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It is between Vin and output pins or pin 1 and 2.Where are you seeing the spikes?
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.It is between Vin and output pins or pin 1 and 2.
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.Is this like the scope trace on pin 3 to pin 2? If this be the case I would expect around 23v under no load ( I know its a different chip, but these all work similar )
Test condition is: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.
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.
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.Please read post no.8 again I add some inf. but if you want another information let me know .
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?My wave form is exactly like in the condition that document says long Ground prob condition!!!
The board is going to be a QC.3 charger which is 9v and 5v ...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?
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?
Pls see last picture of post No.17This 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)