Barkhausen criterion and oscillation

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Manu Varkey - 20 May 2007 12:24 GMT

The usual explanation of a feedback oscillator goes like this. ""We have an
Amplifier 'A(w)', a feedback circuit 'B(w)'. An initial input v=Ke(jwt)
produses an output A(w)*B(w)*Ke(jwt) which in turn produces an output
(AB)^2 * Ke(jwt) ...... => If the signal is to be sustained |AB|=1 and
arg(A)+arg(B)=2*pi then each delayed echo or cycle of fluctuation
will ‘tack itself onto the tail’ of the previous fluctuation with the same
sinusoidal phase leading to oscillation."" I really don't get it . An
Amplifier produces the output based on instantaneous value of input signal
and there is no mechanism which stores an AC signal. Then how is the
oscillation sustained if the initial disturbance is removed ? Please
correct me if I am wrong.

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John Larkin - 20 May 2007 18:32 GMT

>The usual explanation of a feedback oscillator goes like this. ""We have an
>Amplifier 'A(w)', a feedback circuit 'B(w)'. An initial input v=Ke(jwt)
[quoted text clipped - 3 lines]
>will tack itself onto the tail of the previous fluctuation with the same
>sinusoidal phase leading to oscillation."" I really don't get it .

>An
>Amplifier produces the output based on instantaneous value of input signal
>and there is no mechanism which stores an AC signal. Then how is the
>oscillation sustained if the initial disturbance is removed ? Please
>correct me if I am wrong.

Any real amplifier has time lags of various forms, the simplest being
the limits of the speed of light through the amplifier and whatever
path closes the feedback loop. Very high-frequency oscillators are
dominated by this lag. Slower amplifiers have internal capacitive and
inductive elements that slow them down, add phase shift, and
effectively store information within the amp.

The A(w) transfer function would be dimensionless (like '3' or some
such) for an ideal, zero-delay, infinite bandwidth amp, but it never
really is, except maybe in Spice.

You can still build an oscillator from an ideal zero-delay amp, so
long as the B(w) transfer function keeps things happy. If you connect
the input of a very fast amplifier to its own output through a long
hank of coaxial cable (the cable providing a nice time delay), that
will oscillate too. The information storage is then inside the length
of the cable.

John

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Jon Slaughter - 20 May 2007 19:23 GMT

> The usual explanation of a feedback oscillator goes like this. ""We have
> an
[quoted text clipped - 8 lines]
> oscillation sustained if the initial disturbance is removed ? Please
> correct me if I am wrong.

Capacitence and inductanace store charge and delay the signal. Amplifiers
are not perfect... there are also transient behaviors that you are not
taking into account.

Try this,

http://www.st-andrews.ac.uk/~jcgl/Scots_Guide/RadCom/part4/page1.html

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Don Foreman - 30 May 2007 18:35 GMT

>The usual explanation of a feedback oscillator goes like this. ""We have an
>Amplifier 'A(w)', a feedback circuit 'B(w)'. An initial input v=Ke(jwt)
[quoted text clipped - 7 lines]
>oscillation sustained if the initial disturbance is removed ? Please
>correct me if I am wrong.

It is impossible to remove the initial disturbance. All electronic
components, including resistors, exhibit some noise. At frequencies
where Barkhausen's criteria are met, any signal at those frequencies
will grow until limited. Limiting occurs when amplitude reaches a
point where gain is reduced, often due to saturation of some
component.

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Electronics Forum / Basics / May 2007

Barkhausen criterion and oscillation