What keeps the kickback voltage from reaching an infinite voltage?

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We know that the voltage over an inductor is defined by the formula:



$V = L * frac didt $



So in the case where the current flow is suddenly interrupted (like when a mechanical contact is opened), voltage spikes occur in real life.



However, this is not always the case: we don't see arcs happen in small inductive loads. (By small inductive loads I mean a toy car motor, for example.) However, the formula says that the $ fracdidt $ term should approach infinity when mechanical contacts are opened, therefore the $L$ term (which should be small in small inductive loads) shouldn't have a significant effect. Simply, we should be able to see sparks any time we open any inductive load - independent of the inductance.



What are the practical factors that stop the voltage from reaching infinity? Does the current flow actually decrease slower, or is the formula perhaps insufficient for such a "discontinuity"?










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  • 5




    A practical coil has non-zero resistance.
    – filo
    Aug 23 at 14:48






  • 2




    @filo Why would resistance matter if there is no current flow?
    – Çetin Köktürk
    Aug 23 at 14:54






  • 2




    If there's no current flow at the moment the contacts open, why would you expect a spark across the contacts?
    – The Photon
    Aug 23 at 15:52






  • 2




    But the real answer is in Laptop's answer --- interwinding capacitance limits the voltage.
    – The Photon
    Aug 23 at 15:53






  • 5




    Infinity happens when you assume something is zero that is, in reality, not.
    – J...
    Aug 23 at 16:35















up vote
11
down vote

favorite
1












We know that the voltage over an inductor is defined by the formula:



$V = L * frac didt $



So in the case where the current flow is suddenly interrupted (like when a mechanical contact is opened), voltage spikes occur in real life.



However, this is not always the case: we don't see arcs happen in small inductive loads. (By small inductive loads I mean a toy car motor, for example.) However, the formula says that the $ fracdidt $ term should approach infinity when mechanical contacts are opened, therefore the $L$ term (which should be small in small inductive loads) shouldn't have a significant effect. Simply, we should be able to see sparks any time we open any inductive load - independent of the inductance.



What are the practical factors that stop the voltage from reaching infinity? Does the current flow actually decrease slower, or is the formula perhaps insufficient for such a "discontinuity"?










share|improve this question



















  • 5




    A practical coil has non-zero resistance.
    – filo
    Aug 23 at 14:48






  • 2




    @filo Why would resistance matter if there is no current flow?
    – Çetin Köktürk
    Aug 23 at 14:54






  • 2




    If there's no current flow at the moment the contacts open, why would you expect a spark across the contacts?
    – The Photon
    Aug 23 at 15:52






  • 2




    But the real answer is in Laptop's answer --- interwinding capacitance limits the voltage.
    – The Photon
    Aug 23 at 15:53






  • 5




    Infinity happens when you assume something is zero that is, in reality, not.
    – J...
    Aug 23 at 16:35













up vote
11
down vote

favorite
1









up vote
11
down vote

favorite
1






1





We know that the voltage over an inductor is defined by the formula:



$V = L * frac didt $



So in the case where the current flow is suddenly interrupted (like when a mechanical contact is opened), voltage spikes occur in real life.



However, this is not always the case: we don't see arcs happen in small inductive loads. (By small inductive loads I mean a toy car motor, for example.) However, the formula says that the $ fracdidt $ term should approach infinity when mechanical contacts are opened, therefore the $L$ term (which should be small in small inductive loads) shouldn't have a significant effect. Simply, we should be able to see sparks any time we open any inductive load - independent of the inductance.



What are the practical factors that stop the voltage from reaching infinity? Does the current flow actually decrease slower, or is the formula perhaps insufficient for such a "discontinuity"?










share|improve this question















We know that the voltage over an inductor is defined by the formula:



$V = L * frac didt $



So in the case where the current flow is suddenly interrupted (like when a mechanical contact is opened), voltage spikes occur in real life.



However, this is not always the case: we don't see arcs happen in small inductive loads. (By small inductive loads I mean a toy car motor, for example.) However, the formula says that the $ fracdidt $ term should approach infinity when mechanical contacts are opened, therefore the $L$ term (which should be small in small inductive loads) shouldn't have a significant effect. Simply, we should be able to see sparks any time we open any inductive load - independent of the inductance.



What are the practical factors that stop the voltage from reaching infinity? Does the current flow actually decrease slower, or is the formula perhaps insufficient for such a "discontinuity"?







inductor flyback kickback






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share|improve this question













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share|improve this question








edited Aug 26 at 18:34









Peter Mortensen

1,56131422




1,56131422










asked Aug 23 at 14:33









Çetin Köktürk

466315




466315







  • 5




    A practical coil has non-zero resistance.
    – filo
    Aug 23 at 14:48






  • 2




    @filo Why would resistance matter if there is no current flow?
    – Çetin Köktürk
    Aug 23 at 14:54






  • 2




    If there's no current flow at the moment the contacts open, why would you expect a spark across the contacts?
    – The Photon
    Aug 23 at 15:52






  • 2




    But the real answer is in Laptop's answer --- interwinding capacitance limits the voltage.
    – The Photon
    Aug 23 at 15:53






  • 5




    Infinity happens when you assume something is zero that is, in reality, not.
    – J...
    Aug 23 at 16:35













  • 5




    A practical coil has non-zero resistance.
    – filo
    Aug 23 at 14:48






  • 2




    @filo Why would resistance matter if there is no current flow?
    – Çetin Köktürk
    Aug 23 at 14:54






  • 2




    If there's no current flow at the moment the contacts open, why would you expect a spark across the contacts?
    – The Photon
    Aug 23 at 15:52






  • 2




    But the real answer is in Laptop's answer --- interwinding capacitance limits the voltage.
    – The Photon
    Aug 23 at 15:53






  • 5




    Infinity happens when you assume something is zero that is, in reality, not.
    – J...
    Aug 23 at 16:35








5




5




A practical coil has non-zero resistance.
– filo
Aug 23 at 14:48




A practical coil has non-zero resistance.
– filo
Aug 23 at 14:48




2




2




@filo Why would resistance matter if there is no current flow?
– Çetin Köktürk
Aug 23 at 14:54




@filo Why would resistance matter if there is no current flow?
– Çetin Köktürk
Aug 23 at 14:54




2




2




If there's no current flow at the moment the contacts open, why would you expect a spark across the contacts?
– The Photon
Aug 23 at 15:52




If there's no current flow at the moment the contacts open, why would you expect a spark across the contacts?
– The Photon
Aug 23 at 15:52




2




2




But the real answer is in Laptop's answer --- interwinding capacitance limits the voltage.
– The Photon
Aug 23 at 15:53




But the real answer is in Laptop's answer --- interwinding capacitance limits the voltage.
– The Photon
Aug 23 at 15:53




5




5




Infinity happens when you assume something is zero that is, in reality, not.
– J...
Aug 23 at 16:35





Infinity happens when you assume something is zero that is, in reality, not.
– J...
Aug 23 at 16:35











4 Answers
4






active

oldest

votes

















up vote
17
down vote



accepted










A real inductor looks like this (shown below is an inductor with 4 coils) there is a small amount (usually in the pF-fF range) of capacitance between each coil. Each piece of wire also has some resistance associate with it.



Because each coil in an inductor has resistance (or each section of wire if you consider one coil) this impedes the current and reduces the voltage. The small amount of capacitance will also store some of the voltage and prevent an instantaneous change in voltage.



These all soak up energy that prevents the Electro Motive Force (EMF) that has been stored around an inductor from generating an infinite voltage. An inductor can actually be simplified into a circuit such as the one to the left below.





schematic





simulate this circuit – Schematic created using CircuitLab



A superconducting coil would be able to generate much more massive voltages because of much lower losses due to parasitics.






share|improve this answer


















  • 3




    I recommend that you change "impedes the electrons" to "impedes the current". There have been a spate of confused questions regarding electrons in the last few weeks.
    – Transistor
    Aug 23 at 17:28






  • 2




    Yeah, it's not the electrons that are carrying the currentenergy, it's the electric field.
    – laptop2d
    Aug 23 at 17:35






  • 1




    Resonating the capacitance away also allows massive voltages. Then it is a Tesla coil
    – Henry Crun
    Aug 23 at 22:14






  • 1




    Everything is correct, except EMF is not stored inncoils. EMF is Volts, what is stored is magnetic energy, IIL/2, defined by Amperes.
    – Gregory Kornblum
    Aug 23 at 22:36










  • @GregoryKornblum Your right, that should have read "around the inductor" not "in the inductor". It is common to refer to the voltage stored around the coil as EMF. Webers/second = volts
    – laptop2d
    Aug 23 at 22:52


















up vote
7
down vote













Any energy storage system (an inductor) has non-zero size.



Anything of non-zero size has non-zero electric fields, or capacitance. Device junctions are usually a large source of parasitic capacitance. Flyback systems use a diode to transfer energy into a load capacitor.



At peak voltage excursion, all the inductive energy has
(1) been dissipated as heat
(2) been radiated as EM field
(3) been stored in the electric-field of the intentional and the parasitic capacitances.






share|improve this answer



























    up vote
    5
    down vote













    The series resistance matters a great deal with the "kickback" voltage due to the series capacitance of the "switch" when opened. This forms a classic series RLC resonant circuit which has properties of voltage gain by impedance ratio of



    $Q=dfracR = dfracX_LR=dfracomega _0 LR$ at resonant frequency $omega _0= dfrac1sqrtLC$



    For the situation of kickback voltage peak, it can be proven that $|V_p| = Q * V_dc$ for Quality Factor, Q (above) and loop supply voltage Vdc at some resonant frequency.



    When de-energizing a circuit with a contact switch as t goes to 0, V/L=dI/dt, V does not go to infinity due to this parasitic capacitance.



    Example





    schematic





    simulate this circuit – Schematic created using CircuitLab




    e.g. Consider a series circuit, Vdc=1V, L=1uH, R=1 Ohms , Idc= 1A. What is the switch voltage kickback, when just opened, if Csw = 1pF?



    1V , 100V, 1kV, 1e6 V or infinite?




    Now consider the same for a FET switch with 1nF output capacitance with RdsOn << 1% of R=1. What is dV?



    p.s. if you learnt something, then comment your answer.



    The intuitive answer is that the switch goes from a conductor to a tiny stray capacitor which limits the slew rate of the voltage and as does the inductor limit the slew rate of the current and at their resonant frequency the voltage gain, Q at ω0 is inversely proportional to R, so bigger series R dampens the voltage.



    Answer $ V_p= I_dc sqrtdfracLC $ = 1A * √(1uH/1pF)= 1kV



    Misc



    It can be proven the open circuit impedance like a transmission line "characteristic impedance" $ Zo= sqrtdfracLC $



    We see the voltage kickback looks like Ohm's Law. $ V_p = I_dc*Z_0$
    The peak voltage Vp, generated from interrupting an inductive current, $I_dc$.






    share|improve this answer





























      up vote
      3
      down vote













      Just consider a simple example of 100 uH and 1 amp flowing. When the contact in series with the inductor opens, there may be 5 pF of parasitic capacitance left across the inductor and that 1 amp will create a high kick-back voltage but how much?



      $$I = CdfracdVdt$$



      So potentially (no pun intended) the voltage across the 5 pF capacitor could rise at a rate of 200 kV/microsecond. Given that its starting voltage is potentially neglible in comparison, within a few micro seconds a pretty big voltage could develop. However this is mitigated by the lack of energy stored in the inductor: -



      $$W = dfracLcdot I^22$$



      Or 5 micro joules. All this energy will cyclically transfer to the capacitor and we can equate the capacitor energy formula to 5 uJ to give us the maximum voltage: -



      $$W = dfracCcdot V^22$$



      This produces a peak capacitor voltage of 1414 volts.






      share|improve this answer




















      • Thank you for the answer Andy, I was sure that there was a "conservation of energy" answer to this.
        – Çetin Köktürk
        Aug 23 at 20:54










      • No probs dude..
        – Andy aka
        Aug 23 at 21:23










      • @ÇetinKöktürk I would agree that "energy" stored in L's and C's is the best way of thinking about this. It leads directly to a fundamentally correct understanding. (whereas a "circuit analysis" perspective is kind of indirect and somewhat confuses the real issue: energy storage and movement)
        – Henry Crun
        Aug 23 at 23:00










      • @ Andy the fun thing about switches is the variable contact spacing as the switch continues to open further; this reduces the capacitance and lets the voltage become even higher, perhaps once again striking an arc; switches are evil trash generators when energy can be stored in some wiring and then resonated with the switch-contact variable-capacitance.
        – analogsystemsrf
        Aug 25 at 3:20










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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      17
      down vote



      accepted










      A real inductor looks like this (shown below is an inductor with 4 coils) there is a small amount (usually in the pF-fF range) of capacitance between each coil. Each piece of wire also has some resistance associate with it.



      Because each coil in an inductor has resistance (or each section of wire if you consider one coil) this impedes the current and reduces the voltage. The small amount of capacitance will also store some of the voltage and prevent an instantaneous change in voltage.



      These all soak up energy that prevents the Electro Motive Force (EMF) that has been stored around an inductor from generating an infinite voltage. An inductor can actually be simplified into a circuit such as the one to the left below.





      schematic





      simulate this circuit – Schematic created using CircuitLab



      A superconducting coil would be able to generate much more massive voltages because of much lower losses due to parasitics.






      share|improve this answer


















      • 3




        I recommend that you change "impedes the electrons" to "impedes the current". There have been a spate of confused questions regarding electrons in the last few weeks.
        – Transistor
        Aug 23 at 17:28






      • 2




        Yeah, it's not the electrons that are carrying the currentenergy, it's the electric field.
        – laptop2d
        Aug 23 at 17:35






      • 1




        Resonating the capacitance away also allows massive voltages. Then it is a Tesla coil
        – Henry Crun
        Aug 23 at 22:14






      • 1




        Everything is correct, except EMF is not stored inncoils. EMF is Volts, what is stored is magnetic energy, IIL/2, defined by Amperes.
        – Gregory Kornblum
        Aug 23 at 22:36










      • @GregoryKornblum Your right, that should have read "around the inductor" not "in the inductor". It is common to refer to the voltage stored around the coil as EMF. Webers/second = volts
        – laptop2d
        Aug 23 at 22:52















      up vote
      17
      down vote



      accepted










      A real inductor looks like this (shown below is an inductor with 4 coils) there is a small amount (usually in the pF-fF range) of capacitance between each coil. Each piece of wire also has some resistance associate with it.



      Because each coil in an inductor has resistance (or each section of wire if you consider one coil) this impedes the current and reduces the voltage. The small amount of capacitance will also store some of the voltage and prevent an instantaneous change in voltage.



      These all soak up energy that prevents the Electro Motive Force (EMF) that has been stored around an inductor from generating an infinite voltage. An inductor can actually be simplified into a circuit such as the one to the left below.





      schematic





      simulate this circuit – Schematic created using CircuitLab



      A superconducting coil would be able to generate much more massive voltages because of much lower losses due to parasitics.






      share|improve this answer


















      • 3




        I recommend that you change "impedes the electrons" to "impedes the current". There have been a spate of confused questions regarding electrons in the last few weeks.
        – Transistor
        Aug 23 at 17:28






      • 2




        Yeah, it's not the electrons that are carrying the currentenergy, it's the electric field.
        – laptop2d
        Aug 23 at 17:35






      • 1




        Resonating the capacitance away also allows massive voltages. Then it is a Tesla coil
        – Henry Crun
        Aug 23 at 22:14






      • 1




        Everything is correct, except EMF is not stored inncoils. EMF is Volts, what is stored is magnetic energy, IIL/2, defined by Amperes.
        – Gregory Kornblum
        Aug 23 at 22:36










      • @GregoryKornblum Your right, that should have read "around the inductor" not "in the inductor". It is common to refer to the voltage stored around the coil as EMF. Webers/second = volts
        – laptop2d
        Aug 23 at 22:52













      up vote
      17
      down vote



      accepted







      up vote
      17
      down vote



      accepted






      A real inductor looks like this (shown below is an inductor with 4 coils) there is a small amount (usually in the pF-fF range) of capacitance between each coil. Each piece of wire also has some resistance associate with it.



      Because each coil in an inductor has resistance (or each section of wire if you consider one coil) this impedes the current and reduces the voltage. The small amount of capacitance will also store some of the voltage and prevent an instantaneous change in voltage.



      These all soak up energy that prevents the Electro Motive Force (EMF) that has been stored around an inductor from generating an infinite voltage. An inductor can actually be simplified into a circuit such as the one to the left below.





      schematic





      simulate this circuit – Schematic created using CircuitLab



      A superconducting coil would be able to generate much more massive voltages because of much lower losses due to parasitics.






      share|improve this answer














      A real inductor looks like this (shown below is an inductor with 4 coils) there is a small amount (usually in the pF-fF range) of capacitance between each coil. Each piece of wire also has some resistance associate with it.



      Because each coil in an inductor has resistance (or each section of wire if you consider one coil) this impedes the current and reduces the voltage. The small amount of capacitance will also store some of the voltage and prevent an instantaneous change in voltage.



      These all soak up energy that prevents the Electro Motive Force (EMF) that has been stored around an inductor from generating an infinite voltage. An inductor can actually be simplified into a circuit such as the one to the left below.





      schematic





      simulate this circuit – Schematic created using CircuitLab



      A superconducting coil would be able to generate much more massive voltages because of much lower losses due to parasitics.







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Aug 23 at 22:51

























      answered Aug 23 at 15:40









      laptop2d

      21.7k123072




      21.7k123072







      • 3




        I recommend that you change "impedes the electrons" to "impedes the current". There have been a spate of confused questions regarding electrons in the last few weeks.
        – Transistor
        Aug 23 at 17:28






      • 2




        Yeah, it's not the electrons that are carrying the currentenergy, it's the electric field.
        – laptop2d
        Aug 23 at 17:35






      • 1




        Resonating the capacitance away also allows massive voltages. Then it is a Tesla coil
        – Henry Crun
        Aug 23 at 22:14






      • 1




        Everything is correct, except EMF is not stored inncoils. EMF is Volts, what is stored is magnetic energy, IIL/2, defined by Amperes.
        – Gregory Kornblum
        Aug 23 at 22:36










      • @GregoryKornblum Your right, that should have read "around the inductor" not "in the inductor". It is common to refer to the voltage stored around the coil as EMF. Webers/second = volts
        – laptop2d
        Aug 23 at 22:52













      • 3




        I recommend that you change "impedes the electrons" to "impedes the current". There have been a spate of confused questions regarding electrons in the last few weeks.
        – Transistor
        Aug 23 at 17:28






      • 2




        Yeah, it's not the electrons that are carrying the currentenergy, it's the electric field.
        – laptop2d
        Aug 23 at 17:35






      • 1




        Resonating the capacitance away also allows massive voltages. Then it is a Tesla coil
        – Henry Crun
        Aug 23 at 22:14






      • 1




        Everything is correct, except EMF is not stored inncoils. EMF is Volts, what is stored is magnetic energy, IIL/2, defined by Amperes.
        – Gregory Kornblum
        Aug 23 at 22:36










      • @GregoryKornblum Your right, that should have read "around the inductor" not "in the inductor". It is common to refer to the voltage stored around the coil as EMF. Webers/second = volts
        – laptop2d
        Aug 23 at 22:52








      3




      3




      I recommend that you change "impedes the electrons" to "impedes the current". There have been a spate of confused questions regarding electrons in the last few weeks.
      – Transistor
      Aug 23 at 17:28




      I recommend that you change "impedes the electrons" to "impedes the current". There have been a spate of confused questions regarding electrons in the last few weeks.
      – Transistor
      Aug 23 at 17:28




      2




      2




      Yeah, it's not the electrons that are carrying the currentenergy, it's the electric field.
      – laptop2d
      Aug 23 at 17:35




      Yeah, it's not the electrons that are carrying the currentenergy, it's the electric field.
      – laptop2d
      Aug 23 at 17:35




      1




      1




      Resonating the capacitance away also allows massive voltages. Then it is a Tesla coil
      – Henry Crun
      Aug 23 at 22:14




      Resonating the capacitance away also allows massive voltages. Then it is a Tesla coil
      – Henry Crun
      Aug 23 at 22:14




      1




      1




      Everything is correct, except EMF is not stored inncoils. EMF is Volts, what is stored is magnetic energy, IIL/2, defined by Amperes.
      – Gregory Kornblum
      Aug 23 at 22:36




      Everything is correct, except EMF is not stored inncoils. EMF is Volts, what is stored is magnetic energy, IIL/2, defined by Amperes.
      – Gregory Kornblum
      Aug 23 at 22:36












      @GregoryKornblum Your right, that should have read "around the inductor" not "in the inductor". It is common to refer to the voltage stored around the coil as EMF. Webers/second = volts
      – laptop2d
      Aug 23 at 22:52





      @GregoryKornblum Your right, that should have read "around the inductor" not "in the inductor". It is common to refer to the voltage stored around the coil as EMF. Webers/second = volts
      – laptop2d
      Aug 23 at 22:52













      up vote
      7
      down vote













      Any energy storage system (an inductor) has non-zero size.



      Anything of non-zero size has non-zero electric fields, or capacitance. Device junctions are usually a large source of parasitic capacitance. Flyback systems use a diode to transfer energy into a load capacitor.



      At peak voltage excursion, all the inductive energy has
      (1) been dissipated as heat
      (2) been radiated as EM field
      (3) been stored in the electric-field of the intentional and the parasitic capacitances.






      share|improve this answer
























        up vote
        7
        down vote













        Any energy storage system (an inductor) has non-zero size.



        Anything of non-zero size has non-zero electric fields, or capacitance. Device junctions are usually a large source of parasitic capacitance. Flyback systems use a diode to transfer energy into a load capacitor.



        At peak voltage excursion, all the inductive energy has
        (1) been dissipated as heat
        (2) been radiated as EM field
        (3) been stored in the electric-field of the intentional and the parasitic capacitances.






        share|improve this answer






















          up vote
          7
          down vote










          up vote
          7
          down vote









          Any energy storage system (an inductor) has non-zero size.



          Anything of non-zero size has non-zero electric fields, or capacitance. Device junctions are usually a large source of parasitic capacitance. Flyback systems use a diode to transfer energy into a load capacitor.



          At peak voltage excursion, all the inductive energy has
          (1) been dissipated as heat
          (2) been radiated as EM field
          (3) been stored in the electric-field of the intentional and the parasitic capacitances.






          share|improve this answer












          Any energy storage system (an inductor) has non-zero size.



          Anything of non-zero size has non-zero electric fields, or capacitance. Device junctions are usually a large source of parasitic capacitance. Flyback systems use a diode to transfer energy into a load capacitor.



          At peak voltage excursion, all the inductive energy has
          (1) been dissipated as heat
          (2) been radiated as EM field
          (3) been stored in the electric-field of the intentional and the parasitic capacitances.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Aug 23 at 14:51









          analogsystemsrf

          11.6k2616




          11.6k2616




















              up vote
              5
              down vote













              The series resistance matters a great deal with the "kickback" voltage due to the series capacitance of the "switch" when opened. This forms a classic series RLC resonant circuit which has properties of voltage gain by impedance ratio of



              $Q=dfracR = dfracX_LR=dfracomega _0 LR$ at resonant frequency $omega _0= dfrac1sqrtLC$



              For the situation of kickback voltage peak, it can be proven that $|V_p| = Q * V_dc$ for Quality Factor, Q (above) and loop supply voltage Vdc at some resonant frequency.



              When de-energizing a circuit with a contact switch as t goes to 0, V/L=dI/dt, V does not go to infinity due to this parasitic capacitance.



              Example





              schematic





              simulate this circuit – Schematic created using CircuitLab




              e.g. Consider a series circuit, Vdc=1V, L=1uH, R=1 Ohms , Idc= 1A. What is the switch voltage kickback, when just opened, if Csw = 1pF?



              1V , 100V, 1kV, 1e6 V or infinite?




              Now consider the same for a FET switch with 1nF output capacitance with RdsOn << 1% of R=1. What is dV?



              p.s. if you learnt something, then comment your answer.



              The intuitive answer is that the switch goes from a conductor to a tiny stray capacitor which limits the slew rate of the voltage and as does the inductor limit the slew rate of the current and at their resonant frequency the voltage gain, Q at ω0 is inversely proportional to R, so bigger series R dampens the voltage.



              Answer $ V_p= I_dc sqrtdfracLC $ = 1A * √(1uH/1pF)= 1kV



              Misc



              It can be proven the open circuit impedance like a transmission line "characteristic impedance" $ Zo= sqrtdfracLC $



              We see the voltage kickback looks like Ohm's Law. $ V_p = I_dc*Z_0$
              The peak voltage Vp, generated from interrupting an inductive current, $I_dc$.






              share|improve this answer


























                up vote
                5
                down vote













                The series resistance matters a great deal with the "kickback" voltage due to the series capacitance of the "switch" when opened. This forms a classic series RLC resonant circuit which has properties of voltage gain by impedance ratio of



                $Q=dfracR = dfracX_LR=dfracomega _0 LR$ at resonant frequency $omega _0= dfrac1sqrtLC$



                For the situation of kickback voltage peak, it can be proven that $|V_p| = Q * V_dc$ for Quality Factor, Q (above) and loop supply voltage Vdc at some resonant frequency.



                When de-energizing a circuit with a contact switch as t goes to 0, V/L=dI/dt, V does not go to infinity due to this parasitic capacitance.



                Example





                schematic





                simulate this circuit – Schematic created using CircuitLab




                e.g. Consider a series circuit, Vdc=1V, L=1uH, R=1 Ohms , Idc= 1A. What is the switch voltage kickback, when just opened, if Csw = 1pF?



                1V , 100V, 1kV, 1e6 V or infinite?




                Now consider the same for a FET switch with 1nF output capacitance with RdsOn << 1% of R=1. What is dV?



                p.s. if you learnt something, then comment your answer.



                The intuitive answer is that the switch goes from a conductor to a tiny stray capacitor which limits the slew rate of the voltage and as does the inductor limit the slew rate of the current and at their resonant frequency the voltage gain, Q at ω0 is inversely proportional to R, so bigger series R dampens the voltage.



                Answer $ V_p= I_dc sqrtdfracLC $ = 1A * √(1uH/1pF)= 1kV



                Misc



                It can be proven the open circuit impedance like a transmission line "characteristic impedance" $ Zo= sqrtdfracLC $



                We see the voltage kickback looks like Ohm's Law. $ V_p = I_dc*Z_0$
                The peak voltage Vp, generated from interrupting an inductive current, $I_dc$.






                share|improve this answer
























                  up vote
                  5
                  down vote










                  up vote
                  5
                  down vote









                  The series resistance matters a great deal with the "kickback" voltage due to the series capacitance of the "switch" when opened. This forms a classic series RLC resonant circuit which has properties of voltage gain by impedance ratio of



                  $Q=dfracR = dfracX_LR=dfracomega _0 LR$ at resonant frequency $omega _0= dfrac1sqrtLC$



                  For the situation of kickback voltage peak, it can be proven that $|V_p| = Q * V_dc$ for Quality Factor, Q (above) and loop supply voltage Vdc at some resonant frequency.



                  When de-energizing a circuit with a contact switch as t goes to 0, V/L=dI/dt, V does not go to infinity due to this parasitic capacitance.



                  Example





                  schematic





                  simulate this circuit – Schematic created using CircuitLab




                  e.g. Consider a series circuit, Vdc=1V, L=1uH, R=1 Ohms , Idc= 1A. What is the switch voltage kickback, when just opened, if Csw = 1pF?



                  1V , 100V, 1kV, 1e6 V or infinite?




                  Now consider the same for a FET switch with 1nF output capacitance with RdsOn << 1% of R=1. What is dV?



                  p.s. if you learnt something, then comment your answer.



                  The intuitive answer is that the switch goes from a conductor to a tiny stray capacitor which limits the slew rate of the voltage and as does the inductor limit the slew rate of the current and at their resonant frequency the voltage gain, Q at ω0 is inversely proportional to R, so bigger series R dampens the voltage.



                  Answer $ V_p= I_dc sqrtdfracLC $ = 1A * √(1uH/1pF)= 1kV



                  Misc



                  It can be proven the open circuit impedance like a transmission line "characteristic impedance" $ Zo= sqrtdfracLC $



                  We see the voltage kickback looks like Ohm's Law. $ V_p = I_dc*Z_0$
                  The peak voltage Vp, generated from interrupting an inductive current, $I_dc$.






                  share|improve this answer














                  The series resistance matters a great deal with the "kickback" voltage due to the series capacitance of the "switch" when opened. This forms a classic series RLC resonant circuit which has properties of voltage gain by impedance ratio of



                  $Q=dfracR = dfracX_LR=dfracomega _0 LR$ at resonant frequency $omega _0= dfrac1sqrtLC$



                  For the situation of kickback voltage peak, it can be proven that $|V_p| = Q * V_dc$ for Quality Factor, Q (above) and loop supply voltage Vdc at some resonant frequency.



                  When de-energizing a circuit with a contact switch as t goes to 0, V/L=dI/dt, V does not go to infinity due to this parasitic capacitance.



                  Example





                  schematic





                  simulate this circuit – Schematic created using CircuitLab




                  e.g. Consider a series circuit, Vdc=1V, L=1uH, R=1 Ohms , Idc= 1A. What is the switch voltage kickback, when just opened, if Csw = 1pF?



                  1V , 100V, 1kV, 1e6 V or infinite?




                  Now consider the same for a FET switch with 1nF output capacitance with RdsOn << 1% of R=1. What is dV?



                  p.s. if you learnt something, then comment your answer.



                  The intuitive answer is that the switch goes from a conductor to a tiny stray capacitor which limits the slew rate of the voltage and as does the inductor limit the slew rate of the current and at their resonant frequency the voltage gain, Q at ω0 is inversely proportional to R, so bigger series R dampens the voltage.



                  Answer $ V_p= I_dc sqrtdfracLC $ = 1A * √(1uH/1pF)= 1kV



                  Misc



                  It can be proven the open circuit impedance like a transmission line "characteristic impedance" $ Zo= sqrtdfracLC $



                  We see the voltage kickback looks like Ohm's Law. $ V_p = I_dc*Z_0$
                  The peak voltage Vp, generated from interrupting an inductive current, $I_dc$.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Aug 23 at 16:13

























                  answered Aug 23 at 15:31









                  Tony EE rocketscientist

                  58.3k22085




                  58.3k22085




















                      up vote
                      3
                      down vote













                      Just consider a simple example of 100 uH and 1 amp flowing. When the contact in series with the inductor opens, there may be 5 pF of parasitic capacitance left across the inductor and that 1 amp will create a high kick-back voltage but how much?



                      $$I = CdfracdVdt$$



                      So potentially (no pun intended) the voltage across the 5 pF capacitor could rise at a rate of 200 kV/microsecond. Given that its starting voltage is potentially neglible in comparison, within a few micro seconds a pretty big voltage could develop. However this is mitigated by the lack of energy stored in the inductor: -



                      $$W = dfracLcdot I^22$$



                      Or 5 micro joules. All this energy will cyclically transfer to the capacitor and we can equate the capacitor energy formula to 5 uJ to give us the maximum voltage: -



                      $$W = dfracCcdot V^22$$



                      This produces a peak capacitor voltage of 1414 volts.






                      share|improve this answer




















                      • Thank you for the answer Andy, I was sure that there was a "conservation of energy" answer to this.
                        – Çetin Köktürk
                        Aug 23 at 20:54










                      • No probs dude..
                        – Andy aka
                        Aug 23 at 21:23










                      • @ÇetinKöktürk I would agree that "energy" stored in L's and C's is the best way of thinking about this. It leads directly to a fundamentally correct understanding. (whereas a "circuit analysis" perspective is kind of indirect and somewhat confuses the real issue: energy storage and movement)
                        – Henry Crun
                        Aug 23 at 23:00










                      • @ Andy the fun thing about switches is the variable contact spacing as the switch continues to open further; this reduces the capacitance and lets the voltage become even higher, perhaps once again striking an arc; switches are evil trash generators when energy can be stored in some wiring and then resonated with the switch-contact variable-capacitance.
                        – analogsystemsrf
                        Aug 25 at 3:20














                      up vote
                      3
                      down vote













                      Just consider a simple example of 100 uH and 1 amp flowing. When the contact in series with the inductor opens, there may be 5 pF of parasitic capacitance left across the inductor and that 1 amp will create a high kick-back voltage but how much?



                      $$I = CdfracdVdt$$



                      So potentially (no pun intended) the voltage across the 5 pF capacitor could rise at a rate of 200 kV/microsecond. Given that its starting voltage is potentially neglible in comparison, within a few micro seconds a pretty big voltage could develop. However this is mitigated by the lack of energy stored in the inductor: -



                      $$W = dfracLcdot I^22$$



                      Or 5 micro joules. All this energy will cyclically transfer to the capacitor and we can equate the capacitor energy formula to 5 uJ to give us the maximum voltage: -



                      $$W = dfracCcdot V^22$$



                      This produces a peak capacitor voltage of 1414 volts.






                      share|improve this answer




















                      • Thank you for the answer Andy, I was sure that there was a "conservation of energy" answer to this.
                        – Çetin Köktürk
                        Aug 23 at 20:54










                      • No probs dude..
                        – Andy aka
                        Aug 23 at 21:23










                      • @ÇetinKöktürk I would agree that "energy" stored in L's and C's is the best way of thinking about this. It leads directly to a fundamentally correct understanding. (whereas a "circuit analysis" perspective is kind of indirect and somewhat confuses the real issue: energy storage and movement)
                        – Henry Crun
                        Aug 23 at 23:00










                      • @ Andy the fun thing about switches is the variable contact spacing as the switch continues to open further; this reduces the capacitance and lets the voltage become even higher, perhaps once again striking an arc; switches are evil trash generators when energy can be stored in some wiring and then resonated with the switch-contact variable-capacitance.
                        – analogsystemsrf
                        Aug 25 at 3:20












                      up vote
                      3
                      down vote










                      up vote
                      3
                      down vote









                      Just consider a simple example of 100 uH and 1 amp flowing. When the contact in series with the inductor opens, there may be 5 pF of parasitic capacitance left across the inductor and that 1 amp will create a high kick-back voltage but how much?



                      $$I = CdfracdVdt$$



                      So potentially (no pun intended) the voltage across the 5 pF capacitor could rise at a rate of 200 kV/microsecond. Given that its starting voltage is potentially neglible in comparison, within a few micro seconds a pretty big voltage could develop. However this is mitigated by the lack of energy stored in the inductor: -



                      $$W = dfracLcdot I^22$$



                      Or 5 micro joules. All this energy will cyclically transfer to the capacitor and we can equate the capacitor energy formula to 5 uJ to give us the maximum voltage: -



                      $$W = dfracCcdot V^22$$



                      This produces a peak capacitor voltage of 1414 volts.






                      share|improve this answer












                      Just consider a simple example of 100 uH and 1 amp flowing. When the contact in series with the inductor opens, there may be 5 pF of parasitic capacitance left across the inductor and that 1 amp will create a high kick-back voltage but how much?



                      $$I = CdfracdVdt$$



                      So potentially (no pun intended) the voltage across the 5 pF capacitor could rise at a rate of 200 kV/microsecond. Given that its starting voltage is potentially neglible in comparison, within a few micro seconds a pretty big voltage could develop. However this is mitigated by the lack of energy stored in the inductor: -



                      $$W = dfracLcdot I^22$$



                      Or 5 micro joules. All this energy will cyclically transfer to the capacitor and we can equate the capacitor energy formula to 5 uJ to give us the maximum voltage: -



                      $$W = dfracCcdot V^22$$



                      This produces a peak capacitor voltage of 1414 volts.







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered Aug 23 at 20:40









                      Andy aka

                      230k10171391




                      230k10171391











                      • Thank you for the answer Andy, I was sure that there was a "conservation of energy" answer to this.
                        – Çetin Köktürk
                        Aug 23 at 20:54










                      • No probs dude..
                        – Andy aka
                        Aug 23 at 21:23










                      • @ÇetinKöktürk I would agree that "energy" stored in L's and C's is the best way of thinking about this. It leads directly to a fundamentally correct understanding. (whereas a "circuit analysis" perspective is kind of indirect and somewhat confuses the real issue: energy storage and movement)
                        – Henry Crun
                        Aug 23 at 23:00










                      • @ Andy the fun thing about switches is the variable contact spacing as the switch continues to open further; this reduces the capacitance and lets the voltage become even higher, perhaps once again striking an arc; switches are evil trash generators when energy can be stored in some wiring and then resonated with the switch-contact variable-capacitance.
                        – analogsystemsrf
                        Aug 25 at 3:20
















                      • Thank you for the answer Andy, I was sure that there was a "conservation of energy" answer to this.
                        – Çetin Köktürk
                        Aug 23 at 20:54










                      • No probs dude..
                        – Andy aka
                        Aug 23 at 21:23










                      • @ÇetinKöktürk I would agree that "energy" stored in L's and C's is the best way of thinking about this. It leads directly to a fundamentally correct understanding. (whereas a "circuit analysis" perspective is kind of indirect and somewhat confuses the real issue: energy storage and movement)
                        – Henry Crun
                        Aug 23 at 23:00










                      • @ Andy the fun thing about switches is the variable contact spacing as the switch continues to open further; this reduces the capacitance and lets the voltage become even higher, perhaps once again striking an arc; switches are evil trash generators when energy can be stored in some wiring and then resonated with the switch-contact variable-capacitance.
                        – analogsystemsrf
                        Aug 25 at 3:20















                      Thank you for the answer Andy, I was sure that there was a "conservation of energy" answer to this.
                      – Çetin Köktürk
                      Aug 23 at 20:54




                      Thank you for the answer Andy, I was sure that there was a "conservation of energy" answer to this.
                      – Çetin Köktürk
                      Aug 23 at 20:54












                      No probs dude..
                      – Andy aka
                      Aug 23 at 21:23




                      No probs dude..
                      – Andy aka
                      Aug 23 at 21:23












                      @ÇetinKöktürk I would agree that "energy" stored in L's and C's is the best way of thinking about this. It leads directly to a fundamentally correct understanding. (whereas a "circuit analysis" perspective is kind of indirect and somewhat confuses the real issue: energy storage and movement)
                      – Henry Crun
                      Aug 23 at 23:00




                      @ÇetinKöktürk I would agree that "energy" stored in L's and C's is the best way of thinking about this. It leads directly to a fundamentally correct understanding. (whereas a "circuit analysis" perspective is kind of indirect and somewhat confuses the real issue: energy storage and movement)
                      – Henry Crun
                      Aug 23 at 23:00












                      @ Andy the fun thing about switches is the variable contact spacing as the switch continues to open further; this reduces the capacitance and lets the voltage become even higher, perhaps once again striking an arc; switches are evil trash generators when energy can be stored in some wiring and then resonated with the switch-contact variable-capacitance.
                      – analogsystemsrf
                      Aug 25 at 3:20




                      @ Andy the fun thing about switches is the variable contact spacing as the switch continues to open further; this reduces the capacitance and lets the voltage become even higher, perhaps once again striking an arc; switches are evil trash generators when energy can be stored in some wiring and then resonated with the switch-contact variable-capacitance.
                      – analogsystemsrf
                      Aug 25 at 3:20

















                       

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