Electronic oscillator and negative feedback

AI Thread Summary
The discussion focuses on understanding the behavior of an electronic oscillator using an operational amplifier (op-amp) and the role of negative feedback in its operation. It highlights that the voltage difference at the input of the op-amp is ideally zero, but non-idealities can complicate this. The oscillator's operation involves charging a capacitor, with a calculated period of oscillation based on the time constant of the circuit. The charging and discharging of the capacitor creates delays, which are critical for determining the frequency of oscillation. The conversation also touches on the relationship between output and input voltages in negative feedback configurations, emphasizing the importance of time dependence in these calculations.
Myrddin
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Dont know where to start for showing voltage difference at input is zero, suspose its something to do with non idealities of the op amp?

For the oscillator question , kinda know how its works; there's positive saturation giving positive voltage to charge capacitor. When negative there's a delay. However don't know how to calculate period of oscillation?
 
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The first question is basic Opamp theory.
See this
http://en.wikipedia.org/wiki/Opamp

The oscillator is interesting.

Suppose the output has just become positive (+10 V) and the supply is plus 10 V and minus 10 V. This opamp is perfect.

The junction of the two resistors goes to +5V (they are across 10 volts and they are equal).
However the capacitor on the other input (the inverting one) takes time to charge.
It has to charge from -5 volts (as we will see in a minute) to +5 V.
So, it has 15 volts charging it (10 volts - (-5 V)).

You could apply the formula for a capacitor charging through a resistor, but notice that this capacitor is charging to very near 2/3rds of the way to the supply voltage, so you can approximate this as one time constant. (One time constant would be when it charged to 63.2% of the supply voltage but this is 66%, so there is an error.)
That is 10000 ohms and 0.000 000 05 Farads or 0.5 milliSeconds

When the capacitor charges up to very slightly more than +5 volts, the opamp output will switch rapidly to -10 volts.
The junction of the resistors drops to -5 volts.
The capacitor is still at +5 volts and it has to start charging the other way until it reaches -5 volts.

Notice that there are two 0.5 mS delays here so the total delay is 1 mS.
So, you should be able to work out the frequency.
 
For the first question is it just showing the relation ship for any negative feed back op amp circuit, like Vout/ Vin = - R2 /R1 for inverting op amp, for this to be so inputs must equal one another. Dont know what it means by time dependent tho?

Q / Qmax = [1 - exp(t / RC] formula to find time delay for question 2?
 
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