What I understood at least at a basic level the operation boost converter, the inductor supplies output current when the switch is OFF.
When the switch is ON the the load current is taken current by the output capacitor.
Update
Is the classification of DCM and CCM is based on the nature output current or nature of inductor current ?
But during switch ON state, the inductor current has nothing to do on the output current ? or what I understood is wrong? Can anybody please explain this ?
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\$\begingroup\$ You say "we take the inductor current same as output current" - what did you base this on? Do you have a link? Also your graphs show IQ1 etc but these don't have references on the circuit - could you fix for clarity please? \$\endgroup\$– Andy akaJun 5, 2013 at 7:02
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\$\begingroup\$ Your nomenclature makes it difficult to relate the graphs with the circuit. Both inductors and capacitors store energy so for half the time you are putting this in and the other half taking it out. By selecting appropriate values for L and C you can alter the amount of energy available from each. \$\endgroup\$– JIm DeardenJun 5, 2013 at 7:06
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\$\begingroup\$ @Andyaka please see my update ... \$\endgroup\$– noufalJun 5, 2013 at 9:34
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\$\begingroup\$ What is IQ1? What is ICR1? I'd also say discontinuous or continuous current mode etc. What do the nodes a, c and p mean? Why does IL = Ic. Is Io actually Ir (i.e. load current)? - these are all very confusing things that do not help people answer because they seem to be irrelevant to the question. I'm trying to help you phrase a better question btw!! I think the last picture is splitting a sentence - just noticed now. \$\endgroup\$– Andy akaJun 5, 2013 at 9:43
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\$\begingroup\$ @Andyaka Q1 is the switch, and CR1 is the diode, a, c, and p are three nodes used for simplicity.. Please see my question again. I have made few changes as you mentioned. Thanks for your patience... \$\endgroup\$– noufalJun 5, 2013 at 10:36
4 Answers
What actually is the output current of a boost converter?
Hopefully the diagrams below will explain the output current and the difference between continuous and discontinuous operation.
Look at the blue traces in both continuous and discontinuous modes - this is the rectifier or diode current (what you refer to as CR1). The diode current is the current into both capacitor and load resistor.
When the average current is not constant the output voltage would either collapse to zero or rise until something went "bang". In reality, it is the average current that flows into the load because without a load, a non-zero average current would continue to push the output voltage up and up.
Output load resistor current is therefore average diode current
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\$\begingroup\$ is the CCM and DCM operation is a function of load current or is it fixed by the values of passive elements and duty ratio? \$\endgroup\$– noufalJun 5, 2013 at 12:47
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\$\begingroup\$ @noufal High load current and fixed frequency operation mean DCM is more likely because the inductor energy is exhausted before the "charge" cycle restarts. \$\endgroup\$– Andy akaJun 5, 2013 at 13:00
When the switch is ON, current ramps up in the inductor and the load is powered by the output capacitor(s) alone. The rectifier diode is reverse biased.
When the switch is OFF, the rectifier diode is forward biased and the load (including the output capacitor) is powered by the output inductor.
CCM and DCM is based on the behaviour of the inductor current, which is a factor of the load current, the inductor size and the switching frequency.
The load current is the arithmetic average of the diode current waveform, thanks to the output capacitors.
(You can easily simulate a boost converter with any of the free Spice-based simulators out there and get a feel for all of these things by playing with the frequency, inductor size and load current. I like LTSpice for this sort of task.)
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\$\begingroup\$ Why do we bother about DCM and CCM ? what is the necessity of this classification? is DCM operation not desirable in a boost converter ? (or in any converter). \$\endgroup\$– noufalJun 5, 2013 at 14:08
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1\$\begingroup\$ The behaviour of the converter is quite different depending on the mode. A boost can be designed to be 100% DCM, but cannot be designed for 100% CCM - it will always go into DCM at very light load. DCM gives you the advantage of turning on the MOSFET at zero inductor current, so there's less switching loss. However, the peak ripple current is higher than in CCM. Also, in CCM the feedback loop compensation generally needs to have a lower bandwidth than DCM since CCM introduces a right-half plane zero into the transfer function. \$\endgroup\$ Jun 5, 2013 at 14:16
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\$\begingroup\$ thanks for your clear explanation. How do I add a MOSFET with gate driver in LTSpice ? \$\endgroup\$– noufalJun 5, 2013 at 14:43
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\$\begingroup\$ You can simply use a pulsed DC source instead of a driver, directly on the gate (or through a small resistor). \$\endgroup\$ Jun 5, 2013 at 15:57
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\$\begingroup\$ yeah.. i did that. Could you please suggest some values of input parameter to visualize DCM and CCM...? \$\endgroup\$– noufalJun 5, 2013 at 17:19
As your current graphs show, the inductor current is not constant, so the concept of 'the inductor current' being a single (time-independent) value is not valid.
What is valid though, is that the inductor current has no sudden changes (the first derivative is always finite). This is a fundamental property of an inductor, and it is illustrated in your 3d graph: the inductor current is a sawtooth.
When you talk about 'output current' you must be careful what you mean: if you say inductor_current = output_current you mean the current going into the (load + output_capacitor).
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(after question update)
CCM means that there is always a current through the inductor. That means that the current fed into the load+capacitor ramps down to a certain value, and then abruptly drops to 0.
DCM means that at some point(s) in time there is no current through the inductor. That means that the current fed into the load+capacitor ramps down to 0.
Hence DCM/CCM can be observed both at the inductor and at the load+capacitor (and at the input too).
I think the key point is that the average current through the load (which is constant, with a small amount of ripple) is equal to the average current through the diode (which is pulsing). In other words, the area under the ICR1 curve for any given interval equals the area under the IO curve for that same interval.