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waveformI am trying to get more familiar with using a 74HC14, as an experiment I have an ldr set up in a voltage divider configuration and the center tap goes to the input of the 74HC14 input 1. I have all of the other inputs tied to ground. Everything is soldered together on a perf board. I have a .1uF, .01uF, .001uF, .0001uF cap on the power pin (probably overkill). Input voltage is 5v. The voltage divider and the IC are both powered off the same rail

The IC triggers as expected when light is shined on/removed from LDR. Below is the waveform I am getting for the falling edge, it is very consistent from test to test.

What I don't understand is why the signal is dropping to -2.4 or so when triggered then staying below 0.

The rising edge has an overshoot to about 6v but is smoother. Any information/guidance would be appreciated.

https://drive.google.com/open?id=1oG6Eh2Gwj9bf4IXGBJ9i5LoJoLvwybN6

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    \$\begingroup\$ Please post the images into the body of the post. Links can go bad over time. \$\endgroup\$
    – Aaron
    Jan 24, 2020 at 20:48
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    \$\begingroup\$ Expand the trace to show at least a few milliseconds after the transition. Have you compensated the scope probe. How are you grounding the probe, how long is the ground connection? These are almost certainly measurement errors, not anything to do with the device or circuit. \$\endgroup\$ Jan 24, 2020 at 20:56
  • \$\begingroup\$ I will add pic in body of post with an expanded trace when I get home \$\endgroup\$
    – Psubond
    Jan 24, 2020 at 22:52

2 Answers 2

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When trace and wires are inductive at the rate up to 10nH/cm interact with the scope probe coax and standard ground leads, any measurements with rise times faster than what the 20MHz filter allows will have impedance mismatch faulty signals, both due to the board design and the scope probe ground lead.

Either apply the 20MHz DSO filter at all times or learn how to capture proper signals and learn the effects of the [robe tip capacitance. It is no secret that Nch low side drivers are much lower impedance than Pch High side drivers and breadboards without a ground plane closer than trace width the chances of matched impedance on traces is zero. Thus overshoot depends on the degree of how much lower the driver impedance is.

If you want any chance of getting a clean signal; the ground plane, IC decoupling cap and output must all be within 1cm be probed as below Also any long traces or loads examined for reflection effects.

Recall 80% Tr = 0.35/f (-3dB) so Tr=0.35/20MHz =17.5ns

This means any rise times < 18 ns common to all 74HCxx CMOS will be inaccurate on a scope unless you follow this test method... or ignore it and apply the 20MHz DSO filter switch.

enter image description here

Getting rid of noises after RC low pass filter stage

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  • \$\begingroup\$ When I soldered it together I soldered in header pins to attach dupont wires to and after reading your response it is clear that my incorrect way of probing the circuit explains the waveform. I will try again using your method. The good thing is that this isn't for practical use, I made the circuit just as a learning exercise because I got curious. Learning has occured. I will repost with results later tonight when i can get back in my lab. Thank you! \$\endgroup\$
    – Psubond
    Jan 24, 2020 at 22:59
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    \$\begingroup\$ You ought to be able to get textbook waveforms. \$\endgroup\$ Jan 24, 2020 at 23:58
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I used the probing method mentioned by Tony Stewart above. It gave me a much cleaner wave form and I am now confident that my circuit is working properly. I am going to keep experimenting to see if I can clean it up more, but I am working with a lower end scope so I am not sure how much I can improve my signal. I may take the circuit to work and test it on a nice high end scope with active probes and see what I get.

New Waveform

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  • \$\begingroup\$ What is your 10% to 90% rise time and DSO BW? \$\endgroup\$ Jan 25, 2020 at 18:16
  • \$\begingroup\$ I have no idea on the 10% and 90% , I am fairly new at this. The scope I am using is a Hantek 5202P \$\endgroup\$
    – Psubond
    Jan 25, 2020 at 18:24
  • \$\begingroup\$ Take the mean levels as 0 100% then measure the time interval from 10 to 90% . Is that 4ns/div? \$\endgroup\$ Jan 25, 2020 at 18:25
  • \$\begingroup\$ Will get back to you in a bit, have to look this up. This is exactly why I like experimenting on stuff, you never only learn what you set out to learn, there are always a bunch of other rabbit holes to go down. To help with me figuring this out, why are you asking what the 10% and 90% rise times are? \$\endgroup\$
    – Psubond
    Jan 25, 2020 at 18:31
  • \$\begingroup\$ Yes, 4ns per division \$\endgroup\$
    – Psubond
    Jan 25, 2020 at 19:44

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