Quick Fundas: Op-amp# 6 Contd...

In the Op-amp circuit application example "Quick Fundas: Op-amp# 6" we have seen that the output of the Op-amp circuit oscillates between positive and negative saturation voltages when VIN is set to 0V. How the output voltage behave when the input voltage VIN is set (-2V) and when VIN is set (+2V)?

Ans:

Setting VIN (-2V) helps in the capacitor C2 to be charged to the negative threshold voltage faster. At the same time it takes more time for the capacitor C2 to be charged to the positive threshold. Hence the Op-amp output VOUT remains in positive saturation for more time than the time it remains in negative saturation. 

The opposite happens when VIN is set at (+2V). In that case, the Op-amp output VOUT remains in negative saturation for more time than the time it remains in positive saturation. 

Basically this is kind of a PWM (Pulse Width Modulation) circuit using an Op-amp. Pulse width modulation is achieved by varying VIN. In the picture below, the simulated waveform of the output VOUT is shown when VIN is varied from (-2V) to (+2V).

Figure 1: Pulse Width Modulation waveform (for circuit shown in Op-amp# 6)

Quick Fundas: Op-amp# 6

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? 

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:

Figure 2: Simulated waveform of VOUT (VIN = 0V)

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?