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Old 21st March 2003, 02:08 PM   #1
uli is offline uli  Austria
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Lightbulb X-Modul

I made some kinde of discreete modul (1.6" x 2") to x
existing designs or make simple new ones.

Uli
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Old 21st March 2003, 02:09 PM   #2
uli is offline uli  Austria
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Some additional drawings...
U.
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Old 21st March 2003, 02:26 PM   #3
uli is offline uli  Austria
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Default Re: X-Modul

sorry, correct attatchment uploaded...
U.


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Old 25th May 2003, 10:26 PM   #4
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Uli,

This is rather nice work. Have you built/tested it? I have do doubt this will work (since it is true to the original).

Petter
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Old 26th May 2003, 07:44 AM   #5
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Thumbs up works

Hi Petter,
it works, there are 2 errors in the pcb design which
I fixed by 2 wires (forgotten connections)

Uli
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Old 26th May 2003, 09:11 AM   #6
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Looks nice, but I have a hard time recognizing the X in this circuit.

Could you please elaborate a little on this.

Thanks in advance.
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Old 26th May 2003, 09:36 AM   #7
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What people typically have been referring to as the "X resistor" is set to zero ohms. That is probably what you are not seeing. Ordinarily there would have been two current sources in the middle, not one. If you look carefully at current levels (reads the middle bottom resistor at half the value of the two surrounding it), you will see that the one current source does the duty of two such units.

Now what happens when the resistor that has been created so much furore is zero?

a) You get more gain.
b) I question whether there are other effects (other than trimmability)

Since JFET's are used, their current gain is much smaller than that of MOSFET's so I don't have a problem with this. In fact, I have my misgivings about the need for a non-zero resistor in there anyway. How small can a resistor be before we are no longer "X-ing" seems a mute point.

Again, nice work, Uli. I am looking forward to impressions.

Petter
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Old 26th May 2003, 10:32 AM   #8
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I understand that the resistor that partially sets the gain is zero ohm. But how is the output signal from the left half of the amp being fed back to the right half of the amp and vice versa?
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Old 26th May 2003, 10:52 AM   #9
uli is offline uli  Austria
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Lightbulb X idea

Quote:
Originally posted by rtirion
I understand that the resistor that partially sets the gain is zero ohm. But how is the output signal from the left half of the amp being fed back to the right half of the amp and vice versa?
Hi
when you take a look at the AX (or XA?) designs you see this
resistor being zero.
IMHO the X does not depend on this resistor.
NP stated somewhere that this resistor is just to control too
high open loop gain which can cause instability.
The X comes from the double feedback R17 & R18.
When you feedback the signal from drain to gate (what happens
here) the errorsignal can be seen at the source which couples
this errorsignal into the source of the second fet thus creating
a double (in phase) mirror image at the drain of the second fet!
Actually you normally need to scale down this coupling to control
mirror chamber effects as seen in the X-Amps. As the admittance
of the JFETs is fairly lo there is no need to scale down!

This is not a standalone design but meant to be
put in as a replacement of existing gainstages!


This means you have to apply the same kind of external circuitry
as if it were an OpAmp.

Thanks

Uli
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Old 26th May 2003, 03:12 PM   #10
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Another way to "X" things would be to connect regular source degeneration resistors above the "short between current sources" which is the way things are done in ordinary long tailed pairs. This is to a large extent equivalent to using the "magical resistor".

Petter
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