Question about hyestrisis

In this circuit voltage on pin 2 is 5 vd and vin change from 0 to 10 Kvdc so when vin is low vo is low and when vin increase vpin3 start to go up until pass 5 v so in this case vo will be high
so what the value of v pin3 when vo is high and how pin3 changE
when vin change

 

voh=15 and vol=0 so what is my threshold voltage

 

Redraw the circuit for both conditions. When Vout is low, R2 goes to ground - draw it that way and analyze it. When Vout is high, R2 goes to VCC - again, draw it and analyze it.

 

ohms law will work this out, 5V /200k = I , Ix 100M=V, which is about 2500Volts.

 

Not even close. Seat-of-the-pants math doesn't work in this case.
Being a Schmitt trigger, there are two thresholds.
Here's a hint: One of them is negative.

 

For VoL = 0V we need 5V at pin 3 to circuit to switch his high state (to Voh = 15V).
(5V - 0V)/200K = 25μA this current will flow through R3 and will cause voltage drop across R3 resistor equal to 0.5V. So the voltage across R6 is equal 5.5V.
IR6 = 5.5V/200K = 27.5μA so the Vin = ((IR3 + IR6)*R5) + 5.5V = 5.2555KV
Now the output voltage is equal to Voh = 15V. And now voltage at pin 3 need to be lower then 5V to circuit change his state.

Now IR2 = (15V - 5V)/200K = 50μA and IR3 = 50μA So the voltage across R6 is equal 4V.
IR4 =4V/200K = 20μA and IR5 = (IR2 - IR4) = 30μA so the Vin = 4V - IR5*R5 = 4V - 3KV = -2.996KV
So the circuit will switch from LOW to HIGH at 5.25KV and from HIGH to LOW at -2.99KV

 

Yeah, that's what I got. We were trying to get TAKYMOUNIR to show his work.

 

when the voltage is more than 5.2555kv vpin3 will go up or will stay at 5v

 

What do you think, and why?

 

How do you get a negative voltage , the ic has only got a 0 to 15V supply?

 

Well as you can see my answer was not very clear for TAKYMOUNIR.
Simply because he don't understand how op amp work and also my analysis is purposely rather peculiar. Also as you can see Dodgydave also have a problem with this simply circuit.