27MHz colpitts oscillator qestion

[samy555]

This circuit was intended as 27MHz colpitts oscillator:

View attachment 63704
and whem simulated using multisim10 sofware I get:
View attachment 63707

I have some questions:
Any osci. Needs a FB to sustain oscillations, I think that this FB from collector to base path through the parallel two caps; the internal transistor capacitance b/w collector and Emitter, and the C4(=4.7p) cap.
From the Fairchild datasheets of the 2n3904 transistor I get:

View attachment 63708
View attachment 63710
Also, there is a curve related the capacitance with REVERSE BIAS VOLTAGE (V)
My questions:
(1) What is the approxi. value of the internal transistor capacitance b/w collector and Emitter at 27MHz?
(2) How the phase shifted from 180 degree at the collector to 0 or 360 degree at the base? Please give me a step by step illustration.
(3) The freq of the osci. was 25.4MHz while my calculaed value of the resonant freq was 27MHz? what factors affect this?
thanks

 

[WTP Pepper]

My first point would be component tolerances. Many cheap capacitors can be as much as 80% out and 695pF is an odd value. If you have made this capacitor out of combinations of others and measured it, how sure are you of the accuracy of the measuring device and if it is homemade then you need to consider extra capacitance and inductance in this homemade device.

50nH is also a strange value and inductors are also prone to (often) poor tolerances.

 

[ccurtis]

The transistor is connected as a common base amplifier. The input is the emitter and the output is the collector. There is no phase shift from input to output. The current that is feed back from the collector to the emitter through capacitor C4 is in phase with the output. Therefore, oscillation occurs.

Your simulated frequency is probably not what you calculated using the simple LC formula mostly because of the stray capacitance in the transistor. Also, swapping the resistor values of R2 and R3 and reducing the value of C4 to 1pF should do better and give less loading on the LC tank circuit.

Overall, your oscillator is not a classic textbook form of the Colpitts oscillator, but is similar. A transistor datasheet will give various internal capacitance values.

 

[samy555]

The transistor is connected as a common base amplifier. The input is the emitter and the output is the collector.

Yes

There is no phase shift from input to output. The current that is feed back from the collector to the emitter through capacitor C4 is in phase with the output. Therefore, oscillation occurs.

Are the FB is current or voltage? if voltage the caps lag it 90 degree! please explain.

Also, swapping the resistor values of R2 and R3

Why swapping them?

and reducing the value of C4 to 1pF should do better and give less loading on the LC tank circuit.

True, I've seen that actually.

 

[Mikebits]

My first point would be component tolerances. Many cheap capacitors can be as much as 80% out and 695pF is an odd value. If you have made this capacitor out of combinations of others and measured it, how sure are you of the accuracy of the measuring device and if it is homemade then you need to consider extra capacitance and inductance in this homemade device.

50nH is also a strange value and inductors are also prone to (often) poor tolerances.

80% out of tolerance, really? I think the OP has only simulated the circuit and has not actually built it.

 

[WTP Pepper]

Yes. Such tolerances exists. As to whether the OP built or simulated the cct we won't know until he posts otherwise.

 

[ccurtis]

Are the FB is current or voltage? if voltage the caps lag it 90 degree! please explain.

BJTs are fundamentally current controlled devices. The FB in your circuit is the current through the capacitor, which is not phase shifted. There is very little to no voltage gain from the emitter input to the collector output because the low resistance of the inductor allows very little DC voltage swing and also because the input resistance at the emitter is very low. It is the current output at the collector that drives the tank circuit into oscillation and a small portion of that current output is fed back, in phase, to the emitter input through the small FB capacitor. The circuit is a current buffer with a current gain of 1 from emitter to collector. That current gain provides the minimum gain needed to maintain oscillation. In order to use the output of your oscillator, it must be followed by another buffer or a very light load or it will easily get pulled off frequency or stop oscillating.

Why swapping them?

I only suggested that to try to reduce the load on the FB capacitor further. Swapping the values will give a larger value resistor at the emitter so that more of the FB current drives the input and gives you a stronger output at the collector (but also more distortion).

 

[samy555]

Thanks ccurtis. I benefited a lot from this clear explanation.

 

[ccurtis]

You're welcome. I know you only asked about the circuit you have there, but you might like to know that if you add another resistor from the transistor base to ground and move C1 so that it is across the added resistor, your output will be much greater. That is because the transistor is not biased so close to +9V, and the output current can swing further.

 

[MRCecil]

Perhaps you should take a look at the relationships of C1 and C4 of your schematic to the tank. Then consider the phase relationship of their currents to see the entire feedback system of the oscillator. Finally, consider what the proper duty cycle of the 3904 might be, and how that can only be accomplished with the relationships of all the feedback components.

 

[Mikebits]

I still have a problem with the 80% tolerance part. Who in their right mind would purchase a part with 80% tolerance? 20% sure, but 80%?

 

[mvs sarma]

while it is true that electrolytic caps have a tolerance of -20% to +80%,
the disc caps at values such as 560, 620, 680, or small ones like 4p7 etc would not have that level of 80%

all said, i fear that 4p7 ( a total value of below 10p including internal capacitance) feed back is insufficient at 27MHz i fear. Why not increase it to say 47p?

 

[samy555]

You might like to know that if you add another resistor from the transistor base to ground and move C1 so that it is across the added resistor, your output will be much greater. That is because the transistor is not biased so close to +9V, and the output current can swing further.

1-"the transistor is not biased so close to +9V," Where it must be biased so close to +9V? at the Base, collector or emitter?
her is the voltage readings after about 1ms (in real time about 5 minutes:
View attachment 63768

2- "add another resistor from the transistor base to ground and move C1 so that it is across the added resistor" a voltage divider
But you did not tell me what is the ratio of voltage division, is it 1:1?
I mean to put another 10K resistance between the base and ground?
thanks alot

 

[samy555]

Perhaps you should take a look at the relationships of C1 and C4 of your schematic to the tank.

What are the relationships among, Try facilitated it to me?

Then consider the phase relationship of their currents to see the entire feedback system of the oscillator

Again, What is the relationship?

Finally, consider what the proper duty cycle of the 3904 might be,and how that can only be accomplished with the relationships of all the feedback components.

How to know or calculate that "duty cycle of the 3904," and how to relate it with the FB?

 

[ccurtis]

1-"the transistor is not biased so close to +9V," Where it must be biased so close to +9V? at the Base, collector or emitter?
her is the voltage readings after about 1ms (in real time about 5 minutes:
View attachment 63768

The base voltage. See, it is nearly at the supply voltage.

2- "add another resistor from the transistor base to ground and move C1 so that it is across the added resistor" a voltage divider
But you did not tell me what is the ratio of voltage division, is it 1:1?
I mean to put another 10K resistance between the base and ground?
thanks alot

Yeah, a 10K resistor should to the trick, bringing the base voltage nearer to the ideal point of 4.5 volts, halfway between 0 and 9 volts.

 

[MRCecil]

What are the relationships among, Try facilitated it to me?

C1 is tied to one junction of the tank and the base of Q1 and C4 is tied to the other junction of the tank and the emitter of Q1.

Again, What is the relationship?

If the tank is to resonate, will both junctions of the tank be in phase? Think about that and then think of the currents of C1 and C4. Will they be in phase?

How to know or calculate that "duty cycle of the 3904," and how to relate it with the FB?

If you have thought about the first two and came to the correct conclusions, you will understand how Q1 injects energy into the resonator at the correct time (think phase) to sustain the oscillations.

I don't think I should be more direct than that since I believe you may be a student. Am I correct? If I'm in error let me know.

 

[samy555]

C1 is tied to one junction of the tank and the base of Q1 and C4 is tied to the other junction of the tank and the emitter of Q1. If the tank is to resonate, will both junctions of the tank be in phase? Think about that and then think of the currents of C1 and C4. Will they be in phase?

First of all, I do not see it from this view, I think that C1 makes the transistor operates as a common base, this is the main and the only role.
Second thing that every junction to the resonance circuit is 180 degrees Out Of Phase of each other. It seems to me you're trying to tell me that FB is done through C1, but I do not think so. FB is through the C4.

If you have thought about the first two and came to the correct conclusions, you will understand how Q1 injects energy into the resonator at the correct time (think phase) to sustain the oscillations.

If you mean time constant, I can calc Zin (looking into the base) and find Zin*C1, but i can't calc Zout

I don't think I should be more direct than that since I believe you may be a student. Am I correct? If I'm in error let me know.

I'm not a student, I finished my university education as an engineer in communication for years
I like to understand some of the electronic circuits prior to their implementation.

thanks

 

[mvs sarma]

i am a student at my 67 , LoL. We can learn during our entire life, is my belief

 

[MRCecil]

First of all, I do not see it from this view, I think that C1 makes the transistor operates as a common base, this is the main and the only role.
Second thing that every junction to the resonance circuit is 180 degrees Out Of Phase of each other. It seems to me you're trying to tell me that FB is done through C1, but I do not think so. FB is through the C4.


If you mean time constant, I can calc Zin (looking into the base) and find Zin*C1, but i can't calc Zout


I'm not a student, I finished my university education as an engineer in communication for years
I like to understand some of the electronic circuits prior to their implementation.

thanks

Ok, you're not a student so here is the information you need to answer your basic question in the attachment below.

Note the currents of C1 and C4 (I(C1) and I(C4) of the plot) are 180° out of phase and cross zero ma at max amplitude of the output and the state of the collector current at those points marked by the two cursors +/- ~70°. Then look at the collector current ~180° past the left cursor and the differential between the currents of C1 and C4 at that same point. It should be obvious that C1 and C4 in their configuration form the feedback network to trigger Q1 into higher conduction momentarily. That rise in the collector current at that point provides the added energy needed to sustain the oscillations of the resonator.

The plot and cursor info box provided should be all the information necessary for correct evaluation.

 

[WTP Pepper]

i am a student at my 67 , LoL. We can learn during our entire life, is my belief

Mvs sarma... I couldn't agree more. I left University in 1987 and that when I started learning...in all walks of technology.