Discussion in 'Electronic Basics' started by phil, Apr 20, 2004.

1. ### philGuest

Can anyone offer his/her explanation as to how the following
oscillator works:

http://members.tripod.com/~transmitters/begin.htm

I think the tank circuit will oscillate at it resonant frquency. As for
the positive feedback A x beta = -1 requirement I am not sure
how much is the coil so I can't calculate. C1p8 has 500ohms at
100mhz. Intuitively I can see that the feedback is positive because
we have two inversions, one at the collector, and one at the emitter.
How is the modulation performed?

2. ### Kevin AylwardGuest

Its a function of L, Collector to positive supply, C collector-emitter,
and the internal capacitance of the transistor Cbe.

FM is achieved because of the *diffusion* capacitance of the emitter
base. The Cbe is given by

Cbe = Cbeo + gm/2.pi.ft

gm = 40.IC.

Varying IC by varying Vbe is what causes the capacitance to change.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

3. ### philGuest

It is funny you should say "No". why not then ?
I can't see the relevance of capacitances outside of the tuned circuit, the
tuned circuit will always want to settle on it's resonant frequency because
then it has the highest impedance and the amplifier as a whole then has
the highest amplification. (common emitter Av= Rc/Re)

4. ### Kevin AylwardGuest

Because the circuit is effectively a Colpits oscillator. A colpits
oscillator is an oscillator that has an inductance from base to
collector and capacitances from collector to emmiter and emmiter to
basee. Redra the circuit with the large capacitors s/c.

We have an LC tank from collector to base. A cap from collector to
emitter, and the internal Cbe, where it is the Cbe that tunes the
oscillator. Its impossible for this circuit to oscillate unless the LC
tank from base to collector is net *inductive*.
That's because you don't understand how the circuits operates.
This is nonsense. To achieve oscillation the loop gain must have the
correct phase. Maximising the gain is irrelevant. Secondly, the tuned
circuit consists of Cbe and Cce in series, across the inductor.

The condition for oscilation is:

gm*(Zce || (Zcb + Zbe)) = Zbe/(Zbe + Zcb)

Calculated by knowing that the gain is gm.ZL and that Vi=Vo.Zbe/(Zbe +
Zcb)

Either Zcb is inductive with Zce and Zbe capacitive - Colpits
or Zcb is capacitive with Zce and Zbe inductive - Hartley

There are no other solutions for oscillation to the above equation.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

5. ### Rich GriseGuest

Why? The tank circuit, you know, includes the transistor's base-
collector capacitance.

It's just a "tuned-circuit oscillator" - the transistor's running
in common-base mode, and the 1p8 is the feedback cap.
Yes, he already knows it's a function of L. He just said it's a function
of L. He also said that he doesn't know the inductance of L. That's OK,
it's not called out, and figuring it out from the physical construction
of the coil, I'd trust my figures to be within maybe 50% of the actual
value. If this were a construction article, I'd wind the coil the way
the guy said to, and take his word for it.
It explains that right there in the next couple of paragraphs on the
very page you pointed out. Please finish reading the article, and you
This is in direct contradiction to what's printed right in the article.

Have we been dipping into the warden's ale a bit much?

Cheers!
Rich

6. ### Kevin AylwardGuest

I am referring to the *apparent* tank, i.e. the one L and one C.
Secondly, the *most* important capacitance is missing as it is hidden,
i.e. Cbe.
This is misleading, sure its a "tuned circuit", but the impedance that
is topologically connected from collector to base must be inductive. It
is a Colpits oscillator. End of story. I already posted the oscillation
condition:

gm*(Zce || (Zcb + Zbe)) = Zbe/(Zbe + Zcb)

in my other post.
Ho humm... Your point would be?

This is pointless as the guy writing the article does not know what he
Indeed it is. And I care a toss?

The article is wrong. Period. Indeed I emailed the author to point out
this fact.

What part of:

Did you fail to understand? At 1ma, ft=500mhz, Cbe ~12pf. Without Cbe it
is simply impossible for the circuit to oscillate. Go and do the math on
the above equation befor putting you foot in it.

What part of "the small ac signals on the base emmiter has negligable
effect on the large rail to rail Vcb swing, hence Cbe", do you not
understand?

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

7. ### philGuest

I think we need someone to draw the ac equivalent circuit
to clear things up since I am not sure why it would be
a colppitz nor why would transistor capacitances effect
the LC tank circuit (the visible one). Are you all ignoring
the emitter resistor to achieve this outcome or perhaps
the base resistor, (47K is rather large I would say).

8. ### philGuest

....meanwhile I have understood. This is a common base configuration,
as someone did point out, therefore the two colppitz capacitors are
1p8 and Cbe, problem solved.

I suppose we can then treat the C=6p8 and the L=6t as X and use
X in the colpittz formula instead of L, is that the right way?

9. ### philGuest

finally, there is a similar circuit where the 1n capacitor is connected to VCC
instead of ground. Is there any difference ? See circuit with C to Vcc here:

http://www.uoguelph.ca/~antoon/circ/fmt1.htm

Can anyone offer his/her explanation as to how the following
oscillator works:

http://members.tripod.com/~transmitters/begin.htm

I think the tank circuit will oscillate at it resonant frquency. As for
the positive feedback A x beta = -1 requirement I am not sure
how much is the coil so I can't calculate. C1p8 has 500ohms at
100mhz. Intuitively I can see that the feedback is positive because
we have two inversions, one at the collector, and one at the emitter.
How is the modulation performed?

10. ### philGuest

which transistor model is gm used in, the re model ? the h-parameters model
?
I have gm=ICQ/25

11. ### philGuest

which transistor model is gm used in, the re model ? the h-parameters model
?
I have gm=ICQ/25

12. ### Kevin AylwardGuest

Topologically, no.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

13. ### Kevin AylwardGuest

Both.

hie = rbb' + (1+hfe)re.
gm = 1/re = I/Vt = 40.Ic

Vt = KT/q = 25mv at nominal T.

The "real" model. If you slightly change the voltage, dv, across a
diode, its current will change, di, by gm.I. where I is the bias. The
base terminal simply allows one to impress this voltage at the diode
junction, but without taking the full current, i.e. most of it flows
into the collector..

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

14. ### philGuest

Something doesn't add up. You say that Vt=25mV at nominal T.
You also say that gm = I/Vt.
There is should follow that gm= I/25 mV at nominal T for I in mA.
Hence for IQ (operating point) gm should be IQ/25 which is
approximately ICQ/25. Now you are saying 40ICQ, this is not the
same
When you say "real" model, I have a problem with that because in
order to get Vbe you need re, and for re you need beta, and beta
originates in the approximate model which describes a current
controlled current source. So it's a catch 22 situation.

15. ### Kevin AylwardGuest

Ahmmm. Its 25 *millivolts*, 1/.025 = 40.
Nope. re = 1/gm = 1/40Ic. Beta doesn't come into re.
Nope. In order to get Vbe, you need Vbe, i.e. Apply a voltage to the
base emitter. Sure, there is a small drop of rbb'.ib, but it is not that
significant, usually. Its error is ~ rbb'/hfe.re.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.