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