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Flyback diodes

  1. Aug 11, 2013 #1
    So in an application like this:
    oKl0KK8.png
    The coil of the motor will, by Lenz's Law, induce a very large negative voltage across its terminals. Would this voltage be very negative at the left terminal of the inductor and very positive at the right? In this way the diode will conduct, but the negative voltage in this case would be directly connected to the power supply.



    Also, how are the flyback diodes in this way utilized[L293 - H Bridge IC]:

    5G1yQQK.png
    It seems that the two left diodes will always be reversed biased, but what about the right-side two? Why are the left-side diodes even present if they are always reversed bias? It seems that the large negative voltage would forward bias one of the diodes to the right (whichever terminal is induced to a large negative potential), but what benefit does that have? Also, current flow back into the H-Bridge circuit does not appear very practical.

    Perhaps I am looking at this all wrong.

    EDIT: Given this situation:
    -Current flow from top to bottom
    -Pin 3 and pin 6 open circuited (transistor base = 0V) to turn off the motor

    Is it that a a positive voltage will be induced at the motor's "bottom" terminal and a negative voltage induced at the motor's "top" terminal? In this way, it seems that current would flow from the top diode from ground through the motor and then through the bottom left diode back into the source .. is that practical? How is this current limited? Current through a diode is unrestricted.
     
    Last edited: Aug 11, 2013
  2. jcsd
  3. Aug 11, 2013 #2
    Sometimes these diodes are called clamp diodes, and yes you are correct, in the first illustration a large negative voltage is created when the switch is opened. The clamp diode is normally reversed biased but when there is an opposite current flowing it will protect the power supply as well as other components in the circuit. The negative voltage will be connected to the power supply but since the switch is opened on the positive side, no current will flow.


    The second illustration is a bit more confusing. There are two scenarios for where the extreme potential will be. Normally all the diodes are always going to be reversed biased but lets say the voltage collapses in the motor and there is an extreme negative voltage induced on the bottom terminal of the motor. In this case current will actually flow out of ground and through the now correctly biased diode on the bottom right and current will flow out of the top terminal of the motor into the top left diode. The other two diodes will be used the same way if the motor stopped after spinning the other way. This setup protects the chip no matter what situation.

    I believe this is right but I am not certain by any means. It would be nice if someone else could re-assure my thoughts but I hope this helped, and btw if you don't mind could I ask what you are trying to build? Or is this all out of curiosity?
     
  4. Aug 11, 2013 #3
    If you have two sources (24V and -30V, lets say) that are grounded - shorting their terminals together will cause harm to the devices. The only difference here is that the induced voltage generated by the inductor is not grounded - but it seems the logic is still there.

    For the top
    What I think happens is that the left side of the inductor is fixed (at Vs), the right side generates sufficient positive voltage to overcome the diode forward voltage and a current is induced in the diode path (thus kept sufficiently small and spikes are minimized). I suppose that perhaps my confusion has resulted from the fact that given the right-sided (positive) voltage induced by the inductor, the left is (fixed - Vs) is negative with respect to the right-sided reference? Only speculation.

    For the bottom
    I see your point and understand to an extent.
    Given this situation:
    -Current flow from top to bottom
    -Pin 3 and pin 6 open circuited (transistor base = 0V) to turn off the motor

    Is it that a a positive voltage will be induced at the motor's "bottom" terminal and a negative voltage induced at the motor's "top" terminal? In this way, it seems that current would flow from the top diode from ground through the motor and then through the bottom left diode back into the source .. is that practical? How is this current limited? Current through a diode is unrestricted.


    I very strongly oppose using anything unless I completely, or within reason, understand how it works. I recently started working on a robotics project and needed an H-bridge to run the motors. I have an Arduino, so I went ahead and purchased a L298N H-bridge driver board fully assembled to fit nicely on the robot. I understood the concept of an H-bridge, but I was not so sure about the protective circuitry surrounding its application.

    The driver is actually quite simple to use, so I was contemplating assembling my own before playing around with the board I purchased. I did not want to just throw a design together from a schematic I found online. I want to actually understand every part and this question was a derivative of my confusion.
     
    Last edited: Aug 11, 2013
  5. Aug 11, 2013 #4
    Hmm... Well you are definitely right about your first example... but it is only true when the two supplies are grounded. When the supplies are not commonly grounded there isn't a way for any current to flow. For example you have 2 completely isolated supplies connecting their output voltages together will do nothing because a circuit is not formed.

    The diode is crucial when semiconductors are involved in the switching of the motor. After a semiconductor switches the motor off, the high voltage generated can easily damage the chip.

    I don't believe that the voltage on the left side of the inductor is fixed I believe that it is less than 0v and the voltage on the right of the inductor is more than Vs during a spike, this would further explain why the diodes arranged in the way they are in the second illustration.
     
  6. Aug 11, 2013 #5
    Consider the water analogy. How can you have two different pressures inside a single pipe without changing the diameter of it (resistance)? Having 5psi at one instance of the pipe and then 10psi immediately after it is impossible.

    What is the purpose of a negative voltage at the left terminal of the inductor? The magnetic field induces a sufficient voltage to further induce a current that is used to dissipate the energy. The right terminal of the inductor will induce a voltage significant enough to produce a current through the resistor and diode.

    For example, after open circuit:
    Vs = 24V = Left inductor terminal
    Voltage at right inductor terminal = approximately 1V + Vs = 25V

    And a current will exist flowing from the 25V "source" through the diode to the 24V source. Therefore, only a 1V potential exists across the inductor (limited by the diode).

    There is no need for a negative potential to be generated at the left terminal.
     
  7. Aug 11, 2013 #6
    I see what you are saying and I guess that the voltages are all relative and when the switch is open there isn't a common voltage to work off of except the 24V. The second example still is a bit confusing to me I can definitely see that depending on which way the motor was turning before being shut off either the top right and bottom left OR the top left and bottom right diodes become forward biased and will suppress the voltage spike. As far as the current being limited the motor has an internal resistance which is actually what R probably represents in the top diagram. The only thing I have trouble with is that the two different voltage sources would be connected in parallel during a spike which is bad depending on the type of the main PSU.
     
  8. Aug 11, 2013 #7

    jim hardy

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    I think there's a misconception about inductance.

    Recall Lenz's law - induced voltage opposes the change in current(well actually flux, but flux and current are in proportion)

    So the voltage that gets 'induced' when switch opens is not necessarily large.
    Voltage induced will be whatever is necessary to maintain current. No more than that.

    In your top figure that'd be (I X R) + diode drop. Is that necessarily large?

    In your second figure, the motor can return current to the supply by producing voltage equal to (Vc + two diode drops), thereby maintaining current.
    Regardless which end of motor goes negative, there's a diode on each end of the motor for conduction to either Vc or common.

    Does that help?
     
  9. Aug 11, 2013 #8
    I meant to correct that first line. I meant to say a "large" induced voltage when the diode was not in place as it is.

    The voltage at the left terminal would be Vs and the voltage at the right would be Va + (I x R) + Vf, right? It makes sense now. (i.e. (IxR) + Vf voltage induced across the inductor). Also, is this current of the exact same magnitude that it was (for a moment) prior to switching? If that is the case, and the current was significantly large, then the voltage may potentially be quite large as well at the right terminal ( IxR ).

    I think I understand pretty well how the above circuits work, although how practical is it to have current flow back into the supply? What would happen? I thought the point would be to protect the circuitry by providing a pathway for current flow that does not affect the rest of the circuit (as is the case with the first schematic).
     
  10. Aug 11, 2013 #9
    Lenz's law just proves the direction of current. Faraday's law is used to find voltage which is the following equation V= -N * (Δ[itex]\Phi[/itex] / Δt)

    Using this equation the time will be a couple of microseconds and N (number of loops) will be in the hundreds (for a DC motor).

    I'm not totally sure about how much the change in flux would be but say it is less than 1 Wb.

    This would put the back EMF over 100 volts which I thought is why these flyback diodes are needed.
     
  11. Aug 11, 2013 #10

    jim hardy

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    I think you answered it in your first post,

    Sure it's practical, it's not the diodes that limit the current it's Lenz's Law.
    Lenz dictates that initial current will be whatever was flowing through the motor and of course it'll decay as motor's inductive energy is returned to the source, and some of that energy is dissipated in circuit resistances..

    So current just swaps from the driver transistors inside your logic block to the diodes that are in parallel with them, and decays rapidly.
     
  12. Aug 11, 2013 #11

    jim hardy

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    Flux and current are related by reluctance and that's why V=L X di/dt.

    Kirchoff's voltage law dictates that when switch is open, voltage across inductance equals voltage across (resistor + diode) with proper sign of course.
     
  13. Aug 11, 2013 #12
    I was just wondering what effect current flow into a battery or bench supply would have. I guess it would be easier to accept if the current was doing into ground (thereby returning to the source"). But, the current instead flow back into the positive terminal of the supply. The current may not last for very long, but depending on the application it may be quite large.
     
  14. Aug 11, 2013 #13
    Again I think you need to brush up on your Lenz's law... It just defines the direction of current it does NOT state that the Voltage will be the necessary voltage to try to sustain the original. Faraday's law defines the voltage in this situation which is very large.
     
  15. Aug 11, 2013 #14

    jim hardy

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    Remember current flows in complete loops. The motor current flows through a diode into supply, out other end of supply into common, then through another diode into end of motor opposite where it left.

    Now you know you can return current to a battery that's how they are charged.

    A bench supply that won't accept some current could never discharge its output filter capacitor....

    it'd be incumbent on designer to select a supply that's compliant enough for motor duty. Or add filters to his driver to smooth out the current.


    We had some inverters that had a transformer load instead of motor. They returned current to the plant battery at 120hz, about 100 amp peaks causing ripple on the DC bus that upset other DC equipment.... we had to add 62,000 microfarad input filters to reduce their effect on battery bus.

    old jim
     
  16. Aug 11, 2013 #15

    jim hardy

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    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html#c2


    If it opposes a change, isn't that trying to keep value constant?

    I think YOU need to brush up, sir.
     
  17. Aug 11, 2013 #16
    If you scroll down on the page you just linked you will see :

    "When a magnet is moved into a coil of wire, changing the magnetic field and magnetic flux through the coil, a voltage will be generated in the coil according to Faraday's Law."

    No where does it show voltage calculations like the one you posted above.

    In my physics class that I had just last semester we learned Faraday's law along with Lenz's law. First you use Lenzes law to determine the direction of current and then Faraday's to determine voltage. No where in Lenz's law does it mention that the currents will be equal and to calculate voltage that way. With a lot of coils of wire and a small amount of time the voltage will be very large.

    The collapsing magnetic field in the motor causes the same effect the magnet creates when it is moved through a coil therefore wouldn't Faraday's Law apply?
     
  18. Aug 11, 2013 #17

    jim hardy

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    What's inductance? Flux linkages per ampere. L = N X Phi / I
    With constant turns, flux and current are equivalent, constant flux = constant current.

    That's why Lenz by opposing change in flux opposes change in current.
     
  19. Aug 11, 2013 #18
    OK so if you don't calculate it using Faraday's law, what is Faraday's law used for? And can you give me some link where someone is actually doing these voltage calculations your way?

    Also if the back EMF is so insignificant why do good engineers even bother adding a flyback diode to protect chips.

    Here is a spike from a 24 volt solenoid with .1 Ω internal resistance according to this wikipedia page http://en.wikipedia.org/wiki/Flyback_diode

    [Broken]


    According to your method first you would calculate the normal operating current through the coil.
    This would be 24V = I * .1Ω I = 240A

    Now I would multiply by the resistance to get back emf?

    240A * .1Ω = 24V So the back emf will be 24 volts according to you...but the picture shows nearly 300V this is why the diode is needed


    What you are saying as far as I can see is that the back emf will be the same as the normal current used to create the magnetic field in the coil but I can't see how that is correct
     
    Last edited by a moderator: May 6, 2017
  20. Aug 11, 2013 #19
    At the very bottom of the wiki page on flyback diodes there is a derivation of the current in the circuit diagram that is the first image in my original post.

    1XRTWuW.jpg

    Where is shows that the voltage across the inductor is the voltage across the diode and resistor.
     
  21. Aug 12, 2013 #20
    @HHOboy & Jim Hardy

    Guys, you are both right but I think you are misunderstanding each other.

    HHOboy is correctly describing the use of Faraday's law to derive the induced EMF when there is no diode, and this can be huge depending on how fast the switch opens.

    Jim is describing the case where we have the diode and, thus, there is no (or little) change in current, it simply gets diverted through the diode instead of the switch. Little di/dt means little d∅/dt in the inductor and little EMF induced. Of course this is the purpose of the diodes.

    The idea is that a large inductor, with its ends shorted, tries to "hold" current in the same way that a large capacitor, with its ends open, tries to "hold" voltage. If we put a large resistor across a large capacitor current will flow, the amount will be what is necessary to hold the voltage (followed by exponential decay). Similarly with inductors, if I replace the short with a small resistor, voltage will appear across the resistor, the amount will be what is necessary to hold the current (followed by exponential decay).
     
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