Understanding Buck Converters

[kong]

The way I understand Buck Converters is that they 'condition' a PWM signal and basically smooth it out for sensitive electronics (anything that isn't a motor or an incandescent bulb) using an LC circuit and two switches. Am I correct in this way of thinking? The actual power variation comes from the PWM signal; as in, the Buck converter doesn't actually 'buck voltage', it allows the charging of a capacitor so as to smooth out a PWM signal and make the load think it's getting clean DC.

Can someone clear this up for me?

 

[KrisBlueNZ]

Kind of. It's definitely true that without the output capacitor, a buck converter won't work properly.

I would describe it more as an energy converter, using an inductor. During the time that the switch (transistor, MOSFET, whatever) is ON, current flows from the input, through the switch, through the inductor, to the output. Current in the inductor increases, and this creates a magnetic field in the inductor which stores energy. Then when the switch turns OFF, the current drain on the input ceases, and the inductor dumps some or all of this energy into the output, and the magnetic field in the inductor decreases.

So while the switch is ON, some current flows from the input to the output, and the energy in the inductor increases. When the switch is OFF, the inductor supplies the output current and the energy in the inductor decreases. Look at the current waveform in the inductor to see how the energy rises and falls during the switching cycle.

I suggest you get data sheets and application notes from manufacturers of buck converters - Texas Instruments, National Semiconductor (now absorbed into Texas Instruments but you may find some old app notes under the NS banner), Linear Technology, ON Semiconductor, Maxim, Fairchild, STMicroelectronics, Micrel, and Analog Devices are good places to start.

 

[kong]

OH! So really it's taking advantage of an inductor's properties over a resistor's properties to minimize thermal loss but still reduce the voltage at the output, using PWM?

 

[kong]

Seems like the capacitor acts more like a stabilizer and the inductor acts more like the reservoir for energy

 

[KrisBlueNZ]

Yes, that's fair. But, although it's true that feeding a PWM signal through a resistor into a capacitor will smooth it out, that's not a good analogy for the inductor. The inductor does things that a resistor couldn't even dream of doing!

 

[kong]

So the buck converter smooths the output using the inductor and capacitor, providing energy to the load during the off periods of the transistor. The PWM control signal is what prompts the circuit, but how does it 'step down' voltage on the output?

 

[KrisBlueNZ]

It's probably best if you read some manufacturers' application notes. They are normally well written and easy to understand. And there are other sources of information on the Internet as well. But they are not all accurate and clearly explained.

 

[kong]

Okay, thanks for your help!

 

[BobK]

Another way of thinking about the inductor in a buck converter is that it reduces the voltage without loss. The voltage across the inductor is proportional to the rate of change of the current. So If the current is always changing, i.e. the inductor is not saturated, then there is a voltage drop, but no power is lost as it would be in a resistor.

And you can make a buck converter without a capacitor, it is not essential. The ripple will be higher, but that can be fixed by using a bigger inductor.

Bob

 

[BobK]

Here is an example of a buck converter without a capacitor. Note the unreasonably high 10mH inductor. If I go down to 1mH the ripple is about 1.8V p-p. Switching speed is 100KHz.

 

[kong]

I hadn't had inductor voltage drop explained to me that way, that clears some things up, thank you. I never did understand exactly why inductors differed from resistors until now, with respect to current limiting in an AC circuit. So that is how it 'drops voltage', on the output in a buck converter? The shorter the mark to space ratio of the waveform feeding the base of the transistor =the faster change in current across the inductor=the more voltage drop across the inductor= the lower the output voltage? That makes perfect sense to me but I'm not sure I have the theory right.