LM2596 and VCEO ?

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arvinfx

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Hi,
How much is VCEO in LM2596D ?
I asked it because I see 23.5V spikes when the switch goes off! my input voltage is 12.5 and out is 5 .
I mention it on this picture:


 
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Where are you seeing the spikes?

If the spikes are on the input, that could be caused by insufficient decoupling of the Vin pin. You want some ceramic capacitors on the input. Electrolytic capacitors probably have too much inductance to be effective decoupling on their own.

If the spikes are on the output, that could be caused by too much inductance in the wires going to the Schottky diode.


The loop shown in orange is the path of the high power alternating current. Other current paths are low power or low frequency. It is important that you make the area of that loop as small as practical on your layout.

The LM2596 can work up to 40 V, so spikes of 23.5 V isn't going to blow the regulator, but it might be causing unnecessary noise.
 
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Capacitance ESR varies significantly for various technologies :



OSCON is polymer tant, best in class for large bulk bypass caps.


Regards, Dana.
 
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arvinfx said:
It is between Vin and output pins or pin 1 and 2.
Click to expand...
When the switch has just turned off, that voltage should be the sum of the input voltage and the forward voltage of the Schottky diode, so around 13 V in your application. When the inductor current gets to zero, the voltage will fall to the difference between the input and output voltages, so around 7.5 V in your application.

However, as the current in the inductor hits zero, the voltage at that point will oscillate. That is shown on figure 9-14 of the datasheet:- https://www.ti.com/lit/ds/symlink/lm2596.pdf

Figure 9-10 of that datasheet shows the wires that need to be kept short. There is a lot of good information there about layout.

If you are getting spikes of 23.5 V there is something wrong with the decoupling or the layout, but you also have to be very careful that you are measuring the spikes correctly. If you connect the earth lead of the oscilloscope to pin 2 of the LM2596, that is almost certainly going to give incorrect readings as you are expecting the earth of the oscilloscope and the earth of the circuit to have different voltages, and you are changing those voltages very quickly when the switch opens or closes. Both the ground of your circuit and the ground of the oscilloscope will have large capacitances to real ground, so that will mess up the reading badly for high frequency signals.

Can you show the waveform between pin 1 and ground, and the waveform between pin 2 and ground. In each case the oscilloscope ground must be connected to the circuit ground. You should use the ground lead attached to the oscilloscope probe for that.
 
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Yes, they all work in similar ways. That waveform in eevblog is going from just under zero volts to just under 12 V, and the peak voltages are almost exactly what I would expect on any buck regulator converting 12 V to 5 V.

The +ve peak voltage is just a bit less than the supply. The difference is due to the losses in the switch. The -ve peak is at about -0.4 V and it's the voltage across the Schottky diode when it's conducting.

The voltage on pin 2 of an LM2596 should always stay within that range if the buck converter is working correctly. The input voltage should be constant and there are capacitors to keep it from varying much, so the peak voltage difference between pin 1 and pin 2 should only be a tiny bit more than the supply voltage.

However, it's not all that easy to measure.
 
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Diver300 said:
Can you show the waveform between pin 1 and ground, and the waveform between pin 2 and ground. In each case the oscilloscope ground must be connected to the circuit ground. You should use the ground lead attached to the oscilloscope probe for that.
Click to expand...
Test condition is:
Input voltage: 12.6V
current consuming: 1.3A

Output voltage 9V

Ground to Ground
Tip to pin 1 of LM2596





Ground to Ground
Tip to pin2 of LM2596



Ground to pin 2 of LM2596
Tip to pin 1 of of LM2596


 
Last edited:
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There's something wrong with how the circuit is working. The average voltage across the inductor should be close to zero, so the average voltage at pin 2 should be only just above 5 V. The switch should be on for less than 40% of the time, as the output voltage is 40% of the input voltage.

Can you send more details about the circuit. Also, the switching waveforms in more detail at the points when the switch turns on and off, with a timebase of maybe 50 ns/division.
 
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Please read post no.8 again I add some inf. but if you want another information let me know .


Ground to pin 2
tip to pin 1
turn off:



ground to pin2
tip to pin1
Turn on:

 
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arvinfx said:
Please read post no.8 again I add some inf. but if you want another information let me know .
Click to expand...
I had assumed that the output voltage was still 5 V, as that is what you said in the first post. The waveform seems fine for 9 V out and a significant output current.

What is the inductor value?
 
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arvinfx said:
My wave form is exactly like in the condition that document says long Ground prob condition!!!
Click to expand...
It certainly looks like the sort of ringing that I've seen with ground connections being too long. Can you show us what the circuit looks like?

If you have oscilloscope leads with metal ring just a few mm back from the tip, try to make the measurement with a direct link, as short as possible, from that to the circuit ground. I've used a knife blade or pin to make that connection, or soldered a short wire to the ground pin, and bent that so that the ring can touch at the same time as the pin touches the point to be tested.

Also, you should never put a ground wire to pin 2. Post #5 says why. It would effectively be making the ground wire longer as the voltage of the whole oscilloscope chassis will have to change each edge.

Can you try to measure pin 2 on the tip, and pin 3 to ground?
 
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This all measurement error.

10 nH/cm probe ground . Remove the 10:1 Probe ground wire and tip clip and only use pin and coaxial ring of probe between 2 R wire test pins soldered as test points. 5 to 10 mm apart and repeat.

Use 20MHz DSO filter if you can't.

Best practice is AC coupled to 50 Ohm coaxial on DC output with 50 ohms at DSO termination. using OSX or SMB, then textbook waveforms on DC output.

"Ground" is wherever you decide is 0V. (by definition)
 
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Diver300 said:
I had assumed that the output voltage was still 5 V, as that is what you said in the first post. The waveform seems fine for 9 V out and a significant output current.

What is the inductor value?
Click to expand...
The board is going to be a QC.3 charger which is 9v and 5v ...

inductor value is 34uh but I used a toroidal core from an old motherboard.
 
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This is the board:



Ground to pin3
tip to pin 2

in 12.4
out 5v 0.8A


prob location and type of measuring:

 
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Pls see last picture of post No.17
 
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You just knew a breadboard and really poor wiring was going to be involved

Don't use breadboards, and make all connection as short and direct as possible - use a PCB.
 
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