LED voltage measurement question - Page 1
Forum section: electronics / beginners

author: Lucky-Luka
author_rank: Unknown
timestamp: March 29, 2019, 12:52:06 am
content: Hi allI'm starting building simple circuits to better understand the theory related to them.As soon as i've finished my first ever super simple circuit something isn't clear at all...I'm talking about my LED voltage measurement.I have a breadboard with a 5V power supply attached.When the power supply is disconnected from the board and when the button on the breadboard is pressed and the power supply is attached to the breadboard I can assume that what I read in the voltmeter is right but I cannot understand what's going on when the power supply is attached to the breadboard and the button on the breadboard is not pushed. Why do I read -1.308V (power on) and -1.496V (power off)?Thanks for your help.Luca

author: Audioguru
author_rank: Super Contributor
timestamp: March 29, 2019, 01:12:58 am
content: Like you were told on the other website, use a battery instead of a wallwart.

author: jackbob
author_rank: Regular Contributor
timestamp: March 29, 2019, 08:16:18 am
content: It is difficult to tell from the pictures. The board you are using for power seems to have multiple regulators on board likely 5V and 3.3V but i'm not sure. It could be due to some difference between these voltage rails and how that board you are using for power is configured. To simplify it further, use a battery or a simpler power supply with only one rail and not all the voltage regulation on board.

author: Lucky-Luka
author_rank: Unknown
timestamp: March 29, 2019, 08:31:01 am
content: Quote from: Audioguru on March 29, 2019, 01:12:58 amAudioguru I understand that you have seen my question in other forums but I don't understand why feel the urge to tell everybody about it in each forum where you have seen my question.Well, never mind. Can you give me another answer to my problem? What if I don't want to use batteries?Luca

author: Lucky-Luka
author_rank: Unknown
timestamp: March 29, 2019, 08:31:34 am
content: Quote from: jackbob on March 29, 2019, 08:16:18 amThanks for your suggestions.Luca

author: Ian.M
author_rank: Super Contributor
timestamp: March 29, 2019, 08:45:23 am
content: Pull the LED out of the breadboard and connect it directly to the DMM (on Volts). You will see a voltage due to the LED acting as a solar cell albeit a very weak poor one. Wrap the LED in thick black cloth or similar so its totally in the dark and the voltage should drop to zero.You can also try measuring the voltage the LED produces in direct sunlight, facing the sun, and also see how much current it can generate - the uA range of the meter will probably be suitable.

author: jackbob
author_rank: Regular Contributor
timestamp: March 30, 2019, 12:33:43 am
content: Quote from: Ian.M on March 29, 2019, 08:45:23 amI thought the same as well but this does not appear to be the case here. When I first saw this post, this was my first thought and I went to my drawer of LED's then hooked it up to a multimeter to see if I got similar results. What I found is yellow LED's will not produce the voltage seen here without a significant amount of light, far more than seen in these pictures. Also when the LED is acting like a solar cell and producing voltage its polarity remains the same as the voltage applied to the LED to light it up so it would not show a negative voltage as seen here in the picture.

author: Jwillis
author_rank: Super Contributor
timestamp: March 30, 2019, 01:26:22 am
content: Your not getting any AC leakage through the power supply are you?Would still show a DC voltage because the LED is acting like a half wave rectifier.So your seeing the LEDs voltage drop.

author: ArthurDent
author_rank: Super Contributor
timestamp: March 30, 2019, 03:23:59 am
content: It is hard to tell from just the photo, but going by the DMM the yellow LED is on at 1.8VDC and not at 1.3VDC. If you check the Vf of different color LEDs you will see that they require different minimum voltages. The minimum for yellow is about 1.8VDC and white is about 3.3VDC and so on. I'm not sure where the 1.3VDC is coming from with the switch open but the 1.3VDC is not enough to cause the yellow LED to light or conduct.If your meter is 10Mohm input impedance, the switch has some leakage, and the meter is the only real load with 1.3VDC across it, the LED will not light. If you place a 1Mohm across the DMM input and the voltage drops with the switch open, what you are reading is leakage.

author: Wimberleytech
author_rank: Super Contributor
timestamp: March 30, 2019, 04:24:42 am
content: I have replicated your circuit using the same power supply configured just like yours.I observed the voltage across the LED with both an oscilloscope and a Fluke DVM.I see a similar forward voltage when the switch is on.When the switch is off, I see a small voltage due to photons generating electron-hole pairs--aka solar cell, but nothing close to what you see, and opposite the polarity you see. I do see about 500mV p-p voltage at line frequency.Unplug the power to the regulator and tell us what you see.

author: hamster_nz
author_rank: Super Contributor
timestamp: March 30, 2019, 08:08:10 am
content: Hi Lucky-Luka,Electronics works by assumptions and approximations. Things like "oh, this wire just conducts electricity" or "current though a resistor varies linearly with voltage", or "9V batteries put out 9V" and so on. These assumptions don't always hold true, but when people design electronics they work with the simplified models that apply at their level of design, so an auto electrician will use different models than a radar engineer for things like wires.So pretty much everybody will agree that a 1V DC over a 1000 ohm resister does cause 1 milliamp to flow through that resistor (+/- however much that resistor isn't exactly 1000 ohms due to manufacturing errors). However, if that is 1V AC of a WiFi signal rather than DC, then that model will be wrong.So there are "first approximations", more true-to-life "second approximations", "third approximations" and so on, each more true then the previous ones.The first approximations are usually the "ideal" device. For a diode it is "A diode lets current only flow one way".The second approximation for a dioide is "A diode lets current over 0.7V flow one way"The third approximation might be "A diode lets current over 0.7V flow one way, and has a low-valued resistor in series with it"Further approximations might include leakage (where a tiny amount of current can flow the other way), and capacitance (where the diode can store a small amount of charge).Everything you are using in your design - the breadboard, the wires, the switch, the power supply, the LED, even the DVM is like this. Some are more complex than others, but all will behave oddly under unusual operating conditions.Modern DVMs are pretty sensitive devices. They are made to detect really small voltages and really small currents. They are sensitive enough to show things that prove that the assumptions and approximations are just that. This is what you are seeing - it is a phantom reading. I mean it is real, but it is most likely of no actual consequence - like when you just hold the probes up in the air, they will read something but not something that is useful.I really don't know what is going in your case - is the LED acting like a solar cell? is the wall wart (that may be a 'smart' device) constantly probing to see if a load is attached? Is some of the mains frequency leaking through the Wall Wart? Is the breadboard acting like a capacitor, not a bunch of passive wire? Do you have a WiFi router at the edge of your desk that is causing this? Do you have flouro lights?The guess that the power supply has something to do with it is pretty good one, as everything else is pretty simple. So you need a simple power supply that works much like the first approximation for an ideal battery... Batteries are about as simple as they get, they pretty much act as a voltage source with a low-valued resistor in series. So try it with a battery and see if the effect remains.This is the "interesting" part of electronics. Finding weird things, analyzing unexpected problems and learning. However, it is also the frustrating part too!One thing I am pretty sure I know is this: if you put a 10k resistor over the legs of the LED will make the problem go away. It will still light when you press the switch, and it will read (very close to) zero when the switch is open. The 10k resistor will swamp out any of these minuscule effects, and make things look more like the "ideal" LED.Sometimes you will find out really odd things, like I found out can't use my Oscilloscope in the wall socket that was in the same circuit as a baby monitor, because it does something weird and adds quite a bit of RF ripple to all the readings. That took me quite a while to understand that it wasn't my circuit at fault!

author: Wimberleytech
author_rank: Super Contributor
timestamp: March 30, 2019, 01:58:05 pm
content: After further experimentation, I conclude that the issue is simply electrostatics. I can move my hand around the probes and modulate the voltage...kinda like a visual theremin.As suggested, I put a resistor across the LED (a 100k was handy) and the problem went away.

author: ModemHead
author_rank: Frequent Contributor
timestamp: March 30, 2019, 03:39:24 pm
content: I have one of those cheap 9V adapters, possibly even the same one. The scope shot represents the voltage to earth from the 9V output (either terminal). I measure about 75uA leakage current. Nothing unusual for these types of devices.So if one lead of the LED is connected to the power supply and the other one is left floating, you'll get a little positive voltage from the photo-voltaic effect, but possibly a much larger negative voltage based on how much capacitive coupling there is between the free end and earth.

author: Lucky-Luka
author_rank: Unknown
timestamp: March 31, 2019, 11:58:03 am
content: Quote from: hamster_nz on March 30, 2019, 08:08:10 amCan you explain me why?thanksLuca

author: Lucky-Luka
author_rank: Unknown
timestamp: March 31, 2019, 12:31:15 pm
content: To clarify the situation I have discovered that touching the power module the voltage drops to mV values.If I measure the voltage differential at the power module + - pins I read 0 V.SoI have upgraded a little bit my system.The problem is gone: 59mV voltage read, which is ok.Stray voltages seem to be the culprit. Am I right? But what does it mean?Thanks everybody for the contributions: I'm a beginner so everyhing I read here helps me a lot.Luca

author: iMo
author_rank: Super Contributor
timestamp: March 31, 2019, 12:40:50 pm
content: While doing a measurement with your DMM with 10Mohm input resistance and with long DMM cables you have to be aware you may always see such effects.A capacitive/inductive coupling to the AC mains in your voltage adapter, AC pickup from wires around you, high frequency signals from radio sources (ie wifi, gsm, ..), wrong grounding or decoupling, all this stuff may influence your measurements.Those AC signals are rectified by your LED diode and with the high 10Mohm input resistance you see the DC voltages.The signals come from "high impedance" sources, say 1V AC pickup signal comes from a source which has say 100Mohm internal impedance.With your 10Mohm DMM input impedance/resistance it creates a voltage divider so you will see (after DC rectification) something like 90mV.When you wire 10k in parallel to your 10Mohm DMM input you will get around 10k input resistance of your DMM in the result and the resulting voltage will be aprox. 1000x smaller.

author: iMo
author_rank: Super Contributor
timestamp: March 31, 2019, 01:21:48 pm
content: I would recommend you to download the LTspice simulator (it is free and easy to operate) and you may learn to draw schematics and play with your designs

author: Lucky-Luka
author_rank: Unknown
timestamp: March 31, 2019, 01:45:10 pm
content: Quote from: imo on March 31, 2019, 12:40:50 pmWhy?btw I've tried a 10k resistance in parallel to my led and now I read 0V.Luca

author: iMo
author_rank: Super Contributor
timestamp: March 31, 2019, 03:26:14 pm
content: Quote from: Lucky-Luka on March 31, 2019, 01:45:10 pmWhy what?I've taken the 10k as an example. You may take 100k if you wish. Or any other value which is much smaller than 10Meg.Note: when you decrease the input impedance/resistance of your DMM (ie with a parallel resistor) you have to think about how it will influence the circuit in which you measure.What is important for you is to get a feeling for what you are actually doing.With high impedance/resistance circuits you have always to calculate with parasitic phenomena

author: hamster_nz
author_rank: Super Contributor
timestamp: March 31, 2019, 06:56:42 pm
content: Quote from: Lucky-Luka on March 31, 2019, 11:58:03 amThe 10k resistor is low enough to provides a path for whatever the stay voltage is to 'balance out'. Any voltage over the LED will cause current to flow in the resistor, which the heats up a tiny bit, getting rid to the energy.But the resistor is also high enough that it doesn't really change the desired operation on the rest of the circuit in a meaningful way (which a low valued 10 or 100 ohm resistor might).