Let us see a little bit complex Op-amp circuit. Can you figure out how the Op-amp circuit shown below behaves when VIN (V2: DC voltage source) could be set anywhere between -2V to +2V?
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| Figure1: Op-amp circuit# 6 |
Ans:
It might look tough, but it is not! :) Actually the circuit looks a bit clumsy. If you assume VIN is set to be zero, then you will find that the output of the Op-amp oscillating between +ve saturation voltage (around +10V) and -ve saturation voltage (around -10V). When the output of the Op-amp is at +ve saturation, the zener diode D1 (clamps @ 6.2V) and D2 (forward biased) sets +ve reference voltage (~7mV) at the non-inverting terminal of the Op-amp. As the capacitor C2 gets charged through R1, the voltage at the inverting terminal of the Op-amp increases and after time t1, just crosses the voltage present at the non-inverting input of the Op-amp. This turns the output to the -ve saturation voltage. The reference voltage at the non-inverting terminal gets changed to ~ (-7V). As the output turns -ve, the capacitor C2 gets charged through R1 to the -ve voltage. After time t2, the voltage at the inverting terminal of the Op-amp decreases just below the voltage set at the non-inverting terminal and the output of the Op-amp switches back to the +ve saturation voltage. The same continues. The output of the Op-amp remains at +ve saturation for the time t1 and remains at -ve saturation for the time t2. The time period of oscillation is (t1 + t2).
Simulated waveforms are shown below:
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| Figure 2: Simulated waveform of VOUT (VIN = 0V) |
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| Figure 3: Simulated waveform of voltage at the input terminals of Op-amp (VIN = 0V) |
Next: What happens when VIN is set at some voltage other than 0V?
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