Buck Converter Startup sequence

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
Hello

So i am exploring the use of standard buck converter. I see from simulation that during startup, especially during the first few cycles i see very high voltage output before it settles into regulation. This seems failure obvious to me as you have not saturated your inductor yet.
However, with that being said.... If you have high output voltage for a little bit at startup how do you protect your downstream loads from that?

I would of thought about a series switch on the output but obviously unloading the buck converter just makes things worse.
So how is startup normally handled?

Also, with so much capacitance....how does one typically protect or deal with inrush?
I am looking at an open loop design right now and not a closed loop feedback design

Thanks
 

MisterBill2

Joined Jan 23, 2018
18,928
The common means of avoiding damage from short high voltage transients has been both shunt capacitance and series inductive reactance. The combination can be very effective. And because those high voltage pulses are narrow they do not contain a lot of energy, so catching them in a capacitor is practical.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
The common means of avoiding damage from short high voltage transients has been both shunt capacitance and series inductive reactance. The combination can be very effective. And because those high voltage pulses are narrow they do not contain a lot of energy, so catching them in a capacitor is practical.
im Confused on what shunt capacitance is. You already have a capacitor on the output?
 

crutschow

Joined Mar 14, 2008
34,676
So how is startup normally handled?
Typically there is a slow start-up circuit to avoid overshoot in a closed-loop buck converter.
Open-loop converters are seldom used in a real application so the overshoot you observed is not a concern.

Below is the sim of a simple hysteretic-feedback type buck converter with such a slow (≈1ms) start-up.
The top plot shows the output start-up with the reference voltage (yellow trace) having a slow rise-time due to the R3-C3 time-constant.
Note that the output voltage (green trace) basically follows the reference voltage and rises to its regulated 5V value without overshoot.

The bottom plot show the start-up without C3.
Now the output overshoots to nearly 9V.

The red traces show the MOSFET gate control switching voltage (high voltage is off).

1713494662535.png
1713494709466.png
 
Last edited:

eetech00

Joined Jun 8, 2013
4,004
Hello

So i am exploring the use of standard buck converter. I see from simulation that during startup, especially during the first few cycles i see very high voltage output before it settles into regulation. This seems failure obvious to me as you have not saturated your inductor yet.
However, with that being said.... If you have high output voltage for a little bit at startup how do you protect your downstream loads from that?

I would of thought about a series switch on the output but obviously unloading the buck converter just makes things worse.
So how is startup normally handled?

Also, with so much capacitance....how does one typically protect or deal with inrush?
I am looking at an open loop design right now and not a closed loop feedback design

Thanks
There should not be a very high output voltage on startup. There will be a small percentage of "overshoot" but it should not be "vary high", otherwise, your right, it could potentially damage the load. The buck converter is probably not designed correctly.
 

Bordodynov

Joined May 20, 2015
3,182
Typically there is a slow start-up circuit to avoid overshoot in a closed-loop buck converter.
Open-loop converters are seldom used in a real application so the overshoot you observed is not a concern.

Below is the sim of a simple hysteretic-feedback type buck converter with such a slow (≈1ms) start-up.
The top plot shows the output start-up with the reference voltage (yellow trace) having a slow rise-time due to the R3-C3 time-constant.
Note that the output voltage (green trace) basically follows the reference voltage and rises to its regulated 5V value without overshoot.

The bottom plot show the start-up without C3.
Now the output overshoots to nearly 9V.

The red traces show the MOSFET gate control switching voltage (high voltage is off).

View attachment 320277
View attachment 320278
It's a bad idea to shunt the TL431 with a capacitor. Some capacitor values may cause oscillations.
1713508234597.png
 

MisterBill2

Joined Jan 23, 2018
18,928
As has already been stated again, it is important for the design of any converter design to go thru the startup stage without any over-voltage output events. Certainly startup is a separate condition from the running state, requiring additional effort.
The reality is that design of switching regulator circuits is not simple, nor even intuitive for most designers. Unfortunately some presentations ignore that problem.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
I can think of lots of ways to start it up especially if you start looking at closed loop designs. Or better yet add a uC
I just wanted to know what was typical.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
Because it’s 2024 and things like the atiny85 exist which is cheaper then op amps and comparators and offers other benefits like software control
 

crutschow

Joined Mar 14, 2008
34,676
Because it’s 2024 and things like the atiny85 exist which is cheaper then op amps and comparators and offers other benefits like software control
I don't need the snark.
I'm well aware that it's 2024, but that doesn't mean you use a µC for every application just because it's cheap.
Unless you are using it to fully control all the signals to make a switching regulator, a dedicated switching regulator IC that includes all functions such as current-limit, loop compensation, soft-start, etc., is usually the best way to go.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
I don't need the snark.
I'm well aware that it's 2024, but that doesn't mean you use a µC for every application just because it's cheap.
Unless you are using it to fully control all the signals to make a switching regulator, a dedicated switching regulator IC that includes all functions such as current-limit, loop compensation, soft-start, etc., is usually the best way to go.
Its not snark its meant to be literal. Not every application calls for an analog feedback loop. They take up cost and board space and there are certain industries where a uC is the best option because of availability and a dedicated IC is not a thing.

But riddle me this......

I am trying to find an equation that represents how the output voltage of a buck converter (open Loop) changes with the effects of different loads. Say like a resistor sweep.

On my simulation... i see the following.

The lower the load impedance the more current there is and im assuming you are running the inductor dry and into discontinuous mode (i think?) and the voltage regulation eventually falls to ~0V.

ok that makes sense....

But... i am looking for an equation that shows that.
I have taken the following equation that is intended to show minimum inductance requirements for certain parameters. Shown below.

1713544228133.png

Lets say i take that equation and i solve for my Lmin and i get some value.... ok great....
Plug it in the sim, it works fine.

I want to use that equation to assess Vout instead of Lmin

So since Lmin is now a fixed value i solved for previously, and if i rearrange the equation in terms of Vout i get this.

1713544331541.png

Now if i take this equation and hold Lmin fixed and start changing the output current as if i was simulating a load change from high impedance to very low i get a bizzare answer.

Vout behaves conversely to how the simulation works.

With a high current i get a high regulation voltage and with a low current i get a low regulation voltage which is is the opposite of the simulation behavior.

Why is this equation not holding?

If there an equation i can use to see the effects of current loading on my output voltage?

Thanks
 

MisterBill2

Joined Jan 23, 2018
18,928
Relative to the startup short term over voltage condition, one of the unfortunate realities related to closed loop systems is that the loop is not closed until the feedback system is stable. What that means for a switcher type power supply is that there must be, as the default start up condition, that the control portion needs to be fully awake and functional prior to the power portion being active, because otherwise, with nothing telling it when to reduce output, it will go to MAXIMUM. That is the intrinsic flaw of many feedback systems.
To avoid that problem the power section needs to default to the off state until the regulate system asks it to deliver power. That is always more difficult than the other way.
 

crutschow

Joined Mar 14, 2008
34,676
Not every application calls for an analog feedback loop. They take up cost and board space and there are certain industries where a uC is the best option because of availability and a dedicated IC is not a thing
But we are talking about a buck regulator here, not feedback loops in general, and there are many dedicated ICs for that.
Certainly there are many applications where a µC is the best option.
A buck regulator is not necessarily that application.

And telling me its 2024 is snark.
If there an equation i can use to see the effects of current loading on my output voltage?
There likely is, but that tells you nothing about the normal closed-loop operation of a buck converter where the duty-cycle is varied in response to load changes.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
There likely is, but that tells you nothing about the normal closed-loop operation of a buck converter where the duty-cycle is varied in response to load changes.
I think the goal here is to learn the basic operation before trying to look at closed loop operation.
I wish to understand at a fundamental level how the change in load impedance screws up the output voltage regulation
Do you see anything fundamentally wrong with why my equation (which comes from a text book) would be giving me inverse results?
 
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