# Homework Help: Operational Amplifier with positive feedback problems

1. Jan 12, 2007

### esmeco

1. The problem statement, all variables and given/known data

I have some problems determining the tension output value for this circuit and this confusion comes from the presence of positive feedback and an alternate current.Since I'm used to negative feedback circuits and direct current it confuses me a bit,so any help is deeply appreciated!The objective of calculating the output value of the circuit is so we can represent that signal on paper.

2. Relevant equations

I know that the equation for alternate current is v=vout*sinw*t ,where w=angular frequency
The angular frequency is calculated by w=2*pi*f

f=0.5Hz and amplitude=1v;

The equations for the currents in this circuit would be:

I1=I2 + Iout

3. The attempt at a solution

So, using the above equations we have:

(1- Vp)/500=(Vp-Vout)/2000 + (Vp-V0*sinw*t)/R2

What I'm not sure if it's correct is the current that goes through Rin2.Does it still work like a DC circuit where to calculate the currents,we want the difference between the tension in both terminals divided by the resistance?And how the we know what is the value of vp if we want to solve in order to vout so we can represent the signal ?

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2. Jan 12, 2007

### Staff: Mentor

Did you look up the comparator with input hysteresis link that I posted in your other thread? That should get you to the solution pretty quickly, I would think.

3. Jan 12, 2007

### esmeco

Yes,I did but I didn't quite understand it,I'm afraid...I know that when there's positive feedback there's saturation contrary to negative feedback but I really don't have many more ideas besides that..:Do the equations look ok?

4. Jan 12, 2007

### Staff: Mentor

For others who didn't see the thread, here's the link to comparator/hysteresis info:

http://www.maxim-ic.com/appnotes.cfm/appnote_number/3616

Now, your equation is valid, but it is not sufficient in the general case to solve this circuit. In the general case for comparator circuits with feedback, you would assume that the output is pegged one way, say high (at 15V in the ideal in this case, or at whatever practical + saturation voltage is listed in the datasheet for the real part), and use that as Vo to calculate where the Vp voltage sits with the output pegged high. Then repeat for when the output is pegged low. Then calculate where the positive-going hysteresis trip point is (that's when Vp=Vn with the output going from low to high), and then calculate where the negative-going hysteresis trip point is. Then in the case of this problem, you factor in the sine wave, to see if it is able to hit both trip points. If it doesn't, then the output waveform will just stayed pegged. If it does, you figure out what values of Vi hit the two trip points, and that gives you the output rectangular wave solution.

Kind of make sense? Just visualize what is going on, starting with the basic comparator/hysteresis circuit in the link, and then look at this problem in a similar way.

5. Jan 12, 2007

### esmeco

So,if I understood correctly we should introduce the equations:

I3=(0 - Vout)/2000
2000=(15 - 0)/I*2000
2000=-15/I*2000

Does this look correct?

6. Jan 12, 2007

### Staff: Mentor

I don't think so, but I'm not really understanding what you are doing. Instead, I'd solve for the Vp voltage given several different endpoint situations:

-- Vout pegged high, V1= 0
-- Vout pegged high, V1= 0.5V
-- Vout pegged low, V1= 0
-- Vout pegged low, V1= -0.5V

That will help to bracket the solutions a bit. Then you need to figure out where the hysteresis switching points are for V1 increasing from -0.5V up to +0.5V (so that the output switches for some V1 from low to high), and for V1 decreasing back down from 0.5V down to -0.5V (so that the output switches for some V1 from high to low. The output is going to snap between pegged at 15V to pegged at -15V, but it will do it based on a biased version of the V1 input, by virtue of the positive feedback through the voltage dividers and with the added quirk of the V4 offset. I believe that the V4 input bias voltage will offset the two switching points for the V1 input, and result in a non-square output waveform, but I haven't done the math to be sure what the final answer is. I'm relying on you for that!

EDIT -- fixed a typo in the list of initial calcs I'd do.