op amp feedback vs source impedance - Page 1
Forum section: electronics / beginners

author: electrolust
author_rank: Unknown
timestamp: July 27, 2017, 08:46:47 pm
content: For the textbook standard configuration of non-inverting op amp, U1, let's say it's G=2 with R2 = R1 = 10k. G=1+R2/R1=2.Now you connect the output to a second stage inverting op amp, U2, with RIN= 1k and RF= 2k; G=-2.Is the source impedance that U1 sees: RIN= 1k, or is it RIN|| R2 = 909r? Surely it must be the latter? Or, continuing that line of reasoning, wouldn't it be RIN|| R1+R2 = 952r4?[followup question to come]

author: Andy Watson
author_rank: Super Contributor
timestamp: July 27, 2017, 08:54:39 pm
content: Depends. It depends on what the other input of U2 is connected to? If it is connected to ground, you have a "virtual ground" at the negative input of U2. So U1 sees a load of 1k.

author: electrolust
author_rank: Unknown
timestamp: July 27, 2017, 09:03:49 pm
content: Quote from: Andy Watson on July 27, 2017, 08:54:39 pmYes, U2 IN+ terminal connected to ground. Right, "virtual ground" is at U2 IN-. So doesn't that mean that RINand R2+R1 form a parallel resistor network?circuit diagram for clarity:

author: Andy Watson
author_rank: Super Contributor
timestamp: July 27, 2017, 09:15:49 pm
content: Quote from: electrolust on July 27, 2017, 09:03:49 pmYes. I think that's correct. I calculate the result as 952(Sorry, I'd mentally discarded R2 in my previous post)

author: Benta
author_rank: Super Contributor
timestamp: July 27, 2017, 09:29:22 pm
content: QuoteNo. The output of amp 1 (presuming it's an ideal opamp) has zero impedance, so R1 and R2 have no influence on the following stage.You have a circuit with +2 followed by -2 amplification = -4.

author: danadak
author_rank: Super Contributor
timestamp: July 27, 2017, 10:47:41 pm
content: The OpAmp output Z is a function of G and frequency, and is not 0.Look at page 38 for DC calculations.https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/lecture-notes/22_op_amps1.pdfHere is ac analysis (open loop) -http://www.ti.com/lit/an/slyt677/slyt677.pdfRegards, Dana.

author: electrolust
author_rank: Unknown
timestamp: July 28, 2017, 12:12:10 am
content: In my question about parallel resistors, I'm not suggesting any influence on U2. I hadn't thought about that actually, that might be another followup question. But at this point, I'm just considering the effect on U1.Whenever the simple tutorials on op amps discuss it, it's always an op amp in isolation, driving nothing. But it seems to me, once there is a load (in this case, 1k), there must also be an interaction with the feedback network.EDIT: sorry, I see now that I botched the resistor designations in the first post, and indeed made it sound like I was asking about influence on U2. I edited the post to clear it up.

author: Audioguru
author_rank: Super Contributor
timestamp: July 28, 2017, 02:24:04 am
content: Why are the resistors on the second opamp as low as 1k and 2k? Why not 10k and 20k instead? Most opamps cannot drive a load as low as 1k.

author: MrAl
author_rank: Super Contributor
timestamp: July 28, 2017, 10:36:26 am
content: Quote from: electrolust on July 28, 2017, 12:12:10 amHello,Yes there is some interaction with the feedback network, but it will usually be small because the output impedance is so low.For an example, do a DC analysis. Give the ideal op amp an output resistance of say 1 ohm and redo the analysis you did before with the parallel 1k and 20k and see how that changes things. The feedback now senses the output of the 1 ohm resistor and so it should provide very nearly the same output, although not exactly the same with a fixed finite ideal op amp internal gain. Doing that, try different internal gains like 100, 1000, 10000, and 100000, with and without the 1 ohm resistor. If the change is small and you want to see more of the effect, try making the output resistance 10 ohms instead of 1 ohm.

author: electrolust
author_rank: Unknown
timestamp: July 28, 2017, 02:00:57 pm
content: The circuitlab drawing of the circuit is a "live" URL. I should have taken a static png image instead. Anyway I updated it so you can refer back a few posts to see the update where I added R3 = 1r.In that configuration, the total source impedance seen by U1 is 953r, implying it would be 952r without R3. Because R3 is just in series with the rest of the resistance, it simply adds to it and essentially doesn't change how R1+R2 interacts with R_IN. So I don't understand when you say, the interaction is small because of the low output impedance. No matter how small we make the output impedance (down to 0), the R_IN and R1+R2 interaction is the same. If the output impedance were verylarge, then in that case any R_IN vs R1+R2 interaction would still be there but would be irrelevant because the output impedance dominates.I'll note that if I change the output impedance in the U1 model, rather than via the external R3 resistor, there's zero effect on the output voltage or current.The CL simulation seems to confirm my thought, that the total load seen by U1 isn't just R_IN, which is what all the tutorial material on the web suggests. I suppose that the source impedance of U2 is in fact just R_IN, but the source impedance of U2 isn't the entire story when it comes to the load seen by U1.But I'm still curious to understand your statement about how the output impedance of U1 makes the interaction small.

author: Benta
author_rank: Super Contributor
timestamp: July 28, 2017, 02:02:02 pm
content: QuoteYou apparently didn't notice the little note in brackets about an ideal opamp.But it's moot, since the circuit has been changed and the comments no longer apply.

author: MrAl
author_rank: Super Contributor
timestamp: July 29, 2017, 02:56:57 am
content: Quote from: electrolust on July 28, 2017, 02:00:57 pmHi,Are you talking about the output resistance of the first op amp? That's what i am talking about.You start with an ideal op amp. That is where there is no input offset, zero output resistance.You then add a small output resistor in series with the output of the ideal op amp, say 10 ohms.You then take the output of the non ideal op amp to be the open end of the resistor you just added.You then connect the 1k resistor and two 10k resistors to the open end of the small resistor, and that end also becomes your new 'real' output. You then consider how the voltage changes as you change something else like the two 10k resistors, the 1k, or the 10 ohm resistor, at a convenient non zero operating point. The internal gain of the ideal op amp must also be finite, not infinite, such as 100 to 100000.If this isnt clear or doesnt make sense or some other problem, i'll draw up a circuit and throw a few calculations out there for you to examine.If i understand you right you want to understand the loading effect of the 1k and two 10k resistors right?

author: electrolust
author_rank: Unknown
timestamp: July 29, 2017, 08:00:33 am
content: Quote from: MrAl on July 29, 2017, 02:56:57 amNot quite. Thanks for the detailed breakdown. I do understand now what you're saying, and agree.What I really wanted to clarify was, whatisthe load seen by the op amp. The simplified online introductory material would have you believe that it's just RIN.

author: MrAl
author_rank: Super Contributor
timestamp: July 30, 2017, 01:07:37 pm
content: Quote from: electrolust on July 29, 2017, 08:00:33 amHello,The "load seen by the op amp" involves everything connected to the OUTPUT. Any resistors or other amplifier input would be part of that load.IN your reply #2 you show some resistors and a second op amp. The second op amp input resistor is part of the load on the first op amp output, and so are the two 10k resistors, and you can get a measurement of the current by looking at the voltage across that 1 ohm resistor.So i am not sure why you would think any Rin had an effect on the LOADING of the first op amp. The time that is true (more or less) is when you go into the second op amp through the non inverting input terminal. For example, if you change that second op amp to a voltage follower, Rin will be the load due to that op amp now.Thus when you read that Rin is the only load, that's only for a non inverting amplifier as the second stage. An inverting op amp stage like you show in your drawing always has a lower resistance because of the way the input resistor connects to that 'virtual' ground. To reduce that effect, you could use some positive feedback (with care). Positive feedback has the effect of raising the apparent input resistance (with inverting op amp stage mostly).