Timeline for Circuit with two opamps
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 18, 2019 at 20:35 | comment | added | cdummie | It was a typo actually, i meant to say n1 instead of n2. | |
| Jan 18, 2019 at 19:56 | comment | added | Transistor | No. n1 will switch negative. Read the response to your comments in my answer again. I explained that n2 remains at virtual ground. It never deviates more than a mV or so from 0 V as with any negative feedback op-amp configuration. Are you sure you understand how a simple op-amp inverting amplifier actually works? | |
| Jan 18, 2019 at 19:49 | comment | added | cdummie | I know, i need to provide a negative supply too,but is the value of that negative supply voltage actually a voltage at n2 in the moment when Schmitt trigger input voltage is high enough, and it remains that way until significant change happens at the input? Is this right? | |
| Jan 18, 2019 at 18:10 | comment | added | Transistor | Nope. Low is the negative supply voltage, and when the Schmitt (note capitalisation) trigger output goes low it goes as close as it can to that voltage, -12 V in the example I gave in the caption of Figure 1. Your circuit will not work with a positive supply only. | |
| Jan 18, 2019 at 16:45 | comment | added | cdummie | Low should be the lower threshold voltage of schmitt trigger, right? | |
| Jan 18, 2019 at 8:19 | comment | added | Transistor | See the update. | |
| Jan 18, 2019 at 8:17 | history | edited | Transistor | CC BY-SA 4.0 |
Response to comments.
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| Jan 18, 2019 at 6:34 | comment | added | cdummie | also, i think that same thing should happen in n2 since those points are separated only by a single resistor | |
| Jan 18, 2019 at 6:28 | comment | added | cdummie | Well, when the input reaches high enough level we will have logic zero at the output of the schmitt trigger | |
| Jan 17, 2019 at 21:56 | comment | added | Transistor | Schmitt trigger is correct but you haven't generated a pulse train yet. I have given you four questions. What is the answer to the first one? | |
| Jan 17, 2019 at 21:43 | comment | added | cdummie | Well, the first part of the circuit is schmitt trigger, so it is generating a pulse train at its output (node n1) i expect the same waveform at node n2, but i cannot see what will happen with the waveform after integrator in this case | |
| Jan 17, 2019 at 19:49 | history | answered | Transistor | CC BY-SA 4.0 |