DC-DC voltage buck: Why an inductor?

[tylernt]

I'm familiar with the DC-DC voltage boost circuit (SMPS or switch-mode power supply), and I understand why an inductor is used. By using a low voltage to create a magnetic field in the inductor windings, you get a high voltage back out when the field is allowed to collapse. Makes sense.

What I don't understand is why I've seen several DC-DC buck circuits, where Vout is less than Vin, with inductors. (The one on Wikipedia for example, and others.) Couldn't a capacitor be used instead? Whenever Vout falls below spec, start charging a cap. As Vout rises above spec, stop charging the cap. You'll get a little ripple, but surely less than an inductor produces?

I'm sure there's a good reason for using an inductor. I'm just curious what that reason is?

 

[alec_t]

Whenever Vout falls below spec, start charging a cap. As Vout rises above spec, stop charging the cap.

That's what happens in a buck converter: but something has to limit the charging current. An inductor can do this without signicant power loss due to heating.

 

[tylernt]

Ah, so, efficiency. Thanks for the reply!

 

[crutschow]

Ah, so, efficiency. Thanks for the reply!

Yes, the efficiency of such a switched circuit with a capacitor would be the same as a linear regulator.

 

[ronsimpson]

Efficiency Power loss is voltage across the switch X current in the switch. In the case of linear regulation, (10V in 5V out 2A) power loss is 5X2=10 watts. For a switcher. The switch is open part of the time (10 volts 0 current = 0 power) and closed part of the time (0 volts 2amps = 0 power).

 

[Grossel]

Remember that an inductor tries to resist quick changes in the current through it.

 

[crutschow]

Efficiency Power loss is voltage across the switch X current in the switch. In the case of linear regulation, (10V in 5V out 2A) power loss is 5X2=10 watts. For a switcher. The switch is open part of the time (10 volts 0 current = 0 power) and closed part of the time (0 volts 2amps = 0 power).

Which is true when the switch is driving an inductor, of course, but not when it is driving a capacitor. Thus the difference in efficiency.

 

[RCinFLA]

Somewhat simplistic, but one way to think about a buck converter is an input chopper followed by low pass filter to remove the chopping ripple.

For example, if the input chopper is running at 50% duty cycle, the output of the L-C filter will be about half the input supply. This does not take into account resistive losses of the inductor and resistance of the input chopper devices. An open loop (no control feedback) can be made like this but there will be some voltage drop on output as it is loaded. A feedback correction would modified the chopper duty cycle to compensate for circuit losses to maintain a constant output voltage.

 

[simonbramble]

Just to summarise, the inductor and capacitor are often called the filter component. Why not use an RC instead of an LC? Well, an inductor theoretically has zero resistance (hence zero loss). Same is true for a capacitor. If the switch has zero ON resistance and infinite OFF resistance, you have zero loss in the switch. This is the beauty of a switched mode power supply (buck, boost or invert). You can convert one voltage to the other with near 100% efficiency. The controller designers (as well as the inductor and capacitor manufacturers) are constantly striving to get zero impedance to minimise losses and hence get nearer to 100% efficiency (hence longer talk time etc).

Here is a simple guide to buck regulator design if you want to learn more:
http://www.simonbramble.co.uk/dc_dc_converter_design/buck_converter/buck_converter_design.htm

Hope this helps!

Simon