I am revisiting operational amplifiers and negative feedback and having some trouble getting it well. I'd appreciate it if anyone could verify/correct my understanding at the moment.
The way I understand it, negative feedback serves as an adjusting/controlling connection which ensures the the opamp output voltage settles to a definite voltage instead of just hitting the extremes (set by adjusting the fraction of voltage sent from the output to the input using a voltage divider network).
To this end, it drives the voltage difference between the terminals to a very low value (almost but not exactly zero) which is why both the terminals appear to be shorted together in negative-feedback connections.
On the other hand, positive feedback drives the opamp harder in the direction it is already going.
So far so good. To test my understanding I drew up four basic opamp configurations in an attempt to predict the output voltage in each case.
CircuitLab Schematic
According to what I presently understand, I figured the output voltages should be:
(a) The negative feedback would try to minimize the input volt difference which gives, \(V_{out}\) + \(V_{2}\) = \(V_{1}\).
Thus, output would probably be \(V_{1}\)-\(V_{2}\).
(c) I figured the output would be driven to saturation because of the positive feedback. Whether it goes to positive or negative saturation is decided by which is higher \(V_{1}\) or \(V_{2}\). Output voltage goes to positive saturation if \(V_{1}\) is higher, negative otherwise.
(b) The same as (a) above goes for this where \(V_{2}\) would be set to zero giving us the familiar voltage buffer topology.
(d) This one is still a conundrum for me. I guessed it would depend on what the output voltage is at any point of time. Say the output voltage is higher than the voltage at the negative terminal it would be driven to positive saturation and negative otherwise.
The way I understand it, negative feedback serves as an adjusting/controlling connection which ensures the the opamp output voltage settles to a definite voltage instead of just hitting the extremes (set by adjusting the fraction of voltage sent from the output to the input using a voltage divider network).
To this end, it drives the voltage difference between the terminals to a very low value (almost but not exactly zero) which is why both the terminals appear to be shorted together in negative-feedback connections.
On the other hand, positive feedback drives the opamp harder in the direction it is already going.
So far so good. To test my understanding I drew up four basic opamp configurations in an attempt to predict the output voltage in each case.
CircuitLab Schematic
According to what I presently understand, I figured the output voltages should be:
(a) The negative feedback would try to minimize the input volt difference which gives, \(V_{out}\) + \(V_{2}\) = \(V_{1}\).
Thus, output would probably be \(V_{1}\)-\(V_{2}\).
(c) I figured the output would be driven to saturation because of the positive feedback. Whether it goes to positive or negative saturation is decided by which is higher \(V_{1}\) or \(V_{2}\). Output voltage goes to positive saturation if \(V_{1}\) is higher, negative otherwise.
(b) The same as (a) above goes for this where \(V_{2}\) would be set to zero giving us the familiar voltage buffer topology.
(d) This one is still a conundrum for me. I guessed it would depend on what the output voltage is at any point of time. Say the output voltage is higher than the voltage at the negative terminal it would be driven to positive saturation and negative otherwise.
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