K
Ken Smith
Guest
In article <ceEGe.22172$Ag3.17873@newsfe4-gui.ntli.net>,
colin <no.spam.for.me@ntlworld.com> wrote:
Imagine that you have a little black box that contains the function that
creates the distortion. This box has an input and an output. If the
contents of that box produces an output that depends only on the
instantanious value of the input, that box must create balanced side
bands.
The filtering of the sidebands will only happen in the high Q tuned
circuit, assuming that we are trying to make a low drift oscillator. We
can also assume on that basis that the design is done such that the
effects of JFET parameters on tuning have been minimized. When this is
the case, the sidebands will remain balanced.
other, you have phase modulation. When they are in phase, you have
amplitude modulation. Any pair of side bands can be broken down into the
amplitude modulation component and the phase modulation component. Then
you can disregard the amplitude part.
Actually, the 1/F^3 rise continues into the part near the carrier where
the tuned circuit curve flattens. For that matter, there is often a
sudden increase in the slope at that point to well above the cubed factor.
A frequency drift will appear as a 1/F in the graph. A 1/F noise
modulation of a capacitance will appear as a sqrt(1/F)
[...]
frequencies, the desire is to make the ratio higher. To make the ratio
higher you make the impedance even lower. The rule still applies it just
gets harder to follow.
Remember that I suggested that the amplifier section (FET) be one that has
a much higher band width than needed. This and the low terminal impedance
to to prevent the modulation of FET parameters from being a problem.
more normal sense of the word.
If you mean in the more normal sense of the word then I disagree as stated
above. Non-linearities that operate on the instantanious value always
make equal sidebands.
terminals connected to the frequency determining circuit. The measurement
circuit does not.
If we take a perfectly impractical set of cases I think you will see:
Oscillator #1:
We use a simple 2 capacitor divider
Voltage on inductor = 20V RMS
Voltage from gate to source = 10VRMS
Oscillator #2:
We use the 3 capacitor divider
Voltage on the inductor = 10 billion VRms
Voltage from the gate to source = 10VRMS
Both have the same signal to noise at the gate of the FET but the second
one has about 10^12 less ability for the FET to control the frequency.
10fA/sqrt(Hz) * 100K = 1nV/sqrt(Hz) so we are in the same range as the
noise voltage of a low noise JFET. I have run into this fact in practical
circuits.
not pass through the tuned circuit before they hit the non-linearity in
the FET.
Imagine we have an extremely non-linear amplifier. The amplifier is
assumed to be noiseless and the noise is in a generator, added to the
signal just before the amplifier. The input to output function of the
amplifier can be represented by a series. Lets just take the first few
terms:
Y = X + AX^2
Where:
X is the input
A is a constant
Y is the output
Now we consider X as the sum of (S)ignal and (N)oise.
Y = (S + N) + A(S + N)^2
Y = S + N + AS^2 + 2ANS + AN^2
It is the 2NS that does the dirty work. It will mix noise near the second
harmonic down to near the operating frequency. The more nonlinear things
are, the bigger A will be and the more 2nd harmoic noise gets shifted
down. The higher terms bring the higher frequencies down. As a result,
the more nonlinear the circuit is the noisier it is.
[...]
The amplifier's gain does not effect the bandwidth of the system unless we
are taking the case of a poorly designed oscillator where the transistor
controls the frequency.
--
--
kensmith@rahul.net forging knowledge
colin <no.spam.for.me@ntlworld.com> wrote:
I don't think you understood what I said (meant to say) so I'll try again."Ken Smith" <kensmith@green.rahul.net> wrote in message
news:dcel8g$fvl$1@blue.rahul.net...
In article <2UtGe.18947$Hd4.14395@newsfe2-gui.ntli.net>,
colin <no.spam.for.me@ntlworld.com> wrote:
[...]
No, mixing by multiplying makes equal side bands. The amplitude varies
but the frequency remains exactly the same. This is why I was suggesting
making sure that the amplifier is good to a much higher frequency than
needed. A distortion that acts only on the instantanious voltage, always
creates equal sidebands.
You would be correct if the circuit had a flat flat phase and balanced
amplitude response about its operating frequcncy but this isnt necessarily
so, it depends largly on the feedback topology.
Imagine that you have a little black box that contains the function that
creates the distortion. This box has an input and an output. If the
contents of that box produces an output that depends only on the
instantanious value of the input, that box must create balanced side
bands.
The filtering of the sidebands will only happen in the high Q tuned
circuit, assuming that we are trying to make a low drift oscillator. We
can also assume on that basis that the design is done such that the
effects of JFET parameters on tuning have been minimized. When this is
the case, the sidebands will remain balanced.
When the upper and lower side band components are at 180 degrees to eachAlthough I cant say ive gone into the maths of how altering the
amplitude/phase of the sidebands affects the phase of the output in any
great detail, the modulation gets in there somehow, If you look at the
sidebands of a typical good oscilator the close in noise rises at 1/f^3 this
is due to 1/f noise and the 1/f^2 response of the resonant circuit.
other, you have phase modulation. When they are in phase, you have
amplitude modulation. Any pair of side bands can be broken down into the
amplitude modulation component and the phase modulation component. Then
you can disregard the amplitude part.
Actually, the 1/F^3 rise continues into the part near the carrier where
the tuned circuit curve flattens. For that matter, there is often a
sudden increase in the slope at that point to well above the cubed factor.
A frequency drift will appear as a 1/F in the graph. A 1/F noise
modulation of a capacitance will appear as a sqrt(1/F)
[...]
The OP is talking about a lowish frequency oscillator. Even at highI stated this happens and that, this again is why I suggested low AC
impedance nodes at the FET. This reduces the effect of any variation in
capacitance.
The ratio of parasitic capacitance seen by the tank circuit to capacitance
of the tank itself would determine how much this would afect the frequcncy.
If you have a high frequcncy oscilator you cant make this very large,
despite that the impedance would be quite low.
frequencies, the desire is to make the ratio higher. To make the ratio
higher you make the impedance even lower. The rule still applies it just
gets harder to follow.
Once again I think you've misunderstood or perhaps I've misunderstood you.also any noise will efectivly apear as a phase error if it is used in
highly
non linear mode, and a changing phase erroor = change frequcncy.
No, this is not true. Noise that is identical side to side WRT the
carrier does not effect the phase.
If it moves the operating point of the circuit then this may have an effect
on the delay through the amplifier and hence phase.
Remember that I suggested that the amplifier section (FET) be one that has
a much higher band width than needed. This and the low terminal impedance
to to prevent the modulation of FET parameters from being a problem.
Huh? Do you mean non-linearities in the phase or nonlinearities in thealso non linearities
will efectivly make the noise no longer identical side to side of the
operating point.
more normal sense of the word.
If you mean in the more normal sense of the word then I disagree as stated
above. Non-linearities that operate on the instantanious value always
make equal sidebands.
This is a different case. The amplifier in an oscillator has itsNo, this is not true. You can run a larger voltage on the inductor to
keep the voltage at the gate the same as you lower the impedance.
I dont see why you think this is not true, many RF FET datasheets specify
noise performance by using 50 ohm mathcing networks wich also step up the
voltage, this is how they arive at such spectacular noise figures.
terminals connected to the frequency determining circuit. The measurement
circuit does not.
If we take a perfectly impractical set of cases I think you will see:
Oscillator #1:
We use a simple 2 capacitor divider
Voltage on inductor = 20V RMS
Voltage from gate to source = 10VRMS
Oscillator #2:
We use the 3 capacitor divider
Voltage on the inductor = 10 billion VRms
Voltage from the gate to source = 10VRMS
Both have the same signal to noise at the gate of the FET but the second
one has about 10^12 less ability for the FET to control the frequency.
Lets say 10fA/sqrt(Hz) and few 100K impedance.A very high Q tuned circuit has a high enough impedance that the noise
current of the FET starts to matter too.
Noise current of good FETs/MOSFETs is small, realy small, ... down to a few
fa/rt hz.
10fA/sqrt(Hz) * 100K = 1nV/sqrt(Hz) so we are in the same range as the
noise voltage of a low noise JFET. I have run into this fact in practical
circuits.
No, the harmonics are in the noise voltage of the gate of the FET they doThis is certainly not true. ALC style oscillators are what you use if you
want the most stable frequency. Clipping and distortion of any type
allows the noise components near the harmonics of the carrier to be
re-introduced as near carrier components. This reduces the performance a
great deal.
The harmonics will be greatly attenuated by the tank circuit before they can
be re introduced close to the carrier frequency.
not pass through the tuned circuit before they hit the non-linearity in
the FET.
It is fairly straight forward if you take a very simplified case:Although im not too clear of the exact mechanism where they get introduced
Imagine we have an extremely non-linear amplifier. The amplifier is
assumed to be noiseless and the noise is in a generator, added to the
signal just before the amplifier. The input to output function of the
amplifier can be represented by a series. Lets just take the first few
terms:
Y = X + AX^2
Where:
X is the input
A is a constant
Y is the output
Now we consider X as the sum of (S)ignal and (N)oise.
Y = (S + N) + A(S + N)^2
Y = S + N + AS^2 + 2ANS + AN^2
It is the 2NS that does the dirty work. It will mix noise near the second
harmonic down to near the operating frequency. The more nonlinear things
are, the bigger A will be and the more 2nd harmoic noise gets shifted
down. The higher terms bring the higher frequencies down. As a result,
the more nonlinear the circuit is the noisier it is.
Reference left for later look up.on this, look up "Oscillator Phase Noise Reduction Using Nonlinear Design
Techniques"
[...]
Actually the gain must be exactly one if the amplitude is constant.At the fequency of oscillation the gain of the total loop will be slightly
greater than 1 to sustain the oscillation frequcncy, any noise that gets
introduced that is close enough to the center frequcncy so that it is still
within the response of gain >1 will be amplified significantly.
The gain around the loop must be exactly one at the operating frequency.After all the oscilation frequency is in fact just the noise selectivly
amplified many many times by positive feedback, therfore what im saying is
that the close in noise in the output is much much greater than simply the
voltage noise at the input.
Thus keeping the gain as low as posible will ensure the bandwidth with gain
1 is also as low as posible.
The amplifier's gain does not effect the bandwidth of the system unless we
are taking the case of a poorly designed oscillator where the transistor
controls the frequency.
--
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kensmith@rahul.net forging knowledge