# Schmitt trigger

Discussion in 'Electronics Homework Help' started by mrkbuddy, May 2, 2013.

1. ### mrkbuddy

3
0
May 2, 2013

Hello everyone
can any one answer 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?

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.

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Last edited: May 2, 2013
2. ### duke37

5,364
769
Jan 9, 2011

The 7k and 6k resistors do NOT give negative feedback, it is positive.

You may have popped off the red led. There is a limit to the reverse voltage which a led will stand. Since you have plenty of voltage available the best way of protecting them is to put a diode in series (1N4148).

You do not give details of the amp, a standard op-amp will source and sink current but a comparator will normally only sink current. This may be why only one led will light.

3
0
May 2, 2013

4. ### Merlin3189

250
69
Aug 4, 2011
= Right. I'd have guessed nearer +13V, but you are drawing 40mA, so a lower Voltage might be expected.
= But the question was, at what value of Vin? So see below.
= as said before, it is positive feedback, otherwise good.

OK, in a) what is the Voltage on the + input? It is +9.41 * 6/(6+7) the resistors providing a potential divider of the output Voltage. [By the way, can you see why I guessed the output Voltage to be 13V!] Anyhow, that must come to about +4.5V, in my head.
So the input must reach that +4.5V to start to switch the opamp.
When it does, the output will start to fall, which will also lower the Voltage on the + input to less than 4.5V, so the amp will be well and truely switched over.
Then the output is -9.21V, so the + input will be at -9.21 * 6/(6+7) about -4.5V
So now the input must fall to -4.5V to start the switching back.

The opamp switches one way at +4.5V and the other way at -4.5V a wide gap.
If you change the resistors, say make the 6k into 2k, these Voltages become +2V and -2V, closer but still a gap.
If you made the 6k into 0k, then the Voltages become +0V and -0V the same! You've lost your positive feedback and the opamp switches every time the signal crosses 0V.
By choosing the (non-zero) values of the resistors, you set the gap between the switching points - known as the hysteresis.