Hello, I am once again trying to learn more about circuitry and I am on the subject of buck converters and I will have pictures below of my schematics.
From what I believe I know(and please correct me if I'm wrong) about buck converters is that a switch (transistor) opens and closes its contacts at a certain PWM that steps down the voltage to a lower voltage. By switching the circuit at a lower rate, the average voltage decrease and the DC wave is them smoothed out by the inductor and capacitor which together keep the voltage and current from changing.
When the switch is ON the current flows and it is supplying the load with a current. Initially, current flow to the load is restricted as energy is being stored in the inductor and the inductor produces an opposing voltage across the load with the positive on the left side of the inductor and therefore the current in the load and the charge on c1 builds up gradually during the "ON" period. There will also be a large positive voltage on D1 cathode and the diode will be reverse bias and cut off.
When the switch is OFF the energy stored in the magnetic field around the inductor is released back into the circuit. The voltage across the inductor(back EMF) is now in reverse polarity to the voltage across the inductor during the "ON" time and sufficient storage energy is available in the collapsing magnetic field to keep current flowing for at least part of the time the transistor switch is open. The back EMF from L1 now causes current to flow around the circuit via the load and the diode, which is now forward bias. Once the inductor has returned a large part of its stored energy to the circuit and the load voltage begins to fall, the charge stored in the capacitor becomes the main source of current, keeping current flowing through the load until the next "on" period begins.
Is how I explained sound right? You are pretty much controlling the duty cycle of the transistor and by switching it on and off. You average a lower voltage and the capacitor and inductor take the pulsating lower DC voltage and supply current and voltage during the "OFF" time to make a smooth DC voltage?
My last question is why does the inductor when the switch is turned on, have the voltage sign on the left-hand side like shown in the picture?
The last question is the easiest to answer. It's called "Back EMF" for a reason. A collapsing magnetic
field of an inductor is reversed biased and therefore the voltage is indeed "negative" . This is a negative
spike (google Lenz's Law)
I am not an expert on buck converters but I believe you may be incorrect about the current flow in a buck converter. The whole point of the switching speed is to charge the cap faster than the load can discharge it
so nothing is considered to be "equal" during a single cycle. the cap should always be charged to a minimum to maintain the load during the 'off' cycle. In addition, I am almost certain the whole concept depends on the presence of a filter cap to store the energy.
I have made a simple buck converter before and it is not difficult if you have the right component values
and the right code to generate the PWM.
I really like the tutorial linked in the previous post !
Now that i think of it does the buck converter have anything to do with inductive reactance and the inductor at high frequencies offers resistance or doesn't have enough time to establish a full current before it switches? and thats how it "bucks" it down?
but then again im not sure if inductive reactance works with a switching device?
If you think about the basics of inductors and caps as presented in entry level electronics courses:
Inductors are a SHORT at DC and an OPEN at High Frequency.
Capacitors are an OPEN at DC and a SHORT at High Frequency
(There may some caveats in the above statements but they get the idea across)
As far as your question is concerned :
does the buck converter have anything to do with inductive reactance and the inductor at high frequencies offers resistance or doesn't have enough time to establish a full current before it switches? and thats how it "bucks" it down?
I would not want to comment on such an oversimplification.
To ask if a buck converter has any to do with inductive reactance is, at best, ludicrous.
It's like asking if the state of ice has anything to do with temperature...
Just saying...
If you ask a dumb question we are obliged to tell you so you don't make a fool of yourself in front of
others. Is it nice to do that ? Maybe not, nevertheless as experts we are not here to be 'nice' per se
as much as we are here to tell you the truth, the whole truth and nothing but the truth.
The truth about buck converters is that they would not exist were it not for inductive reactance.
That much should be obvious. Could I have been more tactful ? Possibly. Does it matter ? No.
Thanks for reply and i could see how that was a dumb question, its just the more i look into how it works, the more confusing it becomes. Can someone explain in easy to undertsand how a buck converter works?
Think of the inductor as a bucket.
When the transistor turns on the water flows into the bucket.
The diode can be thought of as a one-way valve preventing the water from flowing to the capacitor,
which can be thought of as a storage tank with a faucet attached.
Given a known rate of water flow, the time to fill the bucket (duty cycle) will always be the same.
When the bucket is full, the transistor turns off and the bucket is dumped upside down (back EMF)
into the one way valve funnel which allows the water to flow into the tank.
The time to empty the bucket is also a fixed constant so the once empty , the bucket is righted
and the valve (transistor) is turned on and starts filling the bucket.
The battery is a water tank supplying the bucket.
The capacitor is a tank holding the water received from the diode/inductor.
How the water gets out of the tank (capacitor) is irrelevant to the discussion.
If you are asking what causes back EMF you will have to refer to Lenz's Law
raschemmel:
Think of the inductor as a bucket.
When the transistor turns on the water flows into the bucket.
The diode can be thought of as a one-way valve preventing the water from flowing to the capacitor,
which can be thought of as a storage tank with a faucet attached.
Given a known rate of water flow, the time to fill the bucket (duty cycle) will always be the same.
When the bucket is full, the transistor turns off and the bucket is dumped upside down (back EMF)
into the one way valve funnel which allows the water to flow into the tank.
The time to empty the bucket is also a fixed constant so the once empty , the bucket is righted
and the valve (transistor) is turned on and starts filling the bucket.
The battery is a water tank supplying the bucket.
The capacitor is a tank holding the water received from the diode/inductor.
How the water gets out of the tank (capacitor) is irrelevant to the discussion.
If you are asking what causes back EMF you will have to refer to Lenz's Law
Thanks, i couldn't of asked for a better analogy, but i guess what i am getting down to is, how does that process step down voltage when the inductor and capacitor are supplying the voltage and current when the switch is turned off?
Inductance behaves like mass or momentum, capacitance like a spring. Inductors store energy
proportional to current squared, capacitors proportional to charge (or equivalently, voltage) squared.
In switch mode converters the current in inductors is fluctuating up and down as the switch switches,
fluctuating about a steady average value (due to the load). Energy stored as the current increases
is released back as the current falls again.
There always has to be a path to allow current to continue to flow when a switch device opens, in
the standard buck converter its provided by the diode. (sometimes augmented with a second switching
device).
Since ideal inductors and capacitors don't dissipate energy (only store it and give it back), the ideal converter
is 100% efficient. In real circuits inductors have some resistance, and diodes have some forward voltage
drop, and switching devices aren't perfect, so efficiencies between 80% and 95+% are typically seen.
If the converter is constant current output, no output capacitor is needed as the inductor already regulates
the current, but most converters are constant voltage output requiring an output capacitor.
That makes a little more sense. Would it be correct to say that when the switch is ON current is provided to load, current also charges the capacitor and builds up a magnetic field in the inductor, then when the switch is OFF, the flywheel part of the circuit begins, and voltage is lost due to the forward voltage of the diode and inductor, and like you said components not being 100% efficient which gives a sawtooth-like wave that can be smoothed with the capacitor shown. and looks like the chart below?
Well , probably but without posting the switching frequency it is not that useful.
If , for example the switching frequency is
Independent designs at frequencies of 350, 700, and 1600 kHz will be compared to illustrate the benefits and obstacles. The TPS54317, a 1.6 MHz, low-voltage, 3 A synchronous-buck DC/DC converter with integrated MOSFETs was chosen as the regulator in each example. The TPS54317 from Texas Instruments features a programmable frequency, external compensation and is intended for high-density processor power point-of-load applications.
then you can see that the charging of the capacitor is probably occuring far more rapidly than the discharging of it by the load, especially if there is a large filter further down the line to support the load
during the discharge cycle of the cap.
tjones, your first explanation was remarkably good. But I will jump in with my take on it...
Inductors resist changes in current. If there is a DC current through an inductor, the inductor will try to keep that current going. So when the transistor switches off, the inductor keeps current flowing to the load.
That current must come from somewhere. So the diode sources the current.
There is a lot more required to design a practical buck converter. It is not as simple as the textbook examples.
The sawtooth wave is nothing to do with component imperfections, its the ramping up and down of
current in the inductor depending on the voltage across it.
The load gets is power from the output capacitor, the output capacitor gets its power from the inductor,
the inductor gets its power from the input capacitor (only when the switch is closed).
Each component smooths out the flow of power - the most important part is the inductor
as is has to take its power in pulses and deliver it continuously. The output capacitor principally
acts to keep the output voltage constant (or nearly constant). Remember the load needs
constant voltage, not necessarily constant power.
The control chip has to monitor the output voltage and decide when to switch, and it has to
monitor the current through the switch to protect against inductor saturation (which would destroy
the output). When the inductor saturates there is suddenly nothing to stop the full input voltage
reaching the output and frying the load. Protection circuitry is a vital part of every switch-mode
power converter.
