Have i been thinking about a buck converter wrong this whole time?

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
So in my head i have been trying to understand the basis of why a buck converter has a lower output voltage then input....

Fundamentally.... i have been thinking about it in the following way until very recenty....

So Thought 1......

Because the inductor voltage drop changes polarity when the switch is in the off portion of the period, There is a voltage drop in the opposite direction causing the voltage at the load to be lower......


Then i read this the other day....


Thought 2.....

Lets pretend the inductor isnt there and the fet is replaced with an idea switch.
You are getting a period square wave out....

Fourier analysis tells us that the DC component of any period waveform is simply the average of the waveform given by the integral of course.....
Then it went on to say, the average DC component will however have some harmonics to it that need to be filtered out....

Ok.... so now im thinking that a buck converter has a lower output voltage strictly because the DC component of the periodic signal is the average....
and the inductor and the capacitor are there strictly to act as a filter to actually extract the DC component out of the period signal...


Which way is sounds more correct? Or are they both correct?
 

BobTPH

Joined Jun 5, 2013
9,148
Neither is correct.

The voltage drop across an inductor depends on the rate of change of the current in the inductor. That is the basic inductor equation.

In continuous mode operation, when the switch is on, the current starts increasing which produces a voltage opposing the source voltage, lowering the voltage to the load.

When the switch is off, the current keeps flowing in the same direction, but it is now shunted across the input source by a diode (or MOSFET). The current decreases at the rate which puts the same voltage across the load.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
Neither is correct.

The voltage drop across an inductor depends on the rate of change of the current in the inductor. That is the basic inductor equation.

In continuous mode operation, when the switch is on, the current starts increasing which produces a voltage opposing the source voltage, lowering the voltage to the load.

When the switch is off, the current keeps flowing in the same direction, but it is now shunted across the input source by a diode (or MOSFET). The current decreases at the rate which puts the same voltage across the load.
Im not sure it is fair to say thought 2 is incorrect which comes DIRECTLY from a power electronics book that is used in most universities.

"Fundamentals of Power electronics" Robert E Erickson Page 5
 

BobTPH

Joined Jun 5, 2013
9,148
Im not sure it is fair to say thought 2 is incorrect which comes DIRECTLY from a power electronics book that is used in most universities.

"Fundamentals of Power electronics" Robert E Erickson Page 5
Certainly it s true, but it is not a description of the operation of a buck converter. As soon as you say “pretend the inductor isn’t there”, you no longer have a buck converter, you have a PWM controller.

The easiest buck converter to understand is the “bang bang”, or hysteric buck converter.

It works by using a comparator with hysteresis to control the current. When the current is less than the desired amount, the switch is turned on. When the current reaches the desired current + hysteresis amount, it turns off and the current starts decreasing.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
Certainly it s true, but it is not a description of the operation of a buck converter. As soon as you say “pretend the inductor isn’t there”, you no longer have a buck converter, you have a PWM controller.

The easiest buck converter to understand is the “bang bang”, or hysteric buck converter.

It works by using a comparator with hysteresis to control the current. When the current is less than the desired amount, the switch is turned on. When the current reaches the desired current + hysteresis amount, it turns off and the current starts decreasing.
I guess im trying to understand its operation from two persepctives.
One being the purely electrical one where voltages drops of components are considered and secondly a filter / mathematically approach.

The books speaks of how in an ideal since the voltage is reduced strictly because the DC component of a period signal is simply equal to the average of the signal. The book claims that is the ideal reason why the voltage out is less then the voltage in.

The LC low pass filter is used to essentially form the action of extracting the DC component.
It speaks of how if the corner frequency of the LC filter is sufficiently less then the switching frequency you will obtain the DC component.

This makes kind of sense to me.....

My brain works better thinking in terms of this then it does voltage drops across the inductor and such although i do understand the electrical idea as well...

What am i missing.
 

BobTPH

Joined Jun 5, 2013
9,148
Apply what you are saying to a PWM circuit. Yes you can determine the average voltage, current or power. But there is no voltage drop anywhere. It is simply stating the mathematical definition of average over time. What a PWM circuit is limiting is the on time, not the voltage.

Why do you think that helps you understand a buck converter, where there is an actual voltage drop across the inductor? I thought that is what you were trying to understand.
 

crutschow

Joined Mar 14, 2008
34,684
Yes, you can look at the buck converter output as the average of the PWM voltage which is filtered by the LC output.
That does require that the inductance is large enough so the current doesn't drop to zero when the PWM signal is low (continuous mode operation).
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
hi mike,
Check through these videos by Ben-Yaakov, he explains the operation of Buck converters in detail.

E
https://www.youtube.com/results?search_query=ben-yaakov+buck+converter
Love this guys videos.
I have watched that video over and over again and i understand his math.
Until recently when i was reading this power electronics book i had always thought of the circuit like explained in this video and it becomes clear how the math unfolds UNTIL
I read this part of the book that talks about a buck essentially working as an average PWM signal with a filter on the end....

Then i got all screwed up and now sure which way is more correct.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
Apply what you are saying to a PWM circuit. Yes you can determine the average voltage, current or power. But there is no voltage drop anywhere. It is simply stating the mathematical definition of average over time. What a PWM circuit is limiting is the on time, not the voltage.

Why do you think that helps you understand a buck converter, where there is an actual voltage drop across the inductor? I thought that is what you were trying to understand.
Sooo....
Thinking back to it more electrically in terms of voltage drop....

Ok, yes the inductor has a voltage drop which essentially acts like R1 in a voltage divider only without the power dissipation.
that gets your output voltage lower.... ok cool but.....

If your voltage drop comes from that inductor then why do we care when the polarity of the inductor swaps.

The switch goes off. the polarity switches and you get a negative voltage with respect to ground on the left side of the inductor.
Why does that matter? Why bother with the diode or second mosfet.

If the voltage drop comes from the inductor what happens if you dont but the diode in?

The only thing i can think of is..... All the current being supplied from the cap would flow into the inductor instead of the load?
 

BobTPH

Joined Jun 5, 2013
9,148
The switch goes off. the polarity switches and you get a negative voltage with respect to ground on the left side of the inductor.
Why does that matter? Why bother with the diode or second mosfet.
Forget the capacitor. A buck converter can operate without one, and is easier to understand without one. Once you understand it without the capacitor, you can see why the capacitor will reduce the ripple voltage / current.

Yes, indeed. When the switch is on, the voltage across the inductor subtracts from the voltage of the power source. When the switch is off, the voltage across the inductor is driving the load. So it must invert for the buck converter to work.
 

BobTPH

Joined Jun 5, 2013
9,148

Ian0

Joined Aug 7, 2020
10,032
If the voltage drop comes from the inductor what happens if you dont but the diode in?
The current must continue to flow in the inductor, because it can only reduce at the rate of V/L, so the inductor produces a voltage large enough to keep the current flowing.
In the absence of the diode, the voltage increases until something breaks down so that the current can continue to flow.
That something could be the MOSFET, or some insulation somewhere, whatever happens to fail first.
Needless to say, it's not a good outcome.
 

nsaspook

Joined Aug 27, 2009
13,423
Sooo....
Thinking back to it more electrically in terms of voltage drop....

Ok, yes the inductor has a voltage drop which essentially acts like R1 in a voltage divider only without the power dissipation.
that gets your output voltage lower.... ok cool but.....

If your voltage drop comes from that inductor then why do we care when the polarity of the inductor swaps.

The switch goes off. the polarity switches and you get a negative voltage with respect to ground on the left side of the inductor.
Why does that matter? Why bother with the diode or second mosfet.

If the voltage drop comes from the inductor what happens if you dont but the diode in?

The only thing i can think of is..... All the current being supplied from the cap would flow into the inductor instead of the load?
IMO you need to stop thinking about purely current or voltage absolute values, polarities and voltage drops for your basic mental model. The inductor (and the other reactive components in in the circuit) stores energy and we quantify that energy flow from input to output as a voltage*current expression over time.
https://www.analog.com/en/resources/technical-articles/dc-to-dc-buck-converter-tutorial.html
Control Techniques
From the derivations for the boost, buck, and inverter (flyback), it can be seen that changing the duty cycle controls the steady-state output with respect to the input voltage. This is a key concept governing all inductor-based switching circuits.
inductor-based switching
1713640292919.png
Figure 5. Charging phase: when the switch closes, current ramps up through the inductor. (ENERGY STORED IN THE MAGNETIC FIELD)

1713640366961.png
Figure 6. Discharge phase: when the switch opens, current flows to the load through the rectifying diode. (ENERGY RELEASED FROM THE MAGNETIC FIELD)
 

BobTPH

Joined Jun 5, 2013
9,148
If the voltage drop comes from the inductor what happens if you dont but the diode in?
The diode completes the circuit of the reversed voltage inductor with the load.

Ir is very informative to draw the circuits with the switch on and off and see how the current flows.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
It that were true, a buck converter would have the same efficiency as a PWM. And it does not, it has better efficiency. So yes, that is incorrect.
I dont see how this is possible.

That link specifically shows a PWM signal acting as a DAC where the DAC signal is the DC component or the PWM average signal
The PWM signal is filtered with an LC circuit to obtain the DC bit.

I dont understand how this is at all in any way shape or form different from how the converter runs.
It even has the same parts?

So how is this not the same?
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
Then you need to study it some more.
I dont disagree there

How can a circuit with the same parts not be the same the same concept.

Its a PWM signal with a filter. It provides a DC output.....
How is it different?


Since i clearly dont get it.... im trying to understand PWM at a more basic level to make sure i fully get it.

So lets say i make a simple voltage divider circuit Where Vin = 10V and R1 = 10 ohms. R2 is an IDEAL IGBT driven by a PWM generator
Say 50% Duty Cycle

The gate turns off and on such that the switch opens and closes rapidly.
The current is 1A when closed and 0 amps when open.

With a 50% duty cycle should the average current not be .5A? and therefore the average power equal to 2.5 Watts????

Or am i so screwed up i dont even understand that much?
 
Last edited:

nsaspook

Joined Aug 27, 2009
13,423
Switches (and components in general) are not ideal.

WIth PWM only, there is no energy storage (flywheel) so there is constant switching of X high energy peaks with resistive losses to the PWM duty cycle for X voltage and current output. With inductive energy (storing energy with lower losses in a magnetic field) storage, I can design the circuit to be very efficient with lower peak energy per duty cycle switch pulses, letting the stored (inductor flywheel) energy flow for X voltage and current output.
1713658220899.png
A current level example.
 
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