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Schmitt trigger

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mrkbuddy

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Hello guys
can any one answers me the the questions below with respect to the circuit diagram. thanks in advance.

a) When Vin= -12 the green LED glows. What is the output voltage polarity and magnitude approx. and why does the red LED not glow?

b) if Vin is now altered from -12V towards +12V at what value of Vin will the output voltage alter in polarity to make the red LED glow

c) Explain how the 7k and 6k resistors provide feedback to control the values of input voltage necessary to alternate illumination of the LED's

RedLED.jpg
 
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You seem to have forgotten to include details of your attempt(s) at answering the questions. Try doing the work, then ask for help.
 
a) When Vin= -12 the green LED glows. What is the output voltage polarity and magnitude approx. and why does the red LED not glow?

Output voltage polarity is + 9.41v. I checked the output voltage polarity by using a multi meter to measure the voltage.
The amplifier is connected in inverting mode this changes the polarity from – to + that’s why only the green LED glows because the green LED is connected in forward bias mode. Whilst the amplifier saturates it has to consume some of the voltage for it to operate correctly.

b) if Vin is now altered from -12V towards +12V at what value of Vin will the output voltage alter in polarity to make the red LED glow

Vin is 12V and I have tested the output voltage using a multi-meter and the voltage polarity shows – 9.21 V. the red LED only lights up because its connected in reverse bias in which the amplifier is connected in the inverting mode so this changes the polarity from + to – Value of Vin is One of the characteristic of a diode it has large input impedance only very little current is used for the op amp to work

c) Explain how the 7k and 6k resistors provide feedback to control the values of input voltage necessary to alternate illumination of the LED's

7k and 6k resistors provide negative feedback to prevent switching back to the other state until the input passes through a lower threshold voltage, thus stabilizing the switching against rapid triggering by noise as it passes the trigger point.

i am not sure am i going on the right direction or not.
 
a) When Vin= -12 the green LED glows. What is the output voltage polarity and magnitude approx. and why does the red LED not glow?

Output voltage polarity is + 9.41v. I checked the output voltage polarity by using a multi meter to measure the voltage.
The amplifier is connected in inverting mode this changes the polarity from – to + that’s why only the green LED glows because the green LED is connected in forward bias mode.
Ok. The output polarity is simply the sign of the output voltage. In this case the output polarity is positive w.r.t. ground/0V. The output magnitude you've stated to be +9.41V.

The green LED is forward-biased, i.e. the voltage at its anode is higher than that at its cathode. Because this voltage is more than the forward voltage of the LED (Vf ~ 2V), there will be current flowing through the LED and the LED will be illuminated.

The red LED is reverse-biased, i.e. the voltage at its anode is less than that at its cathode. No current flows and the LED remains dark.

Whilst the amplifier saturates it has to consume some of the voltage for it to operate correctly.
The amplifier will drop some voltage in it's output driver components, yes.

b) if Vin is now altered from -12V towards +12V at what value of Vin will the output voltage alter in polarity to make the red LED glow

Vin is 12V and I have tested the output voltage using a multi-meter and the voltage polarity shows – 9.21 V. the red LED only lights up because its connected in reverse bias in which the amplifier is connected in the inverting mode so this changes the polarity from + to – Value of Vin is One of the characteristic of a diode it has large input impedance only very little current is used for the op amp to work
When the output voltage is -9.21V, the red LED is not reverse-biased; it will be forward-biased.

The question asks at what input voltage the output will change from a positive to negative voltage. This will involve calculations based on the 6k and 7k resistors, as well as the measured output voltage. The output will become negative when the voltage on the -ve input become greater than the voltage on the +ve input.

c) Explain how the 7k and 6k resistors provide feedback to control the values of input voltage necessary to alternate illumination of the LED's

7k and 6k resistors provide negative feedback to prevent switching back to the other state until the input passes through a lower threshold voltage, thus stabilizing the switching against rapid triggering by noise as it passes the trigger point.
There is no negative feedback in that circuit. The positive feedback has the effects you described. There will be two threshold voltages; one when the output voltage is positive and another when it is negative.
 
The circuit damages the LEDs because the maximum allowed reverse-bias voltage for an ordinary LED is only 5V.
Adding a normal diode in series with each LED blocks the high reverse bias voltage that is too high.
 
thanks alot dougy83, you been a big help mate. i really appreciate your help. stay bless ..

thanks to you too audioguru. you guys rocks. :)
 
thanks alot dougy83, you been a big help mate. i really appreciate your help. stay bless ..

thanks to you too audioguru. you guys rocks. :)
You are welcome, Mate. (Shipmate?) We are not from England and have never worked on a boat.
Sorry, my wife is my mate, not you.
 
To find threshold voltage you just need assume that opamp is in saturation, its doesn't mater if its is positive saturation or negative sat.
And then you need to find the value of Vin for which voltage on non-inverting input is equal input voltage.
And this Vin is equal threshold voltage.
Next you repeat the procedure but you change saturation voltage.
 
Jony,
I have to consider two cases: positive saturation (+Vcc) or negative saturation (-Vcc)?
How can you know it is certainly in saturation?
 
Jony,
I have to consider two cases: positive saturation (+Vcc) or negative saturation (-Vcc)?
yes, you have to consider both of this cases.

How can you know it is certainly in saturation?
Because we have a positive feedback here.

As the output voltage increases, the voltage on the (+) input also increases. The output of the op amp will go toward positive direction. This rising potential, through 6K resistor (R2), further increases the potential on the (+) input pin, in our case, the output will drive all the way to its positive saturation limit (positive supply voltage).
So as you can see we have a positive feedback here.
 
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