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Assuming the S is ground, you can protect the opamp's other input from high V by: A) connect to the D with a R and use a low V zener to gnd. The zener will clamp the opamps V to a safe value. Or B) us …

They would, but the feedback via RF overrides it. Say the voltage VF1 was slightly positive (say + 1 mV). Because it has a very high gain, the opamp's gain will drive its output very negative; RF will …

For simplicity, assume R4=R6 and R7=R10. Pin5 of U2B is then (A1 + Temperef)/2. Pin7 of U2B is then 2x this, or A1+tempref. Thus R3 has Tempref across it, no matter what value A1 is, and so acts as a …

In practice for a single supply amplifier, if the input is GND, the current will flown out from the input. However this doesn't affect the design goal - make the effective resistance at each input equ …

R1 doesn't do anything to help the gain -- eliminate it and increase R2 & R3 to compensate. You actually don't even need R2 (replace with a short). It helps control input offset, but the loading of th …

The closed loop output impedance will be quite low -- equal to the open loop output impedance divided by the loop gain. If you want to measure the output impedance at DC, you don't need an input V sou …

A photodiode will produce an AC signal (on top of a DC signal) if exposed to an AC stimulus -- e.g. a signal that is (amplitude) modulated. Even if you aren't using it this way, you still need to ensu …

It's a full wave rectifier. Feedback for positive inputs is via the 2nd opamp; for negative it is via the diode.

This is used to reduce the noise on the + pin; this in turn prevents the opamp from amplifying the noise. There is no precise answer and no component values listed, but a time constant of 10's of ms m …

In addition, C34 acts as a higher resistance at low freqencies (below 1/(2.pi.R.C) = 1/(2 * 3.14 * 680 * 100n) = 2.3 kHz. Thus gain rolls off towards 1 at frequencies below 2 kHz (the 100 nF likely is …

Noise in most circuits depends on bias current in the critical components. It is easy to make bias currents quite independent of supply voltage with an internal current regulator. There is no reason t …

This faulty analysis is commonly because of incorrect placement of the loop 'break'. When you break the loop at the point shown, the Miller effect of the FET is not analysed correctly because the FET' …

Using opamps (which are loop compensated) and diodes to switch loops will always have some type of unnecessary delay as some signal transitions through 1 or 2 diode drops. Instead, use 2 opamps: One a …

Cfb is not really considered a feedback capacitor. At low frequencies, Cfb has little effect and loop gain has a term R3/(R2+R3) in it. At higher frequencies, Cfb 'shorts' R2, eliminating that term (i …

It is meaningless to compare dBs in percentage terms. Decibels are already relative measurements, so you should just refer to the dB difference (0.028 dB in your case, or about 0.3 % in the amplitude) …