please with the numerous references that say it's a current and not a voltage that is proportional to the rate of change in a magnetic field? I don't see how it can be both.
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Show me a law of physics that states such a proportion mathematically. How are you going to get a current in a transformer secondary if it's open? Same thing with a generator or motor - the EMF or back EMF is what we talk about.
-- Best Regards, Mike
Ok. That would have been in reply to someone else, but lemme see... no, I don't see how you could, but I might be brain farting here.
-- Best Regards, Mike
More accurately, capacitors store energy proportional to voltage squared, and inductors store energy proportional to current squared.
For fixed inductances and capacitances, I think these are good generalizations.
If the magnetic path can change (think relay armatures moving, solenoids with moving plungers, etc.) or if the current changes from all the turns to some of the turns (or charges through one winding and discharges through another winding) the current can change.
Likewise, if capacitor plate spacing can change, voltage can change.
There are somewhat unusual conditions but not at all unheard of.
-- John Popelish
Obvious. Should have seen that. My physics is very rusty.
But I asked (3 of my messages back) if I was correct in stating that the snubber can't see a current higher than what the coil carried before it was turned off. What about that?
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