# Current amplifier - Shunt ammeter VS Feedback ammeter

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#### Tueftler

##### Newbie level 4
Dear electronic tinkerers,

I want to measure the current of a photo diode (1 pF, ca. 1 MOhm, 0.1-1 V pos. bias) with a large dynamic range (100 pA - 100 nA), with a high bandwidth (DC - 10 MHz) and, of course, with a good SNR of 60 dB or better.

The following paper summarizes the possible current-to-voltage amplifier circuits pretty well:
https://cdn.testequity.com/documents/pdf/keithley/LowCurtMsmntsAppNote.pdf

I think the shunt resistance amplifier can be excluded, because with a relative small 1kOhm-shunt-resistance (for a negligible voltage drop of about 0.1 %), one would need an op amp with a gain bandwidth of 10 MHz * (100mV output / (100nA*1kOhm)) = 10 GHz. And a such a high GBWP op amp with low-noise voltage specs literally does not exist, right?

I think the best way to go is the traditional transimpedance amplifier (see Figure 3). With a high GBWP op amp (ADA4817 and LTC6268 are the best that exist in my eyes), a feedback resistor of 1 MOhm and a parasitic capacitance of about 0.1 pF one can obtain a bandwidth of almost 10 MHz and a good gain of 100mV/100nA. Additional gain (adjustable 10-100) can be added with a second op amp stage.

We know that for of a good SNR it is (almost) always recommended to realize as much gain as possible in the first op amp stage. Figure 4 shows a nice trick how to add additional voltage gain to the TIA op amp by connecting the feedback resistor to a split resistor between output and ground (or common mode voltage). By this one can obtain E_out = - I * R_F * (1+R_A/R_B). My question is: Does this nice trick increase the SNR? Or is this voltage gain comparable with a second voltage gain op amp stage that amplifies both the singal and the noise by the same factor? My second question is: Is the bandwidth still determined by the feedback resistor and the capacitance in parallel to the feedback resistor (fc=1/(2pi*R_F*C_F))?

Thanks so far!

#### KlausST

##### Super Moderator
Staff member
Hi,

I agree with TIA and a second gain stage, and that the first stage should have the higher gain.

Increasing gain: the circuit increases closed_loop_gain. But in your application I assume the limiting factor is open_loop_gain of the first stage Opamp.
The benefit against a single feedback resistor is that the feedback path can be lower impedance, and there some kind of voltage divider. Both may increase cut off frequency.
If I calculated correct then you need less (parallel) stray capacitance of 16fF across the 1M resistor to achieve 10MHz cutoff frequency.
This is imposiible. Here I see the benefit.
But I don't see a benefit regarding noise, not resistor noise, not voltage noise and current noise of Opamp.

The drawback is, that the offset voltage of the Opamp becomes multiplied.

Klaus

#### FvM

##### Super Moderator
Staff member
Obviously the Keithley application note isn't targeting to MHz Bandwidth, thus you shouldn't expect answers in this regard.

realize as much gain as possible in the first op amp stage
Should be IMHO corrected to as much gain as necessary to make the noise contribution of succeeding stages negligible. Gain of 10 is usually sufficient.

Dropped 2 pi factor in bandwidth calculation has been already mentioned by KlausST.

Rf is directly related to SNR by resistor noise current proportional to 1/√R. See https://en.wikipedia.org/wiki/Johnson–Nyquist_noise

#### Tueftler

##### Newbie level 4
Thanks, I am aware of all the noise contributions. For a 1uA-diode I calculated a total noise floor of about 6 mV rms for an output of 1 V which is acceptable.

Now, let me try to formulate my question more precisely:
Compared to a larger feedback resistor, a voltage divider between output and feedback resistor of the transimpedance amplifier (TIA) does not increase SNR because signal voltage and noise voltage are amplified equally. But, a voltage divider causes slightly less voltage noise than a second op amp stage would cause due to additional input voltage noise. Is that right?
A concern might the output current. Without the voltage divider, the output current of the TIA is exactly the current through the diode. With the voltage divider, only a small part of the output current goes through the diode. Doesn't this reduce the precision and/or response time of the TIA?

This leads to my second question concerning the design:
A bandwidth of 5-10 MHz should be possible for the TIA by using a large 0805 resistor and a ground shield or a guard ring under the resistor to reduce the parasitic capacitance to less than 50 fF (see application note in the LTC6268 datasheet pages 15 and 17). Can you confirm that? If that seems impossible for you, I could reduce the feedback resistor from 1M to, let's say, 200k and add a 1:5 voltage divider with 4k between output and feedback resistor and 1k between common-mode voltage (2.5 V) and feedback resistor. Right?
For possible ultra-low current diodes (<100 nA), I want to add variable voltage gain. Multiplexing one of the resistors of the TIA’s voltage divider would add capacitance (of 50-100 pF) to ground and leakage current (10-1000 pA) which might disturb the functioning of the TIA. So, in this case it might be better to move the additional variable gain to a second op amp stage. Right?

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