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Buck converter inductor current at start up

  1. Mar 8, 2012 #1
    I am trying to understand the inductor current ramp up to steady state in a buck converter.

    I don't understand why the inductor current slope is not steep in the beginning.

    Let's say Vin=12v, Vo desired across the capacitor is 5V.

    Switch turns ON, Inductor current ramps up. Switch goes OFF, Inductor current starts charging the capacitor for a time interval Δt. Now the cap is charged to say, 1v. In the next cycle, the inductor current has increased, it again charges the cap for Δt. Now the cap voltage goes to 2v. This goes on.
    But I don't understand why the current slope becomes steeper.

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  3. Mar 8, 2012 #2


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    Staff: Mentor

    What is the voltage at the input side of the inductor during the OFF portion of the cycle?

    The slope of the OFF portion of the current waveform is determined by the voltage across the inductor during the OFF portion of the cycle...
  4. Mar 8, 2012 #3
    How do you find out the voltage across the inductor during the off portion?
    The ckt is same as - http://en.wikipedia.org/wiki/File:Buck_circuit_diagram.svg
    Load is capacitor.
    At the moment the switch is turned OFF, inductor voltage is 12V.
  5. Mar 8, 2012 #4


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    Inductor voltage = voltage across the inductor
    One side of the inductor is held at either 12v or -0.7v, the other side is the load voltage. Subtract the two to find the difference.
  6. Mar 9, 2012 #5
    Okay. Inductor one side is 12V, connected to a capacitor and a diode in series. The diode voltage is 0.7v, what about the capacitor voltage?
  7. Mar 9, 2012 #6
    3 things to keep in mind:

    1) You can not instantaneously change the current in an indutor. At the moment you "turn on" the inductor sees the full voltage and has 0 current.
    2) There is always some other resistance / impedance - esp from the supply to the circuit, this is usually a critical limit.
    3) Also the pulse width - should be much less than the time constant (over which you will see the effect of the inductor. If you look at the V and I for the inductor over time ( many on-off cycles / pulses) the curve will look more like an typical inductor curve. The model typically used within one pulse is a triangle - not a curve.

    During the off portion - the current wants to stay the same - and decreases as the load absorbs the energy now stored in the inductor.

    A capacitor is used to stabilize voltage ( by being an instantaneous current source) only makes sense here before the switch or after the inductor ( parallel with the load).

    The trick is to realize that the VOLTAGE of the inductor can instantaneously change - so the instant the switch opens - it goes from absorbing energy to releasing energy - so what happens to the voltage across the inductor?? it should be clear - it reverses if there is any load current.

    If the load is a capacitor - when the circuit reaches steady state the I goes to 0 ( cap is fully charged), then the diode is reverse biased and the cap just sits there charged.
  8. Mar 9, 2012 #7


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    When the circuit operates correctly, the capacitor voltage gets pumped up in steps, until it levels out at its final value.
  9. Mar 9, 2012 #8
    Okay. Let me plug in some numbers.

    Lets say input voltage is 12v, L=100mH, C=1uF, switching freq=1KHz ( ignore diode drop).
    Let's say Ton=Toff=0.5ms (is this valid)?

    So final inductor volage after Ton will be 12v.

    V = L Δi/Δt
    12 = 0.1 (di/0.5m)

    Δi= 60mA.

    Toff is 0.5ms.

    capacitor voltage Vc = 1/c ∫I dt

    Vc = 1E6 (60mA*0.5mS)
    Vc = 30V ?
  10. Mar 9, 2012 #9
    The V ind after T on is coming to 0, not 12 V ( Assuming we are at steady state)

    Here is a good reference : http://www.ti.com/lit/sg/sluw001d/sluw001d.pdf [Broken]

    Basic Buck and Boost converter voltages are relative to the duty cycle ( Ton:Toff)
    Last edited by a moderator: May 5, 2017
  11. Mar 9, 2012 #10


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    Staff: Mentor

    At any time while switch is closed, for inductor voltage =12v the load voltage must =0.
    If load voltage is not equal to 0, then inductor voltage cannot be 12v.
  12. Mar 11, 2012 #11
    Okay okay. VL is 0 at the end of Ton.
    But it's limited to (Vo+Vd) when the switch opens, right?

    Also capacitor voltage can only go upto Vo (5V).

    Vc = ΔI ΔT /C

    5 = 60mA (ΔT)/1E6

    ΔT = 5m/60 = 83uSecs

    So the cap gets charged really quickly compared to Toff.
  13. Mar 12, 2012 #12


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    Probably not 0, but certainly heading that way. Dropping away.

    That 60mA is probably an overestimate. You arrived at that value by applying 12v to the inductor and assuming that while the capacitor charges its voltage remains at (or close to) 0.
  14. Mar 12, 2012 #13
    Ok, Initially the load voltage is 0v.
    But then, max VL is (12-Vo)
  15. Mar 12, 2012 #14
    For Vin = 12V and Vout = 6V and RL = 6Ω ( load current = 1A) and F = 1K.
    The duty cycle = 0.5 so Ton = Toff = 500us and the L = 10mH

    At the beginning Vout is 0V.

    When T1 is ON voltage across the coil is equal to 12V.
    At the end of time ON time coil current reaches the value.

    I _(0.5ms) = ( 12V * 500us) / 10mH = 0.6A So to output voltage will reach 3.6V

    So now at time >500us T1 is OFF and we have 3.6V across the coil.
    So the current wail drop to

    I = ( 3.6V * 500us)/10mH = 0.18A

    I_(1ms) = 0.6A - 0.18 = 0.42A and the output voltage will drop to 2.5V

    But now at time 1ms T1 is ON.

    The voltage across the coil is equal to
    12 - 2.5V = 9.5V

    and the coil current will ramp up.

    I = (9.5V * 500us) / 10mH = 0.475A

    So at time 1.5ms Icoil = Iload = 0.42A + 0.475A = 0.895A and the load voltage
    V = 5.37V

    At T = 1.5ms T1 is OFF and the coil current will ramp down.
    The voltage across the coil is now equal to 5.37V.
    And at the end of the OFF time (2ms) coil current drop to :

    I = ( 5.37 * 500us) / 10mH = 0.268A

    I_(2ms) = 0.895A - 0.2685 = 0.626A and the load voltage

    V_(2ms) = 3.759V

    Now T1 is ON up to 2.5ms

    I = ( ( 12V - 3.759V) * 500us)/10mH = 0.412A

    I_(2.5ms) = 0.626 + 0.412A = 1.038A and Vout = 6.228V

    At time 2.5mS T1 is OFF until 3ms

    So the current will drops to

    I = ( 6.228V * 500us/10mH = 0.3114A

    I_(3ms) = 1.038A - 0.3114A = 0.7266A---> Vout = 4.3596V

    T1 --> ON up to 3.5ms

    I = (( 12V - 4.3596V) * 500us)/10mH = 0.382A

    I_(3.5ms) = 0.382A + 0.7266A = 1.10862A ---> Vout = 6.63172V

    And this will continue until a steady state is reach.
  16. Mar 13, 2012 #15


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    Only at the initial moment is voltage across the inductor 12v; as the capacitor charges the voltage across L falls.
    This is a very rough-and-ready estimation. Using your figures, the inductor voltage is only 8.4v towards the end of the ON period, so relying on 12v gives an overestimate.

    Also, you are totally overlooking the fact that some of this inductor current powers the load (the 6Ω resistor) and does not add to the charge on the capacitor. So your figure of 3.6v is going to be way off, and for two reasons you have overlooked/forgotten/not mentioned.
    Current in the inductor doesn't "drop", it smoothly ramps from one value to another. Apart from this, I don't think your figure of 0.18A is right.
    How can determine this without even mentioning the load resistorhttps://www.physicsforums.com/images/icons/icon5.gif [Broken]

    Your attempt to explain the workings of this are a good start, Jony130, but a shaky start, and would not get even half marks on a test paper. https://www.physicsforums.com/images/icons/icon9.gif [Broken]
    Last edited by a moderator: May 5, 2017
  17. Mar 14, 2012 #16
    I know that my analysis is very simplified. I assume RL = 6Ω without output capacitor.
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