Is there a way to tell max volts an LED will take from a CC power supply? - Page 1
Forum section: electronics / beginners

author: Beamin
author_rank: Super Contributor
timestamp: June 01, 2019, 09:54:57 pm
content: I was going through somethings and I found something I built but can't remember anything about it. Its a transistor, two resistors, a photocell, and a 1.5"X3" white LED backlight from ada fruit, can't really make out what values are on the parts.So I want to figure out how bright this will go with out blowing it up, it works pretty well to light up rooms at night. I have no idea what kind of led this is or anyway to look up the data sheet. BUT I do have a power supply that I can set by 1ma and 0.01v. Setting the ma at some arbitrary number then seeing the power supply drop the voltage is there a way to get a close estimate of the LED max current brightness? At 3v the circuit pulls about 1ma to see it start to glow then at 5v with your hand blocking the sensor it pulls 3 ma. Sounds really low, or a backlights supposed to be dim low power leds? Looks like a single LED. I would like to crank up the voltage but not sure. Does the photocell care how much voltage it sees?

author: mariush
author_rank: Super Contributor
timestamp: June 01, 2019, 10:29:53 pm
content: Limit current to something reasonable like 10..20mA (pretty much any led will tolerate 10mA and 10mA is big enough for power supplies to measure)Set voltage to some low value like 1vconnect led in circuit.Slowly increase voltage while monitoring currentWhen you get close to the forward voltage, led will start to conduct and current will rise.Leds have a very narrow region where they behave like partially open, like there's something chocking current flowFor example (using bogus numbers) a white 100mA 3.2v led may be fully closed up to 2.8v and then between 2.8v and 2.9v it may be partially on, letting 1..10mA flow through...and once you go over 2.9v it may be fully open letting hundreds of mA go through if you don't limit current.But, you can't rely on these values as they're dependent on led temperature. it may be 2.8v..2.9v at 25..30c ambient but if you just turned off the led after an hour of running and led is 50-60c warm, that narrow region most likely drifted to something like 2.75v..2.85vIf you limit psu to 10..20mA and ramp voltage slowly you'll see the current jump from nothing to 10..20mA. It's also big enough value to see individual diodes if led is made with multiple diodes in series... ex a 6v led may be made with 2 leds in series inside package.If psu won't let you limit current, you can make your own with a linear regulatorex see the plain LM317 regulator:http://www.ti.com/lit/ds/symlink/lm317.pdfSee page 12, section 8.3.3 ... how to limit current Iout = 1.2 / R1 ... so for example if you want to limit current to 20mA (0.02A) then 0.02 = 1.2/R1 so R1=1.2/0.02 = 60 ohmThis is not astandard E12 or E24 valuebut 56 ohm or 62 ohm is and they're close enough, instead of 20mA you have maybe 18..22mA which is still good enough. If worried, you can recalculate for 15mA and then pick closest resistor standard value.So you can feed 5v..12v to lm317 and you get on the output some voltage with a current limit.Now you can use a second LM317 in standard adjustable configuration (see page 10, 8.2) ... with potentiometer you can adust output voltage from minimum 1.25v to input voltage minus ~ 1..1.5vAnd note that the regulator itself will consume around 5mA so if you limit current to 20mA only with first regulator and 2nd regulator consumes ~5mA+ then led may only get 10..15mA

author: bob91343
author_rank: Super Contributor
timestamp: June 01, 2019, 10:33:51 pm
content: That probably isn't the best way to test. My method is to apply around 10V to the LED through a resistor of around 470 Ohms. If you vary the voltage you will also vary the current. Set the current at 10 mA and measure the voltage across the LED.Connecting an LED directly to a power supply is too risky. You can blow it up. Best to have that series resistor.

author: Beamin
author_rank: Super Contributor
timestamp: June 01, 2019, 10:55:25 pm
content: The post I just typed got deleted grrr..img too big it says. Anyways.Im trying the above method now but how did I wire this? I drew diagram just now from what I can see, I'm assuming the photodiode is to the base(center pin of trans, but I can't read the number on it nor see resistor values) and either C or E, that's why its unlabeled. Shouldn't the photodiode have +Vcc hooked to it then to the base? Looks like I wired it between base and either e or C. Seems backwards? The power supply does 1ma and 0.01v all the way down to zero, so there is also the circuit as protection, I turn the power supply on then block the sensor and watch the ma go up.EDIT: Forgot: Why doesthe LED flash on when I first turn it on the power even though the sensor is in light? (Turns on with dark)

author: schmitt trigger
author_rank: Super Contributor
timestamp: June 01, 2019, 11:16:01 pm
content: What you drew as a "photo diode" is in reality a CdS photoresistive cell.

author: Beamin
author_rank: Super Contributor
timestamp: June 01, 2019, 11:24:30 pm
content: Quote from: schmitt trigger on June 01, 2019, 11:16:01 pmAh yes, it's the thing that is a flat disk with the zigzag brown and white under the clear plastic. So if it's resisting the way I drew the schematic is wrong? Maybe the center of the transistor is not the base? On the other side of that circuit board is a wire that goes from one of those resistors to the center pin of the trans. That's the only wire underneath.So far set the power supply to 10.1 v and putting my hand over the sensor makes it go up to 13ma with the psw set to cut off at 15ma. I have a 30 volt power supply should I go higher or will that eventually burn out the sensor or transistor. Can anyone see the resistor colors from the picture? Then we could do some ohms law action instead of guessing...

author: Brumby
author_rank: Unknown
timestamp: June 02, 2019, 01:15:26 am
content: Quote from: Beamin on June 01, 2019, 11:24:30 pmYou really should look it up as some transistors are E-C-B. Guessing gives you a 1 chance in 6 of getting it right.

author: bob91343
author_rank: Super Contributor
timestamp: June 02, 2019, 04:58:41 am
content: You ought to have one of those magic Chinese boxes that for $15 or $20 tells you the connections to the part. It also measures resistors and capacitors and inductors and diodes and transistors and more. Google MEGA328 and see what ebay has.It's almost the most useful thing on my bench. It won't tell you maximum ratings but it does give ESR of capacitors and much more.Mine came in a little box with a rechargeable battery. I opened it and added test leads so I don't have to try to fit parts into that tiny zip socket.

author: Beamin
author_rank: Super Contributor
timestamp: June 02, 2019, 12:51:36 pm
content: Quote from: bob91343 on June 02, 2019, 04:58:41 amI do actually have one of those but my electronics and huge desktop magnifying screen/camera are packed up. I have vision problems and can't see the part. I don't think the center is base because if it was that circuit I drew shouldn't work. Wouldn't the photo resistor have to send the current to ground through a resistor and when it has light on it start sending it to the base since you want it to turn on in the dark and not the light? I'm not sure though when I look up circuits like this they don't look like this or the way I would think it would work.So far I have tested it to 10 volts where it pulls 13 ma max. Since its pulling so few ma can I keep cranking up the voltage or will that burn out the photo sensor? With these resistors how many ma should I stop at?

author: Beamin
author_rank: Super Contributor
timestamp: June 02, 2019, 04:45:27 pm
content: OK that cd S sensor lowers resistance as it gets light. If it worked backwards this is how the circuit would look(top circuit with the X)? So to fix this circuit we need to give the current an alternate path of lower resistance to ground while it is light. I think the actual circuit I built is reflected in the bottom drawing? Whats strange is the actual circuit I built( that I'm having trouble seeing) pulls 0ma when it is off. My power supply display is accurate to 1ma and so when it says zero it truly is zero. Unless its possible to set the resistor values to pull less then 1ma? That doesn't sound right if the CdS cell goes from hundred of ohm to kohms.

author: bob91343
author_rank: Super Contributor
timestamp: June 02, 2019, 04:46:45 pm
content: Generally the photo resistor would connect from the collector to the base so that when it's illuminated the transistor conducts. Converely it could be between base and emitter and shut the transistor off when illuminated.The main problems with mixing up collector and emitter is that the transistor has lousy gain and voltage limit if backwards. And of course you won't know if it's NPN or PNP.You can try it with a low enough voltage so it won't get damaged.

author: Beamin
author_rank: Super Contributor
timestamp: June 02, 2019, 04:48:45 pm
content: Quote from: bob91343 on June 02, 2019, 04:46:45 pmI think you typed this as I was uploading the img, so you are saying the schematic on the top with the red X is how it should work?

author: bob91343
author_rank: Super Contributor
timestamp: June 03, 2019, 12:03:52 am
content: Yes that should work. The one with the green check cannot work.

author: Beamin
author_rank: Super Contributor
timestamp: June 03, 2019, 12:45:19 am
content: Quote from: bob91343 on June 03, 2019, 12:03:52 amNow that I think about it the circuit I drew at the top of the page should work. When ambient light is bright the ohms of the CdS should go down making the least resistive path through the CdS and around the transistor. When its dark the CdS has a high resistance making the current flow through the base of the transistor. It was the fact that light=lower resistance not higher that was throwing me off. So when this is off in bright light shouldn't it pull at least 1 ma? The meter says zero and if it was 1ma the meter can show that.

author: bob91343
author_rank: Super Contributor
timestamp: June 03, 2019, 02:57:46 am
content: Base current for the transistor is provided by the photocell. When resistance is high in the dark, the base current is small. When illuminated, there is more base current and then we have collector current, through the LED.This is not a good circuit because it depends greatly on the gain of the transistor. A high gain transistor will light the LED with very little illumination. In fact, it could do it with no illumination.A lower gain transistor will work unless its gain is too low in which case insufficient current will flow even in bright light to make the LED glow.