(Damien Teney) wrote in message . com...

And one last thing, could you re-explain how you advise to shield the
FET ? I don't see what you mean :-(

Thanks for your answer ;-)

let me explain this a little more visually.

to begin with, i imagine that you are building this ugly style over a
copper clad board. Now, imagine that you have soldered another piece
of copper clad board (about an inch high and 2 inches across) so that
stands vertically at ninety degrees from the base board.
Now imagine that you have cut out a small mouse hole in this 'wall'.
The size of the mouse hole is just enough to let the FET's body (and
not the leads) pass through.
You take an FET, bend its drain and source to ninety degrees and away
from the FET body. Now you slide the FET into the mouse hole. Bend
down the gate and solder it to the base copper clad board.The source
and drain leads should be on either sides of this wall (the sheild).
this will prevent the source and drain from coupling the energy back
to each other.

- farhan

Thanks I've understood what you mean ;-)

bias the FET for nominal current at about half the Idss.

As I'm not (yet) familiar with such circuits, so could you tell me how to do
this :-$ ? Thanks.

Damien

In article ,
Ashhar Farhan wrote:

let me explain this a little more visually.

to begin with, i imagine that you are building this ugly style over a
copper clad board. Now, imagine that you have soldered another piece
of copper clad board (about an inch high and 2 inches across) so that
stands vertically at ninety degrees from the base board.
Now imagine that you have cut out a small mouse hole in this 'wall'.
The size of the mouse hole is just enough to let the FET's body (and
not the leads) pass through.
You take an FET, bend its drain and source to ninety degrees and away
from the FET body. Now you slide the FET into the mouse hole. Bend
down the gate and solder it to the base copper clad board.The source
and drain leads should be on either sides of this wall (the sheild).
this will prevent the source and drain from coupling the energy back
to each other.

A slight variation on this approach, which I've seen recommended for
use with U310 (TO-52 metal case) is to actually drill a small hole
downwards through the "ugly-style" copperclad, just barely large
enough to admit the body of the JFET. Drop the JFET into the hole -
it's fine if the metal case contacts the copper, as the case and gate
are connected together. Bend the source and drain out sidewise, leave
the gate lead sticking up in the air, put the shielding piece of
copperclad with the "mouse hole" into place, solder it to the board,
and then solder the gate lead to the copperclad on one side of the
shield.

This results in a very short, low-inductance connection of the gate to
ground.

This technique would probably work just about as well with a J310 or
similar TO-92 plastic-package JFET, although it won't give the
additional benefit of grounding/shielding via the metal case.

As to the U310 - anyone know of a convenient source? I haven't seen
anyone selling old-stock U310s on the Net, and there's only one
manufacturer I know of making them (Linear Systems).

--
Dave Platt AE6EO
Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
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"Damien Teney" wrote in message ...
Thanks I've understood what you mean ;-)

bias the FET for nominal current at about half the Idss.

As I'm not (yet) familiar with such circuits, so could you tell me how to do
this :-$ ? Thanks.

Damien

W7ZOI has a quick and dirty way of finding this out.
1) Connect a 1K resistor from the drain of the FET to 12v power
supply.
2) solder a 10K resistor from the source of the FET to ground.
3) ground the gate
4) apply power and measure the voltage on the source. this gives you
the pinchoff voltage (Vp).
5) short the source to the ground as well. Meausre the current flowing
through the drain (you can measure the voltage between the drain and
the 12v supply and divide it by 1000). This current value is IDss.

Now, choose a source resistor of the value Vp/(Idss * 4).

btw, you can simpley chuck all this and try 560 ohms. it should work
. the gain, noise figure etc are not too crucial. the idea is to
buffer the input and output.

- farhan

- farhan

Okay, but that stuff is only used to find out the Vp and Idss I guess ?
So the "amplifier" itself is only made up of the FET, with the gate
grounded, the drain as output, and the source as input via 1 resistor (560
ohms) ?

Damien

PS: that's probably the last question ;-)

W7ZOI has a quick and dirty way of finding this out.
1) Connect a 1K resistor from the drain of the FET to 12v power
supply.
2) solder a 10K resistor from the source of the FET to ground.
3) ground the gate
4) apply power and measure the voltage on the source. this gives you
the pinchoff voltage (Vp).
5) short the source to the ground as well. Meausre the current flowing
through the drain (you can measure the voltage between the drain and
the 12v supply and divide it by 1000). This current value is IDss.

Now, choose a source resistor of the value Vp/(Idss * 4).

btw, you can simpley chuck all this and try 560 ohms. it should work
. the gain, noise figure etc are not too crucial. the idea is to
buffer the input and output.

- farhan

There are a few possible sources of problems
(1) Power supply for the PLL and VCO - noise on these will appear in the
output. use separate supplies with ?zeners rather than IC regulators
(2) Is the amplitude from the buffer enough to drive the prescaler properly?
(3) Is the loop stable - how did you arrive at your calculations for the
loop values?
(4) The VCO tank seems to be heavily loaded - it has an output directly
to the Rx and a large (68p) capacitor to the varicap which is low Q.
This would make it noisier. Why not take the output to the Rx from the
buffer? (is that how it was originally?)
Also could you use a smaller cap to the varicap? (say 5p or less)

Richard


Damien Teney wrote:
Hello everybody,
I have build a VHF receiver for the airband, published in the magazine
Elektor a few years ago, and I've managed to build a PLL to improve it. The
PLL is build on another PCB, that is connected to the main circuit, and that
replaces the original potmeter, in order to choose the frequency. The PLL is
able to lock on the right frequency, but the receiver is a lot more noisy,
and it seems really less sensitive with the PLL. See these sketches,
I think it is better than a long explaination.

http://www.mcarsweb.com/_divers/1sketch.jpg
http://www.mcarsweb.com/_divers/2sketch.jpg
http://www.mcarsweb.com/_divers/3sketch.jpg

I have also included the schematics of the VCO (with the buffering
transistor, T4), and the basic schematic of the PLL (taken from a Motorola
application note). I don't know where the problem could come from. I would
say from the bad buffering of the VCO output, but I don't see how it could
be improved.

http://www.mcarsweb.com/_divers/4VCO.jpg
http://www.mcarsweb.com/_divers/5PLL.jpg

I'm really getting mad with this problem ! So thanks to all of you who will
answer ;-)

Thank you Ashhar, your tip was the best :-) I've added a grounded gate FET
between the VCO and the PLL and now the receiver works as well with as
without the PLL circuit. I still have to adjust the loop filter because it
is not perfect; it can lock on the right frequency and keep it but the
output voltage of the filter doesn't look clean enough on the scope.

Thanks everybody for your advises, and cya ;-)

Damien