If you shorted R3, Vos or the voltage (because of not being a perfect OP amp) between the + and - gets amplified.

The current some 2 pA per ppm gets converted to a voltage e.g. (2e-12 Amps) and multiplied by a big number.

Vos has to be small, so it doesn't swamp out the measured value.

I'm sorry, but i don't understand at all now. I thought that what you have stated is what is true with my circuit already. I thought that the two inputs are practically both the same, give or take a very small amount (The amount given out by the sensor, plus a little from Vos). I'm not getting how i would bias the opamp by 1v by what your saying. Sorry.

 

So, your gain is -1001; I'll use 1000

If Vos is 1 mV and you multiply that by 1000, you get 1V; Here the 1000 is the 1K resistor.

2 pA/ppm * 2*1000*1001*500 ppm aprox~= 1 mV; I used 1000 instead of 1001 to get 1 mV.

So 2 mV at the input translates to 2V at the output for a 500 ppm signal.
1 mV from Vos and 1 mV from the actual signal.

Hopefully I got the numbers right.

 

I'm going to put this down for a couple of days and resume back in the weekend. I can't seem to figure this out despite the helpful, straight to the point, information you guys have provided. I'd like to thank you for attempting to help here.

 

Let your head rest.

1. Amplifying a small voltage and a small signal, Vos gets amplified too.

2. When you use the I-V converter topology. Vos also gets converted into current. With low currents, the FB resistor is high (e.g. 1 M ohm) So, a 1 mV offset gets converted into I=0.001/1e6 or about 1 nA

3. Autozero amplifiers really put a dent into the offset voltage.

4. Vos is temperature dependent. Vos can be nulled in some OP amps.

1mV of offset is not state of the art. I don't want to force a topology on you. Single supply, battery operated adds other complications.

 

I may be completely losing the point here, but would an op amp in a differential amplifier setup work with providing the inverting side an input that would equal 1V on the output?