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I am trying to understand working of the following circuit;

enter image description here It has Input from 0 to 7V and based on the Input value Output A to D will be High or Low but ...

  1. Is it Inverting Schmitt Trigger comparator circuit?
  2. Which Output will be High or Low at first? Out A or D?
  3. How other Outputs will switch accordingly?
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    \$\begingroup\$ If that's not your drawing or design then you need to credit it. (This is site policy.) Can you give some context? What do you think the circuit is for? Hit the edit link ... \$\endgroup\$
    – Transistor
    Feb 13, 2021 at 11:18
  • \$\begingroup\$ Why is R32 6k8 in value? What is the purpose of resistors R11, R20, R25 and R29? There might be some subtlety here that needs details of the chip (U2). Ditto what transistor said. \$\endgroup\$
    – Andy aka
    Feb 13, 2021 at 11:21
  • \$\begingroup\$ R11,20,25 & 29 is used to minimize the offset error. U2(A to D) are general purpose OmAmps. \$\endgroup\$ Feb 13, 2021 at 12:51
  • \$\begingroup\$ Is positive supply P7V0 equal to 7V? \$\endgroup\$
    – G36
    Feb 13, 2021 at 12:55
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    \$\begingroup\$ They all are schmitt triggers. You can recognize them by the positive feedback resistors from the output to the positive input. The initial state is undefined, you should work out the threshold for each amplifier to see how the sequence goes \$\endgroup\$ Feb 13, 2021 at 13:45

2 Answers 2

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There seems to be an original idea in this "ladder" - the comparators are turned on sequentially and each of them allows the next.

Okay, here are some more tips to think about:

Is it Inverting Schmitt Trigger comparator circuit?

Yes, it is as the input voltage is applied to the inverting input.

Which Output will be High or Low at first? Out A or D?

It seems all the outputs will be HIGH at first.

How other Outputs will switch accordingly?

They should switch sequentially.


The voltage dividers are not firmly connected to Vcc and ground; instead their ends are "movable". Let's see what happens when Vin begins increasing above zero.

Vin = 0. In the beginning, the outputs of the all comparators are HIGH. It is convenient to check this by going from back to front.

Output D is HIGH since R31 is firmly connected to 7 V... and regardless of the previous output C, the input voltage of the non-inverting input is positive (> 0.7 V). We continue up and see that, for the same reason, the outputs C, B and A are HIGH.

Vin > VH. When the high threshold is exceeded, the first comparator switches and its output A becomes LOW; the 3 k R21 of the second voltage divider is grounded.

Since R23 is connected to HIGH, the second comparator switches and its output B becomes LOW; so the 3 k R26 of the third voltage divider is grounded.

Then, since R27 is connected to HIGH, the third comparator switches and its output C becomes LOW; so the 3 k R30 of the fourth voltage divider is grounded.

Finally, since R31 is connected to 7 V, the fourth comparator switches and its output D becomes LOW.

Vin < VL. When the low threshold is reached, the fourth comparator switches and its output D becomes HIGH; the 6.8 k R27 of the third voltage divider is connected to HIGH.

Since the 3 k R26 is connected to LOW, the third comparator switches and its output C becomes HIGH and R23 of the second voltage divider is connected to HIGH... and so on...

Conclusion

Like a domino effect, once the input voltage exceeds the upper threshold, the comparators start switching from HIGH to LOW in sequence A -> B -> C -> D. When the input voltage goes below the lower threshold, they start switching from LOW to HIGH in the reverse sequence D -> C -> B - > A.

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  • \$\begingroup\$ Circuit fantasist, you are spot on! So all the Outputs are HIGH at the beginning but which one go low first? A or D. It will switch in sequence from A to D or D to A? \$\endgroup\$ Feb 13, 2021 at 18:49
  • \$\begingroup\$ @Electric Monkey, I think you should answer this question after you read my explanations above? \$\endgroup\$ Feb 13, 2021 at 18:54
  • \$\begingroup\$ i guess it will switch from A to D. \$\endgroup\$ Feb 13, 2021 at 18:58
  • \$\begingroup\$ Why would all the Output will be HIGH despite of having floating/movable reference on non-inverting input? \$\endgroup\$ Feb 13, 2021 at 19:00
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    \$\begingroup\$ Thank you so much! You really have your Analog knowledge in phase! \$\endgroup\$ Feb 13, 2021 at 19:04
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The purpose of Schmitt Trigger hysteresis is to "denoise" the analog input into some binary output according to the spectrum and amplitude of the noise and the signal. Essentially to improve the SNR of the input Analog to Digital output to the fundamental frequency of the input.

Might also beg the question, what was it from? and what was intended? and can it perform any better than a single Schmitt trigger, that is well designed? IMHO, I would say no.

How does this circuit work?

It doesn't.

For Vin >0 and <=7V and R12 = N2V0 , the OUT A will always = N2V0 (i.e. the +ve input is always negative below the -ve = positive input Vin.)

But if R12 is connected to some Positive voltage around 7V, it will work.

I see no methodical filtering or advantage to this design.

Update

Missing information in question "P7V0 = +7V and N2V0 = -2V "

Only makes the duty cycle closer to 55% due to Vin average = 3.5V and Avg{7V,-2V} = 2.5V (assuming correction made per above)

There is some slight LPF effect and denoising by cascading the circuits which could be better achieved by using a better analog "denoise" LPF front end.

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