# oscillator

Discussion in 'General Electronics Chat' started by napos, Apr 13, 2006.

1. ### napos Thread Starter New Member

Mar 3, 2006
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0
Hello

Here is a site about wien oscillator http://sound.westhost.com/project22.htm
and i have this circuit:

The question is: how it is possible that this circuit starts to oscillate when there is no input , I've understand that the meaning of oscillation is that there is some input, lets say a little ac (or radiofrequency), and now when this voltage comes in then the opamp will amplify it and the the positive feedback feeds back the output ac and the opamp again amplifies, and this loop goes around until the maximum gain is obtained. Is this right? And the output oscillation frequency depends on the choosed resistors and capacitors [f=1/(2*pi*RC)], so the input frequency and output frequency are not the same?

Or is it so that the input can be just a dc voltage which makes the capacitors charged and the opamp will discharge them and feeds back the signal, and this charging and discharging will give us a sine wave in output?

I've read many texts about oscillators and their circuits, put there isn't any information about the input, which will make the oscillator circuit to start the oscillation.

And also this circuit, there is +12V supply voltage, but there is no input which would make the BJT to open and go to amplifying regime. If the transistor is not opened then there is no use of the positive feedback in my knowledge.

napos

2. ### kubeek AAC Fanatic!

Sep 20, 2005
4,691
806
I don´t know too much about oscillators, but I think that both these oscillators are in linear regime, so it is sensitive enough (through the negative feedback) that even thermal noise in resistors should be enough for the oscillator to start.

Sep 20, 2005
4,691
806
oops

4. ### hgmjr Moderator

Jan 28, 2005
9,030
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I agree with Kubeek's reply. The inherent amplifier noise is involved in starting up an oscillator such as the one you have posted. The inherent high gain of the opamp and the existence of positive feedback contribute to the kickstarting of the oscillator.

When power is first applied to the circuit, the inherent offset conditions internal to the opamp will dictate whether it rises with the input power or stays at the negative rail. Either way, the rise in power will soon reach a point that forces the opamp's output to switch rapidly to the opposite power rail. This rapid transistion of the opamp's output should be enough to kickstart the oscillator.

In the situation where power is already applied and you momentarily short the positive and negative inputs to the opamp together, the circuit will cease to oscillate as long as the terminals are shorted. The output of the opamp in this case will go to one of the power rails depending on the inherent offset at the opamp inputs and remain there until the short between the two inputs is removed. As soon as the short between the input terminals is removed, the output will switch rapidly to the opposite power rail. This transition of the opamp's output should be adequate to kickstart the oscillator.

hgmjr

P.S. Your command of English appears to be very good.

5. ### PRR New Member

Dec 16, 2003
5
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> this circuit, there is +12V supply voltage, but there is no input which would make the BJT to open and go to amplifying regime.

There is also a -12V supply. It may not be clear, but (without oscillator feedback) the output wants to sit at about -0.7V DC, the first stage idles about 10uA, the second stage idles about 2mA, and the output stage about 10mA. Forward linear gain is pretty high, say 1,000.

> how it is possible that this circuit starts to oscillate when there is no input

There is ALWAYS "input": the random noise of the universe.

Self-Noise at the first Base might be very-very-very small, but the positive feedback will increase the noise over and over again. But only in a narrow bandwidth, converging on a nearly perfect sinewave.

If the loop gain with cold lamp is 1.1, and the random noise is 0.1 microVolts, then 0.1uV *1.1*1.1*1.1*... 200 times brings us over 10 volts output. If the oscillation frequency is 200hz, it takes about a second to start-up: that is typical of this class of oscillator. We design the lamp stabilizer so the lamp will warm-up at a few volts, and reduce loop gain to 1.00000.... exactly. If the gain wanders to 0.999 or 1.001, the lamp resistance changes in the right direction to bring it back to 1.000....

That's the theory. In fact the lamp takes time to catch-on to the fact the output has changed. The oscillator output overshoots, then undershoots. Simple lamp-stabilized oscillators never settle down unless the "linear" amplifier has a little non-linearity, its gain dropping a bit at high output. The Zener-stabilized oscillator has a little more THD but is a lot more stable. Start without the Zener, and adjust the resistor so it just barely starts. Add the Zener with a large series resistor, and reduce that resistor until it just quits, then increase until it starts. That makes a good reliable sine with reasonably low THD.

6. ### JoeJester AAC Fanatic!

Apr 26, 2005
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1,230
A decent reference on oscillators is from Texas instruments. It's a pdf link so you can do the ol' right mouse, save target as ... routine.