# Measuring integrated current

I want to measure the amount of light striking a photodiode in a specific time window (about 1ms). In this window, we get $$\10^6-10^7\$$ photons, which get converted into electrons with about 80% efficiency.

The standard method of performing this measurement is to use a transimpedance amplifier to convert the current into a voltage which is then sampled rapidly throughout the time period and integrated. However, there are lots of places for noise to creep in. Each of the measurements is quite noisy, and this get integrated up.

I was wondering if there was a more accurate method of measuring the total charge produced in this window?

So far I have been considering something like an analog CCD:

• A capacitor is used to store the charge.
• When the time window is over a digital signal can isolate the capacitor from the photodiode
• The charge on the capacitor is then measured, by some method I haven't thought of yet

If anyone has any advice I'd love to hear it. Essentially, I currently have a noisy measurement of the current over time, which I would like to trade for a more accurate measurement of the total charge produced.

Edit: As some context, this question was brought up when I realised that a single-pixel CCD would actually do this job much better than a photodiode. They can have large quantum efficiencies and read noise in the 10s of electrons. Sadly, single-pixel CCD's don't appear to exist, as far as I can tell. (Also, the dark current is a little problematic, but you can always cool them down).

An avalanche photdiode (or photomultiplier) might also be better suited, but we're trying to avoid high voltage.

• Perhaps this thesis would be useful: Design of High-Resolution Photodiode Readout Circuitry for a Bio-Implantable Continuous Glucose Sensing Chip ( repository.tudelft.nl/islandora/object/… ) – crj11 Mar 19 '19 at 20:30
• Integration tends to reduce noise. It's differentiation that tends to enhance noise. – The Photon Mar 19 '19 at 20:31
• The number of photons you need to count is very low, so perhaps the best choice is a photomultiplier: however, operating such devices require several cares since they are very high speed and high voltage devices. – Daniele Tampieri Mar 19 '19 at 20:36
• @ThePhoton is right; since you're integrating this, the noise will be reduced, not worsened. – Hearth Mar 19 '19 at 20:37
• It appears you want to get down to as low as $100\:\text{pA}$ of photogenerated current. What resolution into that do you expect? What noise floor is acceptable? What accuracy do you require (keep in mind that accuracy involves traceability to standards.) What methods of calibration for accuracy do you possess? What repeatability would you require if you built two of these? (How close would you expect their measurements to be, given the exact same measurement situation?) And definitely consider the change in bias currents for opamps over temperature. Scary, not infrequently. – jonk Mar 19 '19 at 21:26

The standard method of performing this measurement is to use a transimpedance amplifier to convert the current into a voltage which is then sampled rapidly throughout the time period and integrated. However, there are lots of places for noise to creep in.

Reducing the places for noise to creep in is exactly why you would want to digitize your signal as soon as possible, and use as few analog processing stages as you can.

Each of the measurements is quite noisy, and this get integrated up.

Integration will tend to reduce noise, not increase it, because integration is inherently bandwidth-limiting.

It's differentiation that tends to enhance noise.

You can certainly build an integrating response into your transimpedance amplifier:

simulate this circuit – Schematic created using CircuitLab

If R1 is very high, then the integrating response provided by C1 will tend to dominate the TIA response, and you will effective have an integrating TIA. In fact, this configuration is often used because the capacitive feedback path can improve the amplifier's stability, even when an integrating response is not desired. The integrating version is essentially just over-compensating this common circuit.

If you need to reset the output, you will need some additional circuitry to discharge C1 (without letting the op-amp go into saturation).