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Coquelicot's answer explains what is going on, so I won't duplicate that. However, one of the sources of confusion is that you are looking at the two waveforms in isolation. Your scope obviously ...
Answer
#2: Post edited
by
Olin Lathrop
·
2020-12-25T20:25:40Z (almost 4 years ago)
- Coquelicot's answer explains what is going on, so I won't duplicate that.
- However, one of the sources of confusion is that you are looking at the two waveforms in isolation. Your scope obviously has at least two channels, since we can see a unused blue line in each picture.
Use both channels. Set everything up like in the first screen shot, meaning to show and trigger off the input signal. Then put the probe for the second channel on the LED cathode. Now you will see both signals together. The fact that your circuit inverts will be immediately obvious.
- Coquelicot's answer explains what is going on, so I won't duplicate that.
- However, one of the sources of confusion is that you are looking at the two waveforms in isolation. Your scope obviously has at least two channels, since we can see a unused blue line in each picture.
- Use both channels. Set everything up like in the first screen shot, meaning to show and trigger off the input signal. Then put the probe for the second channel on the LED cathode. Now you will see both signals together. The fact that your circuit inverts will be immediately obvious.
- <hr>
- <blockquote>mistake on my part, to assume the x-axes would be matched</blockquote>
- Yes. The scope is clearly set up to trigger on a rising edge, when it crosses about 2 V. When you look at each trace separately, each signal will be shown aligned to the trigger point at a rising edge. If the signals happen to be inverted (as they are in this case), then the two signals will be shown phase shifted by 180° relative to each other when viewed separately.
- When a scope shows multiple traces, they are shown together in time. You choose which signal will be used as the trigger. I suggested using the input signal as the trigger, since it's the cleaner and more obvious signal, and what is causing the other signals. When you do that, you will see the LED cathode voltage go down as the input goes up, and vice versa. This would actually be a good lesson.
- Go try it, even though you may already know what you will see. Then you can crank up the time scale and see that the signals are not exactly inverted from each other. When you get to 1 µs/div or less, the delay should be apparent. It would actually be quite educational to see how exactly the LED cathode responds to the input signal at the sub-microsecond level. You will notice some asymmetry in the response too. Think about that for a bit, then come back here and ask another question if you can't figure out what is going on at the detail level.
#1: Initial revision
by
Olin Lathrop
·
2020-12-25T14:58:50Z (almost 4 years ago)
Coquelicot's answer explains what is going on, so I won't duplicate that. However, one of the sources of confusion is that you are looking at the two waveforms in isolation. Your scope obviously has at least two channels, since we can see a unused blue line in each picture. Use both channels. Set everything up like in the first screen shot, meaning to show and trigger off the input signal. Then put the probe for the second channel on the LED cathode. Now you will see both signals together. The fact that your circuit inverts will be immediately obvious.