Electronic – Why isn’t a 1 Farad capacitor equivalent to a wire in LTspice

My only problem with the electrical example is regarding how the system may couple its output back into its input. Is this a simple case of simple feedback? How does simple feedback relate to capacitive coupling?

An 1 Mohm 0805 resistor will have self capacitance of about 0.2pF and this, when applied in the feedback loop of an op-amp will reduce the gain from the 3dB point of 795 kHz. This means that the transimpedance amplifier you are designing will not work well at 10 MHz because there is too much capacitance - and that's just from a single surface mount resistor - imagine what capacitance there is between tracks and if this is not properly catered for by good PCB design, you'll be lucky to get 100 kHz flat operation from the above circuit.

The above is the case of negative feedback causing a signal to become smaller on the output due to parasitic capacitance.

It seems that the output would need some ripple voltage to begin with that would be amplified.

The minutest noise will be enough to trigger oscillation if the feedback is positive. Noise is present in all components at temperatures above absolute zero.

Regards driving tracks having capacitance this is not directly related to the possible cause of oscillations and additionally the rule of thumb in your question doesn't take into account many, many factors such as the power capabilities of the driver, and the track dimensions and whether the track is terminated. At low frequencies there is no real rule at all.

without running your circuit to check, it is usual to expect leakage current in bipolar devices, and silicon diodes are p type and n type, hence the bipolar name. When one diode is reversed biased, impurities in the silicon structure allow a very small current to flow through the device. This current is very temperature sensitive. In a circuit such as this I would expect the reverse bias leakage current to be micro amps or less.