Electronic oscillator and negative feedback

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SUMMARY

The discussion focuses on the operation of electronic oscillators utilizing operational amplifiers (op-amps) and the impact of negative feedback on their performance. Key insights include the calculation of the period of oscillation, which involves understanding the charging time of a capacitor in the circuit. The formula for capacitor charging through a resistor is applied, with specific values such as 10,000 ohms and 0.5 microfarads leading to a calculated delay of 1 millisecond. The relationship between output voltage and input voltage in inverting op-amp configurations is also highlighted, emphasizing the necessity for input voltages to equalize.

PREREQUISITES
  • Understanding of operational amplifier (op-amp) theory
  • Familiarity with capacitor charging equations
  • Knowledge of resistor-capacitor (RC) time constants
  • Basic principles of negative feedback in electronic circuits
NEXT STEPS
  • Study the formula for capacitor charging: Q / Qmax = [1 - exp(-t / RC)]
  • Learn about the characteristics of ideal vs. non-ideal op-amps
  • Explore the design and analysis of inverting op-amp circuits
  • Investigate the effects of negative feedback on oscillator stability and performance
USEFUL FOR

Electronics students, circuit designers, and engineers interested in understanding oscillator design and the behavior of op-amps in feedback configurations.

Myrddin
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http://img707.imageshack.us/i/electronicsoscillator.jpg/

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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