Operational amplifier

Have you tried using a search engine?

Your statement and assumptions aren't true. A voltage-follower has the input tied to the '+' terminal. The key thing to remember about an (idealized) op-amp is that the differential gain is infinite. This implies that the differential input voltage (from the '+' to '-' terminals) is zero. Also, the input impedance is infinite and the output impedance is zero. The rest can be easily derived from there; solve for the currents.

He didn't say 'input', he asked about tieing the 'output' to the '+' terminal.

Tieing the 'output' to the '+' terminal is a recipe for saturating the amplifier.

daestrom

Saturation occurs when the *OUTPUT* (me be bad) is tied to the '+' input because when the output tries to go high the input forces it to go higher (positive feedback) until it hits the power rail.

Yes, I missed that.

Depends on the phase shift. ;-)

True. But a LITTLE positive feedback can be useful in applications where the output is TRUE/FALSE (e.g.: thermostat) and you want a "dead zone" where the output will not swith.

Aka, a "schmitt trigger", being a comparator with positive feedback to set a hysteresis window ("dead zone" as the above called it). One common application being using the hysteresis to debounce on something that switches.

With positive feedback, the output will saturate to one or the other rails with a very small difference in the inputs. As real op-amps are not ideal, their open loop gain is *not* infinite, so, one approximation for output voltage is simply:

Vo=Aol(V+ - V-)

This shows that as Aol increases, the differential voltage needed to saturate the op-amp (since, the output voltage should really be very high or low, but gets clamped to the rails) decreases.

You can bias up an op-amp by providing a very small differential voltage across the inputs such that the output is some value between the rails. This becomes more difficult with a higher open loop gain, since the higher the gain, the smaller the differential voltage required to cause the op-amp to saturate.

that the input stage to the op-amp is a differential pair of transistors with large current mirrors set as an active resistance, providing much more current than just using simple resistors and adapting over temp and such. Basically, the current gets divided between the + and - legs of the diff pair which in turn appropriately biases the output stage of an op-amp, like a common source amplifier. Depending on the amount of current, it changes how well this amplifier can pull up to the positive rail or down to the bottom rail. In real op-amps, there are additional stages, which provide more gain (meaning the output transistors can pull close to the rails with less current from the differential pair).

Positive feedback is not well-suited for amplification purposes.

Funny, my very first shortwave radio used positive feedback, and it worked sorta ok.

Telephones have worked for a hundred years or so too.

One of the early open wire carrier amplifiers was conditionally stable. When turned on in the normal way, it would oscillate. The procedure was to remove one particular tube, turn on the power, and after the remaining tubes had warmed up, insert the missing tube. This avoided the gain level at which the amplifier was unstable.

These were in service for many years!

The regenerative receiver was invented by Edwin Armstrong before he was a full fledged adult. To me, the performance from a single tube device was amazing. I still marvel at it. Moreover, it worked when it was unstable. The oscillating detector enabled listening to cw signals.

In the old days when the microphone and hand-held receivers, it was great fun to place the receiver over the mouthpiece to get howling from the feedback. I just tried it again. It still howls B I had to use two telephones in order get receiver to mouthpiece proximity.

Bill