Question about hyestrisis

Thread Starter

TAKYMOUNIR

Joined Jun 23, 2008
352
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
 

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WBahn

Joined Mar 31, 2012
29,979
You need to analyze the circuit twice.

First, assume that the output voltage is high. What will it be (check the data sheet)?
Now, what value of Vin will result in the voltage on pin 3 being equal to the voltage on pin 2 (i.e., 5V)?

Next, assume that the output voltage is low. Again, what will it be (check the data sheet)? Again, what value of Vin will result in the voltage on pin 3 being equal to the voltage on pin 2?

These two voltages are your switching thresholds.

NOTE: Because of the high resistances you have in this circuit, be sure to compare the currents in your feedback path to the input bias current of the opamp. Unless the bias current is at least 10x smaller (preferably 100x), you will need to take it into account.
 

Thread Starter

TAKYMOUNIR

Joined Jun 23, 2008
352
you need to analyze the circuit twice.

First, assume that the output voltage is high. What will it be (check the data sheet)?
Now, what value of vin will result in the voltage on pin 3 being equal to the voltage on pin 2 (i.e., 5v)?

Next, assume that the output voltage is low. Again, what will it be (check the data sheet)? Again, what value of vin will result in the voltage on pin 3 being equal to the voltage on pin 2?

These two voltages are your switching thresholds.

Note: Because of the high resistances you have in this circuit, be sure to compare the currents in your feedback path to the input bias current of the opamp. Unless the bias current is at least 10x smaller (preferably 100x), you will need to take it into account.
voh=15 and vol=0 so what is my threshold voltage
 

WBahn

Joined Mar 31, 2012
29,979
At what point are you going to start making an effort to answer your own questions?

Do your best, show your work, and we can proceed from there.
 

JMac3108

Joined Aug 16, 2010
348
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.
 

Ron H

Joined Apr 14, 2005
7,063
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.:rolleyes:
Being a Schmitt trigger, there are two thresholds.
Here's a hint: One of them is negative.
 

Jony130

Joined Feb 17, 2009
5,487
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
 
Last edited:

Ron H

Joined Apr 14, 2005
7,063
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.
 

WBahn

Joined Mar 31, 2012
29,979
Yeah, that's what I got. We were trying to get TAKYMOUNIR to show his work.
Why should he? As long as there are people that are willing to do all his thinking and work for him, there is no reason for him to even attempt to learn to think or work things out on his own. That's been demonstrated in thread after thread.
 

Thread Starter

TAKYMOUNIR

Joined Jun 23, 2008
352
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
when the voltage is more than 5.2555kv vpin3 will go up or will stay at 5v
 

WBahn

Joined Mar 31, 2012
29,979
How do you get a negative voltage , the ic has only got a 0 to 15V supply?
When the output is 0V, you have to have a huge positive voltage in order to get the voltage at the non-inverting input up to 5V.

Well, when the output is 15V, you have to apply a huge negative voltage to get the voltage at the non-inverting input down to 5V.
 

Jony130

Joined Feb 17, 2009
5,487
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.
 
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