Oscillator problem, explanation needed

[greg997]

Hi, I've got a few questions about that common base colpitts oscillator. I am trying to simulate it but, I am not sure if the circuit is complete or is it missing anything? I have some formulas but not sure how to use them.I have some assumptions, are they right?

1. Capacitors R3 R4 nad R5 can have any value, but at given frequency must act as short circuit, that means impedance must be small, so the capacitors value must be high. What is the resistance that is considerd short circuit? 1 Ohm? I can't see what I need that C4 for.

2. L resonant Q factor. From formula Q= wL/R. What exactly is that R? Is it like collector resistor in a dc circuit? And when I got Rload , how is it related to R? I've got something called [Total equivalent load resistance across oscillator output at given frequency. I ve got Rload, calculated from output power and peak Ac output.

I have formula Rt= (RdRo)/(RdRo), which does not make sense because the nominator and denominator are the same. Is the Rt= that total equivalent resistance?, then what is Rd and Ro? I've formula for Rd= L/Cr. but now i don't know what is Cr.

3. At what point should I see oscillation? Across the load or transistor? So far I can see only dc voltage equal input voltage. Should I seen pure sinewave?


Appreciate any help and explanation. Those values of resistors and capacitors are random. I just need general answer not calculations. Thanks

 

[vk6kro]

I drew this up in a different simulator (LTSpice) but it may work in yours:

[PLAIN]http://dl.dropbox.com/u/4222062/Colpitts%20Oscillator.PNG

Basically, it is the same as yours except the emitter resistor and one of the bias resistors had to be changed.

Output of 27 Volts p-p is available at the collector of the transistor. Oscillation is at about 3.5 MHz.

1. Capacitors R3 R4 nad R5 can have any value, but at given frequency must act as short circuit, that means impedance must be small, so the capacitors value must be high. What is the resistance that is considerd short circuit? 1 Ohm? I can't see what I need that C4 for.

R3, R4 and R5 are resistors, not capacitors, and I don't understand the rest of this paragraph.
Maybe you could have a read of this:
http://en.wikipedia.org/wiki/Capacitors
to understand what capacitors are used for.

Don't forget to ground the main part of your circuit. The rail along the bottom must be grounded or connected to the negative side of the battery for the circuit to work.

 

[greg997]

HI, thanks for the replay.
Yeah, That was my mistake with the resitor and capaitors names. I meant capacitors C3 C4 C5. I created your circuit in Multisim and there is sinewave, not a nice pure one but it is there :)
In your case, the impedance of the capacitors is 4.5 Ohm, so I guess that is considerd short circuit. I created my own new circuit and it wroks now. Can you tell how exactly I can see the oscillating frequency, I mean how can I confirm that the oscillation frequency is what it is suposed to be.
And I calculated Load Resistror from Pout= Vout^2/2Rload but the power Pout is twice of what I wanted and Vout Ac is nearly twice then the one used for the formula above. So I got stuck again.

I hope somebody can explain my question in the point 2.

Thank you

 

[vk6kro]

To measure the frequency, you need to measure the period first.

Measure the time between two successive peaks on the waveform. Then just take the reciprocal of that time. With my simulator, you just put the cursor on the spot you want to measure and then read it off the bottom of the screen.
You could also measure the time for, say, 10 cycles of output. Then divide the time by 10 to hopefully get better accuracy.

Q= X_L/ R where R is the resistance of the coil wire.

Actually, it is also affected by any parallel resistance. This includes the output resistance of the transistor, the load and, in this case, the emitter resistor.

Read about this in any good book about RF tuned circuits. I have an old book by Terman (Electronic and Radio Engineering) that handles this very well, but any good book on this topic should deal with it.

Yes, a few ohms of reactance doesn't matter in circuits where the other reactances are much higher.

 

[vk6kro]

On a simulator, bigger bypass capacitors will appear to be better.

In real life, capacitors have series inductance and behave like series tuned circuits.

So, bigger capacitors can have a lower series resonant frequency, and above this frequency they behave like inductors.

This also means that if you get the series resonance exactly right, you can have this series resonance giving you perfect bypassing for a very narrow range of frequencies.

 

[greg997]

Hi, thanks for that explanations, that really helps.
One more question. Is that rigth that at resonace the resonant tank acts like resitor due to internal inductor resistance? So it is like collector resistor?
Now, I reading some books about that topic, hope to find what I need.

 

[vk6kro]

If the parallel tuned circuit were perfect, at resonance, it would have infinite impedance and its resonance would be only over a very narrow range of frequencies.

Since it is not perfect (because it is loaded by various resistors, including the coil resistance), then it has less impedance than this and it works over a wider range of frequencies.

It really isn't like a resistor. It has selectivity and it doesn't inherently dissipate power.
It can also produce some surprising effects which magnify voltages and currents within the tuned circuit.

You can think of high impedance a being a bit like high resistance, but beware of thinking the two are really anything like each other.

Also, you should be aware that a transistor or a FET should not have a parallel tuned circuit as its load. Both have low output impedances and they lower the Q of a parallel tuned circuit in their outputs.
These should normally be connected to a low impedance tap on the coil or a low impedance winding placed near the main coil.