# How to bootstrap

A

#### amdx

Jan 1, 1970
0
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

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

P

#### Phil Allison

Jan 1, 1970
0
"amdx"
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

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.

** 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

P

#### Phil Allison

Jan 1, 1970
0
"Jim Thompson"
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.

** Another totally pedantic WANK.

" Jim Thompson = Autistic Pig "

.... Phil

A

#### amdx

Jan 1, 1970
0
"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
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

A

#### amdx

Jan 1, 1970
0
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
Now Jim, if I could do that I wouldn't be asking such a question.
Mikek

P

#### Phil Allison

Jan 1, 1970
0
"amdx"
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.

** Don't invent silly specs.

It's the mark of an utter fuckwit to do that.
I have a High Q LC circuit, I put my bootstrapped measurement device
in parallel,

** FET follower OK.

No bootstrapping needed.

So does the C of my measurement device load the circuit?
Or just change the resonant frequency?

** It will probably do neither.

Cos you circuit is a fantasy wank anyhow.

Piss off.

P

#### Phil Allison

Jan 1, 1970
0
"amdx"
Now Jim, if I could do that I wouldn't be asking such a question.

** 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

G

#### George Herold

Jan 1, 1970
0
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

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

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.

A

#### amdx

Jan 1, 1970
0
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.

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

Mikek

A

#### amdx

Jan 1, 1970
0
"amdx"

** Don't invent silly specs.

It's the mark of an utter fuckwit to do that.

** FET follower OK.

No bootstrapping needed.

Piss off.

http://i395.photobucket.com/albums/pp37/Qmavam/inside.jpg
I didn't design the circuit. Did layout the pcb. I'm looking for a
different input circuit. I'm not happy with the mechanical part that I
put together. I used polystyrene with 3/4" hole. Not good mechanical
strength and it doesn't like the heat of soldering.

Mikek

J

#### Jon Kirwan

Jan 1, 1970
0
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

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

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

+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
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.
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

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

A

#### amdx

Jan 1, 1970
0
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

+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
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.
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

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
Thanks for that Jon,
I'm at work now, but I'll print it and look it over when I'm home.
Mikek

J

#### Jon Kirwan

Jan 1, 1970
0
Thanks for that Jon,
I'm at work now, but I'll print it and look it over when I'm home.
Mikek

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

T

#### Tom Del Rosso

Jan 1, 1970
0
Jim said:
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.

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'.

P

#### Phil Allison

Jan 1, 1970
0
"Phil Hobbs"
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.
(snip)

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.

** 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

T

#### Tom Del Rosso

Jan 1, 1970
0
Jim said:
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.

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?

P

#### Phil Allison

Jan 1, 1970
0
"Tom Del Rosso"
But why do you call it delta? What change does it refer to?

** 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

P

#### Phil Allison

Jan 1, 1970
0
"Phil Allison"
** 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.

** Another sort of bootstrapping involves the output stage of an audio power
amplifier.

See low budget Germanium output stage typical of the late 1960s:

http://1.bp.blogspot.com/_7q93wN0Pq1k/TRcwk1PjNZI/AAAAAAAAAA0/BvJcYYW0-FA/s1600/Picture+012.jpg

The 180ohm collector load of the BFX88 is bootstrapped direct to the
loudspeaker. Doing this improves both drive and drive linearity - as the
speaker signal swings to 5 volts below the ground rail.

Note how if the ( 3 ohm) speaker is disconnected, the OP stage is disabled.

There is a double dose of bootstrapping going on at the input too.

..... Phil

J

#### Jon Kirwan

Jan 1, 1970
0
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.

Very bad wording. What I should have written (and intended to
write) was:

"And the DC load impedance as seen by the input increases..."

I think it's clear if you take the whole context in... but I
wrote too abruptly and it could have changed the meaning.

Jon

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