C
Chuck Harris
Guest
Watson A.Name - "Watt Sun, the Dark Remover" wrote:
RL is the total load, and as such is the parallel combination of the drain
resistor, and the resistance of the FET, rs. In the linear region, rs
changes depending on where you bias the FET. rs is low when Vgs is near
zero, and high when Vgs is very negative. To get maximal gain, you
want rs to be high. Like all things, you have to reach a compromise.
extra isolation capacitor, and gate resistor. The values you used are
fine. I prefer high impedance because it makes for a smaller network.
classic technique of making each section of the network 10x the impedance
of the last. eg:
-----0.0033---o---0.033---o---0.33---o---gate
| | |
10K 100K 1M
| | |
G G G
Then you will only need a gain of 9 or 10 to overcome the loss of the
phase shift network. (This technique works because it reduces
the loading on the previous stage, and gets you closer to the theoretical
stage where there is no loading on the output of any stage.)
value of rs is everything! If I wanted to operate with a higher
voltage supply, I would then reduce Rs from 22K to 12K. That would give
me clean operation at 15V, but sacrifice operation at 9V.
-Chuck Harris
The problem is the actual gain of the stage is: Av = -RL/gm, whereYou have to give the poor little MPF102 a break, and bias it properly.
It needs about -5V Vgs to operate, so you need to up your source
resistor
to be equal to your drain resistor, if you want it to run at 9V.
Well, that was what I mentioned in another followup. According to the
specs, the MPF102 can have a Vgs of anywhere from .5V to 7.5V; that's a
ratio of fifteen to one. In the case of the FET I chose, it seems to be
working pretty good at about 1.5 V. But I'll vary it to see what effect
it has.
RL is the total load, and as such is the parallel combination of the drain
resistor, and the resistance of the FET, rs. In the linear region, rs
changes depending on where you bias the FET. rs is low when Vgs is near
zero, and high when Vgs is very negative. To get maximal gain, you
want rs to be high. Like all things, you have to reach a compromise.
I arranged the phase circuit the way I did to eliminate the need for theThe 330k and 1uF sections of my schem, and the 1M and .33uF sections of
your schem have the same net effect because they are both loaded by the
gate, which is nearly infinite impedance, so the gate is negligible.
And since in both cases the impedance of the RC net is more than ten
times the drain load resistor, the RC net has negligible loading effect
in the drain resistor. Do you agree with this? If so, then it seems
that there should be no need to change from 330k and 1 uF to 1M and .33
uF RC sections.
extra isolation capacitor, and gate resistor. The values you used are
fine. I prefer high impedance because it makes for a smaller network.
Well, if you are really concerned about attenuation, you can use theSo after all this, it comes down to the number of sections, mine with
four, and yours with three. I tried the three sections, and I could
barely tell that there was some overshoot on the output waveform, it was
_very_ damped. So there was a serious need for more gain or less
attenuation. So I went with four sections to get less attenuation.
classic technique of making each section of the network 10x the impedance
of the last. eg:
-----0.0033---o---0.033---o---0.33---o---gate
| | |
10K 100K 1M
| | |
G G G
Then you will only need a gain of 9 or 10 to overcome the loss of the
phase shift network. (This technique works because it reduces
the loading on the previous stage, and gets you closer to the theoretical
stage where there is no loading on the output of any stage.)
Getting the gate biased into the correct region that gives a highAnd I tried the 22k drain load resistor, but it was still too low.
Finally when I got up into the high 20s I started to see some life in
the damped oscillation. At 33k it finally was able to sustain
oscillations. What I need to do now is fiddle with the source bias
resistor to get the waveform to clip symmetrically. So your suggestion
of changing the Vgs looks like it'll be one key to get the waveform
looking reasonable.
value of rs is everything! If I wanted to operate with a higher
voltage supply, I would then reduce Rs from 22K to 12K. That would give
me clean operation at 15V, but sacrifice operation at 9V.
-Chuck Harris