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Input Resistance of Tranresistance Amplifier
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Kirill V.:
Hi every one!
I set out to find the input resistance of the transresistance amplifier taking into account the imperfections of our world. I design a picoammeter as a hobby and that's what I did the calculations for without relying on any tutorials or manuals. Calculations are applicable for DC or very low frequencies, also some parameters of the operational amplifier are not taken into account.
In the attached archive there are documents with calculations and LTSpice file for you to check the convergence of the results of the simulator and my formulas.
Did it prove useful to you? I would be grateful if you leave comments and suggestions here
magic:
--- Quote from: Kirill V. on December 12, 2019, 03:43:24 pm ---Calculations are applicable for DC or very low frequencies
--- End quote ---
Something along the lines of feedback resistance divided by open loop gain?
In parallel with inverting input dynamic resistance, obviously.
That's interesting. It means that those simple LMC662 picoammeters actually have plentiful kiloohms of input impedance.
Of course it makes little difference with such low currents.
On a more practical note, it could probably become a serious headache at higher frequencies and high currents. Say, trying to build a zero burden voltage general purpose AC ammeter.
David Hess:
National Semiconductor application note 242 has some discussion about the practical aspects of the transimpedance configuration for measuring current.
Kirill V.:
Thanks!
Transimpedance/transresistance amp design seems to have become an axiom. It is assumed that the high open loop gain gives solid advantages-just select the desired amplifier, add a feedback resistor and the amplifier is ready for typical use - with photodiodes or something similar.
However, commercial picoammeters have a different topology and I paid attention to the Advantest picoammeter. The first stage is a buffer and it does not take part in loop gain. It allows you to achieve a giant input resistance of the device. And by the user's choice, the gain may even be a unity - it's so unlike the widespread solutions that I decided to do a theoretical analysis. This is what the calculations are made for. Now anyone can use them, the only problem is that the exact value of the input resistance also depends on the insulation resistance, for example.
Kirill V.:
It is interesting to note that the right term in square brackets is, in fact, the transfer coefficient of the voltage divider. Also interesting is that the input impedance of the op-amp is multiplied by (1+Aol), which is typical for a non-inverting amplifier. Feedback is interesting-everything revolves around the same multipliers :)
How will this work with a current source that has a non-infinite internal resistance? Intuitively, this internal resistance is included in parallel with the input resistance of the op amp, and one can add this to the square brackets in the denominator, right?
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