Help understanding buck converters

[Peepsalot]

I am reading this article on wikipedia:
http://en.wikipedia.org/wiki/Buck_converter

They show this schematic:
http://en.wikipedia.org/wiki/Image:Buck_conventions.svg

I know it's a very simple circuit, but it still seems overly complicated to me. I guess part of the problem is that I still don't have a completely clear understanding of the purpose of inductors in general.
I'm wondering if someone could make a step down regulator using only a switch and a capacitor, and leave out the inductor + diode? Leave the switch on until the capacitor is charged to the desired level, turn it off, turn it back on when it gets too low, and so on.
It seems like my idea could be made more efficient than the given schematic, since there will always be losses from resistance in the inductor and voltage drop across the diode. Is there something wrong with this idea?

 

[IMSabbel]

Hm, not being that into that stuff myself, the most sensible idea would be that the inductance acts as a lowpass filter, filtering out spikes/ect due to the switching before that can hit the consumer.

 

[half-watt]

I guess part of the problem is that I still don't have a completely clear understanding of the purpose of inductors in general.

keeping it short and very basic...

one function is to limit (resist) initial current flow, analogous to how a capacitor limits initial voltage. another function is use in filters (e.g., a low pass filter - combined with a capacitor). remember the "switch" is electronic in nature (often a MOSFET) and is switching quickly producing noise and harmonics. diggin' deep into my memory banks, i'd say that it's part of a low-pass filter to allow only the pure DC to pass and to filter out, albeit not perfect, noise from the switching. the diode, aka single-quadrant switch, functions as a second throw position as part of an idealized SPDT switch (often indicated in a simplified schematic of the buck converter) for when the schematic's switch (i.e., typically a MOSFET) is open (off). in some simplified schematics it may illustrated by a second switch instead of being part of a SPDT switch.

look again at the simplified circuit and think on this just a bit.

 

[MrAl]

Hi there,


Dont be dismayed if you find switching regulators a little hard to
understand at first, many folks do, but if you keep with it you
will start to understand it very well.

You dont seem to have a very good grasp of inductors although you
do seem to know capacitors very well, so lets start there.

The inductor is called the electrical 'dual' to the capacitor. This is
because theoretically if you swap current and voltage the inductor
behaves exactly as the cap does. You know that the cap charges
up to a certain 'voltage' and it holds that voltage for a while. Well
an inductor charges up to a certain 'current' and holds that current
for a while. It's this property that makes it work so well in switching
regulators like the buck.

To help understand this circuit better, think of the cap as having a
voltage stored across it and when you try to discharge this voltage with
a resistor it causes a current to flow. With the inductor, think of it
as storing a current level and when you try to discharge it with a
resistor it causes a voltage to appear across the resistor.
Note to get the inductor theory all we had to do was swap the
words 'voltage' with 'current' and 'current' with 'voltage' and of course
'capacitor' with 'inductor'.
Another way of looking at it is that the cap tried to maintain a constant
voltage, while an inductor tries to maintain a constant current.

The switching circuit has two 'states', one is called 'on' and the
other is called 'off'. The on and off refer to the switch as being
either on or off. What this means is that we have two different
circuits to look at: one when the switch is on and the other when
the switch is off.

In the circuit you showed, before the thing is turned on at all, the
voltage across the cap is zero and the current through the inductor
is zero. Note that we say the voltage 'across'
the cap and the current 'through' the inductor. This is only
because voltage appears 'across' something while current flows 'through'
something.
Ok so before turn on everything is zero. Now we turn the circuit on.
The oscillator first turns the switch (shown on the drawing) on, and this
puts a voltage across the inductor. The inductor at first does not
conduct much because that's one of it's properties, but as time goes
on (microseconds) it conducts more and more, and as it conducts more
energy gets stored in it. Also because of the conduction, the cap
gets current through it now, and that causes the voltage to start to
rise more and more, and as it rises the stored energy also rises.
The diode does not conduct because it is reverse biased.
Now the end of the half cycle comes, and the switch turns off. The
inductor doesnt like this idea, and it tries to maintain a constant
current through the cap, and in doing so it causes the voltage across
itself (inductor) to reverse (because the magnetic field *change* changes
level). This reverse in voltage makes the diode conduct.
The diode conducting means current still flows though the cap too,
and that raises the voltage even more. Of course as the voltage
in the cap rises the output load gets more and more voltage which
causes the load to draw more and more current. During this time
the inductor dumps (transfers) some of it's energy (in the form of a current)
into the cap and that depletes the energy a bit, so it needs
more energy to keep supplying the cap.
Eventually the end of the 'off' period ends, and the switch turns
back on. This causes the inductor voltage to reverse back to it's
original state and it draws more current from the battery again.
Eventually the 'on' period again ends and the cycle keep repeating.

Now we take a step back and see what this has done.
1. The inductor kept getting energy from the battery during the 'on'
periods.
2. The cap kept getting energy from the battery during the 'on'
period and from the inductor during the 'off' period.
3. The inductor current ramps up and down as the period changes
from on to off.
4. The inductor current ramps up and down and from this we can
calculate an average value of current. The average value is the
average value of the current that the load draws.

All this means the circuit basically turns on and off causing the
current in the inductor to ramp up and down and the average value
of that ramping current is the average load current.
In a good design for a power supply, the cap voltage doesnt change
too much (maybe a few tens of millivolts) while the current may
ramp up and down by a few amps.
The output voltage gets regulated by controlling the pulse width
on time and its off time. Longer on time means a higher output
voltage (more energy in the inductor) while a longer off time means
less output voltage (less energy in the inductor).

The final note is that during this process you will note that
we talked about energy being 'transferred'. The energy
is transferred from the battery to the inductor and from the
inductor to the cap, and so it changes form twice...once into
current (into the inductor) and then back into a voltage,
and the theoretical efficiency for an inductor and capacitor
is 100 percent (never found in real life however). This
means that while the inductor and cap are storing energy
they dont loose any of it during the conversion.
This means a true energy conversion has taken place where
we converted the voltage and current of the battery into
another voltage and current in the load. This is very
different from using a series resistor to drop voltage
because the series resistor dissipates a lot of energy
while the inductor and cap are usually fairly efficient
(real life devices). The switching circuit could
have worked at 10v input and 0.5 amps and 5v output at 1 amp,
which means the power input equaled the power output.
In real life however, the components are not ideal and
always have some loss, so the input has to draw a little
more power than it puts out in order to satisfy the extra energy
needs of the real life components.

 

[koala]

Hey MrAl, did you copied this from somewhere? Ok all right, it's your's I know.

No words can explain better, *subscribed to this thread*.

To help understand this circuit better, think of the cap as having a
voltage stored across it and when you try to discharge this voltage with
a resistor it causes a current to flow. With the inductor, think of it
as storing a current level and when you try to discharge it with a
resistor it causes a voltage to appear across the resistor.
Note to get the inductor theory all we had to do was swap the
words 'voltage' with 'current' and 'current' with 'voltage' and of course
'capacitor' with 'inductor'.
Another way of looking at it is that the cap tried to maintain a constant
voltage, while an inductor tries to maintain a constant current.

 

[MrAl]

Hey MrAl, did you copied this from somewhere? Ok all right, it's your's I know.

No words can explain better, *subscribed to this thread*.

Hi there koala,

Thanks much. My goal is to help anyone interested enough to understand
the workings of an electrical circuit (usually power supplies, battery
chargers, flashlight electronics, etc.).

One thing i forgot to mention is that the circuit simulators help more
than i ever could. Inside one of these programs you can draw almost
any part you wish (inductors, capacitors, transistors, etc.) and see how
the circuit works. You can then swap parts and see the effect it
has on the output. Oh yeah, and no math required