Hello,

I am struggling to calculate the Schmitt Trigger for this circuit. I got 3.75V for when the switch occurs. Could someone please explain how to do it etc.

R1 is 18K and the input voltage is 5V

Thanks

 

hi 25,
Check this PDF.
E

 

hi 25,
Check this PDF.
E

I get 2V, I need to calculate figure 5 where it says OUTPUT, those two points VCC and VSS. Vcc x R2/R1+R2 can't be correct for my circuit ?

I used this video before posting here and got 3.75V. I don't if it is correct or not

 

hi,
What is the OPA type, the supply voltages.

E

 

hi,
What is the OPA type, the supply voltages and the value of R1.
E

Its an LM741, Supply voltage of 5V and the value of R1 is 18k. It was 8.2k for negative resistance now working on positive resistance by changing the value of R1 to 18k

 

An LM741 won't work on a single supply of 5V.
You need a single supply amp, such as the LM339 or LM393.

 

An LM741 won't work on a single supply of 5V.
You need a single supply amp, such as the LM339 or LM393.

I used a powersupply and connected the positive to the negative etc

 

But your circuit is not a standard Schmitt Trigger. This circuit looks more like a Negative impedance converter.
Also, your supply voltage is too low for LM741.

 

But your circuit is not a standard Schmitt Trigger. This circuit looks more like a Negative impedance converter.
Also, your supply voltage is too low for LM741.

Not sure, the professor gave us this to do lol. By changing R1 to 18k it becomes a positive circuit i believe. The 3rd picture in the initial post is what i need to calculate. The two points, basically when the switch occurs

 

I only need to calculate the Vcc and Vss in the picture shown. any ideas

 

You may have misunderstood the power supply comment. The 741 will not work from +5V and GND. It needs something like +12V and -12V, which is called a bipolar power supply.
Also it seem strange beyond belief that you are taking the output from the Non-inverting Input of the amplifier. Any downstream circuitry will affect the operation. the output should come from pin 6 on your diagram. What you need to do is take a step back and understand what is going on. You do not need to calculate Vcc and Vss; then are the given power supply voltages. what you need to calculate are the threshold voltages along the horizontal axis where the output changes state. Every time the output changes state, it establishes a new and different threshold for the next change of state. Wash, rinse, repeat.

 

You may have misunderstood the power supply comment. The 741 will not work from +5V and GND. It needs something like +12V and -12V, which is called a bipolar power supply.
Also it seem strange beyond belief that you are taking the output from the Non-inverting Input of the amplifier. Any downstream circuitry will affect the operation. the output should come from pin 6 on your diagram.

This is the first thing we did and what it says on the sheet "Use VCC=5V. With a sine wave input, set around 200Hz, obtain the „static‟ transfer curve (Vout/Vin) by using the scope on XY." So you're saying this is not correct? This is my professor asking us to do this, and if he's wrong im going to be very frustrated.

The step i am at is as follows "Change R1 to 18k and note that the effect of this change on the output. Explain what is happening for this value change of R1 and develop a general explanation of the circuit behaviour if R1 exceeds R2 (12K)." now, R1 was 8.2k, changing it to 18k switches it into a positive feedback circuit. That's what we've been told and how we see the circuit via oscilliscope.

 

To do that you need to know the positive and negative output voltage when opamp is in saturation.
For example, if we assumed Vo (opamp output voltage) can reach Vcc and Vee (+5V and -5V).
We can find the threshold voltage:

If we have a +5V at the output we will have +2.5V at inverting input. So, to find the threshold voltage we need to find Vin that will give us +2.5V at the noninverting input.

(5V - 2.5V)/12kΩ = (2.5V - Vin)/18kΩ

And if we solve this for Vin we get: -1.25V

And the second case when opamp output is at -5V (negative saturation)
(-5V - (-2.5V))/12kΩ = ((-2.5V) - Vin)/18kΩ we get +1.25V

And this is the end of ballpark calculations.

 

th

This is the first thing we did and what it says on the sheet "Use VCC=5V. With a sine wave input, set around 200Hz, obtain the „static‟ transfer curve (Vout/Vin) by using the scope on XY." So you're saying this is not correct? This is my professor asking us to do this, and if he's wrong im going to be very frustrated.

The step i am at is as follows "Change R1 to 18k and note that the effect of this change on the output. Explain what is happening for this value change of R1 and develop a general explanation of the circuit behaviour if R1 exceeds R2 (12K)." now, R1 was 8.2k, changing it to 18k switches it into a positive feedback circuit. That's what we've been told and how we see the circuit via oscilliscope.

The original diagram does not show a 741 it has a generic symbol there. How did we start talking about that particular part? I see it was post #5 where you mentioned it. If your professor told you to use a 741 with Vcc=+5V and Vss=GND then he seriously misled you. That part will not work in that configuration. There are many other parts that will work in that configuration, so what exactly is the story and how are we supposed to assume facts not in evidence?

I see in post #12 an unrecognizable symbol in front of "5V" If that symbol was the plus/minus symbol that makes more sense. The 741 may work with Vcc=+5VDC and Vss=-5VDC. So maybe a moderator can explain why the board can no longer render a plus/minus symbol? Hows about it?

 

th

The original diagram does not show a 741 it has a generic symbol there. How did we start talking about that particular part?

Not sure but the op amp we used is an LM741. Seems as though if it is incorrect in somewhat way's it's for us to figure stuff out. Maybe it is done on purpose? i don't have a clue, unfortunately

 

Not sure but the op amp we used is an LM741. Seems as though if it is incorrect in somewhat way's it's for us to figure stuff out. Maybe it is done on purpose? i don't have a clue, unfortunately

See post #14. There was a problem in your earlier post that the plus/minus symbol was not rendered correctly.

 

th

The original diagram does not show a 741 it has a generic symbol there. How did we start talking about that particular part? I see it was post #5 where you mentioned it. If your professor told you to use a 741 with Vcc=+5V and Vss=GND then he seriously misled you. That part will not work in that configuration. There are many other parts that will work in that configuration, so what exactly is the story and how are we supposed to assume facts not in evidence?

I see in post #12 an unrecognizable symbol in front of "5V" If that symbol was the plus/minus symbol that makes more sense. The 741 may work with Vcc=+5VDC and Vss=-5VDC. So maybe a moderator can explain why the board can no longer render a plus/minus symbol? Hows about it?

Oh yeah not sure why it hasnt. Its +-5v. So is this correct or there still an issue between the config you mentioned.

 

hi 25,
This is an LTSpice sim of your circuit but using +/-5V supplies, as you can see it follows @Jony130 calculations.
E

 

hi 25,
This is an LTSpice sim of your circuit but using +/-5V supplies, as you can see it follows @Jony130 calculations.
E

I initially got 2V. i can't figure out how he got 1.25 though from his calculation, unless im missing something?

 

I initially got 2V. i can't figure out how he got 1.25 though from his calculation, unless im missing something?

Look:
(5V - 2.5V)/12kΩ = (2.5V - Vin)/18kΩ

(2.5 - Vin)/18 = 2.5/12

(2.5 - Vin) = (2.5*18)/12

(2.5 - Vin) = (2.5*18)/12

2.5 - Vin = 3.75

Vin = 2.5 - 3.75 = -1.25V