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the output voltage of buck converter at no load and at high load

PG1995

Active Member
Hi,

Please have a look here. You can see that in Figure 2 no load is connected across the output. Don't you think that in such a case the capacitor's voltage would keep on increasing until it equal the input voltage, i.e. 20V. Do I have it correct? Assuming I have it correct, how is this handled for practical buck converters? I understand that a feedback could be used but I think it would require to decrease the duty cycle , or is there any other way to limit the output voltage to around 10 V? Could you please guide me?

Thanks a lot for your help
 

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rjenkinsgb

Well-Known Member
Most Helpful Member
IF the duty cycle was locked at 50%, there is no voltage regulation whatsoever and the output depends on load, component resistance, input supply etc.

In reality, the duty cycle is regulated to hold the output voltage constant or as near as possible. The duty cycle would drop to zero with no load.
 

ronsimpson

Well-Known Member
Most Helpful Member
IF the duty cycle was locked at 50%,
With no load Vout goes to Vin.
if the duty-cycle can go to zero.
Because switches (in the real world) are not perfect, some current might leak through and lift the output. On the other hand the capacitor will leak and pull down a little.

In real power supplies there is a feed back loop that will force the duty cycle to zero if the voltage gets even the smallest amount above what it should be. Also there is a resistor across Vout in most power supplies. It helps discharge the capacitors when the input power is turned off. It also gives the supply a minimum load. If you are having problems thinking about "Vout too high" then put a Zener diode across Vout.
 

PG1995

Active Member
Thank you for your help!

What I have seen in practical buck converter modules is that their switching frequency is fixed and they use an internal feedback loop. So, I don't know if their duty cycle is adjusted using the feedback. Without adjusting the duty cycle I don't think the buck converter would be able to perform properly. Please note that I have never used one of those buck converter modules and had a look on specs of just few ones.

What really confuses me is that when talking about a buck converter's output, no one ever talks about a feedback loop; voltage or current sensors. The voltage sensor is essential for proper regulation of output voltage by increasing or decreasing its duty cycle.

Using a zener diode across the output to regulate the voltage is a good idea but it would waste power and you need to make sure that the zener could dissipate maximum power supplied by buck at no-load condition.

Thanks.
 

crutschow

Well-Known Member
Most Helpful Member
Using a zener diode across the output to regulate the voltage is a good idea
No, that's a very bad idea.
It would dissipate a lot of power, and thus defeats\ the purpose of the converter to generate a voltage with high efficiency.

Virtually all buck regulators control the duty-cycle with feedback to regulate the duty-cycle.
Some types change both frequency and duty-cycle (such as simple hysteretic (bang-bang) feedback regulators.
 

ronsimpson

Well-Known Member
Most Helpful Member
Using a zener diode across the output to regulate the voltage is a good idea
You are so certain the voltage will get too high. I was thinking about a 6.8V Zener on a 5.0 supply. Not for regulation. For runaway conditions.
What really confuses me is that when talking about a buck converter's output, no one ever talks about a feedback loop; voltage or current sensors.
You are in a text book. (theory) They are working on duty cycle to output formulas. You have not seen a complete power supply.

Here is the Switch=FET, Diode, L, C and load.
Added is an error amplifier. In this case it is looking at a fraction of Vout and compariting it to a reference voltage.
The output of the error amp makes a voltage that with a sawtooth wave form creates a duty cycle.
If Vout is not right then the error amp amplifies the error, with delay & phase shift.
The output of the error amp adjusts the duty cycle. (closed loop)
1571704682531.png
 

crutschow

Well-Known Member
Most Helpful Member
Below is an LTspice simulation to show the basic operation of a simple Hysteretic regulator, requiring just a comparator to drive the MOSFET.
When the output drops below the reference voltage Vref, as determined by R1 and R3, it turns the MOSFET on.
When the output goes above the reference voltage, it turns the MOSFET off.

The switching frequency is determined by a combination of the values of the inductor, output capacitor, and load current.

This type of regulator requires no loop compensation, as the one in post #9 does that uses a fixed frequency PWM modulator circuit.

.C3 provides a slow startup ramp for Vref so the output voltage doesn't oveshoot.

In an actual circuit, the voltage reference would likely be provided by a voltage reference to prevent any input voltage variation from affecting the output voltage.

1571706628541.png
 
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PG1995

Active Member
Thank you so much!

You are in a text book. (theory)
Exactly! Now I can see how a feedback loop is an integral part of a buck converter.

Questions:
The following questions are about post #10.
1: What is the role of R2? Is it used to improve switching speed while charging up the gate to turn it off?
2: Is the output voltage of buck going to be 5V or 7V? I'm asking this because Vref at the inverting input is 7V.


Note to self:
In post #10 p-channel mosfet is being used which turns on low voltage. I believe the output voltage is 5V. When output voltage falls below the reference voltage, gate is turned on. As long as the voltage on non-inverting input of comparator is high, the output remains high which means mosfet is off.

Helpful links:
1: /watch?v=ZiD_X-uo_TQ&t=300s (insert youtube.com and watch around 4:20)
 

ronsimpson

Well-Known Member
Most Helpful Member
What is the role of R2?
LM339 only pulls down. There is no pull up on its output. R2 pulls up. (open collector)
Vref at the inverting input is 7V
Please show your work. 5V

Note:
You know this is not a PWM. 'on while low and off while high'
Normally in a PWM the frequency is set and the duty cycle is adjusted by an error amplifier. 'decrease DC while high and increase DC while low' more accurately 'If the error is small change the DC slowly and If the error is large change the DC fast' and 'if no error hold the DC where it is'
 
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PG1995

Active Member
Hi again,

The following quotes made me to search this topic more.

If Vout is not right then the error amp amplifies the error, with delay & phase shift.
This type of regulator requires no loop compensation, as the one in post #9 does that uses a fixed frequency PWM modulator circuit.
I found the following video on buck converter loop gain and phase. In my opinion, it seems pretty good.


This question is about the segment from 4:30 to 5:27. The following is said, "converter loop gain and phase can be measured by injecting a small sine wave signal into the loop and measuring the converter output signal versus the input signal."

It's a DC-DC converter and only a DC load could be connected to its output. So, why worry about injecting a small sine wave signal of different frequencies at the output because no frequencies are involved other than the DC? I understand that it has to do something with the transient response which occurs when a load is connected to the output and different harmonics are generared. The transient response only lasts a very brief time then why a steady small amplitude sinusoid signal is injected to the output; if one really has to inject sinusoids then, I'd say, decaying sinusoids should be used instead.

Helpful links:
1: https://www.digikey.com/en/articles/techzone/2015/jan/the-role-of-slope-compensation-in-current-mode-controlled-voltage-regulators
2: /watch?v=gIZW6jJe8Tc (insert youtube.com, very good, control algorithms for switching controllers)
 

ronsimpson

Well-Known Member
Most Helpful Member
why worry about injecting a small sine wave signal of different frequencies at the output because no frequencies are involved other than the DC?
If the time constants in the error amp are wrong you can make a power phase shift oscillator. (too fast)
If the time constants in the error amp are wrong the power supply will not respond, in time, to changes in load or line. (too slow)

By injecting a known range of frequencies a graph is drawn of gain and phase which tells us if the supply is stable.
1572614327368.png
Here is a step response. mu=6 wants to ring. Much larger and it will oscilate. A mu of 0.1 will respond so slow to be not much good as a supply.
1572614602380.png
 

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