The circuit is a very poor wien bridge oscillator and will not be stable - amplitude adjustments will need to be made and distortion is quite likely to be high and vary with time. There are better sinewave oscillators that can be found so I urge you to do some google image research on this. This is the sort of circuit I'd be considering for a decent sine wave oscillator: -
The basic Wien bridge oscillator has nothing to define or control the sinewave amplitude whereas this circuit uses a JFET to do that. Here is a google image page for sine wave oscillators.
As for the output impedance of the op-amp, because negative feedback is used you can assume the output impedance is less than an ohm when driving loads that a 741 is capable of. This won't give you much power into a transformer so don't expect miracles. The 741 isn't capable of delivering (relatively cleanly) any more than a few handful of tens of mA.
If you want to model this circuit I have no hesitation in suggesting LTSpice (free version) but I think you'll find you'll need a transistor buffer between op-amp and transformer if you expect to drive powers more than a few tens of milli watts.
1) Yes. The usual way to handle this for an AC signal is to construct the feedback so that it has unity gain at DC, and your required higher gain above your minimum signal frequency. That way, only 1x the input offset voltage gets onto the output, rather than gain times.
2) The problem with bias current is that the resistance of the DC path attached to either input must be low enough to source this current, without developing excessive voltage drop. Think of input bias current as being due to current sinks at the input terminals, these are biasing the bases of the input transistors. If the input AC signal is AC coupled, then the R to ground at the input needs to be small enough.
3) Yes, as for input offset voltage, but multiplied by those R to ground input resistors.
You don't want to damage the equipment you are feeding into. It doesn't matter what filtering or feedback you use in your amplifier to control gains and offsets, you must assume that at some point, your amplifier output will hit the rails. If the equipment you are feeding will be damaged by that, then you must take additional steps to protect it.
One solution is to run your amplifier from low enough voltage rails that output saturation will be OK.
Another good solution is to use a pair of anti-parallel silicon diodes to ground after a suitable current limiting resistor from your output. This will clip the output voltage to +/- 0.7v from ground, and allow a signal of +/- 500mV or so with minimal distortion or gain error.
simulate this circuit – Schematic created using CircuitLab
Not all of these AC coupling components are required, but I've shown them all to discuss what they do. Depending on the DC level of your source, the DC input requirements of the thing you're driving, your amplifier offsets and required system gain, none to all of these will be useful.
a) Input AC coupling, R1C1, prevents any input DC reaching the opamp.
b) Feedback AC coupling, R2C2, gives unity gain at DC, and high frequency gain of 11. Any input offsets appear at the output without the gain that the signal has. With a gain as low as 11, typical input offsets are unlikely to be a problem at the output, even if that gain is not rolled off at DC.
c) Output AC coupling, C3R4, removes any amplifier output DC from the final output.
d) The input bias currents will flow through R1 and R3, generating an offset voltage across these resistors. These voltages balance each other and do not create an amplifier output offset. For the amplifier shown which is FET input, these will be very low, and even megohm values for these resistors would be OK. With a bipolar input amplifier, values like this would be more appropriate.
e) The input offset current is the difference between the input currents. The difference in voltage generated on R1 and R3 does cause an output offset.
f) You may for input filtering or gain bandwidth reasons want R1 and R3 to be different values. This mismatch will generate an output offset voltage with the input bias current, but it will rarely be a problem, especially if you can AC couple the output.
g) Output protection of R5D1D2 limits the output voltage to around +/- 700mV, R1 limits the current into the diodes to something they and the amplifier can handle. This form of output protection will only work if +/-700mV is safe for your load, +/- 500mV signal level is adequate (600mV signals will start to get seriously distorted), and R1 is not so big that it limits output current excessively into the load, or its stray capacitance. If all of these are not met, you will need to do something more clever.
Best Answer
Input resistance (1.5T\$\Omega\$ typical) is the change in input current for change in input voltage
\$R_{in}\$ = \$\frac {\Delta V_{in}}{\Delta I_{in}}\$
It is not clear whether this figure is intended to apply to differential input voltage or to common mode voltage or both.
Input current (10pA typical) is the current flowing into or out of the input pin.
If the input was an ideal current source, the input current would be constant so the input impedance would be infinite.
You can model the input (at DC) as a 1.5T ohm resistor to the other input (probably, given the disposition of the input protection diodes) and two +/-10pA current sources, one connected to each input.
Some op-amps (the ancient CA3140 is not one of them) have a rather high input resistance when the two inputs are close to each other in voltage but nonlinear networks across the inputs that turns into k-ohms if you apply more than a diode drop differentially. Not a problem in normal op-amp applications, but problematic if you're using it in applications where it might saturate (precision comparator, some precision rectifier circuits etc.).