Calculating V in an Op-Amp Circuit with Constant Input: Schmitt Trigger Question

[StripesUK]

Homework Statement

FIGURE 2 shows an op-amp circuit with a constant input voltage V connected to the inverting input. If the output
voltage is +2.4 V, calculate the value of V.

Schmitt trigger values:
R1=4.8 ohms
R2=2 ohms
V OUT= +2.4V

Homework Equations

V+=(R1*V SAT)/(R1+R2)

The Attempt at a Solution

V+=11.52/6.8=1.69V

Now this is the Reference Voltage going to the non inverting terminal but I'm lost as to where to go from here, this would indicate that V is less than 1.69?
Any help to point me in the right direction would be much appreciated. This is the first time I've studied the electronics side of things and it's safe to say I'm struggling a little.

 

[gneill]

Hi StripesUK, Welcome to Physics Forums.

Where is "FIGURE 2"? We'll need to see the circuit if we're to check your attempt or offer advice.

 

[StripesUK]

Sorry about that. Please see attached.

 

[gneill]

Why don't you try assuming some finite gain "A" for the op-amp and writing an expression for the output voltage for a given input voltage? Then you can see what happens when the gain is allowed to be very large. Don't assume any non-linear (Schmitt or comparator) qualities at the outset.

 

[StripesUK]

I think I have it now...

So if,

V2/V1=-R2/R1

then,

V1=(R1*V2)/-R2
?

 

[gneill]

You haven't defined what V1 and V2 are. Can you show more of your work in detail?

 

[StripesUK]

V2=2.4V
R1=4.8 ohms
R2=2 ohms

V= \frac{R1V2}{-R2}

V= \frac{4.8*2.4}{-2}=-5.76VI derived all this from the equation, (V1 being the voltage(V) that I am looking for);

\frac{V2}{V1}= \frac{-R2}{R1}

Both sides can be used to find the gain.

 

[gneill]

V2=2.4V
R1=4.8 ohms
R2=2 ohms

Those should be in kilo ohms (although it won't change the voltage results).

V= \frac{R1V2}{-R2}

I don't see where this comes from. Why the negative sign on R2?

In this circuit feedback is provided from the output to the input via a voltage divider (R1 & R2). The result is to "lift" the potential of the + input of the op amp by a fraction of the output voltage. If the output happens to be positive, then this "lift" is also positive. The very large gain of the op amp will ensure that this "lift" will be driven towards matching the potential at the - input of the op amp (enforcing a zero potential difference between the inputs).

You should be able to derive this result by first assuming that the op amp has some finite gain A and writing the circuit equations to solve for $V_{out}$ in terms of $V_{in}$, then taking the limit as A gets very large.

 

[jim hardy]

uhhh,, mr stripes

as drawn
that opamp has positive feedback , so it can't drive its inputs equal unless its gain A were to be negative.

mr gniel has hinted at that

Why don't you try assuming some finite gain "A" for the op-amp and writing an expression for the output voltage for a given input voltage? Then you can see what happens when the gain is allowed to be very large. Don't assume any non-linear (Schmitt or comparator) qualities at the outset.

your calculation for V+ in post 1 is correct, to the digits you showed

 

[boxofissues]

Hi guys, I too am struggling to get my head around this question and was wondering if you could see if I'm on the right path to my final answer and give me a bit of guidance if I am way off.

My attempt so far:

Vout = +2.4V
Rf = 2k ohms
R1 = 4.8k ohms
Vin = ?

First I calculated V+ by using:

V+ = (R1*Vout)/(R1+Rf)
V+ = 11520/6800
V+ = 1.694V

I then calculated the gain, as stated above because it has positive feedback the gain should be negative therefore I have used:

A = -Rf/R1
A = -0.416

I then used the following and have transposed to try and calculate Vin (or V-):

Vout = A(V+ - V-)

Transposing for Vin I came out with:

Vin = V+ - (Vout/A)
Vin = 1.694 - (2.4/-0.416)
Vin = 1.694 - (-5.76)
Vin = 7.454V

So I believe this is my final answer however I am not totally confident as I was expecting Vin < V+ due to Vout being a positive value, my only explanation is that my initial assumption was wrong and Vout is actually its lower limit but again I am not 100% sure and this is where my confusion lies. Any help or guidance would be greatly appreciated as I would love to get my head around this.

 

[gneill]

Your answer is not correct. You've made an assumption about the gain being negative which is not necessarily true, and you've used a formula for the circuit gain that is not correct.

Don't calculate the actual gain of the circuit. Just assume that the op amp has an open loop gain of some value A (where A is a positive number). Then analyze the circuit and determine what happens when A is very large.

 

[boxofissues]

So if I just give the gain a practical value such as:

A = 10^5

Then use this value in:

Vin = 1.694118 - (2.4/10^5)
Vin = 1.694094

Vin is now less than V+ but only by a very small amount but this is still enough for Vout to be at its high limit. Is this more accurate or am I still going in the wrong direction? Apologies if I'm missing the obvious but I'm finding this topic a little difficult to wrap my head around.

 

[gneill]

So if I just give the gain a practical value such as:

A = 10^5

Then use this value in:

Vin = 1.694118 - (2.4/10^5)
Vin = 1.694094

Vin is now less than V+ but only by a very small amount but this is still enough for Vout to be at its high limit.

No, it is not. You have assumed a finite gain for the op amp, so you won't get infinitely magnified differences. What you've discovered is that the input voltage will be very close in value to the potential at the op amp - input, even with a finite gain of 10⁵.

Is this more accurate or am I still going in the wrong direction? Apologies if I'm missing the obvious but I'm finding this topic a little difficult to wrap my head around.

You should analyze the circuit purely symbolically first, without plugging in numbers or making assumptions about the op amp open loop gain or expressions for the circuit's closed loop gain. Give the op amp an open loop gain of A and find the relationship between Vin and Vout. Then see what happens when A becomes very large.

 

[boxofissues]

Give the op amp an open loop gain of A and find the relationship between Vin and Vout. Then see what happens when A becomes very large.

Again, this may just be my stupidity but I thought I had already seen what happens when A becomes large by using the value 10^5? I can see that the closer my value of V- is to V+ then the higher the value of gain A is? I just don't understand how I am supposed to calculate Vin without an actual gain value?

 

[gneill]

The actual circuit gain comes from analyzing the circuit (which I'm hoping to coax you into doing), but your approximation using a fixed value for the gain of the op amp should give you a clue that you can expect the difference between the V+ and V- potentials to tend to zero as the op amp open loop gain grows very large.

 

[jim hardy]

yes, make yourself a table
columns for A and Vin

"A" being gain of the amplifier itself whic you might vary from 0.1 to 10^6 in steps of ten, ie 0.1, 1, 10, 100 1000 etc
you'll see gneill's point that if the circuit is "operating", ie output not saturated, amplifier's inputs must be different by Vout/A .

Now
you mentioned A of Rf/R1
make yourself aware that there's two different gains
one is the gain of the raw amplifier by itself, usually quite high for op-amps, like 10^5

the other is the gain of a circuit encompassing an op-amp which the circuit designer makes whatever he wants by selecting resistors for feedback and input. That one we call "closed loop gain" .
So long as the amplifier is able to push its inputs equal, it can be said to be "operating" that's why we call it an operational amplifier circuit.

In the circuit you presented
the amplifier is unable to push its inputs equal
look - raising V will lower Vout driving the inputs farther apart
so that question is either badly phrased or a trick question,
i'd say an unfair one to give to a beginner
we must rigorously do the arithmetic to get the answers
but if you build that circuit and apply input that you can adjust with a knob
you will be unable to hold Vout steady. It's a math dream only.

That said:
Swap the + and - inputs \ and it'll work beautifully, grab a LM324 and try it.

old jim

 

[Tom.G]

Swap the + and - inputs \ and it'll work beautifully, grab a LM324 and try it.

Then it wouldn't be a schmitt trigger.

(EDIT)
StripesUK
I'm pretty sure your original answer of <1.69V is correct. Reading the other posts, it seems at least some assume an inverting amplifier rather than a schmitt trigger.

 

[jim hardy]

Then it wouldn't be a schmitt trigger.

Right !

what i should have said

That said:
Trying to balance the inputs to get constant output from a Schmitt Trigger is like balancing a pencil on its point.
Swap the + and - inputs \ and it'll work beautifully as a linear ampifier, output will be input X (6.8/4.8), grab a LM324 and try it.

 

[StripesUK]

The only other equation I can find involving Gain, Vin, and Vout is,

Vo=G(V+-V-)

Transposed...

V-=V+ - \frac {Vo}{G}

This would mean that V- is proportional to the Gain. The larger the Gain the larger V- becomes, this would also be the same for Vout.
Am I now just looking for the correct calculation of the circuit gain G?

 

[jim hardy]

This would mean that V- is proportional to the Gain. The larger the Gain the larger V- becomes, this would also be the same for Vout.

Do you see now that circuit is counter-intuitive?
They have fixed Vout at 2.4 volts and ask you to calculate the Vin that would cause that Vout .

That's working the circuit backward. Adding insult to injury, the circuit is unstable .
Very confusing and unfair to treat a beginner that way.

V−=V+−VoGV-=V+ - \frac {Vo}{G}

This would mean that V- is proportional to the Gain. The larger the Gain the larger V- becomes, this would also be the same for Vout.
Am I now just looking for the correct calculation of the circuit gain G?

did you mean "The larger the Gain the more nearly equal V- and V+ must become ? " That's required if the circuit is to not saturate.

The term " gain " for a schmitt trigger circuit is meaningless because its output is not linear but bistable.

http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/positive-feedback/

 

[StripesUK]

did you mean "The larger the Gain the more nearly equal V- and V+ must become ? " That's required if the circuit is to not saturate.

The term " gain " for a schmitt trigger circuit is meaningless because its output is not linear but bistable.

Yes, sorry I can see that is correct from my different calculations of the Gain.

So now because the feedback is positive the calculation of the Gain from the circuit would be G=1+ \frac {R2}{R1}?

 

[jim hardy]

So now because the feedback is positive the calculation of the Gain from the circuit would be G=1+ $\frac {R2}{R1}$?

If gain =$\frac {Δoutput}{Δinput}$

what is the Δin required to cause a one volt change at output ?

G X ( Vout X (4.8/6.8) - Vin ) = Vout

Vin = Vout/G + Vout X (4.8/6.8)
Vin = Vout X (1/G + 4.8/6.8)

so the circuit can exist precariously balanced at any input Vin = Vout X (1/G + 4.8/6.8)
but if Vin moves one single nanovolt away from that value
positive feedback will move the + input AWAY from that balance , causing more imbalance
by the time Δoutput reaches our 1 volt, imbalance at input has grown from our nanovolt to (1/G + 4.8/6.8 +1X10^-9) volts
that's what positive feedback does, instead of balancing things it further unbalances them

so our gain of $\frac {Δoutput}{Δinput}$ has a denominator that's made infinitesimal by positive feedback
that's what positive feedback does, raises gain.

Gain infers linear operation
and a schmitt is not a linear circuit.

I still think this was an unfair problem to give to beginners.

How did you make those nice looking fractions ?

 

[the-brammo]

http://www.electronics-tutorials.ws/opamp/op-amp-multivibrator.html

I think this site explains it pretty well, try to understand the hysteresis curve/plot. The input could be a varying value, maybe you should think about using an inequality?

I think the answer warrants a more qualitative answer.

Edit: Wrong Link

 

[jim hardy]

I think this site explains it pretty well, try to understand the hysteresis curve/plot. The input could be a varying value, maybe you should think about using an inequality?

i don't see a schmitt on that pageTI doesn't even mention "gain" and i think it's a mistake to apply the term to a schmitt, gain is infinite... sum of a divergent series.
http://www.ti.com/lit/an/scea046/scea046.pdf

Best way to learn them is get a 555, tie pins 2&6 together as input and you have a handy schmitt that'll interface with most any logic...

wiki has a decent page
https://en.wikipedia.org/wiki/Schmitt_trigger

What's gain at those vertical transitions ?

 

[the-brammo]

i don't see a schmitt on that page

Did you click on it before I edited the link? I originally posted an incorrect link.

 

[StripesUK]

I'm struggling to see how i calculate the value of G still. Are you saying there is no way of doing so? If that's the case perhaps I've gone drastically wrong somewhere earlier in the question?

I believe I'm correct in saying this is the right equation to find the answer?

Vo=G(V+-V-)

Here's how to create better looking equations.
https://www.physicsforums.com/help/latexhelp/

 

[StripesUK]

The input could be a varying value, maybe you should think about using an inequality?

I think the answer warrants a more qualitative answer.

So my previous answer of <1.69V could be correct?

 

[jim hardy]

Did you click on it before I edited the link? I originally posted an incorrect link.

Aha ! Now it's a multivibrator page , looks perfect.

 

[jim hardy]

Let's make clear the difference between
"An Operational Amplifier"
and
"An Operational Amplifier Circuit"

An "operational amplifier" is an amplifier with high but finite gain perhaps 10^5 or 10^6 volts per volt.
It is represented by that ubiquitous triangle with 2 inputs and one output.
Its gain is given on its datasheet as Avol , acronym for Amplification, volts per volt, open loop.

An "operational amplifier circuit" is such an amplifier wrapped in a circuit that let's it force its inputs equal to one another by means of feedback.
It is the duty of the circuit designer to wrap the amplifier with such a circuit.
The gain of the 'operational amplifier circuit' is set by the circuit with which the designer has surrounded the 'operational amplifier".
Usually that gain is of the form (Zfeedback) / (Zinput) which defines the mathematical operation that the 'operational amplifier circuit ' will perform on an input. It's called "closed loop gain" of the operational amplifier circuit.

We have rather abused G by not clarifying whether we meant open loop gain or closed loop gain.
The operational amplifier at the heart of a schmitt trigger circuit has Avol and the more of it the better.
But how do you define the 'gain' of a bistable circuit ?
I submit that "Gain of a schmitt trigger" is an oxymoron, undefinable or at best infinite. Bistable is a discontinuous math function.

So, once again , your exercise postulates a condition that's mathematically possible but not practically achievable
unless Vsat is 2.4 volts (which was not given)
in which case any Vin less than 2.4 X(4.8/6.8) volts will keep output saturated
and that is this many volts

so, yes, V < 1.69 is a plausible answer ..
but it's not the one i would have inferred from this question:

Homework Statement

FIGURE 2 shows an op-amp circuit with a constant input voltage V connected to the inverting input. If the output
voltage is +2.4 V, calculate the value of V.

because it needn't be constant, just < 1.69etc

was i off track much ?

hope i didnt hurt you on this one.

old jim

 

[jim hardy]

The input could be a varying value, maybe you should think about using an inequality?

He nailed it in post 23 .

@the-brammo - suggestions for a formula?

 

[StripesUK]

was i off track much ?

hope i didnt hurt you on this one.

old jim

Absolutely not, my stress levels have been sitting nice and high but you have been great in terms of helping me understand the subject more. I find my study material never supplies anywhere near enough information so places like these forums are invaluable!

Many thanks!

Just to confirm though, there is no equation for an accurate input voltage of a schmitt trigger?

 

[jim hardy]

Just to confirm though, there is no equation for an accurate input voltage of a schmitt trigger?

Input voltage that'll cause what condition?

It is easy to calculate the voltage at which the schmitt will switch output states.
It is not physically possible to apply an input voltage that will hold its output in between those two states, in its linear region , as i interpreted your original post.
you can calculate such a voltage but it is not practical to measure let alone control to the required precision
because changing the applied voltage in the last digit by just one

will start the positive feedback loop on its merry way to saturation.

I'm not dodging your question just trying to answer it precisely.

 

[StripesUK]

Understood. Thanks for all your help and patience!

 

[jim hardy]

Understood. Thanks for all your help and patience!

Feedback behavior does not come intuitively to some people, myself for instance. I struggled for years.

Pardon the un-academic next paragraph
i learn a lot just watching everyday stuff around me, like burgers on the charcoal grill.
Ever heard the expression "The fat's in the fire now! " ? It means something calamitous is imminent.
Burgers- The fat drips down and burns hot, flames erupt causing more out-juicing which feeds more fat into the flames and you wind up with charcoal-burgers.
That's positive feedback.
Positive feedback usually results in a departure from linear behavior and often destructive failure like that galloping bridge in Tacoma... or a runaway fire in your barbecue grill...
It makes your schmitt trigger circuit nonlinear.
Sometime buy a completely lean round or sirloin steak and have it ground to hamburgers and grill them. You'll see immediately the effect of removing positive feedback. If you add water to those ultra-lean burgers you'll see that negative feedback calms a system..

In early days of radio we had "Regenerative receivers" that employed positive feedback to increase gain of the RF stages. They operated on the verge of unstable oscillation , that's the "whistling" you hear in old movies when somebody is tuning a radio.

Little stupid experiments like that help you work systems in your head, which for a plodder like me is necessary to understand the math.

Work that schmitt in your head. Also conventional negative feedback 'operational amplifier circuits' which lend themselves beautifully to math..

old jim

old jim