Question about switching regulator theory of operation

Discussion in 'Homework Help' started by Saviour Muscat, Jul 17, 2015.

  1. Saviour Muscat

    Thread Starter Member

    Sep 19, 2014
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    Hi
    I need some help about switching regulators step up fig 23-33b I am not fully sure about the text quoted that says
    " when the transistor suddenly cuts off, the magnetic field around the coil collapses and induces a large voltage across the coil of opposite polarity. This keeps the current flowing in the same direction. Further more, inductive kickback voltage(I think back emf) is larger than the input voltage"

    Kindly can someone explain for me the quoted text for me especially the bolded text?

    again thanks
     
  2. WBahn

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    Mar 31, 2012
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  3. WBahn

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    Because the energy in an inductor is stored in the magnetic field, which is the result of the current flowing in the inductor, conservation of energy dictates that you can't make sudden changes in the current. Thus when you open a switch on an inductor the collapsing magnetic field will produce whatever voltage is required in order to keep the current continuous.
     
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  4. crutschow

    Expert

    Mar 14, 2008
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    The statement "induces a large voltage across the coil of opposite polarity" can be confusing unless you think about what is happening.
    Say you apply a positive voltage, left to right, across a coil.
    This will cause to coil current to increase from left to right with a rate determined by the applied voltage and the coil inductance (V = L di/dt), storing an energy equal to 1/2 LI^2.
    When a switch opens in series with the coil the inductance will attempt to keep the current flowing in the same direction (as WBahn stated). This will now cause a plus voltage on the right side of the coil relative to the left side of the coil (thus giving a reverse voltage across the coil as compared to the charging voltage).
    The inductance of the inductor will generate whatever voltage is required to keep the current flowing until all the inductive energy is dissipated. If no path is provided then it will arch across the switch contacts or break down the wire insulation.
     
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  5. Jony130

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  6. Saviour Muscat

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    Sep 19, 2014
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    Hello tankyou for your help
    1 Its a bit difficult for me to grasp the concept, I need help from someone to find me a link or otherwise, a simple diagram for the coil charging and discharging(growth and decay) with battery and resistor and important to show the polarity of the voltage across the coil and the current direction again during charging and discharging,
    2 last question for now Kickback voltage is it called back emf? Am I correct?
    Sorry but I tried searching this things on the internet but what found not all these information were given!
    Very much for your aid
    SM
     
  7. MikeML

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    Oct 2, 2009
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    This shows everything that is relevant to your questions. Think about writing a KCL equation for node d. I show you all the currents in/out of node d. Note the inductor current, the current into the drain of M1, and the current through the diode... Pay attention to V(d), and V(out).

    V(d) shows why I hate switching power supplies. The RFI they radiate is extremely difficult to suppress...

    248.gif
     
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  8. Saviour Muscat

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    Sep 19, 2014
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    Many thanks for your reply but I don't know how to lt spice , I want to tackle the problem step by step, first I want to know how the inductor is working and then applying in the circuit again thanks , In think your post will be useful later on!
    Thanks
     
    Last edited: Jul 20, 2015
  9. Jony130

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    Feb 17, 2009
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    AD1
    http://www.learnabout-electronics.org/ac_theory/dc_ccts44.php
    Also do not forget about this equation VL = L* ΔI/Δt
    This equation indicates that inductance voltage depends not on current which actually flows through the inductance, but on its rate of change. This means that to produce the voltage across an inductance, the applied current must change. If the current is kept constant, no voltage will be induced, no matter how large the current. Conversely, if it is found that the voltage across an inductance is zero this means that the current must be constant but not necessary zero.

    65.png

    Ad2
    Yes, kickback voltage we sometimes called "back emf".
     
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  10. MikeML

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    You do not have to even know that LTSpice was used to make the plots. By looking at the plots, you should be able to answer all of your questions. Everything you asked about is in the plots.
     
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  11. Saviour Muscat

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    Sep 19, 2014
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    Dear Jony130
    Please find attached (growth and decay) What I have in mind Am I correct? If no correct me

    Dear MikeML
    I will find your plots very later on

    million thanks
     
  12. Jony130

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    Looks good to me.
     
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  13. MikeML

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    As the dark blue trace in my plot shows. Note that the left end of the inductor is tied to +5V. When the FET turns on, the right end of the inductor is grounded. Current (from left to right) linearly increases per di/dt=5/L. When the FET turns off, the voltage at the right end of the inductor jumps immediately to ~+25V (clamped by the forward voltage drop of the schottky rectifier) and the previous voltage stored on the output capacitor. Now current continues to flow through the inductor and the rectifier until all of the energy previously stored in the inductor is transferred to the output capacitor...

    At the instant that the FET turns off, the voltage across the inductor (right end referenced to left end) jumps from -5V to +25V, a net 30V step. It would go much higher if it were not for the clamping voltage effect of the rectifier.
     
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  14. Saviour Muscat

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    Sep 19, 2014
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    Many thanks dear Jony130 and MikeML

    I am seeing some light at the end of the tunnel now,
    I think using Bjt transistor from my first post, during saturation the coil is charging(energized) and during cutoff the coil is discharging (denergized) and the output voltage is controlled by pulsewith modulation on the input of the transisitor VL=L*deltaI/deltat. please tell if Am I there?
    Last question the diode is it used to block the charge of the output capacitor to go back to ground during when the transistor is ON ?again tell me if Am I there?
     
  15. Jony130

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    Yes, you are right.
    Yes, but diode also is needed when transistor is OFF, to provide a path for a coil "discharge" current.
     
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  16. Saviour Muscat

    Thread Starter Member

    Sep 19, 2014
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    Yes OK I understood the function of the circuit I appreciate much much your help thanks to all especially Jony130 and MikeML !
    Goodnight to all
     
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