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I am measuring both the 10 MHz (5V pk-pk) input signal, and the output signal of the above mentioned inverters. Why does the output signal looks so much different than the input? I have decoupling capacitors on the supply, near the chips, the supply is stable, there are really no differences between the measurements, other than that I am probing a different pin.

The images below are the result of the CD 4069, but the CD 40106 schmitt trigger version produces very similar results as well.

Input, which looks exactly as I expected: enter image description here

Output: enter image description here

I forgot to adjust the probe scale on the scope, this is why you see these high numbers, but I am doing it with 1:1, and the levels are good.

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    \$\begingroup\$ Well, don't use 1:1 probes, try again with 10:1 probes and see if it works better. \$\endgroup\$ – Justme Sep 19 at 10:16
  • \$\begingroup\$ That does not make any sense. It is not about the levels, but the shape. The exact same measurement configuration should result in exactly the same signal (just inverted) on the output. For some reason, my signal is clipped at the output. \$\endgroup\$ – Gábor DANI Sep 19 at 11:32
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    \$\begingroup\$ It makes a lot of sense. A 1:1 probe has significant capacitance (often 100 pf or more) that will affect the shape of the signal. That is why 10:1 probes are used much more often since their capacitance is at least an order of magnitude lower. Also, digital logic gates are not linear amplifiers. The output signal will not and does not need to look exactly like the input. It only has to satisfy the levels for high and low and be able to drive another gate input. \$\endgroup\$ – Barry Sep 19 at 12:05
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    \$\begingroup\$ eetimes.com/oscilloscope-mistakes-part-2 this article gives some insight into measurement issues with oscilloscopes \$\endgroup\$ – mhaselup Sep 19 at 12:12
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    \$\begingroup\$ @GáborDANI The output does not seem like clipped as you say, it only seems like it is low pass filtered, and probes at 1x setting have low bandwidth compared to 10x setting which would explain the output. Also the CD4000 series chips have quite low output drive capability, so it would explain the waveform too. And the device at the input simply has stronger output drive. \$\endgroup\$ – Justme Sep 19 at 12:24
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What you see here is just the switching speed of the IC. If you check the datasheet, the transition time between high and low is given as 80 ns at a supply voltage of 5V. Actually, given this 'typical' value from the datasheet, your version of the IC seems to be even faster.

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  • \$\begingroup\$ The Op is commenting on the shape of the waveform rather than the transition time. The output is clearly rounded (integrated) and missing the ringing of the input so it's down to capacitance/ drive capability limitations. \$\endgroup\$ – mhaselup Sep 19 at 13:29
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    \$\begingroup\$ @mhaselup The shape is a direct result from the transition time. \$\endgroup\$ – asdfex Sep 19 at 13:43

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