Operational Amplifier with positive feedback problems

In summary, the objective of this homework is to calculate the output value of the circuit in order to represent that signal on paper. The equations for alternate current are v=vout*sinw*t,where w=angular frequency. The equations for currents in the circuit would be I1=I2 + Iout. However, the equation is not sufficient in the general case to solve this circuit. In order to solve the circuit, you would need to introduce the equations for comparator circuits with feedback. Using the equations, you would calculate Vp given several different endpoint situations.
  • #1
esmeco
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Homework Statement



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.

Homework 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

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
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
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
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
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
Does this look correct?

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.
 

Q: What is an operational amplifier with positive feedback?

An operational amplifier (Op-Amp) is an electronic circuit component that amplifies an input signal. Positive feedback is when the output signal is fed back into the input, causing the amplifier to increase its output. Therefore, an operational amplifier with positive feedback is an amplifier that uses this feedback mechanism to increase its output signal.

Q: What are the common problems associated with operational amplifiers with positive feedback?

Some common problems with operational amplifiers with positive feedback are oscillation and instability. These can occur when the feedback loop is not properly designed or when the amplifier is overloaded.

Q: How can I prevent oscillation in an operational amplifier with positive feedback?

To prevent oscillation, it is important to ensure that the feedback loop is properly designed. This includes selecting appropriate resistors and capacitors and ensuring that the phase shift around the loop is less than 360 degrees. Additionally, using decoupling capacitors and keeping the amplifier away from any external electromagnetic interference can also help prevent oscillation.

Q: What is the gain of an operational amplifier with positive feedback?

The gain of an operational amplifier with positive feedback is theoretically infinite. This is because the feedback loop continuously amplifies the output signal, causing it to increase without limit. However, in practical applications, the gain is limited by the power supply and the internal components of the amplifier.

Q: How do I troubleshoot problems with an operational amplifier with positive feedback?

If you are experiencing problems with an operational amplifier with positive feedback, start by checking the connections and components in the feedback loop. Make sure that the feedback loop is properly designed and that there are no loose or damaged components. You can also try changing the values of the resistors and capacitors in the loop to see if that resolves the issue.

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