CFA's have been around forever. It's possible to find versions of them in Vacuum Tube designs from the 1930's. Later, there were versions as Chip and Wire Hybrids. Then Comlinear came out with monolithic IC versions, some time in the early 1980's. If you needed a low gain, high slew rate, low distortion amplifier there was no comparison. Parts like the LH0024 from National Semiconductor (NS) were instantly obsolete.
In self defense perhaps, NS bought out Comlinear and subsumed their product line, which was almost exclusively CFA's, into the LMHxxxx line. That's where you will find what's left of Comlinear's line. Some of the more specialized parts like the CLC501, an amazing output clamping CFA, were instantly obsoleted by NS, leaving some customers desperately scrambling for life time buys. Of course TI bought NS in another take over the competition move. Although suppliers have consolidated, Analog Devices and Linear Tech still make them as well as TI.
CFAs are great for things like:
- Video amps (for focal plane array buffering or some such).
- Driving cables.
- Trans Impedance Amplifiers, for wide band photo detectors. This one of the few applications where it makes sense to use a feedback capacitor (in parallel with the feedback resistor), usually this is a real no no--but the right value of feedback capacitance compensates for the detector capacitance.
- Any kind of low gain pulse amplification.
- Best as non-inverting amplifiers, because of the low impedance of the inverting input. This is forced by the restricted values of gain resistors relative to feedback resistors.
They're not great for:
- Driving Capacitive loads. Reference your question Figure 6., note the heavy duty snubber at the FET gate.
- High gain amplifiers.
- Differential amplifier.
Although the gain equations for the CFA look superficially like a VFA, they are different. The feedback resistor (\$R_f\$) sets the stability of the amplifier, and will be some small value like \$500 \Omega\$. Less than the specified amount will make the amplifier unstable, and greater values will reduce bandwidth. The resistor connected to the inverting input is the gain resistor (\$R_g\$), and since \$R_f\$ is a low value, \$R_g\$ has to be low too. That can make it hard to drive an inverting stage, and is why non-inverting stages are more common.
CFA's aren't going away, they'll be around in some fashion. It's interesting that some of the new VFAs from TI and Linear Tech appear to have a CFA rolled into them. If you look, for example at LM6171, you'll see what looks like a CFA with a buffer added to the inverting input to make a VFA with very wide bandwidth and high slew rate.
The trick you missed (I think) is that, since this opamp is in a negative feedback loop, you can assume that the voltage between the + and - inputs of the opamp is zero.
You can theoretically explain this by assuming that the opamp has an infinite gain so Vout = (Vin+ - Vin-) * infinite. This results in that the output voltage can have any value for input voltage = zero.
With that assumption (voltage at Vin- = zero, because Vin+ is grounded) and realizing that no current flows into or out of the opamp's inputs it is now much easier to calculate the currents and then the transfer function.
Best Answer
You can reasonably assume that currents can flow from an input (Va in your diagram) and from the op-amp output (Vo in your diagram). This means you can assume current does not flow into the two op-amp inputs and these can be regarded as high impedances.
Additionally, you can assume the op-amp open-loop gain is very high and the impact of this is that for an output voltage that is reasonable (i.e. somewhere within the bounds of the power supply rails), the difference between V1 and V2 is zero volts.
All the above assumes that there is an overall negative feedback that keeps the output in the linear region i.e. within the bounds of the power rails.
If the overall feedback is zero or positive then the output will likely be end-stopped against one of the power rails and there may well be currents taken by the inputs AND, you can no-longer assume V1 = V2.