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inverting OpAmp Current to volt, when non inverting input is not at GND??

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settra

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hello forum. i am trying to make a current to voltage converter, that will convert , current produced from a photocell, to voltage.
i know that the simple way is like this :
classic_450.jpg

the thing is, that in my circuit, the non-inverting circuit, is connected to a veriable voltage source (-10, to 10 v) , and so , the inverting outpout, is also used, to BIAS the photocell...
in that case, how can i find the Vout of the OpAmp?? thanks allot!!
 
You need to subtract Vbias from Vout of the I-V amplifier. It takes another opamp to do this.
 
There was a nice article:
"Understand and apply the the Transimpedance amplifier (part 1 of 2)"
by David Westerman, Staff Applications Engineer, National Semiconductor Corp.
This covers mostly what you want.
In general, it is a transimpedance amplifier which is used in CD/DVD players to read the optical output.
Hope this helps.
 
If you apply a voltage to the non-inverting input, the transfer function becomes Vbias-(I*Rf)

You don't have to bias it there. You can bias with an isolated supply

Let's at least establish what I max has to be and some idea of the photocell's characteristics.

Make sure we are talking about the right device: photocell, photodiode, or solar cell?
 
its a photocell. its an experimental apparatus, for photoelectric effect!! the currents i will be dealing with, is from some pA to max 100pA !!
the bias, is my teachers opinion, so i cant do much about it. but the reason for that, is , we have to take measures, for the noise in our signal, so everything has to be guarded, by extremely low input current op-amps (Which are expensive, and we cant use many), and so on...

the output voltage, has to be 1mV per 1pA!! i have also concluded to Vbias-(I*RF) , but i cant seem to be able to use it, since Vbias, will always dominate the output??

i will look on the documentation posted!!
the circuit so far, looks like this : (the current source, represents the photocell... couldn't find the correct symbol!
opDSMXe.jpg
 
You didn't take the diff amp advise, so here it is in more detail:

For the sim, the current source is varied from -10uA to +10uA as the bias is stepped from -10V to +10V. I put a gain of 10 into the diff amp. You could also use an instrumentation amp to subtract out the bias voltage, and get more gain in the second stage.

Putting gain in the diff amp will give you an overall better noise figure than trying to get all the gain you need in the transimpedance amp.
DF50.jpg
 
I'm just curious. The op amp is likely to have some input bias current, which is probably much higher than 1pA (and may be negative too). It is also very likely to change dramatically when you change its "+" input between -10 and +10V. Are you going to calibrate your device to somehow account for the input bias current and its changes over the scale?

You can avoid the influence the cell bias on the output if, instead of biasing the end of the cell connected to the op amp input, you bias the other end of the cell which is currently connected to the ground.
 
Deleted
 
That particular range is going to be very tough. The dominant term is Vos and Vos is VERY temperature dependent.

There is a low current applications book available for download on www.keithley.com.

You need to add a zero check/zero correct function. It's relatively easy to turn an I-V converter to measure Vos.

Your bias circuit really needs to be low Z.

There is a "T-circuit" that reduces the size of the feedback resistors needed.

Connecting the non-inverting input to a Teflon standoff may be necessary. You may not want to run those traces on the board.
 
Havent looked at the project yet, cause i'v been busy. but:
NorthGuy, the Op-amp used, is application specific, ULTRA low bias. in the range of fA. so no problems there.
KeepItSimpleStupid, i am planing to use guard anyway ;)
 
ok. what do you guys say about that circuit??
iOPfUVt.jpg


its supposed to give 1mV per 1nA... my only problem is, if i have bias the photocell correctly....
 
That will certainly bias the cell. I don't see how you are going to distinguish between Vout changes due to cell current and those due to bias voltage change (noise, ripple) :confused:.
 
MM... the bias i Supposed to be noise free... but u think that appart from the cell current, the cell will have current by the bias itself?? (as if the cell was a huge ressistor in serries? )
 
the cell will have current by the bias itself?
Presumably, unless the cell has infinite resistance. Why else would the cell need bias?
 
i see. and i suppose, there is no good way, to be aware of that current, without a second op-amp?? (cause if i where to use 2 op-amps, i would go MikeML's way )
 
there is no good way, to be aware of that current
You haven't told us the cell spec, so we don't know if the cell current due to the bias is predictable.
Try MikeML's method; but I still think errors/drift in bias may mask the wanted cell output.
 
i see. and i suppose, there is no good way, to be aware of that current, without a second op-amp?? (cause if i where to use 2 op-amps, i would go MikeML's way )

There's no way to account for that current at all. You cannot tell if the current you get is leakage current caused by your bias or some other form of current produced by the cell. If you want to distinguish them, you need to turn the cell off (you don't tell us what is the cell, so I don't know if that's possible) and measure leakage current alone. And even then, there's no assurance that the leakage current with cell off is the same as with cell on. If you cannot turn it off, you cannot separate leakage and cell currents.

Why do you need the bias anyway?
 
i dont know anything about the cell... i need the bias, because its a experimental apparatus for the photoelectric effect... i suppose the way to turn the cell "off" would be to stop radiating it with light?? (at least this is what we do with the current apparatus )
 
If you're building the apparatus for someone else, you probably need to follow their specs. Although it would be a good idea to ask them about leakage current.

From the cell viewpoint, it doesn't matter if the current measurement is done from the biased end or from the ground end. It's the same current both ways. Measuring the current from the biased end and then subtracting the bias voltage is likely to be less accurate because the bias voltage is 1000 to 10000 higher than the voltage you're trying to measure - you may get nothing but noise.
 
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