bootstrapped opamp is bothering me...probably useless buildingblock..

[Johan]

two things I can't figure out.

first, I realize that as drawn, it would probably oscillate or do something unpredictable since the bootstrap node is at 1:1 gain. but ignoring that for a moment. lets say the R is 150 Ohm, how do I calculate the reflected input Z of this circuit?  if the gain at the node was 1 : 0,9 or 1 : 0,8..is there a formula?..or is it a "build it and measure" kind of thing..

and second. since the R is strapped across the + and - inputs, is the resistor self noise canselled or amplified?

the simple fact that I can't recall seeing a opamp bootstrapped like this suggests to me it wouldn't have any benefits as far as the second Q goes, and that would make the first Q sort of redundant, but I just cant figure it out and it bothers me..

does anyone have any idea?..
J

[R.G.]

The resistor between the inputs won't do much.

Your schemo doesn't show it, but I'm assuming that the input with the cap connected is the + input and the one with all the resistors is the (-) input.

The fundamental way an opamp works is that the output voltage is the open loop voltage times the voltage difference between the + and - inputs. Another way to say that is that the difference between the inputs is 1/(open loop gain) of the output voltage. For smallish gains (compared to the open loop gain of over 100,000 usually) then the difference between the inputs is millivolts. The + input is a high impedance point, and the - input is a low impedance point, but they are very close in voltage, generally negligibly small. So a resistor between two points with negligibly small voltage across it will have a negligibly small current through it, and hence be itself negligible.

That's not really a bootstrap.

I think you meant to run the bootstrapping resistor from the output to the + input. As long as this is a higher resistance than the source resistance driving the input, and as long as the loop gain from + input to - input is small enough to not make it oscillate, then the bootstrap increases the apparent input impedance.

[earthtonesaudio]

Take a look at a graphic EQ pedal circuit.

The (-) input is a virtual ground, so your input resistance = R.  No bootstrap action occurs.

I'm not entirely sure, but I think the noise generated in R should be canceled.

[Johan]

Your schemo doesn't show it, but I'm assuming that the input with the cap connected is the + input and the one with all the resistors is the (-) input.

thank's R.G.
yes thats the way I meant it..It was just obvious to me, but for clearity, I should have drawn the + at the input.

the way it is set up now, it doesnt matter if the gain is 3db, 20 db or 40 db. the signal ratio between input and node is always 1:1..(as long as no clipping occurs..or that is the idea.. ) tie-ing R to the output would surely couse oscilation at every gain seting greater than 1
the idea is obviously to have a circuit that represent a high Z, but being able to use an opamp that may need a lower Z input, such as a 5532 instead of a TL072. but how do I calculate actuall Z?   and does the R selfnoise cansel or gets amplified?.

J

[R.G.]

the way it is set up now, it doesnt matter if the gain is 3db, 20 db or 40 db. the signal ratio between input and node is always 1:1..(as long as no clipping occurs..or that is the idea..

I don't follow you. If you mean "the signal ratio between + input and - input node is always 1:1", yes, that's true. But this has almost  no effect on the input impedance because the voltage difference between them is always essentially zero. The resistor can't do anything because essentially no current through it. It can't provide any current to the + input to buck out any input current because the (-) input is already being hammered into place following the (+) input.

The idea behind bootstrapping is that the place signal touches the circuit "eats" some current from the signal source. Bootstrapping provides some of that current from another place - the output in almost all cases - to make the input eat less signal current, and appear to be a higher impedance.

Let me try another way. The opamp inside the feedback loop is driven by the voltage difference between the two inputs. The two inputs have two resistances which let current flow from the signal source. One is (conceptually at least) a resistance to ground. The other is the apparent resistance between + and - inputs. Putting a resistor between the two points doesn't change the resistances to ground at all, but it does decrease the resistance between the two. It can't decrease the input current, so it can't make the apparent input impedance higher.

tie-ing R to the output would surely couse oscilation at every gain seting greater than 1

Yep. Quoting Jerald Graeme in "Applications of Operational Amplifiers", "bootstrapping is the controlled application of positive feedback". He goes on to describe why and how to control it to keep the bootstrap loop gain under unity.

the idea is obviously to have a circuit that represent a high Z, but being able to use an opamp that may need a lower Z input, such as a 5532 instead of a TL072. but how do I calculate actuall Z?

Probably the best way is with a high impedance buffer in front of the opamp, or a bootstrap from a second opamp which merely follows the input.

and does the R selfnoise cansel or gets amplified?.

It is amplified by the open loop gain of the amplifier and then reduced by the feedback factor.