Hi,
As you said i analyzed my loop.. i encountered a different behavior of the circuit.
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there must be some reason for this..
thanks.
The phase shift oscillator is finicky to get it going.
Recommendations:
Increase power supply to 4V or more.
Add a bit of bias current, adjusted through a potentiometer.
Reduce emitter resistor to a very low value. You may not need it at all.
I guess the loop gain is too small. Please try to add more 2 stages of amplifier.
My pdk transistor supports upto 1.8V only..
Does this mean you cannot have a power supply greater than 1.1V (referring to post #1)? This is very hard to believe regarding any simulator.
and what is ebb and how it affects oscillations...
C1 (in my schematic) must discharge an equal amount as it charges.
This happens every cycle.
'Ebb and flow' is a phrase we use for cyclical give-and-take action.
You must adjust the transistor bias so that it turns on sufficiently to discharge C1 a certain amount. Then the transistor must shut off (or almost shut off), so that C1 can charge an equal amount as it discharged.
There is only a narrow range of adjustment, where you can get the circuit to oscillate forever.
It is not difficult to make it ring for a few cycles at startup. There is a wide range where you can do that.
I guess the loop gain is too small. Please try to add more 2 stages of amplifier.
No - of course you can provide enough loop gain with one single stage only
However - kenambo, do you know the definition of loop gain? It is the gain of the COMPLETE loop (after opening it at a suitable point) including RC highpass and gain stage.
My pdk transistor supports upto 1.8V only..
Does this mean you cannot have a power supply greater than 1.1V (referring to post #1)? This is very hard to believe regarding any simulator.
It is pdk dependent and i will try with your supply and share the results.
and how to give initial conditions for this circuit.
Because after that only i can fix the charging and discharging am i right.
thanks
Yeah i knew that. and what exactly you mean by suitable point?
Because i opened my circuit after RC and given input to the transistor. Does this exhibit correct loop ?
That means: In your case, you should do the following:
As a first step, open the feedback loop and determine the small-signal (dynamic) input resistance r,in at the opening.
As a second step, load the passive part with this resistance r,in and perform the simulation.
As an alternative, you could load the passive part with a copy of the circuitry which you have disconnected.
Hi
While i am reading about RC oscillator, i came up with the line that specifies the RC network attenuation is 1/29th of the gain.
Where does this 1/29 come from?
and to avoid this our amplifier gain must be sufficient.
but how does it cause this attenuation?
thanks
The number is only valid for a specific RC circuit, unfortunately you forgot to mention it. I'm sure the number becomes obvious if you calculate the transfer function of the circuit.Where does this 1/29 come from?
Possible yes, but not necessary. More likely the amplifier phase will be compensated by a respective RC phase shift, resulting in 0 overall loop phase.that means my amplifier phase shift is zero in that frequency and my RC also gives zero phase for that frequency.
Is that even possible
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