Opamp Feedback Loop Input - how does it work

[seedlings]

I am trying to understand this example.

Input 1 looks like a standard inverting opamp stage with a feedback resistor value of R3+R4, and a gain of (R3+R4)/R1

Input 2 ... how does this work?  Is the feedback loop of input 2 still R3+R4, but the gain is (R3+R4)/(R2+R3)?

What is the input impedance for each input?

Thank you!
CHAD

[R.G.]

I think you're injecting an error current into the feedback loop. That's one way to look at it. I'll have to do more thinking to do better than that.

[Digital Larry]

It could be solved in a straightforward way using Kirchoff's voltage and current laws.  I started writing it down and quickly got to 4 equations with 4 variables.  Well at least, I could see it was heading in that direction.  Then I recalled why I got into management about 15 years ago.  Is there a real audio application for this or is it just a quiz question?

Another way I look at things like this is:

As R4 goes to zero, R2 will become attached to the output and there will be zero voltage gain from R2 to the output.  Input impedance (assuming the output has "zero" impedance) would be R2.  This also suggests that R4 is a standalone multiplier of the gain formula for input 2.

As R3 goes to zero, R2 gets attached to the (-) input and then it's like any other inverting input.  Since the (-) input is (probably) a virtual ground (you didn't show the (+) attached to anything), the input impedance would again be R2.

If you don't have anything connected to input 1, no current flows through R3 (as the op-amp has "infinite" input impedance).  The gain would be -R4/R2.  R3 is in essence in series with the (-) input and is rendered negligible by the infinite input impedance.  If something IS connected to input 1 then there is a current path out of the R3/R1 junction.

Input  1's impedance is R1 regardless of other conditions.

So, unlike most op-amp circuits you'd normally see, the voltage gain at either input depends on the source impedance of the other driving source.

[Digital Larry]

gahhh dual post sorry

[seedlings]

It's not a quiz.  On the PT2399 datasheet, pins 15 and 16 are an opamp, and this is the resistor arrangement to return feedback (input 2) to the input signal (input 1). I don't understand how it works.

CHAD

[R.G.]

DL's comments are correct. The right way to do it is to write the loop and node equations and solve for input current to the (-) input node, then compute what output voltage has to be to make the net current at the (-) node be zero.

Opamps work by the large gain forcing enough current through the feedback network to force any current going into or out of the (-) node to balance to zero.  Having a resistor there, somewhere along the feedback path, either adds or subtracts current from the current fed back to the input. This means the output has to adjust its voltage to cancel this out too.

The resistor values scale how much the currents are for a given voltage.

I'd probably solve this with superposition - if I had to. 

[Digital Larry]

OK I see what you mean.  Assuming we are talking about this:

http://3.bp.blogspot.com/-YR-nTU81BIM/TrhHUVLtyhI/AAAAAAAAAJ4/BLCfN0kw4Hc/s1600/20091227262420268.gif

It is a little different than you have shown.  Both inputs (the 15k/1 uF sections) join to the junction of R3 and R4.  From that perspective (DC analysis only) the 10k connecting to the (-) input should have almost no effect.  But once you put the cap between pins 15 and 16 it introduces a frequency sensitivity.  All that said, it's still pretty bizarre and I'd do a SPICE analysis sooner than I'd solve an equation for 4 variables.  

[seedlings]

OK I see what you mean.  Assuming we are talking about this:

http://3.bp.blogspot.com/-YR-nTU81BIM/TrhHUVLtyhI/AAAAAAAAAJ4/BLCfN0kw4Hc/s1600/20091227262420268.gif

It is a little different than you have shown.  Both inputs (the 15k/1 uF sections) join to the junction of R3 and R4.  From that perspective (DC analysis only) the 10k connecting to the (-) input should have almost no effect.  But once you put the cap between pins 15 and 16 it introduces a frequency sensitivity.  All that said, it's still pretty bizarre and I'd do a SPICE analysis sooner than I'd solve an equation for 4 variables.  

I was looking at the stock schematic from the datasheet:


Why not use one feedback resistor instead of R3  and R4, and connect both R1 and R2 to pin 16?

CHAD

***edit: and now I see that my original sketch is wrong.  Both input resistors go to the junction of R3 and R4... but I still don't get it.

[Digital Larry]

Why not use one feedback resistor instead of R3  and R4, and connect both R1 and R2 to pin 16?
CHAD

I agree with you completely.  I don't know why they have done it this way.

[slacker]

You need to look at the whole thing including the caps, it makes what's called a Multiple Feedback lowpass filter. Have a look about half way down this page http://en.wikipedia.org/wiki/Multiple_feedback_topology or google MFB for more info. I've no idea how they actually work, they seem to be pretty mysterious beasts.

[Digital Larry]

You need to look at the whole thing including the caps, it makes what's called a Multiple Feedback lowpass filter. Have a look about half way down this page http://en.wikipedia.org/wiki/Multiple_feedback_topology or google MFB for more info.

I agree with you completely, and now I realize why they have done it this way!

[seedlings]

You need to look at the whole thing including the caps, it makes what's called a Multiple Feedback lowpass filter. Have a look about half way down this page http://en.wikipedia.org/wiki/Multiple_feedback_topology or google MFB for more info. I've no idea how they actually work, they seem to be pretty mysterious beasts.

Excellent!  "Multiple Feedback' -thank you!  It's a 2nd order low pass filter.  Some things are very hard to learn if you don't know what to ask.  I looked through tons of opamp images trying to find one that corresponds.

Much obliged!



CHAD