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Comments on Design rules for opamp bootstrapping

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Design rules for opamp bootstrapping

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Here is the bootstrapping technique for opamps, as exposed in the Art of Electronics:bootstrap.

This technique is supposed to increase considerably the input impedance of the opamp for an AC input source (usually passed through a cap). The Art of Electronics considers it somewhat outdated, as it is now possible to use extremely low bias input current opamps.

I disagree with him: I see no reason to buy a somewhat expensive femtoamp opamp when you can obtain the same result with a more banal one (as long as we deal with AC sources).

Actually I do have an application where I use the self capacitance of a rotating half cylinder to measure the ambient electric field, which is perfectly tailored to bootstrapping.

Now, to make my question precise:

Assume

  1. I have an AC sinusoidal source of known frequency w;

  2. I want the output signal (voltage) be at least n percents of the input signal (e.g. 99%);

  3. I want the RMS input current be equal to at most I_max.

Question: how should I choose the values of the two resistors and of the capacitor in the schematic above?

Edit: Here is a simulation done with LT spice. I've more or less given random values, following nothing but some intuitive guess:

Simulation with bootstrap: bootstrap2

Result (current through the sine generator):bootstrap3

Result of the same simulation but without cap C1:

bootstrap4

Edit 2:

Perhaps the schematic will make more sense if we pass the input signal through a capacitor, as is usually the case. Here is the schematic for the simulation:

bootstrap5

And here are the results: the first trace is the current through cap C2, and the second trace is the current at the in+ of the opamp: bootstrap6

So, we see that the amplitude of the current through C2 is about 2nA, and the amplitude of the AC current at in+ is 6nA, more than I_C2.

Without cap C1, the current through the sine generator (and hence through C2) is the same as previously

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General comments (1 comment)
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I answer to myself because I believe this subject is interesting and not very well understood, even to professionals. Also, I want to point out that this answer is the product of my own investigations and simulations, and is not coming from reliable sources or knowledge.

First of all, I want to point out that the bootstrapping technique is mainly used in the context of transistor based amplifiers, mainly BJT. In the following image, R1, Rc, RE and R2 form the usual topology of a BJT amplifier. The bootstrap is formed by resistor R3 and cap "Cboot": bootstrap8

In this context, the bootstrap technique is alive and well! Excellent related questions and answers can be found here. I will not enter into this matter since it is well known and the original question is related to opamp bootstrapping. I only point out that the main point is that the bootstrapping technique avoids the loading of the AC signal by the biasing resistors.

Now, let me return to the main question: opamp bootstrapping.

Despite what the "Art of electronics" says (and also, to some respect, what Olin said above), this technique, if correctly understood, can be still useful in some circumstances.

Here is the main point of bootstrapping: often, an AC signal needs to be passed through a cap, and then buffered or amplified. There may be several reasons to pass the input signal through a cap, the most obvious one being you want to annihilate some DC bias present in the input signal.

Doing so, it is mandatory to follow immediately the cap with a resistor called "a return path to ground". Here is an illustration of what is meant:
boot7

Without a return path to ground, the output of the opamp would be undefined: for example, if there is no input signal at all, you expect that the output of the opamp be 0. But without a return path, the opamp would charge cap C, and its output would stabilize somewhere between its supply voltages, most often near Vcc+ or Vcc-. Usually, the return path R has to be chosen in such a way that the RC time constant is well below the lowest frequency of interest present inside the input signal.

This being said, for very very sensitive input signals, even a return path of 10 Mega ohm may load the signal heavily: it may be necessary to make the return path to ground at least 1Gohm or more, which is not convenient. This is where the bootstrap technique may be useful (see schematic inside the question): it virtually annihilates the the return path at the AC level, and ensures the signal loading is minimal, that is, no more than the input current of the amplifier.

A last question remains: what values should be chosen for the resistors R1=R2=R and C in the schematics of the question. Well, this is again function of the RC time constant. More precisely, the cut off frequency 1/2πRC has to be chosen well below the lowest frequency of interest present in the signal. For example, R = 1 Mega ohm and C = 100 nF fits most signals > 10Hz.

To sum up:

  • The opamp bootstrap technique is used to prevent the return path to ground to load the input signal;
  • It can be advantageously used for extremely sensitive input signals to ensure the signal is loaded as few a possible, that is, no more than the input current of the amplifier itself;
  • Hence, it allows using standard resistor (in the range of 1 M ohm), and in fact opamps with higher input current, since they surely works with return path of 1 M ohm.
  • Perhaps most importantly, this technique allows not worrying about the value of the return path to ground: it ensures the circuit works in the optimal way anyway.
  • finally, regarding what values to choose for the resistor and the bootstrap cap, it appears that R = 1 Mega ohm and C = 100 nF works in most of the cases, whenever the frequency of the AC signal is > 10Hz.
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General comments (3 comments)
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Olin Lathrop‭ wrote over 3 years ago · edited over 3 years ago

The circuit you show has nothing to do with any "bootstrap" technique. Its a R-C low pass filter followed by a unity gain buffer. The rolloff frequency is 1/2πRC, which is 1.6 Hz for the 1 MΩ and 100 nF part values you cite.

coquelicot‭ wrote over 3 years ago

@Olin. The circuit I showed was only an illustration of what is a usual return path to ground (see the ":" before the image). The bootstrap technique is already illustrated by the images in the question and in your answer. I'll probably take into account your remark and edit this answer though.

Skipping 1 deleted comment.

Circuit fantasist‭ wrote over 3 years ago

‭@coquelicot‭, I read with interest what you wrote. Definitely it is very enthusiastic and brave... Keep going boldly in this direction! Now I want to ask you, "Do you want to know what the simple truth about bootstrapping is?" To do this, you only need to forget about the existence of the "ground" in the circuit. Sounds weird, doesn't it? But it is very simple...