super dim blue LED - Page 1
Forum section: electronics / general

author: grifftech
author_rank: Frequent Contributor
timestamp: February 23, 2017, 04:02:14 pm
content: I built this circuit but my dmm can't measure the current (less than .5 microamps). can someone who has the right equipment for measuring low currents build and test this circuit. the led is blue

author: Zbig
author_rank: Frequent Contributor
timestamp: February 23, 2017, 04:05:49 pm
content: 10M? But why?

author: grifftech
author_rank: Frequent Contributor
timestamp: February 23, 2017, 04:07:16 pm
content: Quote from: Zbig on February 23, 2017, 04:05:49 pmbecause I can

author: daveshah
author_rank: Unknown
timestamp: February 23, 2017, 04:49:17 pm
content: Two optionsGuestimate the forward voltage of the LED (probably a bit under 3V), obtain the voltage drop over the resistor and use Ohm's law to work out current.Alternatively if your multimeter has an input impedance close to 10Mohm then replace the 10M resistor with your multimeter in voltage mode. Then you can work out current again using Ohm's law.

author: Seekonk
author_rank: Super Contributor
timestamp: February 23, 2017, 05:10:34 pm
content: While a LED can be driven with just microamps, I think that is a bit too low.

author: Fgrir
author_rank: Regular Contributor
timestamp: February 23, 2017, 05:22:09 pm
content: Add a smaller resistor in series (10K-100K) and measure the voltage across that. But yeah, why?

author: macboy
author_rank: Super Contributor
timestamp: February 23, 2017, 06:10:00 pm
content: At such a low forward current, the voltage across the LED will be lower than nominal, but still not insignificant. If you have voltmeter that can do high impedance (>G ohm) voltage measurements at 5 V then measure the voltage across the 10 M resistor and calculate current. many bench meters should be able to do this, but very few handhelds. Alternatively, if your meter tops out at cay 2 V for the high impedance measurements, then you could measure the voltage across the LED, then calculate the voltage across the resistor, then calculate current.

author: bktemp
author_rank: Super Contributor
timestamp: February 23, 2017, 06:16:37 pm
content: The voltage drop across the LED is probably somewhere between 2-3V, therefore the current is between 200-300nA.Using a white LED I could see it glowing at about 10nA. With a cheap photomultiplier tube, I could detect the light at <500pA LED current.

author: Someone
author_rank: Super Contributor
timestamp: February 24, 2017, 02:56:43 am
content: Quote from: grifftech on February 23, 2017, 04:02:14 pmI can build this circuit and measure the result but it won't mean much for you, blue LEDs vary in their IR curves not only between different products and manufacturers but also part to part within the same batch. At these very low currents leakage through the diode (that is not producing light) or the package will make the measurements confusing, there is some already on the forum:https://www.eevblog.com/forum/blog/eevblog-869-counting-led-photons!/

author: Electro707
author_rank: Contributor
timestamp: February 24, 2017, 04:28:04 am
content: If you have the capability to modify the circuit why not PWM the LED instead of putting a 10M Ohm resistor.

author: FrankBuss
author_rank: Unknown
timestamp: February 24, 2017, 04:44:39 am
content: I did a quick test with a RGB LED from my parts bin. I bought it some time ago from eBay, I think this is the datasheet:http://www.media.highlight-led.de/products/documents/pdf/76600026.pdfQuite bright at even 10 mA. It's amazing what the eye can do, with 10 M ohm I can still see a very tiny dim dot when on, if the room is dark, but can't say if it is blue. I guess this is the scotopic vision of the rod cells of the eyes, which are more sensitive than the cone cells required for color vision. I can measure 270 nA with 5 V. This means the voltage drop at this current is about 2.3 V. It is about 3 V with 10 mA.

author: John B
author_rank: Frequent Contributor
timestamp: February 24, 2017, 06:27:31 am
content: Quote from: Electro_Dynamic on February 24, 2017, 04:28:04 amThe power dissipation in a 10meg resistor at these voltages will be so trivial that it's not really worth worrying about more circuitry.

author: grifftech
author_rank: Frequent Contributor
timestamp: February 24, 2017, 06:05:39 pm
content: Quote from: John B on February 24, 2017, 06:27:31 amyes, the resistor with 5v across it dissipates 2.5 microwatts