Is that now correct?

Yes, It is correct. But in real life, the positive and negative saturation voltage will not be the same.
So, real-world measurement will show a different result.

 

Yes, It is correct. But in real life, the positive and negative saturation voltage will not be the same.
So, real-world measurement will show a different result.

This is what i got in the real world.





the forumulas and steps provided in my previous response is that correct.

 

This pictures are way too small to read anything from there.

 

This pictures are way too small to read anything from there.

 

This pictures are way too small to read anything from there.

I think this is the correct one.

 

This pictures are way too small to read anything from there.

 

Do you have any more questions about the circuit?

 

Do you have any more questions about the circuit?

I gave you the pictures to have a look? The real world figures/analysis.

 

Is this blue waveform a voltage at the opamp output? Because it looks more like it was a voltage at the non-inverting input.

 

Is this blue waveform a voltage at the opamp output? Because it looks more like it was a voltage at the non-inverting input.

yes

 

Look:
(5V - 2.5V)/12kΩ = (2.5V - Vin)/18kΩ

(2.5 - Vin)/18 = 2.5/12

(2.5 - Vin) = (2.5*18)/12

(2.5 - Vin) = (2.5*18)/12

2.5 - Vin = 3.75

Vin = 2.5 - 3.75 = -1.25V

Could you ease explain each step for me please. Like Equation 1 = and so on

 

In your circuit we have this situation:


From KCL we know that:

I2 = I1 (1)

And

I2 = (Voh - Vp)/R2 (2)

I1 = (Vp - Vin)/R1 (3)

Aditional form opamp "theory" we know that "switching" occurred when Vp = Vn and Vn = Voh/2 (4)

Therefore we and up with this:

(Voh - Voh/2)/R2 = (Voh/2 - Vin)/R1 (5)

So, we can plug the numbers and solve it for Vin (V_LTP):

(5V - 2.5V)/12kΩ = (2.5V - Vin)/18kΩ

2.5V/12kΩ = (2.5V - Vin)/18kΩ ---> multiply both sides by 18kΩ

2.5V/12kΩ *18kΩ = (2.5V - Vin)

3.75V = 2.5V - Vin ---> subtract 2.5V

3.75V - 2.5V = -Vin

1.25 = -Vin ----> multiply both sides by -1

Vin = -1.25V

But this is just elementary school math.

 

In your circuit we have this situation:

View attachment 190140
From KCL we know that:

I2 = I1 (1)

And

I2 = (Voh - Vp)/R2 (2)

I1 = (Vp - Vin)/R1 (3)

Aditional form opamp "theory" we know that "switching" occurred when Vp = Vn and Vn = Voh/2 (4)

Therefore we and up with this:

(Voh - Voh/2)/R2 = (Voh/2 - Vin)/R1 (5)

So, we can plug the numbers and solve it for Vin (V_LTP):

(5V - 2.5V)/12kΩ = (2.5V - Vin)/18kΩ

2.5V/12kΩ = (2.5V - Vin)/18kΩ ---> multiply both sides by 18kΩ

2.5V/12kΩ *18kΩ = (2.5V - Vin)

3.75V = 2.5V - Vin ---> subtract 2.5V

3.75V - 2.5V = -Vin

1.25 = -Vin ----> multiply both sides by -1

Vin = -1.25V

But this is just elementary school math.

I thought it's not 1.25 and 2.5v as the Vout is 4v. So if its 4v which it is, houd have 1.v = - Vin? (Multiply both sides by what ?) Sorry.

 

1.v = - Vin? (Multiply both sides by what ?) Sorry.

Multiply or divide both sides by -1.

 

Multiply or divide both sides by -1.

for the negative saturation is this correct following the KLC as above
((-4V-(-2V))/12kΩ=((-2V)-Vin))/18kΩ=((-2V)-3V))/18kΩ=+1V

or this

2.0V/12kΩ = (2.0V - Vin)/18kΩ ---> multiply both sides by 18kΩ

2.0V/12kΩ *18kΩ = (2.0V - Vin)

3.0V = 2.0V - Vin ---> subtract 2.0V

3.0V - 2.0V = +Vin

1.0 = +Vin ----> multiply both sides by 1

Vin = 1.0V

 

In your circuit we have this situation:

View attachment 190140

Sorry for coming late with a doubt.

What about the output taken from the non-inverting input? Is that justified / explained?

 

Sorry for coming late with a doubt.

What about the output taken from the non-inverting input? Is that justified / explained?

We take it from Vp

 

What about the output taken from the non-inverting input? Is that justified / explained?

Because the original circuit was a negative impedance converter (NIC) "converted" to Schmitt trigger (when positive feedback wins the fight).
And of course, you can take the output signal from the opamp output as we usually do.

 

We take it from Vp

Because the original circuit was a negative impedance converter (NIC) "converted" to Schmitt trigger (when positive feedback wins the fight).
And of course, you can take the output signal from the opamp output as we usually do.

Thanks. Have to learn about that NIC to understand this.

 

Thanks. Have to learn about that NIC to understand this.

Look here:
http://www.analogzoo.com/2018/01/negative-impedance-converters/