The first thing you need to do is to understand how "ideal" non-inverting Schmitt trigger work.
Maybe this will help
For example if
Vcc = 5V; Vee = -5V and Vref = 0V ; R1 =10kΩ; R2 = 100kΩ
The VH = 10V * 10k/100k = 1V and the threshold voltage are
V1 = Vref - VH/2 = 0V - 1V/2 = -0.5V
and
V2 = Vref + VH/2 = 0.5V
In your circuit, you want the threshold voltage to be around V1 = 3.11V and V2 = 3.15V. Therefore Vh = V2 - V1 = 40mV and because you are using single supply voltage (Vcc = 12V and Vee = 0)
And now we can easily solve for R1 and R2.
Vh = Vcc*(R1/R2)
You need to shift the Vref voltage up to:
Vref = V1 + VH/2 = 3.11V +40mV/2 = 3.13V ( exactly speaking Vref = R2/(R1 + R2) * V2)
Because with a single supply V1 and V2 are equal to :
V1 = Vref - (Vcc - Vref)* R1/R2
V2 = (R1 + R2)/R2 * Vref = Vref + (Vref -Vomin)* R1/R2
Where Vomin - comparator output voltage at low state
Maybe this will help
For example if
Vcc = 5V; Vee = -5V and Vref = 0V ; R1 =10kΩ; R2 = 100kΩ
The VH = 10V * 10k/100k = 1V and the threshold voltage are
V1 = Vref - VH/2 = 0V - 1V/2 = -0.5V
and
V2 = Vref + VH/2 = 0.5V
In your circuit, you want the threshold voltage to be around V1 = 3.11V and V2 = 3.15V. Therefore Vh = V2 - V1 = 40mV and because you are using single supply voltage (Vcc = 12V and Vee = 0)
And now we can easily solve for R1 and R2.
Vh = Vcc*(R1/R2)
You need to shift the Vref voltage up to:
Vref = V1 + VH/2 = 3.11V +40mV/2 = 3.13V ( exactly speaking Vref = R2/(R1 + R2) * V2)
Because with a single supply V1 and V2 are equal to :
V1 = Vref - (Vcc - Vref)* R1/R2
V2 = (R1 + R2)/R2 * Vref = Vref + (Vref -Vomin)* R1/R2
Where Vomin - comparator output voltage at low state
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