Absolute dumb question about high impedance op-amps

Janis59

Joined Aug 21, 2017
1,863
By theory, always without of any exception, op-amp amplifier is made by negative loop resistor between the output and negative input feet and then other resistor is going from the same negative feet to the signal source. Math is clear, works in zillion of cases without of any problem.

But then I met the femtoampere technics having typically say 10^-18 Ohm inner impedance, few femtoampere scale current and bit more nanovolts voltage what must be amplified becoming measured. For that reason exists a bunch of special operationals like AD7721, LMC662, ADA4530, LTC6268-10, AD549 and many more.

But as more I read the know-how things, more I see they many are switching op-amp in crazy way (but that are serious sci publications, thus the brutal typo error is near nil probability) - how sad they never explain WHY they do so - they switch source (for example Faraday Cup) via some 10 Meg resistor to the negative input, BUT the 1 Gig to10 Gig loop resistor they connect BEFORE the 10 Meg - straight on the contacts of F.Cup.

1) What is that the strange idea behind it?
2) How then I can calculate the K(u) and why it havent plainly diminishing value?
3) If they want to degrade the (for example AD549) toward prost transimpeance cascade with K(u)=1, then why to use ANY resistor at all if ever both are needed for nothing?

Janis59

Joined Aug 21, 2017
1,863
Bit read more articles using such articles - seems this semi"crazy" circuit is more wisdom like mistake. Firstly, 10 Meg makes the input guard functionality against the overvoltage in input. Secondly, the main goal is to get the CURRENT amplification factor K(i)=1 Gig / 10 Meg so the loss of K(u) is thing no-one-cares. Third, there is sth about Johnsson noise what is worsened in this version. See page 13 of https://www.analog.com/media/en/technical-documentation/data-sheets/AD549.pdf

Audioguru again

Joined Oct 21, 2019
6,783
High input impedance opamps use Jfets or Cmos at the input that have no input current. The resistors in a circuit determine the input impedance. Most opamps are voltage amplifiers.
A high value resistor in the circuit develops a fairly high thermal noise. Most datasheets show that the noise from a high value resistor is higher than the noise from the opamp.

Janis59

Joined Aug 21, 2017
1,863
Seems You didnt realized the question content. Firstly, however with larger FET amplifiers, if the 9...14....30 Amperes is "just about nil current" <snip> Secondly, question wasnt about current but about nullification of K(u) in the case of realized circuit topology. I may agree, tat 1 GOhm thermal noise MAY be larger than from IC itself, but then WHY to annihilate the negative loop??

Moderator edit: deleted racist remark,

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MrChips

Joined Oct 2, 2009
31,083

Janis59

Joined Aug 21, 2017
1,863
In Your picture, Chips, there is one more resistor between Your circuits soldering near inverting input toward the inverting very input. In the transimpedance amplifier theory nothing such may be tolerated. For the very basics, this resistor reverts the K(u) toward 1/K(u), i.e. much smaller than 1. Such strange "signal annihilation" is very salt of my surprize.

MrChips

Joined Oct 2, 2009
31,083

What you see is a charge sensitive pre-amplifier with open loop gain. Charge is accumulated on the 22pF capacitor. Normally, charge will build up on the capacitor and the voltage on the output will keep rising. The 1GΩ resistor is a discharge path to bring the charge and hence the voltage back to zero.

The input resistance of AD549 is of the order 10000GΩ. The 10MΩ plays no role in the gain of the circuit. Its purpose is to protect the input of the AD549.