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enter image description here

This circuit is a Schmitt trigger, and the supply voltage Ucc.​ is 15V. I want to calculate whether the transistors T1 and T2 are conducting or not when the state is LOW or HIGH(U(A)). These transistors are NPN bipolar junction transistors (BJTs). I suspect that in the LOW state, T1 blocks and T2 conducts, while it's the opposite in the HIGH state! For this, I used Kirchhoff's Voltage Law. Is the result correct?

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  • \$\begingroup\$ It's not a Schmitt trigger; it's plain ordinary negative feedback. You need to provide a citation for the image. Use a simulator if you want to check this out for yourself. \$\endgroup\$
    – Andy aka
    Commented Apr 27 at 19:42
  • \$\begingroup\$ @Andyaka - er, Andy - that's positive feedback from T2 emitter to T1 emitter. They are both NPN's. \$\endgroup\$
    – Kevin White
    Commented Apr 27 at 20:16
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    \$\begingroup\$ @Circuitfantasist I cannot agree with your perspective as being better. It's worse. To my eye. But that takes nothing away from whatever works for you. So long as the perspectives work equally well for either of us, and lacking any specific, narrowing context, then better is simply in the eye of the beholder. \$\endgroup\$
    – periblepsis
    Commented Apr 28 at 9:36
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    \$\begingroup\$ @Circuitfantasist Oh. Sure. And on that point I agree. It's single-ended input and single-ended output. But that's true for so many circuits -- uncountable -- that are also very different from each other. So different that knowing they are single ended I/O doesn't really help much. On the other hand, if someone tells me that this is a diff pair with one of the diff inputs tied to the other's collector output -- then that tells me a great deal about the circuit because I can leverage what I know about diff pairs and apply the ideas right away. \$\endgroup\$
    – periblepsis
    Commented Apr 28 at 10:03
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    \$\begingroup\$ @Circuitfantasist I have used this circuit where T2's collector goes straight into the base of another bipolar (VAS). But there is an important purpose to T1's collector resistor. It converts the teeter-totter current to a managed voltage signal toward's T2's base. A current source wouldn't perform that function and T1's collector voltage would be in poorer management. (Haven't thought about it until you just now asked, though, because if the idea had flitted across my mind I would have discounted it as an unhelpful complexity, worried about the collector voltage, and then moved on.) \$\endgroup\$
    – periblepsis
    Commented Apr 28 at 10:41

1 Answer 1

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Here are some initial simulations.

schematic

simulate this circuit – Schematic created using CircuitLab

To simplify the schematic, collector resistors (Rc1 and Rc2) are represented by real ammeters with the same internal resistance (5 kΩ and 2 kΩ); the emitter resistor Re is represented by a voltmeter with 1 kΩ resistance.

Vin = 0V -> 4V

schematic

simulate this circuit

Vin = 4.1V -> 10V -> 2V

schematic

simulate this circuit

Vin = 0V -> 10V -> 0V

schematic

simulate this circuit

STEP 3

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  • \$\begingroup\$ Strange thing - "Simulate this circuit" disappeared... \$\endgroup\$
    – Circuit fantasist
    Commented Apr 28 at 16:23
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    \$\begingroup\$ Aha... it's already appeared... \$\endgroup\$
    – Circuit fantasist
    Commented Apr 29 at 21:36

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