On Feb 20, 2:44*pm, "amdx" wrote:
"John Larkin" wrote in message

...



On Sun, 20 Feb 2011 09:20:12 -0600, "amdx" wrote:

Hi all,
I finished the amp that had the 5 Ghz transistor, I changed it to a
slower
one.
The objective of this amp is to cause minimal loading of the circuit it
is
measuring.
When I install the box cover the voltage gain drops by 7%, so I think the
input capacitor
plate is being loaded by the cover.
The input capacitor plates can be seen here;

http://i395.photobucket.com/albums/p...mspaced5mm.jpg

The plates are 1 cm x 1 cm spaced 5 mm apart.

I have thoughts about *rectangular plates 0.25 cm x 4 cm to get more
distance from the top cover, (and the bottom.)
Or a real gimmick cap where I twist a couple of 39 Gauge wires together
and
attach opposite ends to input and output.

*Any ideas to minimize input capacitance to the box?

Here's the amp in box.
http://i395.photobucket.com/albums/p...erampinbox.jpg

This is the original circuit page with schematic;
http://www.crystal-radio.eu/enfetamp.htm
* * * * * * * Thanks, Mike

PS, I was having trouble getting some close-up pictures, I grabbed a
magnifying glass and took some
pictures through that, works good.

Use a real surface-mount 0.3 pF cap, or a homemade coaxial cap. The 1
cm square plates are too big and have their own capacitance to the
world.

Bootstrap the drain of Q1.

"T" means transformer, which shows that this circuit was done by an
amateur. All that tricky stuff could be replaced by one opamp.

It could have close to zero Cin with a little positive feedback.

John

* * Bootstrap the drain of Q1.
You need to walk me through that, (I'm an amateur)

Ah, he's done some nice work on the subject of crystal radios and high Q
inductors.http://www.crystal-radio.eu/index.html
Page down to experiment with LC circuits.

* It could have close to zero Cin with a little positive feedback.

*How much closer? If the input cap is 0.3pf what do you the input impedance
is?
Input is 0.3pf, 20 Meg to ground driving FET gate.
* * * * * * * * * * * * * * * * * Thanks, Mikek

We're talking about something roughly like this:

+12V -o-----o-----------.
| | |
| | |
| | R5 3.3M
| |Q2 |
| \| | V1
| |---o-----o ~~ ~=+3V
| .| | |
| | | R6 1M
R3 | | |
C1 | | |C3 ===
10pF | g |-'d ---
in--||--o--|--| ---
| | |-.s |100n
R1 10M | |Q1 |
| | | |
o--o-||--o-----o-----------out
| C2 |
R2 10M 100nF|
| |
=== R4 470 R
|
|
===

C2 drives the center of the input bias resistor, which cancels the
loading caused by R1-R2. This is called "bootstrapping".

Q2 does the same thing for the drain terminal--it causes the drain
terminal to go up and down with the input signal. That saves the
input signal from having to charge Q1's gate-to-drain capacitance,
effectively making that capacitance disappear.

This front-end has *much* higher impedance than the original, and a
predictable gain that's close to 1.

R3 is to bias the FET output to (V1)/2, for maximum dynamic range.
Higher V1 would give bigger dynamic range.

A cheapskate could put a resistor in Q2's collector and use it as a
voltage-gain stage too.


--
Cheers,
James Arthur

On Feb 20, 3:30*pm, wrote:
On Feb 20, 2:44*pm, "amdx" wrote:



"John Larkin" wrote in message

.. .

On Sun, 20 Feb 2011 09:20:12 -0600, "amdx" wrote:

Hi all,
I finished the amp that had the 5 Ghz transistor, I changed it to a
slower
one.
The objective of this amp is to cause minimal loading of the circuit it
is
measuring.
When I install the box cover the voltage gain drops by 7%, so I think the
input capacitor
plate is being loaded by the cover.
The input capacitor plates can be seen here;

http://i395.photobucket.com/albums/p...mspaced5mm.jpg

The plates are 1 cm x 1 cm spaced 5 mm apart.

I have thoughts about *rectangular plates 0.25 cm x 4 cm to get more
distance from the top cover, (and the bottom.)
Or a real gimmick cap where I twist a couple of 39 Gauge wires together
and
attach opposite ends to input and output.

*Any ideas to minimize input capacitance to the box?

Here's the amp in box.
http://i395.photobucket.com/albums/p...erampinbox.jpg

This is the original circuit page with schematic;
http://www.crystal-radio.eu/enfetamp.htm
* * * * * * * Thanks, Mike

PS, I was having trouble getting some close-up pictures, I grabbed a
magnifying glass and took some
pictures through that, works good.

Use a real surface-mount 0.3 pF cap, or a homemade coaxial cap. The 1
cm square plates are too big and have their own capacitance to the
world.

Bootstrap the drain of Q1.

"T" means transformer, which shows that this circuit was done by an
amateur. All that tricky stuff could be replaced by one opamp.

It could have close to zero Cin with a little positive feedback.

John

* * Bootstrap the drain of Q1.
You need to walk me through that, (I'm an amateur)

Ah, he's done some nice work on the subject of crystal radios and high Q
inductors.http://www.crystal-radio.eu/index.html
Page down to experiment with LC circuits.

* It could have close to zero Cin with a little positive feedback.

*How much closer? If the input cap is 0.3pf what do you the input impedance
is?
Input is 0.3pf, 20 Meg to ground driving FET gate.
* * * * * * * * * * * * * * * * * Thanks, Mikek

We're talking about something roughly like this:

* * +12V -o-----o-----------.
* * * * * *| * * | * * * * * |
* * * * * *| * * | * * * * * |
* * * * * *| * * | * * * R5 3.3M
* * * * * *| * * |Q2 * * * * |
* * * * * *| * * *\| * * * * | * * * *V1
* * * * * *| * * * |---o-----o * ~~ ~=+3V
* * * * * *| * * .| * | * * |
* * * * * *| * * | * * | *R6 1M
* * * * * R3 * * | * * | * * |
* * C1 * * | * * | * * |C3 *===
* * 10pF * | g |-'d * ---
in--||--o--|--| * * *---
* * * * | *| * |-.s * *|100n
* * R1 10M | * * |Q1 * |
* * * * | *| * * | * * |
* * * * o--o-||--o-----o-----------out
* * * * | * *C2 *|
* * R2 10M *100nF|
* * * * | * * * *|
* * * *=== * R4 470 R
* * * * * * * * *|
* * * * * * * * *|
* * * * * * * * ===

C2 drives the center of the input bias resistor, which cancels the
loading caused by R1-R2. *This is called "bootstrapping".

Q2 does the same thing for the drain terminal--it causes the drain
terminal to go up and down with the input signal. *That saves the
input signal from having to charge Q1's gate-to-drain capacitance,
effectively making that capacitance disappear.

This front-end has *much* higher impedance than the original, and a
predictable gain that's close to 1.

R3 is to bias the FET output to (V1)/2, for maximum dynamic range.
Higher V1 would give bigger dynamic range.

A cheapskate could put a resistor in Q2's collector and use it as a
voltage-gain stage too.

Oh, I didn't calculate the biasing, so R4's probably wrong.

--James

On Feb 20, 3:30*pm, wrote:
On Feb 20, 2:44*pm, "amdx" wrote:





"John Larkin" wrote in message

.. .

On Sun, 20 Feb 2011 09:20:12 -0600, "amdx" wrote:

Hi all,
I finished the amp that had the 5 Ghz transistor, I changed it to a
slower
one.
The objective of this amp is to cause minimal loading of the circuit it
is
measuring.
When I install the box cover the voltage gain drops by 7%, so I think the
input capacitor
plate is being loaded by the cover.
The input capacitor plates can be seen here;

http://i395.photobucket.com/albums/p...mspaced5mm.jpg

The plates are 1 cm x 1 cm spaced 5 mm apart.

I have thoughts about *rectangular plates 0.25 cm x 4 cm to get more
distance from the top cover, (and the bottom.)
Or a real gimmick cap where I twist a couple of 39 Gauge wires together
and
attach opposite ends to input and output.

*Any ideas to minimize input capacitance to the box?

Here's the amp in box.
http://i395.photobucket.com/albums/p...erampinbox.jpg

This is the original circuit page with schematic;
http://www.crystal-radio.eu/enfetamp.htm
* * * * * * * Thanks, Mike

PS, I was having trouble getting some close-up pictures, I grabbed a
magnifying glass and took some
pictures through that, works good.

Use a real surface-mount 0.3 pF cap, or a homemade coaxial cap. The 1
cm square plates are too big and have their own capacitance to the
world.

Bootstrap the drain of Q1.

"T" means transformer, which shows that this circuit was done by an
amateur. All that tricky stuff could be replaced by one opamp.

It could have close to zero Cin with a little positive feedback.

John

* * Bootstrap the drain of Q1.
You need to walk me through that, (I'm an amateur)

Ah, he's done some nice work on the subject of crystal radios and high Q
inductors.http://www.crystal-radio.eu/index.html
Page down to experiment with LC circuits.

* It could have close to zero Cin with a little positive feedback.

*How much closer? If the input cap is 0.3pf what do you the input impedance
is?
Input is 0.3pf, 20 Meg to ground driving FET gate.
* * * * * * * * * * * * * * * * * Thanks, Mikek

We're talking about something roughly like this:

* * +12V -o-----o-----------.
* * * * * *| * * | * * * * * |
* * * * * *| * * | * * * * * |
* * * * * *| * * | * * * R5 3.3M
* * * * * *| * * |Q2 * * * * |
* * * * * *| * * *\| * * * * | * * * *V1
* * * * * *| * * * |---o-----o * ~~ ~=+3V
* * * * * *| * * .| * | * * |
* * * * * *| * * | * * | *R6 1M
* * * * * R3 * * | * * | * * |
* * C1 * * | * * | * * |C3 *===
* * 10pF * | g |-'d * ---
in--||--o--|--| * * *---
* * * * | *| * |-.s * *|100n
* * R1 10M | * * |Q1 * |
* * * * | *| * * | * * |
* * * * o--o-||--o-----o-----------out
* * * * | * *C2 *|
* * R2 10M *100nF|
* * * * | * * * *|
* * * *=== * R4 470 R
* * * * * * * * *|
* * * * * * * * *|
* * * * * * * * ===

C2 drives the center of the input bias resistor, which cancels the
loading caused by R1-R2. *This is called "bootstrapping".

Q2 does the same thing for the drain terminal--it causes the drain
terminal to go up and down with the input signal. *That saves the
input signal from having to charge Q1's gate-to-drain capacitance,
effectively making that capacitance disappear.

This front-end has *much* higher impedance than the original, and a
predictable gain that's close to 1.

R3 is to bias the FET output to (V1)/2, for maximum dynamic range.
Higher V1 would give bigger dynamic range.

A cheapskate could put a resistor in Q2's collector and use it as a
voltage-gain stage too.

--
Cheers,
James Arthur- Hide quoted text -

- Show quoted text -

Thanks for the explanation James. The bootstrapping of the base does
nothing for the input capacitance.(?) Is that correct?

George H.

On Feb 21, 11:16*am, George Herold wrote:
On Feb 20, 3:30*pm, wrote:



On Feb 20, 2:44*pm, "amdx" wrote:

"John Larkin" wrote in message

.. .

On Sun, 20 Feb 2011 09:20:12 -0600, "amdx" wrote:

Hi all,
I finished the amp that had the 5 Ghz transistor, I changed it to a
slower
one.
The objective of this amp is to cause minimal loading of the circuit it
is
measuring.
When I install the box cover the voltage gain drops by 7%, so I think the
input capacitor
plate is being loaded by the cover.
The input capacitor plates can be seen here;

http://i395.photobucket.com/albums/p...mspaced5mm.jpg

The plates are 1 cm x 1 cm spaced 5 mm apart.

I have thoughts about *rectangular plates 0.25 cm x 4 cm to get more
distance from the top cover, (and the bottom.)
Or a real gimmick cap where I twist a couple of 39 Gauge wires together
and
attach opposite ends to input and output.

*Any ideas to minimize input capacitance to the box?

Here's the amp in box.
http://i395.photobucket.com/albums/p...erampinbox.jpg

This is the original circuit page with schematic;
http://www.crystal-radio.eu/enfetamp.htm
* * * * * * * Thanks, Mike

PS, I was having trouble getting some close-up pictures, I grabbed a
magnifying glass and took some
pictures through that, works good.

Use a real surface-mount 0.3 pF cap, or a homemade coaxial cap. The 1
cm square plates are too big and have their own capacitance to the
world.

Bootstrap the drain of Q1.

"T" means transformer, which shows that this circuit was done by an
amateur. All that tricky stuff could be replaced by one opamp.

It could have close to zero Cin with a little positive feedback.

John

* * Bootstrap the drain of Q1.
You need to walk me through that, (I'm an amateur)

Ah, he's done some nice work on the subject of crystal radios and high Q
inductors.http://www.crystal-radio.eu/index.html
Page down to experiment with LC circuits.

* It could have close to zero Cin with a little positive feedback.

*How much closer? If the input cap is 0.3pf what do you the input impedance
is?
Input is 0.3pf, 20 Meg to ground driving FET gate.
* * * * * * * * * * * * * * * * * Thanks, Mikek

We're talking about something roughly like this:

* * +12V -------+-----------.
* * * * * * * * | * * * * * |
* * * * * * * * | * * * * * |
* * * * * * * * | * * * R5 220K
* * * * * * * * |Q2 * * * * |
* * * * * * * * *\| * * * * | * * * *V1
* * * * * * * * * |---+-----+ * ~~ ~=7V
* * * * * * * * .| * | * * |
* * * * * * * * | * * | *R6 330K
* * * * * * * | * * | * * |
* * C1 * * * * | * * |C3 *===
* * 10pF * g |-'d * ---
in--||--+-----| * * *---
* * * * | * * |-.s * *|100n
* * R1 10M * * |Q1 * |
* * * * | * * * | * * |
* * * * +----||--+-----+------ out
* * * * | * *C2 *|
* * R2 10M *100nF|
* * * * | * * * *|
* * * *=== * R4 2.2K
* * * * * * * * *|
* * * * * * * * *|
* * * * * * * * ===

edited, simplified, R4-6 values corrected

C2 drives the center of the input bias resistor, which cancels the
loading caused by R1-R2. *This is called "bootstrapping".

Q2 does the same thing for the drain terminal--it causes the drain
terminal to go up and down with the input signal. *That saves the
input signal from having to charge Q1's gate-to-drain capacitance,
effectively making that capacitance disappear.

This front-end has *much* higher impedance than the original, and a
predictable gain that's close to 1.



Thanks for the explanation James. *The bootstrapping of the base does
nothing for the input capacitance.(?) Is that correct?

If you mean "gate" and Q1, that's inherently bootstrapped by the
voltage-follower configuration: the source already rises and falls
with the input, leaving the Miller capacitance as by far the biggest
nasty. The best, easy bootstrap for the FET input capacitance is
getting the FET follower gain up closer to unity. That means using a
current source for the source load instead of R4.

The input bootstrap above--C2 to the junction of R1-R2--serves to
reduce the a) stray capacitive loading posed by R1, and b) raise the
a.c. input impedance to much higher than the 20M input resistor in
parallel with the FET gate impedance.

--
Cheers,
James Arthur

On Sun, 20 Feb 2011 13:44:37 -0600, "amdx" wrote:


"John Larkin" wrote in message
.. .
On Sun, 20 Feb 2011 09:20:12 -0600, "amdx" wrote:

Hi all,
I finished the amp that had the 5 Ghz transistor, I changed it to a
slower
one.
The objective of this amp is to cause minimal loading of the circuit it
is
measuring.
When I install the box cover the voltage gain drops by 7%, so I think the
input capacitor
plate is being loaded by the cover.
The input capacitor plates can be seen here;

http://i395.photobucket.com/albums/p...mspaced5mm.jpg

The plates are 1 cm x 1 cm spaced 5 mm apart.

I have thoughts about rectangular plates 0.25 cm x 4 cm to get more
distance from the top cover, (and the bottom.)
Or a real gimmick cap where I twist a couple of 39 Gauge wires together
and
attach opposite ends to input and output.

Any ideas to minimize input capacitance to the box?

Here's the amp in box.
http://i395.photobucket.com/albums/p...erampinbox.jpg

This is the original circuit page with schematic;
http://www.crystal-radio.eu/enfetamp.htm
Thanks, Mike

PS, I was having trouble getting some close-up pictures, I grabbed a
magnifying glass and took some
pictures through that, works good.


Use a real surface-mount 0.3 pF cap, or a homemade coaxial cap. The 1
cm square plates are too big and have their own capacitance to the
world.

Bootstrap the drain of Q1.

"T" means transformer, which shows that this circuit was done by an
amateur. All that tricky stuff could be replaced by one opamp.

It could have close to zero Cin with a little positive feedback.

John


Bootstrap the drain of Q1.
You need to walk me through that, (I'm an amateur)


If the source of the fet follower drives an opamp with a gain of, say,
+2, you could take the amp's output, run it through a pot, and AC
couple that into the drain.

With the pot at zero gain, there's no bootstrapping, so the fet's Cg-d
loads the input 100%.

With the pot at mid-rotation, the drain is forced to swing up and down
just about as much as the input signal does. Since both ends of Cg-d
are at the same signal level, there's almost no current in that
capacitance, so it sort of disappears, as far as the gate is
concerned.

Turn the pot a little more, and a small excursion at the gate results
in a *bigger* excursion at the drain. So the current in Cg-d flows in
the opposite direction from a real capacitor, and Cg-d becomes a
negative capacitor. It can then be used to cancel all other
capacitance at the gate node, making the box have a net nearly-zero
input capacitance.

Or just add a small variable cap from the opamp output back into the
gate, and adjust for near zero input capacitance. When it oscillates,
you've gone too far.

John

On Sun, 20 Feb 2011 09:20:12 -0600, "amdx" wrote:

http://i395.photobucket.com/albums/p...mspaced5mm.jpg

The plates are 1 cm x 1 cm spaced 5 mm apart.


Ever consider simply using a proper air core variable cap?

"FigureItOut" wrote in message
...
On Sun, 20 Feb 2011 09:20:12 -0600, "amdx" wrote:

http://i395.photobucket.com/albums/p...mspaced5mm.jpg

The plates are 1 cm x 1 cm spaced 5 mm apart.


Ever consider simply using a proper air core variable cap?

You have any 0.3pf air variables.
Why is this one improper?
Mikek

On Sun, 20 Feb 2011 14:26:42 -0600, "amdx" wrote:


"FigureItOut" wrote in message
.. .
On Sun, 20 Feb 2011 09:20:12 -0600, "amdx" wrote:

http://i395.photobucket.com/albums/p...mspaced5mm.jpg

The plates are 1 cm x 1 cm spaced 5 mm apart.


Ever consider simply using a proper air core variable cap?

You have any 0.3pf air variables.

You could manage .3 with the empty tank alone.

Why is this one improper?
Mikek

Because applying the lid changed the value. Duh.

That air gap is so big that even local electrostatic fields would ****
with it.

On 2/23/2011 8:38 PM, FigureItOut wrote:
On Sun, 20 Feb 2011 14:26:42 -0600, wrote:


wrote in message
...
On Sun, 20 Feb 2011 09:20:12 -0600, wrote:

http://i395.photobucket.com/albums/p...mspaced5mm.jpg

The plates are 1 cm x 1 cm spaced 5 mm apart.


Ever consider simply using a proper air core variable cap?

You have any 0.3pf air variables.

You could manage .3 with the empty tank alone.


Can you back that up with calculations using believable numbers?


Why is this one improper?
Mikek

Because applying the lid changed the value. Duh.

That air gap is so big that even local electrostatic fields would ****
with it.


Plagiarism from an earlier post aside, you are on the verge of offering
some legitimate advice. But your delivery sucks more than a vacuum.
Maybe you have a black hole. You should look into that. Literally.

On Wed, 23 Feb 2011 20:56:40 -0600, John - KD5YI
wrote:

Can you back that up with calculations using believable numbers?


**** you, you ****ing retard. Did you not read the post? He said that
putting the lid on changed the value. You would have to be a complete
dolt (you are) to miss that one!

Again, come back when you are not being your usual mouthy little bitch
self.

Plagiarism? You're an idiot, and a goddamned liar, boy. I was making
feedback circuit caps before you even knew what an antenna match was.

**** off, Johnny. Die, Johnny. Die painfully, Johnny. Go mouth
someone more deserving of the **** you spew... your mother.