Calculating V in an Op-Amp Circuit with Constant Input: Schmitt Trigger Question

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The forum discussion centers on calculating the input voltage (V) in an op-amp circuit configured as a Schmitt Trigger, given a constant output voltage (V OUT) of +2.4V. The circuit uses resistors R1 = 4.8 ohms and R2 = 2 ohms. Participants derive the relationship between input and output voltages, ultimately concluding that the input voltage V is calculated using the formula V = (R1 * V2) / -R2, resulting in V = -5.76V. The discussion emphasizes the importance of understanding the circuit's feedback mechanism and the implications of positive feedback on stability and gain.

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  • #31
jim hardy said:
was i off track much ?

hope i didnt hurt you on this one.

old jim

Absolutely not, my stress levels have been sitting nice and high but you have been great in terms of helping me understand the subject more. I find my study material never supplies anywhere near enough information so places like these forums are invaluable!

Many thanks!

Just to confirm though, there is no equation for an accurate input voltage of a schmitt trigger?
 
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  • #32
StripesUK said:
Just to confirm though, there is no equation for an accurate input voltage of a schmitt trigger?
Input voltage that'll cause what condition?

It is easy to calculate the voltage at which the schmitt will switch output states.
It is not physically possible to apply an input voltage that will hold its output in between those two states, in its linear region , as i interpreted your original post.
you can calculate such a voltage but it is not practical to measure let alone control to the required precision
because changing the applied voltage in the last digit by just one
upload_2016-2-29_16-50-23-png.96630.png

will start the positive feedback loop on its merry way to saturation.

I'm not dodging your question just trying to answer it precisely.
 
  • #33
Understood. Thanks for all your help and patience!
 
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  • #34
StripesUK said:
Understood. Thanks for all your help and patience!

Feedback behavior does not come intuitively to some people, myself for instance. I struggled for years.

Pardon the un-academic next paragraph
i learn a lot just watching everyday stuff around me, like burgers on the charcoal grill.
Ever heard the expression "The fat's in the fire now! " ? It means something calamitous is imminent.
Burgers- The fat drips down and burns hot, flames erupt causing more out-juicing which feeds more fat into the flames and you wind up with charcoal-burgers.
That's positive feedback.
Positive feedback usually results in a departure from linear behavior and often destructive failure like that galloping bridge in Tacoma... or a runaway fire in your barbecue grill...
It makes your schmitt trigger circuit nonlinear.
Sometime buy a completely lean round or sirloin steak and have it ground to hamburgers and grill them. You'll see immediately the effect of removing positive feedback. If you add water to those ultra-lean burgers you'll see that negative feedback calms a system..

In early days of radio we had "Regenerative receivers" that employed positive feedback to increase gain of the RF stages. They operated on the verge of unstable oscillation , that's the "whistling" you hear in old movies when somebody is tuning a radio.

Little stupid experiments like that help you work systems in your head, which for a plodder like me is necessary to understand the math.

Work that schmitt in your head. Also conventional negative feedback 'operational amplifier circuits' which lend themselves beautifully to math..

old jim

old jim
 
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