This AoE circuit I cannot make sense of

[Moriambar]

Hi, while reading tAoE 3rd edition (fig 4.66, page 259).
This "T" network should make the feedback resistor seen by the op amp as large as 10M (as advertised). But to me it's just 100k + (100k//1.2k). Where am I wrong? Also, why does the noninverting input have a 53k resistor to GND?

Boy sometimes tAoE is frustrating…

Thanks

[langwadt]

it sets the gain as if the feedback resistor was 10M

the 51K is there so that  the resistance to ground for both inverting and non-inverting input is similar, the reduces offset due to bias currents

[TimFox]

These TEE networks are common when the large required resistance is not practicable.
A simple way to consider it is that the 100 k resistor (at op amp output) and 1.02 k resistor feed about 1/100 of the amplifier output voltage to the other 100 k resistor at the input, making the current through that resistor to the "virtual ground" 100 times smaller than that through a 100 k resistor from the output.  Therefore, the feedback current equals the current through a large resistor (100 x 100 k = 10 megohm).
This has other problems, though, including that the "noise gain" (voltage gain for the amplifier input noise and offset voltages) is increased from unity to x100.  Also, the thermal (Johnson) noise current in the circuit is that of the 100 k resistor, not the equivalent 10 megohm resistor, so the noise increases (when driven from a high-impedance source such as a photosensor).
This is an example of the general "Y-Delta" transformation, which is very powerful for circuit analysis:  https://en.wikipedia.org/wiki/Y-%CE%94_transform

[Moriambar]

These TEE networks are common when the large required resistance is not practicable.
A simple way to consider it is that the 100 k resistor (at op amp output) and 1.02 k resistor feed about 1/100 of the amplifier output voltage to the other 100 k resistor at the input, making the current through that resistor to the "virtual ground" 100 times smaller than that through a 100 k resistor from the output.  Therefore, the feedback current equals the current through a large resistor (100 x 100 k = 10 megohm).
This has other problems, though, including that the "noise gain" (voltage gain for the amplifier input noise and offset voltages) is increased from unity to x100.  Also, the thermal (Johnson) noise current in the circuit is that of the 100 k resistor, not the equivalent 10 megohm resistor, so the noise increases (when driven from a high-impedance source such as a photosensor).
This is an example of the general "Y-Delta" transformation, which is very powerful for circuit analysis:  https://en.wikipedia.org/wiki/Y-%CE%94_transform

Cool! Thanks by your message and a bit of thinking I exactly got how to deal with this kind of circuit/feedback. Y-Delta analysis has always escaped me, but that's another topic for another day