# why does buck converter need an inductor?

#### Santa klaus

Joined Nov 16, 2014
36
I don't understand. doesn't it ondulate the current?

#### Santa klaus

Joined Nov 16, 2014
36
So current can keep flowing in the load. Thanks.

#### crutschow

Joined Mar 14, 2008
34,680
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.

#### Santa klaus

Joined Nov 16, 2014
36
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.
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.

#### NorthGuy

Joined Jun 28, 2014
611
To 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.

#### takao21203

Joined Apr 28, 2012
3,702
To 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.
At time zero, inductor current is zero. It is rising exponentially

http://en.wikipedia.org/wiki/RL_circuit

#### NorthGuy

Joined Jun 28, 2014
611
It is rising exponentially
Here you're answering the next question - what the FET is for.

#### crutschow

Joined Mar 14, 2008
34,680
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.
By asking "what is ondulate" I meant what does ondulate mean in your language since it is not a standard English word.

The value of the inductance does affect the output ripple voltage of the converter but that is not the inductor's primary function. It is energy storage, as previously stated.

#### sirch2

Joined Jan 21, 2013
1,037
Possibly the OP meant "undulate", I do like the sound "ondulate" though.

#### MrAl

Joined Jun 17, 2014
11,565
Hello,

Simply put, the inductor is there to act as an energy storage element which is necessary in order to obtain true power conversion. True power conversion means low loss of energy even when the input output differential voltage is large.

For the buck circuit with ideal circuit elements for example, if we have 10 watts in then we have 10 watt out, if we have 100 watts in then we have 100 watts out, the only difference is that the output voltage will be lower than the input voltage. So if we had 20v in and 10v out we would get 100 watts out at 10v with 100 watts in at 20v, so no loss of power (100 watts in and 100 watts out). This means we have an efficiency of 100 percent.

For a linear regulator (does not use an inductor) if we had 100 watts in at 20v we could only get 50 watts out at 10v because we would loose 50 watts in the series dropping element used to get 10v out. So we would have 100 watts in and only 50 watts out, which is an efficiency of only 50 percent.

Comparing the two methods with inductor and without inductor, with the inductor we get 100 percent efficiency (no loss of energy), and without the inductor we only get 50 percent efficiency (large loss of energy). Now you know the real reason for using an inductor.

Inductors are used for filtering too though but that could be with any type of regulator. In that case we just want a smoother output and/or reduce strain on some of the other components (like diodes, capacitors).

#### Santa klaus

Joined Nov 16, 2014
36
sorry for my french influenced english i meant oscillate

#### crutschow

Joined Mar 14, 2008
34,680
sorry for my french influenced english i meant oscillate
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.

#### MrAl

Joined Jun 17, 2014
11,565
Hi,

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.

#### takao21203

Joined Apr 28, 2012
3,702
ONDULATOR is a device used for making curly hairs- mainly by women. You can google it.

#12

#### crutschow

Joined Mar 14, 2008
34,680
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.
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.
This can be readily seen if you place the LC at the output of an op amp and do the negative feedback from the junction of the inductor and capacitor (as is done to regulate the switching regulator's output voltage).

#### MrAl

Joined Jun 17, 2014
11,565
Hi,

What form of compensation are you talking about here? Do you mean like a lead or lag network?

You'll have to specify a particular circuit here because it's not fair to generalize every buck circuit into an oscillator and then state that it oscillates. For example the buck plant is not an LC circuit, it's an RLC circuit, and even a RRLC circuit where we have at least two major resistances working inside the plant.

I've made several buck circuits without the need for any extra compensation, and they work very well. So i have to ask what kind of compensation are you using? I am thinking that a lot of op amps are already compensated for unity gain.

#### NorthGuy

Joined Jun 28, 2014
611
If you replace the inductor with a resistor, which doesn't store energy, but does limit current, it'll still work (much less efficiently though). If you do not use anything that limits current, it'll stop working. The primary goal is to limit current. Why? Because this is the only way to limit the charging speed of the output capacitor.

#### MrAl

Joined Jun 17, 2014
11,565
Hi,

That's an interesting view too.

I think the primary goal however is not to just limit current, because a linear can do that, but rather the primary goal is to CONVERT ENERGY which means an energy storage device has to be used (induct0r) and it has to be used correctly.