Question about hyestrisis

Discussion in 'General Electronics Chat' started by TAKYMOUNIR, Nov 9, 2012.

  1. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    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
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    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.
     
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  3. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    voh=15 and vol=0 so what is my threshold voltage
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    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.
     
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  5. JMac3108

    Active Member

    Aug 16, 2010
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    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.
     
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  6. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    ohms law will work this out, 5V /200k = I , Ix 100M=V, which is about 2500Volts.
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    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.
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    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: Nov 10, 2012
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  9. Ron H

    AAC Fanatic!

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

    Moderator

    Mar 31, 2012
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    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.
     
  11. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    when the voltage is more than 5.2555kv vpin3 will go up or will stay at 5v
     
  12. Ron H

    AAC Fanatic!

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    What do you think, and why?
     
  13. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    How do you get a negative voltage , the ic has only got a 0 to 15V supply?
     
  14. WBahn

    Moderator

    Mar 31, 2012
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    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.
     
  15. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    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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