op amp biasing... why are voltage sources reversed biased +/-?

Question

I'm still trying to understand the theory of audio signal processing, namely guitar amplifiers. The main thing that got me started down this path was a broken amp that I have. I have since figured out the problem and I know how to fix it but I've also been looking at the schematics for it as an example circuit to help me learn the answers to other questions. I will attach, as long as it is ok for educational purpose, two cutouts from schematics that go with the amp (or close to it) and ask a question. I have many questions but I'm focusing on just one: What is the purpose of the two voltage sources (in both schematics, which are different sections of the overall amp circuit) that are both connected via diodes that are reverse biased? I've marked the areas with red. I think I have a guess, but I don't know if it is right.

I included both of these sections to show the similarity, they are nearly the same circuit. The main difference is the input source is expected to be different as the first is a guitar in jack and the second is an effect return jack. Let's focus on the first from here on out. I will try to explain how I think the circuit works, please correct anything that is wrong. The signal comes in at JS1. It seems a little complicated, my guess is it is dealing with the option of "balanced TRS" input, but anyway it goes through a little section I think is a filter and maybe has something to do with impedance matching with the coil looking thing (On the real amp). I'll ignore the details about this section because it doesn't pertain to my real question. And then the signal line hits a resistor R50 and a capacitor C28. I'm not sure what R50 does except maybe just to be part of the voltage divider effect with R49. The C28 capacitor is probably like a DC filter to ensure clean AC. Then there is a split to a 1M resistor R49, which I think is the other half of a voltage divider to set the input impedance and get a high voltage/low current signal to move on to the op amp + input. Then there is a leg to a capacitor C57 which I think is an AC filter to remove unwanted frequencies (not in audio spectrum) IE noise. Lastly, before going to the op amp's noninverting input, there is a connection with +/- 15V. These are my question. Both connections go through a diode that is reverse biased, D1, D2. This seems at first odd and ostensibly useless, but my guess [WAS WRONG!]. The correct reason is given below by several helpful answerers, it's a voltage clamp to protect the op amp, disallowing any voltage that exceeds +-15 in case it somehow was supplied from the external input source. Thanks in advance for your help!

Answer 1

The name of such circuit is a diode clamp or diode limiter.

Assuming the dioded are ideal, they do nothing until they get forward biased, and they only get forward biased if input voltage exceeds the range of power supplies, i.e. goes above positive supply, or below negative supply.

It limits the voltages at the op-amp input to be within supply voltages to protect the op-amp input from getting damaged from accidental overvoltages fed in via audio input.

Answer 2

These diodes are there for OPAMP input protection. If e.g. the voltage at the input is more than +15V, the diode conducts a fault current from the input to the +15V supply, so that the OPAMP not gets more than approx. +15.7V at its inputs. Same for -15V.

Answer 3

These diodes are supposed to protect the operational amplifier. If you look at the NJM072 datasheet, you'll see that it's maximum input voltage is +/- 15 V.

If we assume that D1, D2, D9 and D10 are ideal diodes, they will conduct as soon as they are directly polarized, that is, the anode voltage is greater than their cathode voltage. Suppose that IC2 non inverting input voltage is 16 V (which would, in theory, damage the operational amplifier). D2 would be directly polarized and start conducting current, becoming a short-circuit (remember that we're supposing that the diodes are ideal). After D2 becomes directly polarized, its anode voltage would be clipped at +15 V, protecting the amplifier. You can make a similar analysis for D1 (supposing that its cathode voltage would be -16 V before it starts conducting). The same analysis is valid for D9 and D10.