I've got a question: how should I solve this?
I've seen a thread were it was solved, but I don't know why the V+ is equal = V+ = Vout+(Vin−Vout)*R/(R+Rs)
If someone could guide me how it is calculated then I would be glad.
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\$\begingroup\$ This is a comparator with hysteresis. It doesn't act like a normally operating opamp. \$\endgroup\$– jp314Commented Aug 1, 2022 at 18:23
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\$\begingroup\$ Have you checked specifically this answer to the question that you linked: electronics.stackexchange.com/a/112482/194393? \$\endgroup\$– devnullCommented Aug 1, 2022 at 18:23
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\$\begingroup\$ Perhaps you are right. But I saw that someone could calculate it but I don't know how to calculate here V+. \$\endgroup\$– user331990Commented Aug 1, 2022 at 18:24
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1\$\begingroup\$ No - it is a nonlinear device (comparator) only when the posive feedback factor (k+) is greater than the negative feedback factor (k-). When Rs/R<R1/R1=1 the opamp works in its linear region. \$\endgroup\$– LvWCommented Aug 1, 2022 at 18:55
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1\$\begingroup\$ @user331990 Vx applies to both inputs i.e. when using an op-amp with net negative feedback, the op-amp seeks to keep both inputs at the same voltage i.e. Vx \$\endgroup\$– Andy akaCommented Aug 1, 2022 at 20:03
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1 Answer
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The calculation for this circuit is straightforward:
1.) For operating as a linear amplifier, we must ensure that the positive feedback factor Hr+ is smaller than the negative feedback factor Hr-, that means: Hr+ < Hr-.
2.) Then we can follow the classical procedure: Closed-loop gain Acl=Hf/Hr (assuming an infinite open-loop gain Aol).
With
Hf=Forward factor= R/(Rs+R) and
Hr=Feedback factor= (Hr+ + Hr- )= Rs/(Rs+R) + (-R1/(R1+R1)
Comment: The mentioned closed-loop gain formula Acl=Hf/Hr follows directly from the classical feedback expression:
Acl=Hf[Aol/(1-loop gain)]=Hf[Aol/(1+Aol*Hr)]=Hf/[1/Aol)+Hr] ......=Hf/Hr for 1/Aol=0 .
As you can see, for the loop gain I have used the expression (-AolHr), which means that the net feedback factor Hr=(Hr+ Hr-) is (and must be) negative when Hr=-R1/(R1+R1) dominates.
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\$\begingroup\$ Yes but in the other thread the author of the best answer said that If I'm not sure whether the Op Amp works in linear. Then He used this equation Vout = (V+ - V-)*Av. And he calculated V+ and V- in terms of Vin and Vout. The only thing is that I don't know how the author calculated V+. I see the final result of the equation but I don't know how he got to that point : V+=Vout+(Vin−Vout)f1, f1 = R/(R+Rs). I think there is a method but I don't remember how it was calculated. \$\endgroup\$ Commented Aug 1, 2022 at 19:25
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\$\begingroup\$ @user331990 It's just a potential divider. It could also be expressed as V+ = Vin + Rs/(Rs +R)(Vout-Vin). You assume that no current flows into the op-amp inputs. \$\endgroup\$– FinbarrCommented Aug 2, 2022 at 7:17
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\$\begingroup\$ @LvW May I ask you a question ? 1. You said that Hf had to be negative but here is positive. 2. Why I have to check which feedback factor is stronger ? Isn't V+ = V- the most basic thing to check ? For example if Rs > R then the gain Acl is negative. But it is a certain number which says that V- is equal V+. So why the math here is wrong ? V+ = V- says it is correct so what is the problem here ? 3. "I have used the expression (-AolHr)" , I don't see it, where did you use it ? \$\endgroup\$ Commented Aug 8, 2022 at 0:24
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\$\begingroup\$ @user331990 (1) No, I did not say that "Hf had to be negative". Where do you read this? (2) The equation V+ = V- applies only when a negative feedback keeps the opamp within its linear operation range,. As a counter example: When operated as a comparator, this equation does NOT apply. Therefore, before applying all the rules and equations for a linear amplifier, we have to be sure that the device really can operate in the linear amplification range, \$\endgroup\$– LvWCommented Aug 8, 2022 at 5:16
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1\$\begingroup\$ Yes - thats right. With positive feedback and injecting the signal into this positive network (lets assume that - by mistake - we do that), the CALCULATION with V+ = V- will give a gain value. However, the gain will be negative, which indicates that something is wrong because an input in the non-inv. input must result in a positive gain. So this contradiction shows that our assumption (linear amplification) was wrong. Of course, in reality (practice) the circuit will never work (saturation). \$\endgroup\$– LvWCommented Aug 8, 2022 at 14:12