# Low Level Photodetector opamp LTC1150

I found below schematic from LTC1150 datasheet (PG11) I want know what is the purpose of 10R resistor to ground.

when I add this resistor to my circuit it reduce the noise (or peaks).how? After some research I found Out they call this t network base on this website

but how this help the noise

• It increases the gain 1000 times (in combination with the 10k resistor). – evildemonic Jun 27 at 22:58
• But how does this reduce the noise (or spikes) on the vout. Also could you please let me know how did you get 1000 times – Shahreza Jun 27 at 23:09
• Any more questions? – Tony Stewart Sunnyskyguy EE75 Jun 28 at 5:28

You may need a high V/I transimpedance amplifier (TIA) gain using this configuration, such as this example $$\V_o/I_{in}=10^9 \Omega=[\mu V/\mu A]\$$ This implies a feedback $$\R_f = 10^9\Omega \$$.

This creates problems for; reduced BW, DC bias current offset gain and stray EMI very high impedance induced noise.

If that R was used instead, the input pF and stray capacitance pF, might only be a few pF in total, but this RC product results in 1000 times slower rise time and Bandwidth, BW than the offered T network feedback.

Also, any input bias current even in the low nA range could be amplified to >1Vdc output as an error voltage.

To prevent this all these effects, the T feedback ratio reduces the negative feedback voltage by this divider ratio from the output by 1000:1. That, automatically causes any forward voltage gain to be 1000x.

The feedback is further reduced to a current source by the large 1M to act as a current to voltage conversion to the input for error feedback.

The forward gain is now $$\ Vout/Iin= 1M\Omega * 1000 = 10^9\Omega \$$ .

The shunt cap across the 1M has an RC value =15us chosen to match the RC value of the sensor and cable capacitance That is just a scope probe 10M input impedance matches the cable low capacitance to the scope 1M//xx pF input impedance to flatten the tuned probe frequency response.

Except now it does a better job than $$\10^9*C\$$ because stray capacitance would exceed the 0.015 pF requirement, which cannot be achieved.

Theoretically, the following circuit would achieve the same- The practical problem arrise due to the extreme value of R1. This kind of impedance is difficult to implement on a PCboard (requires shielding and often some kind of silicon rubber). Also, you could not get a 0.015 pF capacitor, it's totally impractical. Instead, feeding the feedback signal through an attenuator such as the 10k-10ohms will maintain a low impedance circuit while achieving equivalent attenuation level.