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Let me talk in general first: I have two LEDs of the same wavelength (940 nm) placed opposite to each other. Between them is a 140 mg/dL glucose solution (only glucose + water). One LED is used as emitter, the other as detector. Of course, the signal generated from detector will be amplified later. But the problem is:

Choosing an opamp depends on how low the current from detector is. Is it in nanoampere, picoampere, or even lower?

How can I know the range of the current generated by the detector? Since every opamp has an input bias current that influences the result.

Peter Mortensen
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Totally New
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    Try it. It's an off-label use of the LED so there are no specs to design to (and even if there were you would have to do some work to figure out how much light falls on the die). It won't be good compared to a proper photodiode or phototransistor. – Spehro Pefhany May 01 '17 at 10:07
  • Since LED's are cheap, it might be worth trying to play around with sanding down the LED's, assuming they are the 3mm/5mm through hole type. This could make them more directive, which is a good thing in this application. Depending on the speed of the circuit, JFET/MOSFET input opamps could be interesting as they have far lower input currents. – Joren Vaes May 01 '17 at 10:10
  • Still dont understand . How can i try ? I want to know the current through led-detector is in which range – Totally New May 01 '17 at 10:14
  • You can try it by getting it set on a breadboard and measure the current. That will give you an idea of what you need in terms of performance. – Joren Vaes May 01 '17 at 10:17
  • Still confused . Currently there is no device which can measure small current below uA – Totally New May 01 '17 at 10:20
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    Yes there is - it's called a microammeter. You can "zero out" the input bias current by measuring what level you get in complete darkness. – pjc50 May 01 '17 at 14:16
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    There are such devices. In fact, if you have a few thousand USD to spare, you can get the Keithley 6482, which can measure currents with a resolution of one femtoampere. More affordably, the μCurrent can measure down to a resolution of a picoampere in combination with a multimeter. – Hearth May 01 '17 at 14:21
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    @TotallyNew, in your other questions you've asked at least 5 times about a circuit that is perfectly capable of measuring micro-amp currents. Buy some parts and build it, see what happens. – The Photon May 01 '17 at 15:27
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    Also, this question, which I'm pretty sure I already linked to in comments on one of your other questions, tells you pretty much exactly what you're asking about. – The Photon May 01 '17 at 15:28

5 Answers5

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The amount of current that is produced by a photodiode is related to the light power it receives in watts and the "yield factor" in converting watts to amps. All recognized photodiodes have a specification for this.

So, you begin by analysing the emitter: if an emitter is producing "so many" milli watts from (ostensibly) a point source, you can calculate the power density (watts per sq metre) at any distance.

For an isotropic emitter, light power is emitted in all directions so, the power density at a given distance (R) is related to the surface area of a sphere (\$4\pi R^2\$) and the power originating from the "point source".

Given that a recognized photodiode will have a specified active surface area you can converts watts per sq metre back to incident power (watts). This then converts to amps via the "yield factor".

So you need to know how much light power is produced by the LED and how this power is concentrated in a particular direction. For instance, it is very unlikely that it will be isotropic and much more likely you can assume that at least 50% of the power produced is in a 3D angle of maybe 60 degrees. The LED spec should tell you that.

This "concentration factor" will give you watts per square metre at the distance your receiver is placed from the emitter. The active surface area of your receiver converts watts per sq metre to watts and then on to amps and you are done. Of course, the glucose solution will absorb some power so you also need to factor that in but, I guess that's the whole point of your experiment.

However, in your scenario you don't appear to know any of this so, unfortunately, you cannot know how much current to expect and, as others have said in comments, your only option is to try it - find an op-amp with low bias currents and try it out.

I would also recommend using a light source that is pulsed because then you can cancel out drift by sampling before light is emitted and then sampling whilst light is being emitted. This also allows you to get rid of the effects of ambient light changing the reading.

Andy aka
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Since your problem seems to be how to measure a small current - you can typically measure a tiny current with an inexpensive multimeter. You need to know the input resistance of the multimeter.

For example, I have an inexpensive Uni-T M890G multimeter. The lowest current range is 2mA (1uA resolution). So we're scuppered, right? Nope. The input resistance of the meter on the 200mV voltage range is 20M, so the resolution when used as a current range is 100uV/20M = 5pA! For higher currents (>10nA in this case), don't use a different voltage range, shunt the input to get a lower input resistance so that the ammeter voltage drop ('burden') is not excessive.

A better longer term solution is to breadboard a transimpedance amplifier but as you point out you need to have an idea of the order of magnitude before you can pick an op-amp (hint: LMC6001 is pretty good but you would need to take some heroic measures such as PTFE standoffs to really get full value from it).

Spehro Pefhany
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How can i know the range of the current generated by detector ?

By measuring it. That really should have been obvious.

LEDs aren't usually characterized as receivers, so you don't have the necessary information to calculate it. Build a test setup and see what you get.

You could use a 940 nm photodiode instead of a LED for the receiver. That should come with enough data to calculate the signal. However, even then, there are things in the setup that are hard to know. The photodiode will probably give you a better signal than the LED, but to really know what the level of that signal will be, you need to measure it.

Olin Lathrop
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  • just wonder how to measure , since the current is so small. – Totally New May 01 '17 at 11:08
  • @Tot: The same way you were planning to measure small currents in the first place. Start with some circuit, and adjust depending on what you find. For small to medium adjustments, you can probably change some resistors to change the gain. For large adjustments, you may need to change topology and get parts with different specs. Again, just try it already. – Olin Lathrop May 01 '17 at 11:14
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    This really is the most practical answer. Though I would add 1 thing, which is that, depending on what equipment you have available, it's probably easier to do the initial current measurements with a handheld or benchtop DMM. Even if the current is below the lower limit of your best DMM, that at least gives you an upper boundary without you having to order any parts and breadboard. – Dave May 01 '17 at 15:55
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You get one point for using the LED to get a matched wavelength detector however a 'proper' photodiode will get you more gain and most likely more repeatable results.

In this instance I would also seriously consider using an off the shelf optical output controller to drive the LED to achieve a constant low output on the detector. This would let you use the LED drive current to determine the path attenuation and make the rest of your circuitry simpler. Sort of like the circuit shown here . - https://www.maximintegrated.com/en/app-notes/index.mvp/id/1769

Here is some interesting reading as well - http://www.mdpi.com/1424-8220/8/4/2453/pdf

Here is also another nice application note from Linear Technologies. - http://www.analog.com/media/en/technical-documentation/technical-articles/Optimizing-Precision-Photodiode-Sensor-Circuit-Design-MS-2624.pdf

And here is a neat experimental setup. - http://www.ece.rice.edu/~jdw/243_lab/exp6.1.html

KalleMP
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The current will depend on the led in use.

Due to inconsistency and temperature sensitivity, it is best that this being a digital sensor - detecting the presence of light, rather than intensity of light.

Mitsubishi lab has a paper on this.

dannyf
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    The whole point of the question is, as far as I can tell, detecting how much the fluid absorbs so, based on that, the intensity of the light is vital to detect. – Andy aka May 01 '17 at 13:55
  • An accurate value comparator and variable light source might be an option but perhaps not what OP wanted. – KalleMP May 02 '17 at 13:28