Common Collector Colpitts Oscillator

Discussion in 'Homework Help' started by s15kid, Oct 17, 2011.

1. s15kid Thread Starter New Member

Oct 16, 2011
10
0
Hey guys,

Ive been asked to design a Colpitts Oscillator in a common collector configuration using a single BJT. Im really new to this so help is greatly appreciated

Vcc = 12v
Frequency has to be 4K Hz
Vout = 3V

Now i know that the formula for frequency is

fo= 1/[(2*pi)*sqrt(LCt)]
and Ct = c1*c2/c1+c2

How would i work out the values for L and C ??

Ive been looking around the forum and I cant seem to find any information about Colpitts Oscillator in a common collector configuration.

I know this is the basic layout of a common collector colpitts oscillator but what do i need to add to the circuit to make it functional? Does Vin go to the collector or the emitter?

Any help would be greatly appreciated

Cheers guys.

Last edited: Oct 17, 2011
2. Hi-Z Member

Jul 31, 2011
157
17
You'll need to provide base current to get the transistor operating properly, and you can do this using a potential divider (between Vcc and gnd): use resistor values as high as possible (bearing in mind the loading effect of the base current) in order not to load the LC circuit too much, and aim for a base voltage of about half the supply.

Once you've done this you'll need to isolate the LC circuit from the base, dc-wise, and you'd do this using a capacitor (whose value is much higher than the tuning capacitors).

You could take the output from the emitter, but don't load it too much...

3. s15kid Thread Starter New Member

Oct 16, 2011
10
0
would this be correct?? dnt mind the values of the components as i havent worked them out yet

4. Hi-Z Member

Jul 31, 2011
157
17
No, the supply must connect directly to the collector, as well as the divider - there needs to be a path for standing current in the transistor. So delete the top capacitor.

The divider should connect directly to the base, as you've shown, but both the L and the Cs need to be coupled to the base via a capacitor. If you don't have this capacitor, the L will short out the bottom resistor in the divider chain, and you'll not be biasing the transistor correctly.

s15kid likes this.
5. s15kid Thread Starter New Member

Oct 16, 2011
10
0
Ok so get rid of the top capacitor and connect it straight to the collector and ive added a coupling capacitor which i think is right.

why does it say that the capacitor loop has no resistance? do i need 2 place a resistor somewhere there?

6. Hi-Z Member

Jul 31, 2011
157
17
No, remove the leftmost capacitor. Disconnect the L and "top" C from the base. Connect them both to a capacitor, and connect the other side of the capacitor to the base. Hope that's clear!

7. s15kid Thread Starter New Member

Oct 16, 2011
10
0

Like this? Also once the circuit is complete what formula would i use to find the values of C1 and C2?

8. Audioguru New Member

Dec 20, 2007
9,411
896
Now the base has no bias voltage. Connect the base to the voltage divider.
Change the resistors in the voltage divider to 18k ohms each.
Change the emitter resistor to 1k ohms.

The frequency is the resonant frequency of the inductor in parallel with the two series capacitors. The formula for capacitors in series and for a parallel LC resonance is in Google.

There is nothing to stabilize the output level so the transistor will clip the signal and it will be almost a square-wave.

s15kid likes this.

Dec 26, 2010
2,147
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The value of Ct should be chosen large enough so that its reactance Xc = 1/(2πfC) is many times less than the input impedance of the transistor circuit.

Don't go too wild with the "large enough": think microfarads, not millifarads. Capacitors suitable for resonating can be inconvenient in large values: electrolytics won't suit this job.

If you get your oscillator to work, but with a rotten waveform, a small resistor in the lead between the emitter and the junction of the two capacitors forming Ct may help. Do not make this too big, or the oscillator will not run any more.

Last edited: Oct 17, 2011
s15kid likes this.
10. s15kid Thread Starter New Member

Oct 16, 2011
10
0
how do i work out the value for the coupling capacitor?

ive worked out these values

c1: 7uF
c2: 4uF
l: 10mH

which gives f= 1KHz which is what ive wanted. now what values would i change to get 3V at the output???

11. Audioguru New Member

Dec 20, 2007
9,411
896
You provided missing details:
3VDC?
3V RMS (AC)?
3V peak-to-peak?

12. s15kid Thread Starter New Member

Oct 16, 2011
10
0
3v peak, so 6 V peak-to-peak

13. Audioguru New Member

Dec 20, 2007
9,411
896
Since you have an extremely low value resistor divider feeding the base of the transistor and shorting the tuned circuit, and you don't bias the base of the transistor, andyou don't power the collector of the transistor then of course the circuit won't work.

14. s15kid Thread Starter New Member

Oct 16, 2011
10
0

this is my current circuit, sorry should of uploaded it 1st.

15. s15kid Thread Starter New Member

Oct 16, 2011
10
0
from the divider i know that there is 6V after the divider, but i dnt understand why my output is not stable. Also using these values, for L and C, ive managed to get 1kHz as i wanted initially, but how do i get a 3v peak output??

16. Audioguru New Member

Dec 20, 2007
9,411
896
The base of the transistor still does not have a bias volltage.
The resistance of the coil still shorts the voltage divider to ground.

Why is your schematic a negative picture with a black background?
Why are the wires covered in dots?
Why doesn't it show a power supply voltage?
What is 1mF? How many uF?

I think the ratio of the capacitors at the emitter determines the output level.

17. Audioguru New Member

Dec 20, 2007
9,411
896
Try a schematic like this (minus the dots on the wires):

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18. s15kid Thread Starter New Member

Oct 16, 2011
10
0
oh ok thanks what is that red wire for? or is that jst to signify the bais voltage from the divider?

the schematic isnt negative, its jst the program that i am using and the dots are just the current flow. 1mf = 1*10^-3 so that would be 1000uF

the output still isnt stable. im using a beta value for 1 for my transistor, is that correct?

19. Audioguru New Member

Dec 20, 2007
9,411
896
Yes.

How did you calculate the value to be about 100 times too high? Try 10uF.

Do you mean its amplitude bounces up and down? Or does the frequency slide around?

A little transistor has a beta (hfe) of about 220. Its alpha is about 0.98.

20. s15kid Thread Starter New Member

Oct 16, 2011
10
0
well i changed the values and im gettin about 987mV at the output. would changing the value of the capacitor which is connected directly to the base effect the output? or what would i need 2 change in order to get 3v peak output