1

I've got a question: how should I solve this?

enter image description here

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)

Like here: How are positive and negative feedback of opamps so different? How to analyse a circuit where both are present?

If someone could guide me how it is calculated then I would be glad.

ocrdu
  • 8,705
  • 21
  • 30
  • 42
user331990
  • 183
  • 1
  • 9
  • Have you checked specifically this answer to the question that you linked: https://electronics.stackexchange.com/a/112482/194393? – devnull Aug 01 '22 at 18:23
  • I think I saw this one but I got lost there because I didn't know how he get some results. I know that there was a way to calculate it and it was lim Av->infinite (Vout/Vin). also Vout = Av(Vp-Vn). I don't know only how Vp is calculated and he also calculated Vx two times and I don't know if it is Vx for +Input or -Input. – user331990 Aug 01 '22 at 18:33
  • Oh and I checked the most voted answer – user331990 Aug 01 '22 at 18:44
  • This is a comparator with hysteresis. It doesn't act like a normally operating opamp. – jp314 Aug 01 '22 at 18:23
  • Perhaps you are right. But I saw that someone could calculate it but I don't know how to calculate here V+. – user331990 Aug 01 '22 at 18:24
  • 1
    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 – LvW Aug 01 '22 at 18:55
  • 1
    @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 – Andy aka Aug 01 '22 at 20:03
  • You mean I can't just make an equation for V+ and V- and use this equation : Vout = Av(Vp-Vn) ? Then use this equation lim Av_>infinite (Vout/Vin) and see if it's fin or infinite ? If it's infinite then it is positive feedback and if it's fin then negative feedback ? It is more universal. Using this lim equation. – user331990 Aug 01 '22 at 20:17
  • @user331990 Of course, you can use the equation Vout=Av(Vp-Vn) which is identical to Vp=Vn for Av approaching infinity. HOWEVER, this is allowed only if the circuit works in the linear amplification range. That means: As a first step you must check if stability of the circuit is ensured. That is the background of the requirement Hr+ – LvW Aug 02 '22 at 07:35
  • I am not to confident with Hr :D I've seen something like this for this circuit ti see which feedback is the dominant one : The negative feedback is just the output multiplied by R4/(R3+R4) and the positive feedback is the output multiplied by R1/(R1+R2). I don't know why it works like that. I know only the basics of Op Amp only for single feedbacks. – user331990 Aug 02 '22 at 21:33

1 Answers1

1

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.

LvW
  • 24,857
  • 2
  • 23
  • 52
  • 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. – user331990 Aug 01 '22 at 19:25
  • @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. – Finbarr Aug 02 '22 at 07:17
  • @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 ? – user331990 Aug 08 '22 at 00:24
  • @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, – LvW Aug 08 '22 at 05:16
  • @LvW I think I might got it. Because when I have only Op Amp with positive feedback (without negative feedback) then I could also calculate V+ = V- and I could get a solution but I know it is positive feedback so this equation could get me wrong even though I have a solution V+ = V- for positive feedback. So I believe it works the same with this one with 2 feedbacks. – user331990 Aug 08 '22 at 13:16
  • 1
    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). – LvW Aug 08 '22 at 14:12
  • @LvW Well the gain is negative and you mentioned that it is wrong because input is in non-inv. Why does it say it is wrong ? I guess input doesn't determine what is the output because the difference between V- and V+ says what is in the output though. And not what I add in the input. For example if I had input on non-inverting 6V and input on inverting 4V and negative feedback then output is positive. Even though I have input in non inverting. – user331990 Aug 10 '22 at 14:30
  • Did you notice the contradiction in your last two sentences? Yes - you are right that it is the difference (V+ - V-) which determines the output. But in my example this difference is positive - and the output signal is shifted by 180deg. That is a logical error. – LvW Aug 11 '22 at 07:03
  • @LvW Yes you are right now I've noticed. But there is one more thing. In your example V- was grounder. Which was more understandable why output shifted by 180 deg is an error. But when there are 2 feedbacks and it's hard to tell which gain is correct because non of them are grounded. Because input is connected to non inverting but the gain is decided not by voltage connected to the input but by the difference in V+ and V- – user331990 Aug 12 '22 at 21:13
  • Tha answer is simple: When the input signal is connected - either directly or through the feedback circuitry - to the V+ (resp. V-) input, the closed-loop gain must have a "+" (resp. "-") sign. If this is not the case, the calculated gain cannot be correct because the following requirement was not met: Negative feedback must govern (negative feedback factor must be larger than the positive feedback factor - if existent). Therefore, the fulfillment of THIS requirement should be checked at first ! By the way: When both are equal, this is another form of the oscillation condition. – LvW Aug 13 '22 at 07:34
  • Even thought the voltage source is not directly connected to the input V- and V+ I have to assume that when I have voltage source connected to V+ with resistor then the gain must be positive ? Even though I don't know what is in the V- and in the V+ Like in the picture from the post and I can;t confirm it using Vout = (V+ - V-) equation to say if the gain is correct or not ? I mean it's pretty complicated because I know that voltage input on V- or V+ doesn't determine the Vout. – user331990 Aug 13 '22 at 15:21
  • I must admit that I do not get the problem description in this last comment. Do you realize that the feedback factor is independ on the node where the input signal Vin is connected? We set Vin=0 for finding the feedback factors. In short: Wenn the negative feedback governs, the circuit works as an amplifier (linear mode). Then, the gain is positive (negative) when Vin drives (directly or through any network) the non-inverting (inverting) opamp input pin. – LvW Aug 13 '22 at 15:44
  • @LvW What I meant is that Output depends on the difference between V- and V+. So I thought that Vin is not connected directly to the V- or V+. Also V- or V+ isn't directly connected to the ground. So I don't know what is the difference in V- and V+ so how can I know that Vout is correct thus the gain is correct ? For example if I had inverting op amp Vin is in inverting side with negative feedback V+ is not a ground but for example has V+ > Vin. So the Gain is not negative but positive. although I gave Vin on the inverting then the feedback should be negative but it's positive. – user331990 Aug 13 '22 at 16:52
  • So how can I know that the gain there with two feedbacks is also correct ? When I don't know the V+ and V-. In other words when I calculate closed loop gain. Why can it be wrong when I don't know what is in V+ and V- ? I can't confirm the closed loop gain sign ("+","-") with the Vout ("+","-"). In the single feedback for positive example it was easier to see because Vout was supposed to be positive but closed loop gain was negative. – user331990 Aug 13 '22 at 17:03
  • Example: The circuit as shown in your detailed contribution. Pos. fedback factor is Rs/(Rs+R). Neg. feedback factor is R1/(R1+R1)=0.5. When Rs – LvW Aug 13 '22 at 18:56
  • @LvW I mean if I didn't know about feedback factor. Then how could I prove that the gain I calculated was wrong ? The closed loop gain I have calculated was wrong, for Op amp with single positive feedback it was easy to prove because from Vout = V+ - V- was positive because Vin is in V+ and 0V was in V-. But with two feedbacks it's hard to say if vout is incorrect with the closed loop gain. – user331990 Aug 14 '22 at 12:58
  • @LvW In single feedback (positive feedback without negative feedback) and the Vin is in V- so the closed loop gain will be postive but the Vout is negative so it's easy to say it's wrong. And it's easy to say because at the beggining V+ was 0V so Vout = 0 - 5V = -5V but the closed loop gain is positive so its obvious that this gain is incorrect. It was proved that the gain is wrong but in both feedbacks I don't know what V+ is the value and what V- is the value. So how can I confirm that the closed loop gain is wrong comparing to Vout ? – user331990 Aug 15 '22 at 18:08
  • Why do you think that in your example the "closed loop gain will be positive"? When Vout is negative, I think it is quite logical that the gain is also negative - because THIS is the reason for a negative output. – LvW Aug 16 '22 at 07:13
  • @LvW Because there is a positive feedback that's why I calculated the closed loop gain that says it is positive but the output is negative. So the equation is wrong so the output will saturated. You said that the calculation for closed loop gain can be fixed for positive feedback but are incorrect in other ways. – user331990 Aug 16 '22 at 12:41
  • Maybe in other words. how can I assume that the closed loop gain is depended on where I place Vin ? You said that connecting Vin in inverting input will result in negative gain. But the output is calculated by V+ - V-. So how can I know that for double feedbacks the closed loop gain is wrong ? not knowing the V+ and V- to confirm if the closed loop gain is wrong ? You said it yourself here : "However, the gain will be negative, which indicates that something is wrong because an input in the non-inv. " One input doesn't determine the output, both inputs determine the output as I remember. – user331990 Aug 16 '22 at 13:11
  • Regarding your last sentence: No - one input (receiving Vin) does ALONE determine the SIGN of the output!! When there is no contradiction (pos. output for Vin at the non-inv. terminal, and vice versa) , then the circuit works in linear amplification mode. Very simple rule! I cannot see where and why do you have such problems. – LvW Aug 16 '22 at 13:20
  • @LvW I heard that one input doesn't determine the output. If I had Vin in the inverting input but V+ was not 0V but 6V. Then for Inverting input that is 5V then the output will be also positive even though you said that the gain should be negative. And for both inputs I don't know what do I have for V+ and for V- so how do I know if the closed loop is incorrect if I don't know if the output is correct. I want to create a chat to show you in pictures because I have a small doubts. But I don't know how to create chat – user331990 Aug 16 '22 at 14:01
  • I spoke about the SIGN of the output - and I wrote it in capital letters!! – LvW Aug 17 '22 at 07:11
  • @LvW But I don't get it. I heard a lot about Vout and closed loop gain. That Vout is determined from two inputs like in the equation Vout = V+ - V-. So to tell that the closed loop gain is wrong is when Vout has the oposite sign than the closed loop gain. I want to talk this one but I can't share the image. How can I create a chat because I want to show you exactly what do I mean. Because many people tell me that one input doesn't determine the output and there must be value of another input to determine the output. – user331990 Aug 17 '22 at 12:33
  • @LvW. Also for double feedback I know the Vin for V+ but how can I know what is the value of V- when it is not grounded ? So how can I confirm that the gain calculated is incorrect. I know that you compared closed loop gain to Vout. If the sign of the gain is incorrect with the Vout then it is going into saturation. But I know only Vin for V+ and V- is unknown and I don't know how to confirm that the sign of the gain is incorrect with the Vout because Vout is equal Vout = (V+ - V-)*Aol. But V- isn't grounded. – user331990 Aug 17 '22 at 13:21
  • I do not know what you are speaking about. How can you say "not grounded"? What does this mean? Open circuit? Why don`t you stick to the example circuit you have shown in your original contribution? It shows everything which is necessary - I have covered this circuit already in my comment 13. August, 18:56. Thats all one can say to this circuit with double feedback. – LvW Aug 17 '22 at 14:12
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/138580/discussion-between-user331990-and-lvw). – user331990 Aug 17 '22 at 14:52
  • @LvW I made a chat with pictures. Because it's hard for me to tell you what do I mean. It is connected really with double feedbacks, but I needed to show you also single feedbacks. I'm sorry for annoying you but I just don't understand one part. Please if you have a time check the pictures I made in the chat. Thank you for everything you did here. – user331990 Aug 17 '22 at 15:08