I don't understand. doesn't it ondulate the current?
i said ondulate because i read that an inductor is used to smooth the output current, i can see it now.. from this formula Delta Il = (Ve-Vs)aT/L, if we increase the inductance in output we can make the output smoother. So we need an inductor to discharge but also we can add additional inductors or increase inductance to have a smoother current in output.What is "ondulate"?
To convert voltages efficiently, the buck converter needs a mechanism to store the energy while converting it from a higher voltage to a lower voltage. This function is performed by the inductor. When the converter switch is on, it applies the input voltage to the inductor, causing its current to increase and storing energy as determined by the equation LI^2 where L is the inductance and I is the current. When the switch is off the inductor current continues to flow but with decreasing current as the energy is transferred to the lower output voltage.
You could use a resistor to transfer from the higher voltage to a lower voltage, which is how a linear regulator works, but that is not as efficient.
At time zero, inductor current is zero. It is rising exponentiallyTo figure out why do you need something, try removing it (mentally). Remove inductor. What do you see? Very high current. FET blown up. Load over-voltaged (killed). Conclusion: inductor is there to limit current and thereby to make sure that your output voltage doesn't get as high as the input voltage.
Here you're answering the next question - what the FET is for.It is rising exponentially
By asking "what is ondulate" I meant what does ondulate mean in your language since it is not a standard English word.i said ondulate because i read that an inductor is used to smooth the output current, i can see it now.. from this formula Delta Il = (Ve-Vs)aT/L, if we increase the inductance in output we can make the output smoother. So we need an inductor to discharge but also we can add additional inductors or increase inductance to have a smoother current in output.
So to your original question --The inductor does form a resonant circuit with the output filter capacitor and that must be compensated in the switching regulators feedback loop, otherwise the regulator will indeed oscillate.sorry for my french influenced english i meant oscillate
It's true that when the switcher control loop is simulated as a linear loop the switching frequency can be ignored if it's sufficiently higher than the LC resonant frequency. But you still need to compensate the control loop for that resonant frequency or the loop will oscillate.What i find is that the inductor cap combination frequency is almost always much lower than the switching frequency, so it is treated almost like it was not there at all. For example, a 1kHz resonant frequency vs 50kHz switching frequency, the control circuit sees only a very slow variation due to the LC combination, which makes it look almost like a simple slowly varying load change...which it is able to handle no problem.
It's only when the switching frequency is too close to the LC frequency that trouble might begin because then the control circuit might see an increase in output with a decrease in control signal, which means it's out of control.
There's also the damping to consider, if the inductor has higher resistance it wont oscillate even under some bad conditions.
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