What is the maximum achievable transimpedance gain with LTC6268. I wish to design a TIA to amplify a current pulse whose magnitude is around 100 pA and FWHM (full width half maximum) of the pulse is around 0.2 ms for which the rise time expectation is around 5 to 10 us. I wish to achieve the transimpedance gain of around 10e9.
Hi M Arivalagan,
If I understand you correctly, you says about transfer coefficient - Volts/Ampere. In your case it value is possible because 100 pA multiplied by 10^9 V/A equal 100 mV output voltage - this voltage can be easily provided by an amplifier.
Open-loop gain defines the input impedance of a transimpedance amplifier that is visible to the input source. Due to this, the input impedance can be obtained very low. You can estimate the value of the input impedance like this: R_feedback divided by Aol.
Thanks for the response.
1. From the data sheet I see the minimum open loop voltage gain Av is 125V/mV. I cannot connect how this Av affects the closed loop transimpedance gain.
2.Second, my source capacitance is unknown. My sensor is a Faraday plate whose source capacitance is unknown to me. Literature says its value is very low and negligible. My worst case rise time requirement is about 1us. When I try to calculate the feedback capacitance value at this transimpedance gain is in the femto farad range which is way below the practically available 0.1pF value. How do I tweak my design accordingly?
1. The transimpedance gain is determined by the value of the feedback resistor. There is an effect of the amplifier's own input resistance, but in your case this value is at least 1 TOhm that is three orders of magnitude greater than the feedback resistor (you will use a 1 GOhm resistor, right?). Then this influence can be neglected. The open loop gain only reduces the input resistance of the TIA.
2. Take a look at this article. I haven't studied it in detail yet but I'm sure there will be something useful for you4xp3_TIA.zip
Will it help you?