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?

enter image description here

After some research I found Out they call this t network base on this website

but how this help the noise

  • 2
    \$\begingroup\$ It increases the gain 1000 times (in combination with the 10k resistor). \$\endgroup\$ – evildemonic Jun 27 at 22:58
  • \$\begingroup\$ 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 \$\endgroup\$ – Shahreza Jun 27 at 23:09
  • \$\begingroup\$ Any more questions? \$\endgroup\$ – 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.

Short Answer

The rule of thumb to reduce stray EMI is try to lower the input impedance on an unbalanced source. Now it is only 1M.

So you might think, why not go further with a million to one attenuation in the T, to increase the BW further using 1k for RF. Maybe, but then we must consider noise currents and other details, which we can discuss another time.

  • \$\begingroup\$ This answered my question, just want know when I calculate the Cf do I use Rf or the 1M \$\endgroup\$ – Shahreza Jun 28 at 15:18
  • \$\begingroup\$ Use actual cct resistance for Rf*Cf= 1Meg Cf but I should have said Rf equiv = Rf ' = 1e9 and Cf equiv or Cf ' = 0.014 pF . ciao . \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 28 at 15:23

Theoretically, the following circuit would achieve the same-enter image description here 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.


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