Requirements for inductor to operate in continuous conduction mode?

  1. Question about continuous conduction mode converter

    I have a few quesitons about CCM converters

    I have posted a basic schematic of the circuit in question as an attachment.

    The circuit is a series LC circuit in which the inductor and capacitor charge during the on time. During the off time the inductor energy dumps into the cap, charging it to 2x the supply voltage.

    My questions about the circuit:

    - What would be required to get the circuit to operate in CCM?
    -Since the charging/discharging impedances are equal could a 50% duty cycle be used?
    -Would the circuit only operate in CCM when the pulsing frequency is equal to the resonant time period between the inductor and capacitor?

    Any help would be greatly appreciated! Thank you.

    Attached Files:

    Last edited: Mar 20, 2013
  2. jcsd
  3. jim hardy

    jim hardy 5,460
    Science Advisor
    Gold Member
    2014 Award

    no help yet?

    see if there's anything helpful here.

    one of the major manufacturers had tutorials on switching regulator design , I think it was National before TI bought them.
    Try a search on phrase "Simple Switchers"
  4. That site has lots of good info but not exactly what I am looking for.

    I have the simulation circuit (attached to my first post). All I want to understand is what the requirements are to get the inductor to operate in continuous mode when the input is a 50% duty cycle pulse.
  5. berkeman

    Staff: Mentor

    The average output current has to be over half of the ripple current.
  6. berkeman

    Staff: Mentor

  7. So, the average output current has to be over half the ripple current.

    Can you explain why this is?
  8. berkeman

    Staff: Mentor

    Sure. Look at the 2nd plot down in the picture I posted. That shows the output current as a function of time. The ripple current is the peak-to-peak variation of the current, and the average output current is the DC average of that graph.

    For continuous output current mode, you do not want to drop to zero output current at the bottom of the ripple current waveform. So if the average output current is at least half of that ripple current, the graph of the output current waveform never reaches the bottom axis (I=0).

    The main problem with allowing non-CCM is that it complicates the feedback equations, and makes it harder to design a stable DC-DC converter circuit. Not impossible, of course, but it makes it harder, and constrains the performance of the circuit (either you add more circuitry to make it more expensive, or you settle for reduced regulator performance).
  9. Ok, that makes sense.

    Thanks! I appreciate the help.
  10. Looking at the circuit that hobbs125 gave in his original post, isn't the determination of whether the inductor current is continuous or discontinuous dependent on the resonant frequency of L1-C1 and the frequency of the square wave?
    (This is a guess on my part.)

    This would be the ideal case for a spice simulation. Is there anyone here that can do a spice simulation?
  11. berkeman

    Staff: Mentor

    I don't think it depends on C1, as long as C1 is big enough to limit the output ripple voltage to a reasonably small value.
  12. Quote from original post "The circuit is a series LC circuit in which the inductor and capacitor charge during the on time. During the off time the inductor energy dumps into the cap, charging it to 2x the supply voltage."
    The preceding is only partially correct, the capacitor is only charged to 2x the supply voltage if the capacitor is charged to zero volts at start of pulse.
    In "normal" operation the capacitor's average voltage would be slightly less than the square wave pulse peak voltage.

    Using a 1.13 mHy inductor, a 1N4148 diode, 5 each 2.2 uf capacitors -1.12 uf measured and a 1000 ohm resistor + a function generator and oscilloscope.
    With a 20 volt peak to peak square wave input, the start of the CCM was measured at between 500 kHz and 1.0 MHz. (Depending on how the oscilloscope wave form was interpreted.)
    It would be interesting if someone would go through the math and say if the measure frequency for CCM is the same as the calculated CCM.
  13. Carl Pugh,

    This is exactly what I am wondering....My question, is why? What is different at Fres that causes the circuit to operate in CCM?

    Then again, I think berkeman has a good point. I tested the circuit in multisim and it seems to operate in CCM at all frequencies..But the resonant thing really has me wondering still, not sure why? It just seems like we would see something different occuring at that freq, even though it's not an AC circuit.
    Last edited: Mar 23, 2013
  14. Carl

    Can you or anyone here show me how you can calculate the frequency at which CCM occurs?
  15. What is a square wave?

    What is the definition for a square wave?
    Whether the circuit is CCM or not depends at least partially on the definition of a square wave.

    Attached Files:

  16. Well, those are all square waves as far as I know. However, the circuit I posted contains a rectifier and is meant for unipolar pulses.

    I do know that if you breakdown the square wave you find out it's made of multiple harmonic AC waves.....IS that what your getting at?
    Last edited: Mar 24, 2013
  17. berkeman

    Staff: Mentor

    But you do understand that the simulation circuit that you posted is non-physical? At least, no boost converter that I know of operates the way that you posted. Where did you get that circuit from?
  18. Yeah, I have noticed in multisim that results vary and sometimes change when I didn't make any changes to the circuit....

    This circuit is more of an inductive charging circuit than a boost converter, meant to charge the cap to 2x Vs. Problem is, the load drains the cap too fast and causes the voltage to drop. That's why I want to know about how to get the cirucit to operate in ccm, if there's a certain frequency required and how to calculate that frequency?

    I still don't really understand it.
  19. Using the circuit and component values originally posted, assuming that all components are ideal and that the square wave goes between 0 and +10 volt.

    At high frequencies, the inductor acts as an open circuit to AC and the capacitor only charges to the average voltage of the square wave, or 5 VDC.

    The current through the inductor goes continuous when the frequency of the square wave increases to a high enough frequency that the capacitor voltage drops to 5 VDC

    For positive portion of square wave, di/dt = (voltage of square wave-C1 voltage)/Inductance
    di/dt = (10-5)/0.0017 = 2940 amp/second

    For 0 voltage portion of square wave, di/dt = (voltage of square wave-C1 voltage)/Inductance
    di/dt = (0-5)/.0017 = -2940 amp/second

    The current through the resistor is 5 volt/1000 ohm = 0.005 amp

    For the average current through the inductor to be 0.005 amp, the peak current has to be twice this or 0.01 amp.

    0.01 amp/2940 amp/second=3.40 microsecond that the square wave is at 10 volt.
    Since the di/dt is the same for when the square wave is at 0 volt, the period of the square wave is 2 X 3.4 microsecond or 6.8 microsecond.

    f = 1/period = 1/6.8 microsecond = 147 kilohertz

    At frequencies below 147 kilohertz, the current through L1 is discontinuous.
    At frequencies above 147 kilohertz, the current through L1 is continuous.

    C1 never charges to a higher voltage than the peak voltage of the square wave.
    When L1 is CCM, the output voltage is one half the square wave peak voltage.
    (Tested this circuit using component values I gave previously and the two preceding statements are correct)
    Last edited: Mar 25, 2013
  20. Just out of curiosity, what is the difference between an inductive charging circuit and a boost to Vout=2Vs?
  21. jim hardy

    jim hardy 5,460
    Science Advisor
    Gold Member
    2014 Award

    I've been trying to work that circuit intuitively in my head for days now.

    I believe that Berkeman's circuit in post #5 is the only one that can work.

    For this reason:

    The capacitor can only be charged to [itex]\frac{1}{2}[/itex] CV2 joules from the supply.

    To charge it to 2V takes it to [itex]\frac{1}{2}[/itex]C X(2V)2 = 2CV2 joules

    Which means the inductor has to deliver the remaining [itex]\frac{3}{2}[/itex]CV2 joules.

    So the inductor must be charged to current where
    [itex]\frac{1}{2}[/itex]LI2 = [itex]\frac{3}{2}[/itex] CV2

    or I = V[itex]\sqrt{3C/L}[/itex]
    That works out to 939 milliamps for 10 volts ?

    During time you are attempting to charge the inductor, capacitor's voltage is approaching Vsupply reducing the differential voltage available across the inductor to charge it.
    That's why Greg's circuit will work and the original cannot.

    EDIT: Greg's keeps constant voltage across the inductor while it is being charged so inductor current can rise as long as it needs to and reach substantial value.

    Original settles at V/R amps., That's 10 milliamps for 10 volts.
    It has a race between charging the capacitor and charging the inductor.
    Surely some simple algebra will resolve whether that race can be won.

    That's as far as I got last night.
    I would approach it from energy , di/dt and dv/dt , myself .
    Times should fall out that'll define your frequency and duty cycle (I hope.)

    old jim
    Last edited: Mar 25, 2013
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?

Draft saved Draft deleted