Attenuation of a signal using a hex inverter

Sakux

Joined Mar 1, 2021
25
Hello,

I encounter this part of a circuit that made an attenuation of a signal coming from a crystal. If I measured this sinus signal I saw an attenuation in the different stages of the path. I really don´t know how it works, mostly for the part of the IC E. Here is what I can measure with the oscilloscope in each point:

Amplitude (V):
• 3 ---- 5,81
• 4 ---- 1,91
• 11 ---- 2,85
• 10 ---- 4,10​

Can anyone be to kind to explain that or give me a clue of what is happening with the amplitude? Thanks in advance!

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BobTPH

Joined Jun 5, 2013
8,108
What is the supply voltage for the gates? A CMOS gate high output should be very near the supply rail.

Your voltage increase between 3 and 4 is not possible. It would indicate a current of 5mA from the input (!) of E to the output of B.

How are you measuring this? You need to use a scope.

Last edited:

Alec_t

Joined Sep 17, 2013
14,009
What does that wire exiting at top left connect to?
Is your 'scope probe AC-coupled or DC-coupled?
Are your measured voltages DC, AC peak-to-peak or AC rms?

Papabravo

Joined Feb 24, 2006
20,607
Measuring the amplitude of an AC signal of that frequency with a voltmeter is a profoundly misguided thing to do. You cannot possibly interpret the results in a meaningful way.

Sakux

Joined Mar 1, 2021
25
Hello,

Thank you all for your responses. This line on the left leads to a pin of a micro to control the output. It puts the signal High impedance, 3.3 or 0. Of course I used an oscilloscope to measure this.It is in DC coupling. And the signal is measured using the amplitude quick measurement of the oscilloscope.

Alec_t

Joined Sep 17, 2013
14,009
It puts the signal High impedance, 3.3 or 0.
What state was the pin in when you took the measurements?

MrChips

Joined Oct 2, 2009
29,850
Remove the connection from the MCU.
Show us screen shots of the oscilloscope.

Sakux

Joined Mar 1, 2021
25

Sakux

Joined Mar 1, 2021
25
Remove the connection from the MCU.
Show us screen shots of the oscilloscope.
Here you can see an example of what you asked. Names are refered of the pins of the hex gates.

Regards and thanks for your time.

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Alec_t

Joined Sep 17, 2013
14,009
If the micro pin was putting out 3,3V, how can pin 3 possibly be at 5.81V?
Have your scope probes been properly calibrated?

schmitt trigger

Joined Jul 12, 2010
772
Some probes can place a 10 pF load, even on the 10X setting. Much worse at 1X.
Even with a good probe, the load will be around 1500 ohm.

Sakux

Joined Mar 1, 2021
25
If the micro pin was putting out 3,3V, how can pin 3 possibly be at 5.81V?
Have your scope probes been properly calibrated?
No idea... Maybe I was wrong and it was in high impedance when I measured it?

Sakux

Joined Mar 1, 2021
25
Some probes can place a 10 pF load, even on the 10X setting. Much worse at 1X.
Even with a good probe, the load will be around 1500 ohm.
Yes, you are right. But how is this afecting with so much the amplitude of the signal in the different points? What I am missing?

Sakux

Joined Mar 1, 2021
25
Yes, you are right. But how is this afecting with so much the amplitude of the signal in the different points? What I am missing?
Moreover I don´t really understand the second structure of the hex gate with this realimentation with the 47K.

Alec_t

Joined Sep 17, 2013
14,009
I don´t really understand the second structure of the hex gate with this realimentation with the 47K.
The 47k provides negative feedback to keep the DC operating point of the gate at about mid-rail voltage. The gate then acts as a more-or-less linear amplifier over a limited range.

MrChips

Joined Oct 2, 2009
29,850
What I see on pin-10 as an example, is a square-wave (digital) of about 3.3V amplitude.
I see overshoot and undershoot caused by the oscilloscope probe coupling.
74HCU04 is a digital gate. Don't expect to see a sine-wave signal.

Sakux

Joined Mar 1, 2021
25
The 47k provides negative feedback to keep the DC operating point of the gate at about mid-rail voltage. The gate then acts as a more-or-less linear amplifier over a limited range.
So this gate is acting like something as an OA?

MrChips

Joined Oct 2, 2009
29,850
That's make sense, but how is this achieved the saturation of the original sine wave of the crystal?
The sine wave you see on pin-3 is the oscillation of the crystal.
Pin-3 input impedance is very high and does not load down the crystal oscillation.

Alec_t

Joined Sep 17, 2013
14,009
So this gate is acting like something as an OA?
Not really. It is an unbuffered gate. It does not have a non-inverting input.