Electronic – Measuring integrated current

amplifiercurrentcurrent measurementmeasurementtransimpedance

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.

Best Answer

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:

schematic

^{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).

Related Solutions

Electronic – Measuring 0 – 1MHz ( 0.25Hz resolution) Squarewave using an MCU

If possible I'd suggest selecting a microcontroller that supports a counter operation using the timer inputs; rather than manually incrementing a counter inside an ISR (which at high frequencies quickly ends up saturating the microcontroller activity) you allow the hardware to handle the counting. At this point your code simply becomes a matter of waiting for your periodic interrupt then calculating the frequency.

To extend the range and make the frequency counter more generalised (removing the need for multiple ranges at the expense of a little more work for the MCU) you could use the following technique.

Select a periodic interrupt rate that allows for measurement accuracy at the highest input frequency; this should take into account your counter size (you need to select the timer period such that the timer counter will not overflow at the maximum input frequency). For this example I'll assume that the input counter value can be read from the variable "timer_input_ctr".

Include a variable for counting periodic interrupts (should be initialised to 0 at startup); for this example I'll refer to this variable as "isr_count". The interrupt period is contained in the constant "isr_period".

Your periodic interrupt should be implemented as (C pseudo-code):

void timer_isr()
{
  isr_count++;
  if (timer_input_ctr > 0)
  {
    frequency = timer_input_ctr / (isr_count * isr_period).
    timer_input_ctr = 0;
    isr_count = 0;
  }
}

Obviously this rough example relies on some floating point math that may not be compatible for low-end microcontrollers, there are techniques to overcome this but they are outside of the scope of this answer.

Electronic – Measuring current with multimeter

Your meter is correct. With a 3.7 V battery and a 10 ohm resistor, Ohm's law (\$V = I\cdot R\$) tells you that the current should be 0.37 A.

The 1260 mA you mentioned is probably 1260 mAh (milliamp hours). This tells you the capacity of the battery, not the current draw. For example, if you drew 100 mA from the battery continuously, it would last 1260 mAh / 100 mA = 12.6 hours.

Best Answer

Related Solutions

Electronic – Measuring 0 – 1MHz ( 0.25Hz resolution) Squarewave using an MCU

Electronic – Measuring current with multimeter

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