Voltage spikes when turn-on step-down module (LM2596S)

Thread Starter

xchcui

Joined May 12, 2014
272
step down module.png
Hi.

This is a step down module LM2596S.
The input voltage are: 5.5-35 VDC.
The output voltage are: 0-30 VDC.
Working frequency: 65KHz
(input voltage must be 1.5 V higher than output voltage).
Are those modules making any voltage spikes at the output,towards the connected device,when turn-on/turn-off?
I would like to connect it to electronic device and i wonder if those modules
are making voltage spikes when turn-on/off,so i need to add a protection?

Thanks.
 
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Thread Starter

xchcui

Joined May 12, 2014
272
Can't really tell without seeing the module circuit schematic.
Otherwise, you would just have to purchase and measure one to see.
It is just a photo that i took from the web,i don't have any data sheet and it is not referred to a specific one,but in general to LM2596S modules.
I just would like to know,in general,whether those kind of modules produce voltage spikes at the OUTPUT when turn on/off?
 
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Thread Starter

xchcui

Joined May 12, 2014
272
No one here used a simple step-down module
like the LM2596S?
If someone used it,Did you experience any voltage spike on the output of the module when you turn-off/on the power supply ?
 

LadySpark

Joined Feb 7, 2024
124
No one here used a simple step-down module
like the LM2596S?
If someone used it,Did you experience any voltage spike on the output of the module when you turn-off/on the power supply ?
on those, if you connect the on/off to ground it can do that. But usually the zener diode that is placed afterwards takes care of that if the load is not highly inductive. But the way I create that circuit would be using a time delay on the on/off pin with a voltage divider with a cap in series to the source voltage like this:
Screenshot_2024-02-25_05-27-40.jpg
 

LadySpark

Joined Feb 7, 2024
124
If the module is working correctly, it will not cause voltage spikes.
I would have to say, if the module was designed correctly, it shouldn't have this issue.
After all, some of these are generic copies from datasheets instead of specifically built for an application and they are not always going to work for every situation. That is why its better to build your own stuff instead of using someone's board they are marketing.
 

Thread Starter

xchcui

Joined May 12, 2014
272
on those, if you connect the on/off to ground it can do that. But usually the zener diode that is placed afterwards takes care of that if the load is not highly inductive. But the way I create that circuit would be using a time delay on the on/off pin with a voltage divider with a cap in series to the source voltage like this:
View attachment 316142
It is a useful information,LadySpark,but i think that i didn't phrase my question well.
I wasn't referred to the on/off of the chip,i meant to say that at startup,when you power(turn on)the module by providing it 12VDC(for example)to the IN+ IN- terminals,
  1. Will a voltage surge/spike happen in the output(OUT+ OUT-)in that fraction of second?
  2. If the output voltage is adjusted to 5VDC,Is it possible that on startup(fraction of second),the output voltage will be higher than 5VDC until the module will start the regulation?
  3. I don't have oscilloscope and i guess that i can't check it with my DMM,is there another way,maybe indirectly,to check that?
 

LadySpark

Joined Feb 7, 2024
124
It is a useful information,LadySpark,but i think that i didn't phrase my question well.
I wasn't referred to the on/off of the chip,i meant to say that at startup,when you power(turn on)the module by providing it 12VDC(for example)to the IN+ IN- terminals,
  1. Will a voltage surge/spike happen in the output(OUT+ OUT-)in that fraction of second?
It can when the on/off pin is grounded. That is why the normal practice is using a soft start circuit that I described on the on/off pin.
Without an oscilloscope, it is going to be hard to see, unless you are using an analog meter or a dmm with a relative change function that will capture the voltage spike, but inrush current is the most concern with these circuits than a voltage spike.
 

Thread Starter

xchcui

Joined May 12, 2014
272
I understood your explanation about the on/off pin,but as you may see at the photo,this is a finished module circuit,which you are supposed to just adjust the voltage and connect it to your device.I assume that the on/off pin is already grounded(apparently not as you mentioned).My main issue is if i really can just connect it to my electronic device(after the adjustment)and turn the module on(and after that off)without any issue.(surge,spike,inrush current...in the output to device).
When you say inrush current,did you mean in the output?Isn't the coil resist to the change on the startup current and delay the current to the capacitor which prevents the inrush current to output,as the capacitor is charged slowly?
 
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MisterBill2

Joined Jan 23, 2018
18,442
The sad fact is that not all modules deliver the performance implied in the description. And performance not specified is often not available.
 

LadySpark

Joined Feb 7, 2024
124
When you say inrush current,did you mean in the output?Isn't the coil resist to the change on the startup current and delay the current to the capacitor which prevents the inrush current to output,as the capacitor is charged slowly?
The inductor is for switching filtering and not for limiting inrush current. Delaying the on pin allows the regulator to be bypassed temporarily (determined by the capacitor used in the on delay circuit) so the load capacitor can charge up enough that the regulator doesn't have so much current demand instantly when power is applied.
 

ronsimpson

Joined Oct 7, 2019
3,037
SPICE has a hard time with PWM, inductors, charging large capacitors at startup. And when inductors have 0 ohms and capacitors have 0 ohms.
I do not use a voltage source to power the PWM. Use a voltage source that starts out at 0V and ramps up to 24V in a mS. That really helps the startup problems.
 

Thread Starter

xchcui

Joined May 12, 2014
272
I have noticed that you made this simulating for a module with the addition of the capacitor and resistor as LadySpark mentioned before,so the results are fine,while my main interest was to know how may the standard LM2596s finished module(without that addition)will act.(on the output to the device)
The inductor is for switching filtering and not for limiting inrush current. Delaying the on pin allows the regulator to be bypassed temporarily (determined by the capacitor used in the on delay circuit) so the load capacitor can charge up enough that the regulator doesn't have so much current demand instantly when power is applied.
If i got it right,the addition of resistor+capacitor are used actually to protect the chip itself from inrush current,am i right?but what about the output without them?
If we won't add them(install the finished module as it is)there will be an inrush current to the chip(as you mentioned),but will an inrush current also go out of the module to the device?or the other capacitor in the end will reduced before it go out the module?
ronsimpson said:
SPICE has a hard time with PWM, inductors, charging large capacitors at startup. And when inductors have 0 ohms and capacitors have 0 ohms.
I do not use a voltage source to power the PWM. Use a voltage source that starts out at 0V and ramps up to 24V in a mS. That really helps the startup problems.
The issue is that i must use a regular switch to turn it on,i can't switch it on as you have described.Anyway my main concern is to understand what happens in the output,after that i will see what can i do next.
 
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LadySpark

Joined Feb 7, 2024
124
If i got it right,the addition of resistor+capacitor are used actually to protect the chip itself from inrush current,am i right?
The capacitor with the voltage divider on the on/off pin is used to delay powering up the chip because these regulators need a load and bias in the ic before its turned on to operate. If either one of these conditions are not met, then a current surge happens through the whole circuit including the load.
If we won't add them(install the finished module as it is)there will be an inrush current to the chip(as you mentioned),but will an inrush current also go out of the module to the device?or the other capacitor in the end will reduced before it go out the module?
The standard practice for this device is to be integrated on a board of the device itself instead of a separate module that could accidentally be powered without a load. There is equipment out there that ignored standard practices and I was hired along with another engineer to redesign the boards that went into one of these equipment. The US distributor of this equipment tried to address this issue with the manufacturer with no resolution, so he hired real experts. The distributor still buys the robotic quilter machine, but puts the re designed boards in so that there is no field failures from accidentally powering up the unit without everything connected or connecting parts of it wile other parts are powered on. That catastrophically causes damage to boards that amount to $2500 in board replacements from the manufacturer since the lands that burn up are inside multi layer boards.

So if you want to build your unit like the $30,000 robotic quilter that self destructs, that is your problem.
 

Bordodynov

Joined May 20, 2015
3,179
I have noticed that you made this simulating for a module with the addition of the capacitor and resistor as LadySpark mentioned before,so the results are fine,while my main interest was to know how may the standard LM2596s finished module(without that addition)will act.(on the output to the device)

If i got it right,the addition of resistor+capacitor are used actually to protect the chip itself from inrush current,am i right?but what about the output without them?
If we won't add them(install the finished module as it is)there will be an inrush current to the chip(as you mentioned),but will an inrush current also go out of the module to the device?or the other capacitor in the end will reduced before it go out the module?

The issue is that i must use a regular switch to turn it on,i can't switch it on as you have described.Anyway my main concern is to understand what happens in the output,after that i will see what can i do next.
The capacitor always has ESR. I added 0.3 Ohm as for a cheap capacitor. A high-quality 100 uF capacitor has 0.1 Ohm or less. The battery resistance is of course less than 1 ohm. For example 0.1 Ohm. I took such a high resistance of the power supply to push out the harmful effects.
 

seanstevens

Joined Sep 22, 2009
252
Its supposed to be a switch mode regulator with voltage feedback from its output to ensure the voltage does not go higher than what it was designed for. So as it has been said here already, if whatever you connect to it is not doing anything silly/faulty, then the regulator's output should remain within the design spec.
 

MisterBill2

Joined Jan 23, 2018
18,442
Its supposed to be a switch mode regulator with voltage feedback from its output to ensure the voltage does not go higher than what it was designed for. So as it has been said here already, if whatever you connect to it is not doing anything silly/faulty, then the regulator's output should remain within the design spec.
Given that we are not very familiar with the source of the regulator module, consider "then the regulator's output should remain within the design spec." that "should be" and "definitely are" are quite different.
 

seanstevens

Joined Sep 22, 2009
252
Given that we are not very familiar with the source of the regulator module, consider "then the regulator's output should remain within the design spec." that "should be" and "definitely are" are quite different.
True, but I was referring to the regulator semiconductor, i.e. LM2596S or any genuine buck converter. Not particularly the module the TS pictured. Although I have had my share of buck & boost converters from AliEx without any issues so far.
 
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