Questions about colpitts oscillator design?

(1) How much the ratio C2/C1 should be in order to start oscillations?

(2) If the output signal is taken from emitter through a suitable capacitor, how will be the feedback path to the base if the transistor?

(3) We all know that there is a stray capacitance came of external connections and the internal capacity of the transistor itself. If I use a 2n3904 transistor which has the following data:

My question is how much capacitance I should add to the series C1 and C2?

(4) Are there any effects from Cc cap on the oscillation frequency?

Thanks

I don't think there is a set ratio rule. You can see **broken link removed** which explains rather well. The aim is an impedance transform, to get the impedance at the tap (low impedance) to become a
higher impedance at the top of C1 (connected to the base) for the frequency in question.

Depends on the gain of the transistor but generally you want C3 to swamp Cbe by factor of 10 since it changes the most with drive and temp. If you are going to load output at emitter then you want C2 to swamp out variation in load capacitance.

If you are AC grounding collector it is actually better to tie ground side of C2 to collector. You then don't go through the collector bypass cap.

With selected C2 & C3 then Cc is selected to maximize Q of tank for the given coil inductance/Q while maintaining enough loop gain for reliable startup.

sky_123 said:

I don't think there is a set ratio rule. You can see here which explains rather

well. The aim is an impedance transform, to get the impedance at the tap (low impedance) to become a
higher impedance at the top of C1 (connected to the base) for the frequency in question.

Click to expand...

Attached file appears to be good, but after navigation I did not find it a lot
I understand from your words that we want to make some feedback from the low impedance emitter to the high

impedance base so we use a impedance transform in the form of a capacitive divider, Is this what you want to say to me?

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RCinFLA said:

Depends on the gain of the transistor but generally you want C3 to swamp Cbe by factor of 10 since it changes the most with drive and temp. If you are going to load output at emitter then you want C2 to swamp out variation in load capacitance.

If you are AC grounding collector it is actually better to tie ground side of C2 to collector. You then don't go through the collector bypass cap.

With selected C2 & C3 then Cc is selected to maximize Q of tank for the given coil inductance/Q while maintaining enough loop gain for reliable startup.

Click to expand...

There is no C3, there are C1, C2 and Cc
I'm a little confused
Please repeat your explaination with those caps
thanks

I have a simulation going.

samy555 said:

(1) How much the ratio C2/C1 should be in order to start oscillations?

Click to expand...

The ratio is not critical. Changing one or the other may cause you to have to adjust resistor values, however.

(2) If the output signal is taken from emitter through a suitable capacitor, how will be the feedback path to the base if the transistor?

Click to expand...

It should be okay if your output is high impedance.

Oscillations are healthy in this circuit (once they get going, because it's tricky to get them going). The tank loop has tens or hundreds of mA going around it. To get an AC output, I believe you can take an AC output off the tank loop, yet the sine wave will not be distorted. (Again your load should be high impedance.)

(3) We all know that there is a stray capacitance came of external connections and the internal capacity of the transistor itself. If I use a 2n3904 transistor which has the following data:
...
My question is how much capacitance I should add to the series C1 and C2?

Click to expand...

The tank loop oscillates on its own and is not affected substantially by transistor capacitance.

(4) Are there any effects from Cc cap on the oscillation frequency?

Click to expand...

All three capacitors contribute to the oscillation frequency. All 3 are in the tank loop.

I do not want to understand howo scillator works , because dozens of books and hundreds of articles talking about it and I'm sick of them
I want to design one
I want clear answers and specific, and not the words of not fattening nor of hunger.
Here is my design

Firstly i designed an emitter follower of Q-point (IC= 1mA and VCE = 1,5 V)
then I calculate the tunning circuit:

But when I stand in order to explain it to my colleagues, I expect to ask me
1 - Why did you choose this ratio for C2/C1 = 0.5?
2 - How will feedback path from emitter to base?
3 - How was stray capacitance value calculated using the 2n3904 transistor datasheet?
4 - How the capacity divider works? i.e how it matching the impedances between low emitter and high base?
5-What relationship of loaded Q to this ratio (C2/C1)?
So I need your help so I do not feel overwhelmed
thank you

First comment: Operating the oscillator near transistor fT limit makes the oscillator calculation rather difficult and less intuitive. To understand the basic circuit relations, a medium frequency oscillator (e.g. 1 MHz) is better suited.

In a low and medium frequency oscillator, C2 is mainly determing the resonance frequency and C1 can be larger by a factor of 50 or 100. As in your circuit, Cc will be set to a large value that doesn't affect the oscillation frequency.

Besides C1 to C2 ratio, the characteristic impedance of the resonant circuit is the other parameter that can be varied. It should be considerably larger than the emitter follower output impedance as well as smaller than the base node load impedance.

FvM said:

First comment: Operating the oscillator near transistor fT limit makes the oscillator calculation rather difficult and less intuitive. To understand the basic circuit relations, a medium frequency oscillator (e.g. 1 MHz) is better suited.

In a low and medium frequency oscillator, C2 is mainly determing the resonance frequency and C1 can be larger by a factor of 50 or 100. As in your circuit, Cc will be set to a large value that doesn't affect the oscillation frequency.

Besides C1 to C2 ratio, the characteristic impedance of the resonant circuit is the other parameter that can be varied. It should be considerably larger than the emitter follower output impedance as well as smaller than the base node load impedance.

Click to expand...

I always just listen to the words of impractical
I can not find where any answer to my questions
only it should be and it not should be

Can you tell me how to calc the characteristic impedance of my resonant circuit?
please I need a mathimaticale expressions.

I thought, characteristic impedance is a commonly known term: Z0 = ω0L = 1/ω0C

I don't think that there are simple answers to all of your question. I even wonder about the reasonability of some questions. E.g. what is loaded Q of an oscillator? By effect of the feedback loop, oscillator Q is raised to infinity.

I preferred to partly change the question and the answers in return.

1 - Why did you choose this ratio for C2/C1 = 0.5?

Click to expand...

Fair enough. Why did you choose that ratio?

2 - How will feedback path from emitter to base?

Click to expand...

Are you asking about feedback in common-base configuration? That is the process at work in your circuit.

A current through the emitter resistor creates a volt level at its upper end.

This has the effect of reducing volt differential across B-E, which reduces bias, thus reducing C-E current.

Now the emitter resistor has LESS current through it, reducing volt level at its upper end.

This has the effect of increasing volt differential across B-E. Increasing C-E current.

Equilibrium is reached almost instantly and I imagine it takes place too fast for us to see it happen.

3 - How was stray capacitance value calculated using the 2n3904 transistor datasheet?

Click to expand...

Your component values are small. Oscillations will be in the RF range.

High internal capacitance in a transistor can limit high-frequency response. You must determine how to choose a transistor on that basis.

4 - How the capacity divider works? i.e how it matching the impedances between low emitter and high base?

Click to expand...

A smaller capacitor exhibits greater voltage swings, with less current (as compared to a large capacitor).

As to impedance matching between C1's reactance and resistance through B-E, they may or may not be equal at the oscillation frequency.

5-What relationship of loaded Q to this ratio (C2/C1)?

Click to expand...

The overall Q is affected by all components in the tank loop. Resistance is small, therefore Q is high.

------------------------

Now for some other questions that at least one of your colleagues will ask:

'Is this a Colpitts type or a Clapp type?'
What is your answer?

'Does your circuit start oscillating on its own reliably at power up, or does it need a kick, or shock, or some instigation?'
What is your answer?

'How much time goes by until oscillations are continuous?'
What is your answer?

---------------------------

I do not want to understand howo scillator works , because dozens of books and hundreds of articles talking about it and I'm sick of them
I want to design one
I want clear answers and specific, and not the words of not fattening nor of hunger.

Click to expand...

Perfectly understandable. You asked specific questions. I tried to answer them.

But I believe there's something more you want.

I think you want to see how well your design works, in simulation or in practice.

Hi samy,

I don't know if it helps - however. from your postings i derive that you need a basic understanding of the circuit and, in particular, identification of the feedback path.
Here are some points that came into my mind:

1.) Contrary to BradtheRad, I think your circuit is working in common collector (CC) configuration, and not in common base (CB) configuration.

2.) This leads to a certain problem during explanation of the oscillation condition which requires a loop gain equal to or slightly larger than unity. Normally, the gain of the active device is larger than unity and, thus, only a portion of the output signal is fed back in order to fulfill the loop gain requirement. However, here in CC configuration the transistor gain is slightly smaller than unity. Therefore, the signal that is fed back to the base must be larger than the signal at the emitter node. Therefore, something similar to "signal up-transformation" takes place: It is easy to recognize that the voltage at the top of the tank circuit (base node) will be larger than between both capacitors (which is the input of the feedback path). But this signal at the base node must be only slightly larger than at the emitter node (because the transistor gain is only slightly lower than unity).

3.) Therefore, FvM's recommendation in post#7: Make C1 much larger than C2 (e.g. factor 50). In this case C2 mainly determines the frequency .

4.) For a better understanding of the circuit's function it is sometimes convenient to use the "virtual ground" principle to modify the visual appearance of the circuit - without modifying its basic properties.
In this case, create a pure ac equivalent circuit by suppressing all elements (resistors) which are necessary to dc bias the transistor only but do not contribute to the oscillation frequency.
Then, ground the base and all elements which were connected to the base (and, of course, lift the ground at the collector). Now you have an equivalent circuit in CB configuration. Now you easily can identify the feedback path between collector and emitter and you can simulate the loop gain (watch the restrictions and requirements for loop gain measurements!).

Here is a document posted by another EDABOARD member which will help explain most of the design values in an optimum Colpitts osc design.

BradtheRad said:

Are you asking about feedback in common-base configuration?

Click to expand...

I think it is a common emitter, What is your guide on it common-base configuration?

Your component values are small. Oscillations will be in the RF range. High internal capacitance in a transistor can limit high-frequency response. You must determine how to choose a transistor on that basis.

Click to expand...

Good point

A smaller capacitor exhibits greater voltage swings, with less current (as compared to a large capacitor). As to impedance matching between C1's reactance and resistance through B-E, they may or may not be equal at the oscillation frequency.

Click to expand...

Can you help me or at least the the guideline method of calculating resistance through B-E of my circuit? Are you mean (1/gm) or
{R2//R4// (beta+1)*((1/gm)+R1)} <<<<< note the R's from my circuit

Now for some other questions that at least one of your colleagues will ask:
'Is this a Colpitts type or a Clapp type?'
What is your answer?

Click to expand...

Of course it is Clapp type

'Does your circuit start oscillating on its own reliably at power up, or does it need a kick, or shock, or some instigation?'
What is your answer?

Click to expand...

No, it starts oscillating on its own reliably at power up.

'How much time goes by until oscillations are continuous?'
What is your answer?

Click to expand...

According to multisim.10 simulator it takes less than 2ms.

Thank you BradtheRad for your help

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LvW said:

1.) Contrary to BradtheRad, I think your circuit is working in common collector (CC) configuration, and not in common base (CB) configuration.

Click to expand...

I'll leave BradtheRad respond to you.

2.) This leads to a certain problem during explanation of the oscillation condition which requires a loop gain equal to or slightly larger than unity. Normally, the gain of the active device is larger than unity and, thus, only a portion of the output signal is fed back in order to fulfill the loop gain requirement. However, here in CC configuration the transistor gain is slightly smaller than unity. Therefore, the signal that is fed back to the base must be larger than the signal at the emitter node. Therefore, something similar to "signal up-transformation" takes place: It is easy to recognize that the voltage at the top of the tank circuit (base node) will be larger than between both capacitors (which is the input of the feedback path). But this signal at the base node must be only slightly larger than at the emitter node (because the transistor gain is only slightly lower than unity).

Click to expand...

Although your analysis was terrific and entered my brain, but I want you to know that you can use voltage-divider common emitter configuration which has a voltage gain much larger than 1. All you have to do is to designe this configuration and then add to it the resonance circuit with little of settings it will oscillate soon.

4.) For a better understanding of the circuit's function it is sometimes convenient to use the "virtual ground" principle to modify the visual appearance of the circuit - without modifying its basic properties.
In this case, create a pure ac equivalent circuit by suppressing all elements (resistors) which are necessary to dc bias the transistor only but do not contribute to the oscillation frequency.
Then, ground the base and all elements which were connected to the base (and, of course, lift the ground at the collector). Now you have an equivalent circuit in CB configuration. Now you easily can identify the feedback path between collector and emitter and you can simulate the loop gain (watch the restrictions and requirements for loop gain measurements!).

Click to expand...

Here you come back and say it is common base (CB) configuration as BradtheRad said!!!!!!

thank you LvW

In the circuit in post 6 the transistor is connected in common collector configuration, also known as emitter follower.

Input is to the base, output is from the emitter. There is feedback from emitter to base to make it oscillate.

Common emitter means input to base, output from collector.
Common base means input to emitter, output from collector.

In this circuit, the collector is connected directly to the battery. There is no output there.

The arrangement of components resembles common collector operation.
Just about all the distinguishing characteristics point to CC.

If it were an amplifier it would be CC operation. I will go along with that.

When watching scope traces, however, I have seen a mode where the transistor shuts off (or nearly shuts off) as much because the lower capacitor has a couple volts on it, as because the base is low.

This means the control signal is (to some degree) applied at the emitter, which is a feature of common base operation. In this arrangement anyway.

Is it feedback? Is it a merging of output and input?

In some regards, oscillators don't always fit in one or the other column of our definitions. Amplifiers are easier to classify

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Right now I am looking at the article linked by sky_123 (post #2). It shows a common-base Colpitts.



In view of this I must concede that sammy555's configuration is a common collector.

In view of this I must concede that sammy555's configuration is a common collector.

Click to expand...

Fortunately the text books don't need to be rewritten.

samy555 said:

Here you come back and say it is common base (CB) configuration as BradtheRad said!!!!!!

Click to expand...

Hi samy, it seems you misunderstood my explanation.

Without any doubt, the given circuit is in CC configuration.
However, in order to investigate the properties of the oscillator - in particular, identification of the loop gain to allow verification of the oscillation condition - it is sometimes more convenient to have a circuit that resembles the classical feedback model: An active unit with a gain larger than unity and a passive frequency-determining feedback path.
For this purpose you can (on the paper, not in reality) transfer the CC circuit into an equivalent CB arrangement.
The principle behind this procedure is the "virtual ground" concept.
Sometimes the same procedure is applied to transfer a negative-resistance oscillator into a feedback topology in order to simplify the analysis.

FvM said:

Fortunately the text books don't need to be rewritten.

Click to expand...

I can accept a bit of good-natured ribbing.

godfreyl said:

In the circuit in post 6 the transistor is connected in common collector configuration, also known as emitter follower.

Input is to the base, output is from the emitter. There is feedback from emitter to base to make it oscillate.

Common emitter means input to base, output from collector.
Common base means input to emitter, output from collector.

In this circuit, the collector is connected directly to the battery. There is no output there.

Click to expand...

Mistake I wrote ce, but I mean cc, SORRY

What do you think of this design,
I borrowed it from a book "RF Circuit design- Chris Bowick"
Initially I'll put the pages quoted






My key design depends on the grounds that I'll put RE (emitter resistance) = Rs and Rs' = RB (base resistance).
I my upper emitter follower design I have the following values:
IC = 1mA, beta = 230, re = 25.887 ohm, so:
RE(total) = RE + re = 1500+26 = 1526 ohm
RB = R2//R4//(beta*[re+RE])= 14.5 kohm.
Now I'll go to design the resonant circuit using mathCAD 7 professional:




Please put your comments and your opinions
thanks