BJT common collector Colpitts oscillator

In summary, the conversation is about finding the amplifier gain for a BJT common collector colpitts oscillator. The speaker has searched for hours but has not been able to find the amplifier gain for this specific circuit. They have found it for similar colpitts oscillators with a collector resistor, but the output of this circuit is the collector voltage. They are seeking help and clarification on the difference between the load in the open loop and closed loop configuration and the different types of gain (voltage, current, power). They also mention a video they are watching for more information and offer to provide an LTspice.asc.txt file for further analysis.
  • #1
Helena Wells
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TL;DR Summary
I have trouble finding the amplifier gain of this colpitts oscillator.
1613059990052.png


This is a BJT common collector colpitts oscillator.I have found how to find the feedback fraction from this site:http://fourier.eng.hmc.edu/e84/lectures/ch4/node12.html but I have searched for hours and haven't found the amplifier gain of this circuit.

I have found the amplifier gain for similar colpitts oscillators with a collector resistor and the output of the circuit is the collector voltage but not for this . Help appreciated!
 
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  • #2
Helena Wells said:
This is a BJT common collector colpitts oscillator.I have found how to find the feedback fraction from this site:http://fourier.eng.hmc.edu/e84/lectures/ch4/node12.html but I have searched for hours and haven't found the amplifier gain of this circuit.

I have found the amplifier gain for similar colpitts oscillators with a collector resistor and the output of the circuit is the collector voltage but not for this . Help appreciated!

Well - the output node of the gain stage in common-collector configuration is - of course - the emitter. But for answering your question one must know which "gain" are you interested in? I suppose, you know that the "open-loop gain" (without the feedback network) is simply Aol=gm*R3/(1+gm*R3), which is very close to unity.
But this is just the most simple part of the problem.
Most important is the loop gain (gain around the closed-loop), which gives you the condition for oscillation and the oscillation frequency.
 
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  • #3
LvW said:
Well - the output node of the gain stage in common-collector configuration is - of course - the emitter. But for answering your question one must know which "gain" are you interested in? I suppose, you know that the "open-loop gain" (without the feedback network) is simply Aol=gm*R3/(1+gm*R3), which is very close to unity.
But this is just the most simple part of the problem.
Most important is the loop gain (gain around the closed-loop), which gives you the condition for oscillation and the oscillation frequency.
@LvW I know the Barkhausen criteria .I am just interested in the amplifier gain.
 
  • #4
Is the load different in the closed loop configuration than the load in the open loop configuration and if so how is it different?
 
  • #5
Helena Wells said:
Is the load different in the closed loop configuration than the load in the open loop configuration and if so how is it different?

Helena Wells said:
Is the load different in the closed loop configuration than the load in the open loop configuration and if so how is it different?
Under open-loop conditions the load is given with the emitter resistor only.
 
  • #6
LvW said:
Under open-loop conditions the load is given with the emitter resistor only.
With the assumption that you've opened the loop between the resistor and the tank, of course. Then when you calculate the closed loop gain you have to account for the extra loading of the amplifier (as defined). This is what you see done in your attachment where they use the Thevenin model with the amp output impedance. In practice, at the resonant frequency, the tank impedance is very high (roughly Q⋅Zc) so it is usually ignored.

BTW, thanks for posting with a nice schematic. It makes it so much easier to understand and communicate about this stuff.
 
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  • #7
Helena Wells said:
Summary:: I have trouble finding the amplifier gain of this colpitts oscillator.

View attachment 277764

This is a BJT common collector colpitts oscillator.I have found how to find the feedback fraction from this site:http://fourier.eng.hmc.edu/e84/lectures/ch4/node12.html but I have searched for hours and haven't found the amplifier gain of this circuit.

I have found the amplifier gain for similar colpitts oscillators with a collector resistor and the output of the circuit is the collector voltage but not for this . Help appreciated!
The transistor cannot be said to be in Common Collector.
Look at the AC circuit ignoring DC. The collector load is half the tuned circuit using C2, and it is in parallel with R3. The base-emitter has C1 and the other half of L1.
 
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  • #8
The capacitor you show as a high-pass filter is there to block the DC. You can reduce the value of that capacitor to get improved stability. That is because it reduces the amplifier load on the tuned circuit.

Helena Wells said:
I know the Barkhausen criteria .I am just interested in the amplifier gain.
The input impedance differs from the output impedance, so do you want the voltage gain, the current gain or the power gain?

If you post your LTspice.asc.txt file, I can break the loop or parallel your circuit to show model the amplifier gain.
 
  • #9
I am watching a video about this .What is she talking about at 4:28?

 
  • #10
Baluncore said:
The capacitor you show as a high-pass filter is there to block the DC. You can reduce the value of that capacitor to get improved stability. That is because it reduces the amplifier load on the tuned circuit.The input impedance differs from the output impedance, so do you want the voltage gain, the current gain or the power gain?

If you post your LTspice.asc.txt file, I can break the loop or parallel your circuit to show model the amplifier gain.
The voltage gain obviously.
 
  • #11
Helena Wells said:
The voltage gain obviously.
Assuming high transistor gain, and ignoring second order stuff, the voltage gain of an emitter follower is 1. As in post #2.
 
  • #12
DaveE said:
Assuming high transistor gain, and ignoring second order stuff, the voltage gain of an emitter follower is 1. As in post #2.
ok but what is the impedanve of the load?It isn't just the emitter resistor is it?How C1 and C2 contribute to the impedance of the load?
 
  • #13
Helena Wells said:
ok but what is the impedanve of the load?It isn't just the emitter resistor is it?How C1 and C2 contribute to the impedance of the load?
At the resonant frequency the tank impedance is very high (since there are essentially no losses shown). So, at that frequency, we would normally assume there is no additional load across the emitter resistor. At other frequencies, or to get an exact solution, is an algebraic nightmare, you have to include lots of parts. You can't just look at C1, or C1 & C2, because all of those parts have similar magnitudes at the frequencies of interest.

For the resonant tank part, if you are far away from resonance, then you will find that either the inductor, or the capacitor can often be ignored because their magnitudes are very different.
 
  • #14
DaveE said:
At the resonant frequency the tank impedance is very high (since there are essentially no losses shown). So, at that frequency, we would normally assume there is no additional load across the emitter resistor. At other frequencies, or to get an exact solution, is an algebraic nightmare, you have to include lots of parts. You can't just look at C1, or C1 & C2, because all of those parts have similar magnitudes at the frequencies of interest.

For the resonant tank part, if you are far away from resonance, then you will find that either the inductor, or the capacitor can often be ignored because their magnitudes are very different.
What about the parasitic resistance of the inductor/capacitors?
 
  • #15
tech99 said:
The transistor cannot be said to be in Common Collector.
Look at the AC circuit ignoring DC. The collector load is half the tuned circuit using C2, and it is in parallel with R3. The base-emitter has C1 and the other half of L1.
The transistor is in Common Emitter. The base-emitter circuit is formed by part of the tuned circuit, and similarly with the collector-emitter circuit. The transistor gain is approx hfe x R3. The Unfortunately, this circuit is very bad because it shunts R3 across the tuned circuit. The Hartley version avoids this issue. There may also be gain due to a step up action by the transformer action of the inductor. It is undesirable to reduce the collector capacitor because this places resistance originating in the collector decoupling resistor across the tuned circuit which lowers the Q. It also creates NFB via R3. If wanting to reduce coupling between tuned circuit and transistor this can be done by tapping the base down the coil (capacitively if wished) or reducing C3.
 
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  • #16
tech99 said:
The transistor is in Common Emitter.
No. This is a common-collector (or emitter-follower) configuration. I think you have your terms mixed up:
https://www.electronics-tutorials.ws/transistor/tran_1.html
 
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  • #17
tech99 said:
It is undesirable to reduce the collector capacitor
What collector cap? The 100uF bypassing Vcc to Ground? It sounds like you are looking at a different schematic than me.
 
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  • #18
Helena Wells said:
What about the parasitic resistance of the inductor/capacitors?
Yes, those are real in the practical world, also losses due to radiation. But at 5MHz they shouldn't be very significant, if the components are selected appropriately.

Also the DC bias resistors R1 & R2 are 6Kohm of parallel load on the tank. Since the tank characteristic impedance Zc = √(L/C) = 300 ohm, the 6K in parallel will limit the Q = R/Zc to less than 6K/300 = 20. Also the output impedance of the amplifier (R3), etc. The exact solution is complex.

These circuits are almost always analyzed by making some assumptions to simplify things. Like ignoring the loading of R1 & R2 because Q=20 is almost the same as Q=∞. We know those assumptions are approximations, once you get a simplified model looking OK, then you can go back and see if you really do care about some of those details we skipped over.
 
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  • #19
Helena Wells said:
The voltage gain obviously.
The BJT is an emitter follower with a voltage gain slightly less than one.

Model another tank circuit, identical to L1, C2 and C3. Connect the lower end of L2 to ground. Connect the centre of the two capacitors to the same emitter drive terminal on the existing modeled oscillator. The simulation should run with the tank circuit slaved in parallel without any big change to the simulation.

The top of the slave, L2 is disconnected from the base terminal. Look at the sine wave on that terminal and compare it to the emitter voltage. You will see that for C1=C2, the voltage ratio is exactly two.

The L1-C1-C2 tank acts like a centre tapped transformer winding = an autotransformer. That provides the voltage gain of two required to sustain oscillation.

The BJT is a buffer amplifier that converts a high impedance at the base terminal, to the low impedance at the emitter terminal. In effect it has a current gain.
 
  • #20
The emitter resistor can be increased to 15 kΩ, and there is less distortion with a higher base bias impedance and a smaller feedback capacitor = 10 pF.

The base impedance is then about 40 kΩ. The output impedance of the emitter follower is about Re / β ≈ 100 Ω.

Here is the schematic with the plot of base and emitter voltages.

schematic.png


Note that Ve with offset is plotted in blue, over Vb/2 in red.
slave_plot.png
 
  • #21
I have attached a diagram showing how the circuit may be regarded as common emitter.
 

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  • #22
tech99 said:
I have attached a diagram showing how the circuit may be regarded as common emitter.
I am not convinced. The transistor is operating as an emitter-follower, which really makes it a common collector configuration. The base is the input, the emitter is the output.
 
  • #23
Baluncore said:
I am not convinced. The transistor is operating as an emitter-follower, which really makes it a common collector configuration. The base is the input, the emitter is the output.
I think people are assuming emitter follower because the circuit is drawn with the collector grounded for RF. That does not make it common collector. The ground can be placed anywhere on an oscillator circuit.
 
  • #24
tech99 said:
I think people are assuming emitter follower because the circuit is drawn with the collector grounded for RF. That does not make it common collector. The ground can be placed anywhere on an oscillator circuit.
No - we are not "assuming" something. The circuit is indeed an emitter follower - the collector is at signal ground and the emitter potential is "following" (with a very small error) the varying base potential. What else do we need to call it correctly "emitter follower"?
And yes - you are right. We can make use of the "virtual ground" concept as proposed by Alechno long time ago. This is a good method to convert the shown circuit into a common emitter stage (gain>1) with a frequency-dependent positive feedback path. In some cases, such a conversion can help to better understand the function of some circuits.
But without such ground shifting the given circuit remains what it is: A common collector stage (emitter follower).
 
  • #25
tech99 said:
I have attached a diagram showing how the circuit may be regarded as common emitter.
Nope. Lots of incorrect network manipulations between steps 2 and three.
 
  • #26
I can't buy it either. I'm surprised anyone would attempt to claim such a thing.
 
  • #27
I believe that treating this oscillator circuit as a common emitter configuration is not completely unreasonable. At least for now, I can't think of a reason to absolutely deny it. Although I really don't know whether this view would violate basic theory or principle. 🤔

In any case, whether we treat it as CE or CC, the working process of this circuit, the current and voltage of each part in the circuit will not change as a result. :smile:

Circuit 3.jpg
 
  • #28
Yes - I agree. The above contribution demonstrates the "virtual ground principle" (explained by Alechno in a corresponding article).
 
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  • #29
alan123hk said:
I believe that treating this oscillator circuit as a common emitter configuration is not completely unreasonable. At least for now, I can't think of a reason to absolutely deny it. Although I really don't know whether this view would violate basic theory or principle. 🤔

In any case, whether we treat it as CE or CC, the working process of this circuit, the current and voltage of each part in the circuit will not change as a result. :smile:

View attachment 277878
Are you saying for the input that VL = VC1? I don't think can be correct.
 
  • #30
tech99 said:
Are you saying for the input that VL = VC1? I don't think can be correct.

Of course, the voltage values of L and C1 in the circuit should be different. What I mean is that as long as the input and output are defined correctly as shown on the diagram I posted, then it seems that it is no problem to call it CE or CC.

This is just an expression of humor. Since the voltage and current of each component in the oscillator circuit will not change due to the different names we give to the oscillator circuit, it seems that we don’t have to care too much about whether its name is CE or CC. :smile:
 
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  • #31
alan123hk said:
Of course, the voltage values of L and C1 in the circuit should be different. What I mean is that as long as the input and output are defined correctly as shown on the diagram I posted, then it seems that it is no problem to call it CE or CC.

This is just an expression of humor. Since the voltage and current of each component in the oscillator circuit will not change due to the different names we give to the oscillator circuit, it seems that we don’t have to care too much about whether its name is CE or CC. :smile:

I think, we should not mix different things.
* The shown circuit is an oscillator with a BJT in CC configuration. No doubt about it.
* However, we can use the "virtual ground" method - and such a new view allows us to consider the BJT now in CE configuration (of course, no change for passive elements).
* However, during calculation of the loop gain (verification of the oscillation condition) it is of course important to know which gain formula we have to apply. Hence, it is important to know if the BJT is in CC or CE configuration.
 
  • #32
LvW said:
I think, we should not mix different things.
* The shown circuit is an oscillator with a BJT in CC configuration. No doubt about it.
* However, we can use the "virtual ground" method - and such a new view allows us to consider the BJT now in CE configuration (of course, no change for passive elements).
* However, during calculation of the loop gain (verification of the oscillation condition) it is of course important to know which gain formula we have to apply. Hence, it is important to know if the BJT is in CC or CE configuration

If we define the collector of the transistor as ground, the base as the input, and the emitter as the output, and the circuit is designed correctly, then this must of course be a common collector configuration. I'm just saying I won't deny that other people use another way to deal with it.

Regarding the derivation of the oscillation conditions, the calculation methods of the CE configuration and the CC configuration can of course be different, and even the calculation methods based on the same configuration (CE or CC) can be different. However, whether we find the oscillation conditions based on CE or CC and use any different calculation methods, the basis should be the same, that is, the closed loop gain is equal to 1, and the results of oscillation conditions should also be the same, because they are actually dealing with the same oscillator circuit. :smile:
 
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  • #33
'Moving ground around' does not change anything about the behavior of the oscillator. The tank circuit is still the two capacitors and inductor in series. Currents still circulate around that loop. If you choose to connect your scope in such a way to be able to say that the output is being fed back 180 degrees out of phase and maintain oscillation since that is what is required with common emitter, I guess that's your business. But I think it's a silly thing to do.
 
  • #34
I guess the point is that if you remove all of the biasing and bypass stuff (create the AC small-signal model), and then redefine what ground and the output is, you can consider this as any "amplifier" configuration. CE, CC, CB, are primarily defined by the inputs and outputs. But there is no input, and everything in the circuit is oscillating, you could choose to take the output from anywhere. So, those definitions become meaningless.

However, the OP defines the analysis approach as an amplifier with feedback, so it is reasonable to choose a canonical form that is easy to understand. Often (for good design documentation), the way the schematic is drawn will guide the analysis to follow the designer's understanding or intent. A good schematic is typically drawn so that the canonical forms or common structures are readily apparent.

I fail to see the broader point. This all seems like a pedantic diversion.
 
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  • #35
alan123hk said:
However, whether we find the oscillation conditions based on CE or CC and use any different calculation methods, the basis should be the same, that is, the closed loop gain is equal to 1..
Just a small correction: It is the LOOP GAIN which must be "1"-
 

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