Is the source permanently connected to power? If so do all transistor in a circuit connect to a shared power rail?
Looking at the TL084 schematic, the JFET source is fed from another transistor (a current source in a current mirror). Some of the transistors are connected to V++ and some are connected to V-- but many of them are not connected directly to any power rail.
Must voltage always be present on both the source and gate in order for current to flow to the drain?
There has to be a voltage differential between the source and the drain for current to flow. JFETs are "normally on" and get pinched off as the voltage differential between the gate and source increases.
Lastly when the transistor is switched "on" does voltage leaving the drain then become input for a gate of some neighboring transistor in the circuit?
Voltage doesn't really leave the drain, but the drain is connected to the base of another transistor, part of the second state of the op amp, where most of the amplification happens. In other words, the output of the differential pair is the input to the next amplifier stage.
Would these last two points also apply to MOSFETs?
In the case of a MOSFET there obviously wouldn't be a "next amplifier stage" but would the output voltage be input to the next transistor's source or the next transistor's gate maybe?
You can put these things one after the another to get more complex behavior.
Roughly speaking, common collector/common drain amplifiers have a gain of approximately 1, but has low output impedance, which makes it suitable for driving big loads (like the outside world). Common emitter/common source amplifiers have high output impedance but tons of gain.
The classic op-amp topology is a differential amplifier (long-tailed pair, in the wikipedia template) feeding into a common emitter/source gain stage, then into a common collector/drain follower output stage.
To answer your last question, the output of the gain stage is the drain of the transistor, that feeds the gate of the common drain output stage transistor.
The gate is usually the input terminal, but sometimes the gate is at a fixed voltage and the drain is the input terminal (cascode-y circuits are like this, aka common base or common gate).
Comments
Looking at the TL084 schematic, the JFET source is fed from another transistor (a current source in a current mirror). Some of the transistors are connected to V++ and some are connected to V-- but many of them are not connected directly to any power rail.
There has to be a voltage differential between the source and the drain for current to flow. JFETs are "normally on" and get pinched off as the voltage differential between the gate and source increases.
Voltage doesn't really leave the drain, but the drain is connected to the base of another transistor, part of the second state of the op amp, where most of the amplification happens. In other words, the output of the differential pair is the input to the next amplifier stage.
Thank you for the detailed reply.
Would these last two points also apply to MOSFETs?
In the case of a MOSFET there obviously wouldn't be a "next amplifier stage" but would the output voltage be input to the next transistor's source or the next transistor's gate maybe?
For MOSFETs the story is about the same.
For the tree transistor horsemen (BJTs, MOSFETs, JFETs), they can all be fitted into similar transistor amplifier building blocks:
https://en.wikipedia.org/wiki/Template:Transistor_amplifiers
You can put these things one after the another to get more complex behavior.
Roughly speaking, common collector/common drain amplifiers have a gain of approximately 1, but has low output impedance, which makes it suitable for driving big loads (like the outside world). Common emitter/common source amplifiers have high output impedance but tons of gain.
The classic op-amp topology is a differential amplifier (long-tailed pair, in the wikipedia template) feeding into a common emitter/source gain stage, then into a common collector/drain follower output stage.
To answer your last question, the output of the gain stage is the drain of the transistor, that feeds the gate of the common drain output stage transistor.
The gate is usually the input terminal, but sometimes the gate is at a fixed voltage and the drain is the input terminal (cascode-y circuits are like this, aka common base or common gate).
Thanks I found the classification in that link quite helpful. Cheers.