Why do I get this waveform at the output after power down of Buck Converter

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
Hi All,

I am having a buck converter (TPS54260 - http://www.ti.com/lit/ds/symlink/tps54260.pdf) which steps down 8-16V input to 5V output. Load current = 0.6A

My schematic is similar to the one in the datasheet's application diagram on page 29. Similar input output capacitors, inductors and resistors are used.
But upon removing the input power, I get the attached waveform.
Please let me know why I get this waveform upon removing the input voltage

Why does the output not go to 0V the instant the input voltage is turned to 0V? We can see from the waveform that the output is gradually decreasing and suddenly peaks and then gradually decreases and peaks till it reaches 0V? Why the sudden peaks? What is this phenomenon?

Thanks
 

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danadak

Joined Mar 10, 2018
4,057
A converter has energy stored in various reactive components which
has to decay, and in that process control loop loses its ability to function
predictably.

If I switch off power source to a fan it does not instantly stop because
energy is still present due to rotating mass.


Regards, Dana.
 

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
A converter has energy stored in various reactive components which
has to decay, and in that process control loop loses its ability to function
predictably.

If I switch off power source to a fan it does not instantly stop because
energy is still present due to rotating mass.


Regards, Dana.
I agree. I understand that is the reason why the 5V decreases gradually rather than dropping to 0V instantly when the power is removed. My question is why the sudden peaks appear while the output voltage is decreasing gradually.
 

BobaMosfet

Joined Jul 1, 2009
1,117
I agree. I understand that is the reason why the 5V decreases gradually rather than dropping to 0V instantly when the power is removed. My question is why the sudden peaks appear while the output voltage is decreasing gradually.
Because the buck converter isn't passive. It's still trying to do the DC : DC conversion, because it doesn't know power's been removed. It's a little bit like poking a hole in yourself. You don't stop breathing and moving because blood is pouring out until there's too little to operate on.

The field energy stored by the inductor has to go somewhere proportional to impedance.
 
Last edited:

MrAl

Joined Jun 17, 2014
7,659
Hi,

Actually it could be due in part to a larger than usual impedance somewhere in the input circuit.

If there is sufficient impedance, then when the buck is turned off and goes undervoltage the current through that impedance decreases and thus the full voltage may get to the input of the converter again thus starting it up again. Once it turns on, same thing happens. It repeats until the cap discharges more.

Just one possibility.
 

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
Seems that works "repeatable restart" function while input capacitor is not fully discharged.

"The SS/TR pin is discharged before the output powers up. This discharging ensures a repeatable restart after an over-temperature fault, UVLO fault or a disabled condition."

http://www.ti.com/lit/ds/symlink/tps54260.pdf , page 11.
I am unable to understand. Can you explain in simple terms please
 

Danko

Joined Nov 22, 2017
934
I am unable to understand.
It is a pity...
Can you explain in simple terms please
After input voltage becomes OFF, voltage on input capacitor V(C) gradually drops.
When V(C) drops to under voltage lockout (UVLO) threshold, then:

1. UVLO detects low V(C) and shuts system down. Converter is OFF. Only restart function (SS/TR pin) still works. Out voltage gradually drops.
2. After some time, when capacitor on SS/TR pin becomes discharged, system starts. Capacitor on SS/TR pin charges. Converter is ON, out voltage instantly increases (spike).
3. Go to point 1.

This cycle repeats while V(C) still enough for system functioning.
 

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
It is a pity...

After input voltage becomes OFF, voltage on input capacitor V(C) gradually drops.
When V(C) drops to under voltage lockout (UVLO) threshold, then:

1. UVLO detects low V(C) and shuts system down. Converter is OFF. Only restart function (SS/TR pin) still works. Out voltage gradually drops.
2. After some time, when capacitor on SS/TR pin becomes discharged, system starts. Capacitor on SS/TR pin charges. Converter is ON, out voltage instantly increases (spike).
3. Go to point 1.

This cycle repeats while V(C) still enough for system functioning.

Thank you. When you say, after SS/TR pin discharges, how does after this, the system starts and when you say, that after this, the capacitor on SS/TR pin charges, from where is the charge supplied to that capacitor?
 

Danko

Joined Nov 22, 2017
934
When you say, after SS/TR pin discharges, how does after this, the system starts
"The SS/TR pin is discharged before the output powers up. This discharging ensures a repeatable restart after an over-temperature fault, UVLO fault or a disabled condition." (TPS54260 datasheet, page 11)
I believe it will restart (start).
Chips like TPS54260 contain inside hundreds transistors, sophisticated monitoring and logic system,
so, if you will provide me detailed schematic diagram of TPS54260, I will give you more detailed answer.

and when you say, that after this, the capacitor on SS/TR pin charges, from where is the charge supplied to that capacitor?
"A capacitor on the SS/TR pin-to-ground implements a slow-start time. The TPS54260 has an internal pullup current source of 2 μA that charges the external slow-start capacitor." (TPS54260 datasheet, page 16)
 
Last edited:

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
"The SS/TR pin is discharged before the output powers up. This discharging ensures a repeatable restart after an over-temperature fault, UVLO fault or a disabled condition." (TPS54260 datasheet, page 11)
I believe it will restart (start).
Chips like TPS54260 contains inside hundreds transistors, sophisticated monitoring and logic system,
so, if you will provide me detailed schematic diagram of TPS54260, I will give you more detailed answer.


"A capacitor on the SS/TR pin-to-ground implements a slow-start time. The TPS54260 has an internal pullup current source of 2 μA that charges the external slow-start capacitor." (TPS54260 datasheet, page 16)
Thank you for the answer
 
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