How to bootstrap

I see the term bootstrapping, used when the designer wants a high impedance and usually low capacitance. I know it involves feedback but that's all I know.

I want to learn enough to build a bootstrapped input with 10Meg/2pf impedance. Those are ballpark numbers, the end use would be used to measure voltage on a high Q coil and NOT load it. Frequency 100khz up to 2Mhz, but 10 Mhz adds more uses. (crystal radio stuff)

I would prefer a transistor circuit, that way I'll learn something, but if there is an obvious IC circuit, I would like to know. Mikek

Reply to
amdx
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"amdx"

** I think you have been told this already - but a single JFET ( wired as a source follower ) is ideal.

BTW: a 2pF cap has an impedance of 8000 ohms at 10MHz.

... Phil

Reply to
Phil Allison

Just analyze an ideal amplifier of gain +(1-delta) where delta is small, but non-zero Apply a feedback resistor from output to input, then calculate input impedance. ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

Don't recall being told about the JFET source follower. With your reminder, I do recall the circuit I built before had 0.3pf. So forget the 2pf, need 0.3pf or less.

Can you give me incite to this;

I have a High Q LC circuit, I put my bootstrapped measurement device in parallel, the C of the circuit adds to the C I used to resonate the L. So does the C of my measurement device load the circuit? Or just change the resonant frequency? Assuming a high Q C in my measurement circuit. Mikek

Reply to
amdx

Now Jim, if I could do that I wouldn't be asking such a question. I'll Google delta later. Mikek

Reply to
amdx

"amdx"

** What the smug with delusions of grandeur is attempting to point out is that if YOU organise things so that the same voltage appears at both ends of a resistor - it passes no current.

Makes an input bias resistor virtually disappear.

.... Phil

Reply to
Phil Allison

The easiest way (for me) to make a bootstrap is an opamp unity gain buffer the output drives a shield 'capacitance'. But as Phil said the jfet follower is a classic. A bootstrap in a High Q LC circuit is hard to understand. What are you doing?

George H.

Reply to
George Herold

I hear that delta is a babe >:-}

Math is your friend. It's mostly really all about reading comprehension... what does the problem say?

The bootstrap...

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In my late 20's I taught Algebra to disadvantaged youth... otherwise known as South Phoenix thugs.

I had about a 90% success rate.

I hired my best student as my technician and he stayed with me for 27 years. (Jim Foster, the JF initials seen on many of my ancient hand-drawn schematics.) ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
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Reply to
Jim Thompson

As Phil A. says, the idea of bootstrapping is that if the input admittance of your circuit has zero swing across it, it draws zero current.

A BF862 JFET follower has an output resistance near 30 ohms. (That is, assuming that the source current is somewhere near I_DSS.) If the current sink on its source is a 1k resistor, the voltage gain of the source follower stage is set by the voltage divider ratio:

A_V = 1k/(1k + 30) ~ 0.97.

Thus the voltage swing across the G-S capacitance is reduced by a factor of 30. The G-S current thus goes down by the same factor, so effectively the input impedance goes up by the same factor of 30.

If you hang an emitter follower on the source of the JFET, and use that to move the JFET drain up and down to follow its gate, the contribution of C_DG goes down by about the same factor.

This is not a free lunch, because the SNR stays more or less the same, but it does give you a much nicer frequency response in general.

Cheers

Phil Hobbs

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Reply to
Phil Hobbs

Maybe a jfet follower then an opamp or emitter follower. The second follower would be AC coupled into the jfet drain, so it goes up and down to track the ac gate voltage. That cancels most of the d-g capacitance. AoE has a bunch about bootstrapping.

Bootstrapping can be used for DC applications, like constant-current sources.

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Here's a 1940-ish Blumlein circuit that uses a bootstrap to reduce the effective capacitance of an early TV picture tube:

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I'm not sure how exactly to define "bootstrap".

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Reply to
John Larkin

Not really part of the circuit, it would me an instrument. Several things. Measure Q (3db points), signal strength meter that doesn't load the coil. As a start. Here's the circuit I have built, just curious about a bootstrapped input.

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Mikek

Reply to
amdx

Okay, mikek. Here's your bog standard bootstrapped degenerative amplifier using a single BJT. I'll try and look over it according to my poor hobbyist mind. I will assume you understand much of it, already. But ask about stuff, where I'm wrong about that.

+V | +V | | \ | / Rc | \ \ / / R1 | \ | / +----Out | C1 | | || | In----------------||--, | | || | | | R3 | |/c Q1 +----/\/\-----+--| | |>e | | | C2 | | || +------, +---------||-------+ | | || | | | | \ | | / Ra \ | \ / R2 \ / \ / Re | / \ | | / | | | --- Ca | | --- gnd | | gnd | gnd

(The simpler pieces missing from the above diagram are RS, the source resistance, and RL, the load resistance. Ignore them for now.)

As you probably know already, the biasing pair of resistors R1 and R2 (in a non-bootstrapped case) are supposed to be stiff enough for the task of keeping Q1's bias point from moving much. It's common to read suggestions that the current through R1+R2 should be about 1/10th of Iq, which is the current through Rc in the quiescent state. Making them that stiff also means that they load the source (by their shunting effect.)

I liked the description I saw from Phil Hobbs. It's exactly how I learned to see this, as well. He said that if you can make the swing across a resistor (he said admittance) to be identical (or as close as possible to that), that this means no current is drawn.

Look at R3 for a moment. Assume that there is a DC bias across R3 providing the necessary Q1 bias current for its base. Now imagine keeping that DC bias, but asking yourself what would happen if the biasing pair node can be made to move up and down (AC wise) in exact lock-step with the base of Q1. If that could happen then the bias current would remain intact, but there would be no "change" in the current with AC changes. So no AC loading. And the DC loading can be increased because, although R1 and R2 still have their Thevenin equivalent shunting effect, now you can add R3 straight away to that, so that the DC loading is much lighter than before despite a stiff bias pair.

So more degrees of design freedom.

What C2 does is to implement that requirement that the biasing pair node moves up and down in concert with Q1's base. At AC, the emitter is "following" the base (with very slightly less than 1 gain, Q1's alpha.) Assume the gain is 1 for now. If C2 is designed to be a 'short' at AC frequencies of interest and if the emitter of Q1 is a good, low impedance "source" of these AC changes (it is such a good source), then C2 will bypass emitter fluctuations directly to the biasing pair and it will do so at low impedance, easily driving the biasing node up and down in concert with the base signal. What's neat about this is that it uses an existing low impedance replica of the input signal that is isolated (mostly) from the input by the beta of Q1. And it uses it to force the biasing pair node up and down in phase with the AC signal. Since it is obvious from the circuit that Q1's base itself is moving also up and down by the same amount (almost) then it follows that the AC signal on both sides of R3 is the same (almost.) So at AC signals, R3 "looks" like an infinite impedance. And at DC, R3 greatly reduces the bypass loading.

Okay. So some analysis. Miller's theorem says that the impedance between two nodes can be resolved out into two different components: z/(1-k) and z*k/(k-1). In this case, k is the voltage gain, k = Av = Vo/Vi, and R3 is z. Since we are tapping off of the emitter, not the collector, Vo/Vi is nearly 1. For example, if Av=.99 and R3=200k then the effective resistance is 20M Ohm.

Full analysis is, of course, more algebra and work. But I thought I'd just get this out there for you to get the main point across. I'd use the term bootstrapping anytime this particular kind of approach is being taken.

Jon

Reply to
Jon Kirwan

No problem. The basic idea boils down to this:

If you can keep both sides of R3 moving around in lock step then in effect R3 has infinite impedance (from the AC point of view) and completely isolates the Q1 base from the divider network. So the biasing pair (via R3) gets to perform its DC function of biasing Q1 but, at AC, R3 works to decouple the Q1 base from the biasing pair.

Of course, in reality it's not perfect and so the isolation is imperfect as well.

Jon

Reply to
Jon Kirwan

What delta - just a change on the input? If gain is a function of that, even with the minus sign, I'd think it was an exponential amplifier rather than 'ideal'.

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Reply to
Tom Del Rosso

GAIN = (1-delta) , delta small, but non-zero; so say that GAIN = 0.99, for example.

Draw yourself a picture/schematic of what I wrote in words. ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

"Phil Hobbs"

(snip)

** One issue with bootstrapping input resistors is that it can dramatically increase the noise compared to simply using a large value resistor. 30 odd years ago I attempted to build a JFET pre-amp for a condenser mic capsule and not having any 1Gohm resistors handy tried bootstrapping a 10Mohm one. The pre-amp tested fine, with an effective input resistance close to 1Gohm - ie response was flat across the audio band when driven via a 22pF cap simulating the capsule.

When the capsule was tried, the background noise ( hiss) was about 20dB more than with a commercial pre-amp and quite unacceptable for studio work.

Thing is, with a 1Gohm ( gate bias) resistor, 22pF is enough to shunt nearly all the audio frequency ( Johnson ) noise away - not so with 10 Mohms and a bunch of positive feedback in place.

... Phil

Reply to
Phil Allison

Yup. Bootstrapping doesn't help the SNR, it just flattens out the frequency response. It's the same with photodiodes--the noise floor due to the amplifier's voltage noise rises steadily with frequency.

Cheers

Phil Hobbs

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Reply to
Phil Hobbs

Ok I can solve for that, if I treat delta like a constant of, say, 0.01.

But why do you call it delta? What change does it refer to?

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Reply to
Tom Del Rosso

"Tom Del Rosso"

** One is reminded of Humpty Dumpty's declaration to Alice:

" When I use a word ... it means just what I want it to mean, neither more nor less"

... Phil

Reply to
Phil Allison

Just call it "mary" instead ;-) ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

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