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

1. ### TAKYMOUNIR Thread Starter Active Member

Jun 23, 2008
351
1
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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2. ### WBahn Moderator

Mar 31, 2012
17,716
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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
351
1
voh=15 and vol=0 so what is my threshold voltage

4. ### WBahn Moderator

Mar 31, 2012
17,716
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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
349
66
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
4,969
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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
7,050
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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.

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
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!

Apr 14, 2005
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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
351
1
when the voltage is more than 5.2555kv vpin3 will go up or will stay at 5v

12. ### Ron H AAC Fanatic!

Apr 14, 2005
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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
17,716
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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
3,957
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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.