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enter image description hereI am trying to designthe rc phase shift oscillator .I designedfor 1khz frequency using BJT. But when i try to convert the same circuit by changing the R and C values fot 90 khz I am getting the damped sinusoidal oscillation. Can some one please say what was the issue and also wether can we generate 90khz signal with RC phase shift oscillator.

And can some one suggest what is the easiest way to design the RC phase shift oscillator for higher frequency ranges above 100 khz.

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    \$\begingroup\$ Can some one please say what was the issue No because 1) You did not include a schematic 2) "trying to design" is not possible, either you know how and design or you don't know and then you're just "messing with components". Either you find a ready-designed circuit for your needs and build that or you learn what makes an oscillator oscillate so that you can really make a design. I could tell you that probably the loopgain in your circuit is too small but then of course you'd ask how to fix that. So learn what loopgain is and try to increase it. \$\endgroup\$ – Bimpelrekkie Oct 28 '17 at 13:42
  • \$\begingroup\$ can you please see the screeshot of the circuit \$\endgroup\$ – Suneeldatta Kolipakula Oct 28 '17 at 14:19
  • \$\begingroup\$ More or less the circuit I expected to see. I can already see that this will never work with those 100 ohm resistors. You really need to do your research on how there kinds of circuits work. with these component values it is obvious that you don't. That's OK but then don't change a circuit which works. \$\endgroup\$ – Bimpelrekkie Oct 28 '17 at 15:13
  • \$\begingroup\$ This is ridiculous! We have now a screenshot, but no info what is the difference between the working one and the non-working 90kHz model. You should show both or write the different component values. Otherwise the details of the error will remain secret. I bet this is the non-working version. Try to keep the resistors as they were at 1kHz, but make the capacitors smaller. - Divide them by 90. \$\endgroup\$ – user287001 Oct 28 '17 at 15:26
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In order to oscillate, the loop gain of the circuit has to be greater than 1 at the oscillating frequency. That is to say the gain produced by the amplifier has to be greater than the in-phase loss in the feedback path. If the loop gain is >1 then the signal will get stronger each time it goes around the loop, but if it is <1 then any oscillation will get weaker and die out. So the reason your 90kHz circuit doesn't oscillate is that its loop gain is less than 1.

But if it worked at 1kHZ, why does it not work at 90kHz? Because when you changed the component values to increase the frequency, you also lowered the loop gain. If you changed the values proportionally then the feedback loss should not have changed, so it must have somehow lowered the amplifier's gain. To understand how this may have happened you have to know what determines the gain of a transistor amplifier. Here's your amplifier with the feedback load represented by C1 and R1:-

schematic

simulate this circuit – Schematic created using CircuitLab

A bipolar transistor is a current amplifier. In Common Emitter configuration the Base I/V characteristics are similar to a diode, and the Collector sinks a current which is relatively insensitive to Collector voltage. In practice this means that the transistor acts like a voltage controlled current generator, and its voltage gain is proportional to the Collector load impedance. With a Collector load of 1K this circuit has a voltage gain of about 300 (49.5dB).

However any external load impedance will be effectively in parallel with R6 and so will reduce the gain. At high frequencies C1 is a virtual short circuit, so R1 is in parallel with R6 which reduces voltage gain by about 10 times to 30 (29.5dB). Will this lesser gain be enough?

With the feedback component values shown in your circuit (R1-3 = 100Ω, C1 = 470nF, C3-4 = 100nF) LTspice says the oscillation frequency is 4.9kHz. At this frequency the filter has a loss of ~20dB. Since the total loop gain (29.5dB-20dB = 9.5dB or 3) is greater than 1 the circuit should oscillate. However with such low gain LTspice thinks it will take about 5ms for the oscillation to build up to full strength.

Now what happens when you try to increase the frequency? If you reduce the values of R1-3 then their lower impedance will cause amplifier gain to be further reduced, perhaps dropping loop gain below 1 and preventing the circuit from oscillating. If you only reduce the capacitor values then the impedance at the oscillating frequency should stay the same, so provided the transistor still has sufficient gain at the higher frequency it should work.

To get 90kHz in LTspice I changed C1 to 26nF and C2-4 to 5.5nF. Changing all the capacitors by equal proportions should ensure that the loss remains constant. However In a circuit like this which has marginal gain, many factors can affect its ability to oscillate. Increased load on the output, a transistor with slightly lower gain, parasitic capacitance, a different bias point or lower supply voltage, even capacitor and resistor tolerances could be enough to stop it from working.

To get stronger and more reliable oscillation you could increase the values of R1-R2 to reduce loading, and make C1 closer in value to C2-4 to increase its contribution to phase shift and reduce loss at the oscillating frequency. The ultimate limit on frequency may be when the capacitors are so small that parasitic capacitance takes over.

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  • \$\begingroup\$ I liked the approach here. It isn't easy to reach 90 kHz with a single BJT (though it often seems all too easy to reach well over 1 MHz from parasitics when you don't want it!) +1 \$\endgroup\$ – jonk Oct 28 '17 at 21:32
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I actually think Bruce's answer here is good and you should learn from it. I'm not going to replace it here, but instead offer some quick thoughts and a specific circuit to try out.

But keep in mind that \$90\:\textrm{kHz}\$ with a single BJT probably means taking some care in construction and care in selection of the BJT and care in whatever you load it down with. The load will affect the frequency! Do not forget this! So everything matters!

If you use a protoboard, for example, I'd expect trouble. So don't do that. If I were attempting this, I'd probably wire it up "dead bug" so that I could be a bit more sure about managing strays.

To push up towards this frequency, I'd need to use smaller capacitor values. But I don't want them too small because then the BJT capacitances and other factors start loading things down. So I'd just pick something more than \$100\:\textrm{pF}\$ and less than \$270\:\textrm{pF}\$ as the capacitor value. In this case, I decided on \$180\:\textrm{pF}\$ as a reasonable all-around compromise. The rest derives off of that.

Here's a schematic to try. The transistor is important. If you replace it, do so with the knowledge that it may not work. I'd recommend using something with low capacitance and a decent corner frequency, but with good general purpose gain and low saturation current otherwise. In this case, I selected the 2N4124.

schematic

simulate this circuit – Schematic created using CircuitLab

I've shown a load there that is "significant." That load will definitely "pull" the frequency upwards a little. But the point here is that you need to concern yourself with what you will be driving. That's going to be very important, as well. You have to take everything into account both in determining if it will work at all as well as in what frequency you are likely to get. I suppose another point is that you could "tune" the frequency up a little with a different load. So if you get a slightly lower frequency with your circuit, just load it down a little as shown to pull upwards as needed.

The above circuit is a concept diagram. I think it will work, if wired up well and if you select the right BJT for it. I don't know exactly what frequency you will get. But it should be in a rough neighborhood of where you were asking to get.

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