Purpose of inductor in buck converter

[Rusttree]

In a boost converter, the inductor is necessary to achieve a higher voltage on the output than the input. I was thinking about a buck converter and I realized I don't understand why an inductor is necessary to step down the voltage. It seems it'd be much simpler to just have a capacitor bank at the output and a switching transistor that keeps the capacitors at the desired voltage using a feedback loop. I threw one like that together in a SPICE model with a periodic load and it seemed to work just fine. What's the advantage to the inductor in the circuit? Is it an efficiency thing?

 

[MikeMl]

Think about the difference between an RC low-pass filter (1 pole) vs an LC low-pass filter (2 poles).

Think about peak current required to charge the capacitor with and without the series inductor.

 

[Bob Scott]

If you switch a transistor on in order to "fill up" a filter capacitor and that transistor is directly directly connected between the source of B+ and the capacitor, theoretically infinite current will flow. An inductor will reduce the peak current to a manageable level. What is the peak current in your SPICE model?

Or are you using a linear regulator?

 

[Rusttree]

Ah, right, the 2-pole LC filter will attenuate the higher frequencies more aggressively. I wasn't thinking about filtering when I asked the question. That makes sense now.

So there's nothing "wrong" with taking out the inductor. You'll just have to deal with more ripple voltage. Which, now that I'm looking at my SPICE model a little more closely, is definitely evident.

 

[Rusttree]

If you switch a transistor on in order to "fill up" a filter capacitor and that transistor is directly directly connected between the source of B+ and the capacitor, theoretically infinite current will flow. An inductor will reduce the peak current to a manageable level. What is the peak current in your SPICE model?

True, the Rdson of the MOSFET is the only appreciable resistance, which is very tiny. The model does predict high current spikes (10s of Amps) for very short durations every time the transistor turns on.

 

[MikeMl]

...The model does predict high current spikes (10s of Amps) for very short durations every time the transistor turns on.

And where are you going to get currents of hundreds of A? Maybe from a car battery. Not likely from a practical upstream power supply. Just what will you do to the regulation of the upstream supply by loading with huge currents with periods of no load?

 

[dougy83]

In a boost converter, the inductor is necessary to achieve a higher voltage on the output than the input. I was thinking about a buck converter and I realized I don't understand why an inductor is necessary to step down the voltage. It seems it'd be much simpler to just have a capacitor bank at the output and a switching transistor that keeps the capacitors at the desired voltage using a feedback loop. I threw one like that together in a SPICE model with a periodic load and it seemed to work just fine. What's the advantage to the inductor in the circuit? Is it an efficiency thing?

The inductor isn't neccessary to step down the voltage; as you have noted, pulsing the voltage through an RC circuit (which is what you are doing) will provide a lower voltage. The problem is that half of the energy from the input is lost in the resistor while being transferred to the output.

The use of an inductor does allow the peak current to be reduced, but this is not it's main function in a buck converter. Its purpose is to store energy while the switch is ON and transfer that stored energy to the capacitor when the switch is OFF. A resistor is not capable of storing energy and just loses it as heat.

If you switch a transistor on in order to "fill up" a filter capacitor and that transistor is directly directly connected between the source of B+ and the capacitor, theoretically infinite current will flow. An inductor will reduce the peak current to a manageable level. What is the peak current in your SPICE model?

There is an inherent resistance between the source and the capacitor within the wiring, the transistor and the capacitor itself - all of which will reduce the current. The inductor has more effect than just a current limiter.

So there's nothing "wrong" with taking out the inductor. You'll just have to deal with more ripple voltage.

You will get much decreased efficiency (worse than a linear regulator), and increased stress on all parts in the circuit (possibly some explosions due to the high peak I²R). You will get a higher ripple voltage as the current is only supplied for a fraction of the cycle, whereas with an inductor the current can be supplied continuously.

 

[Bob Scott]

There is an inherent resistance between the source and the capacitor within the wiring, the transistor and the capacitor itself - all of which will reduce the current. The inductor has more effect than just a current limiter.

I didn't say it was just a current limiter. Please don't talk down to me. It's the OP that asked a specific question.

 

[dougy83]

I didn't say it was just a current limiter. Please don't talk down to me. It's the OP that asked a specific question.

I'm sorry that you took it to be condescending (if that's what you meant by 'talk down'); it certainly wasn't meant to be.

 

[panic mode]

inductor is used as energy storage element. it is essential for operation of SMPS.

 

[MrAl]

Hi,


Yes you guys are getting to the main point here and that is that the most important role of the inductor is energy storage, and that energy storage is what allows a buck converter to be called a "true" power converter, whereas a resistive divider (transistor or not in there) is not.

A true power converter actually converts energy from one form to another and that's the mechanism that makes it work. It takes a higher voltage and lower current and converts it to a lower voltage and higher current, similar to how a transformer transforms power. For a 100 percent theoretical system, the power in equals the power out but what changed was the voltage and current, so that makes it a true power converter.

Take out the inductor though and what we see is a pulse with modulator (you can add resistance to reduce current) but that only means that somewhere there has to be resistance eating up power. It is true that the average current will go down, even the current for each pulse, but the current pulse will still be high enough where any resistance (necessary resistance to limit current peaks) eats up power.
If you analyze this you find that as you lower the resistance to try to reduce losses, the current peak goes up so you end up loosing a lot of power anyway.

So without the inductor your efficiency drops quite a bit except in the case where the output voltage is almost the same as the input voltage. When the output voltage is almost the same as the input voltage then the loss in any resistance starts to equal the loss in a real life buck converter where it can not be exactly 100 percent efficient. With a typical efficiency of 80 percent, this means somewhat small but not insignificant differences between input and output voltages. For example, if you want to drop only 1 volt with an input of 10 volts and output of 9 volts you may never see a buck that can equal the efficiency of just a series resistor, depending on the current requirement. But for 10 volts in and 3.3 volts out for example you'd never get the efficiency anywhere near as good with anything but a true buck circuit even at 80 percent efficiency.
We could look at more specific examples.

 

[crutschow]

They do make switched-capacitor IC voltage converters without inductors that are reasonably efficient but they are generally low current (100mA or less) and step the voltage in discrete multiples of the input voltage (unless you add an inefficient linear regulator to the mix).

 

[MrAl]

Hi there,


I think the capacitor switched converters are only good for developing output voltages that are integer multiples of the input voltage. This includes the negative integers too so for example we could use one for an 'inverter' that converts say 10v in to -10v out at reasonable efficiency for relatively low current levels. But that works because the voltage differentials that appear across the several series resistances are low. Once the voltage differential becomes significant (as in a 2:1 buck converter) the efficiency will drop off. So we might be able to do a 10v to 20v converter, or a 10v to -10v converter, but as soon as we try to do a 10v to 5v converter we see a big drop in efficiency because there is always a relatively large voltage difference between any voltage trying to charge a capacitor and the charging source (input or another cap).
Maybe adding pulse width modulation would help, but there's always going to be that largish voltage differential and along with the current means we loose more power.

 

[MikeMl]

To build a 10 to 5V converter, how about charging two identical capacitors in series, and then switching them in parallel to the load.? You could have two sets of two, and multiplex the two sets so that the load is not disconnected from it's source. Never tried it, but it should work...

 

[MrAl]

Hi Mike,

Sounds like it could work. We'll have to try it. This config would require some complex switching, but who cares. At least the voltage differentials would be low. Three in series to three in parallel might give us 1/3 of the output or 3.3v out. The limitation is that they would all have to be the same value and pretty close or else we'd get some strange results.

 

[Brevor]

The inductor is usually the most expensive component in a SMPS, If there was a way to eliminate it someone would have done it.

 

[panic mode]

exactly

 

[simonbramble]

Have a look at this:
http://www.simonbramble.co.uk/dc_dc_converter_design/buck_converter/buck_converter_design.htm

I have done a lot of thinking about buck converters and the role of the inductor - and indeed their simulation in LTSpice.

Hope it helps