Op-Amp Upper and Lower Thresholds

Hello guys, once again, i'm here to ask for your help.

I have an Op-Amp which Vin is driven through a R1 resistance to the inverting input.

What i need to calculate are the 2 thresholds of this circuit, the resistors and the Vreference.

I managed to calculated them with Vref=0 and the equations : Vutp=Voh*(R2/[R2+R3]) etc
and with Vref!=0 but with no success. I dont understand how to keep it going. Can you give me a clue from where to start calculating? Thank you in advance!

This is the circuit :

To find threshold voltage you just need assume that op-amp is in saturation, its doesn't mater if its is positive saturation or negative sat.
And then you need to find the voltage on non-inverting input simply by use superposition.

[LATEX]Vt1 =Vcc* \frac{R2}{R2+R3}+ Vref* \frac{R3}{R2+R3}[/LATEX]

Vt2 = ??

i see. and what if i need to find the resistors for example Vutp=4Volts and Vltp=-3Volts?

can you calculate this with Vref=0 and Vref!=0 ??

If Vref = 0 then for this circuits it is impassible to achieve this condition (Vutp=4Volts and Vltp=-3Volts).
In order to achieve this you need to add extra resistor into the circuit.

But for Vref>0 we can solve for R2; R3 and Vref to get Vutp=4Volts and Vltp=-3Volts.
I get this result

Vref = 15/23 = 0.652; R2 = 3K ; R3 = 10K

Hm, i still cant make the calculations.

For Vref!=0 : Vutp=Vref* [R3/(R2+R3)]+Vcc*[R2/(R2+R3)] , where Vcc=+15Volts

Vltp=Vref* [R3/(R2+R3)]+Vee*[R2/(R2+R3)] , where Vee=-15Volts

and also Vhys=Vutp-Vltp=2(Vcc-Vee)* [R2/(R2+R3)]

right???

What do i miss???

Hmm,
So you have two equations

4V = Vref*[R3/(R2+R3)] + 15V*[R2/(R2+R3)]

-3V = Vref*[R3/(R2+R3)] + (-15V)*[R2/(R2+R3)]

But we have two equations and three unknowns.

So to solve this we can assume additional R3 = 10K and solve for R2 and Vref.

Hmmm... I'm not sure where's the missing point, but i get the result : Vref=0,5Volts and R2=2,33kOhms.

I explain

Vcc=+15Volts, Vee=-15Volts
Vref=?
R1=?
R2=?
R3=10KOhms
Vutp=4Volts
Vltp=-3Volts


Vutp= Vref*[R3/(R2+R3)]+Vcc*[R2/(R2+R3)]=Vref*[10k/(R2+10k)]+15[R2/(R2+10k)] <=> Vref= {[4-15*[R2/(R2+10k)]}/10k]}*[R2/(R2+10k)] (1)

Vltp=Vref*[R3/(R2+R3)]+Vee*[R2/(R2+R3)]=(1)={[4-15*[R2/(R2+10k)]}/10k]}*[R2/(R2+10k)]*[R3/(R2+R3)]-15*[R2/(R2+R3)] <=> R2=2,33kOhms (2)


Replacing R2 with the (2) result in equation (1) i get : Vref=0,5Volts

Very strange

4V = Vref*[10K/(R2+10K)] + 15V*[R2/(R2+10K)]

4 = (Vref*10K + 15*R2) / ( 10K + R2) ---> multiply by ( 10K + R2)

4*( 10K + R2) = (Vref*10K + 15*R2)

40K + 4R2 = Vref*10K + 15R2 ---> subtracted 15R2

40K - 11R2 = Vref*10K ---> divided by 10K

(40K - 11R2)/10K = Vref

Vref = (40K - 11R2)/10K (1)

OK now second equation

-3V = Vref*[10K/(R2+10K)] + (-15V)*[R2/(R2+10K)]

-3 = (Vref *10K - 15R2) /(R2 + 10K)

-3*(R2 + 10K) = (Vref *10K - 15R2)

-30K - 3R2 = 10K*Vref - 15R2

12R2 - 30K - 10K*Vref

12R2 = 10K*Vref + 30K

R2 = (10K*Vref + 30K) / 12 (2)

so we have

Vref = (40K - 11((10K*Vref + 30K) / 12))/10K

Vref = (40k - 11/12*(10K*Vref +30K))/10K

10K*Vref = 40k - 11/12*(10K*Vref +30K)

And after some several additional steps i get

Vref = 15/23 = 0.652174V

And

R2 = (10K*15/23 + 30K) / 12 = 70/23 = 3.04347826KΩ

And simulation confirms my result.

And for Vref= 0.5V and R2 = 2.33K

Vt1 = 3.2V
Vt2 = - 2.42V