# LED current rising

Discussion in 'The Projects Forum' started by aguaman99, Dec 23, 2012.

1. ### aguaman99 Thread Starter New Member

Dec 23, 2012
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Im building an LED array and relitivly new to electronics. I thought i understood the basics of led's in series however my logic is being challanged.....

im powering a string of 8 leds with a bench power supply in voltage control mode. the leds are 3W high output cree, have fv of 3.00V and are able to run up to 1500 mA. when i connect 8 of them with a 24V fixed supply voltage and a current limiting resistor the current through the string is not fixed and it rises. Shouldnt the current be a constant value based on the limiting resistor value and the supply voltage.

I suppose i just dont understand something correctly....

appreciate any insight helping me to understand what im missing

2. ### panic mode Senior Member

Oct 10, 2011
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Q point may change due temperature rise. also if Vf=3.0V per LED and you have 8 of them, that is 8x3V=24V which is fine for voltage drop for LEDs, but where is the overhead voltage for resistor? what is the value of your resistor? what is it's power rating? what is the rating of the power supply?

say you want to run them at 0.8*1500mA=1200mA
say supply voltage is 32V (need that overhead for resistor).
32-8*3.0=8V, got 8V overhead
8V / 1.200A = 6.666 Ohm

say we choose 6.8 Ohm as a common value

8*8/6.8=9.4W
so our 6.8 Ohm resistor is 15 or 20W component.

i doubt you would still have problem with this setup.

if you are using industrial 24V power supply, they will have slight adjustment of the output.
if you can get 26V you will be lucky. this only gives you 2V overhead which is very little for 8 LED string.
you would proceed same way and hope for the best:
(26-24)/1.2=1.666 Ohm. Suppose you pick 1.8 Ohm as common value
P=V^2/R=2*2/1.8 = 4/1.8=2.22W so you may try 5W resistor.
note this is really pushing it.

Last edited: Dec 23, 2012
3. ### aguaman99 Thread Starter New Member

Dec 23, 2012
12
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i was being general with the values because i am away from the shop and dont recall exact values however i used a supply voltage slightly higher than the 24 volts needed for 8 leds with VF of 3 V. I then used ohms law to calculate the resistor required to drop the remaining voltage. The issue is that when i hooked it the the power supply and held the voltage at a constant (lets say 24.8 ) voltage the current was rising.

What i think understand is that if i have a constant voltage applied to a series of led's with a resistor shouldnt i get a fixed current draw through the circuit. what would make it just keep slowly rising?

4. ### aguaman99 Thread Starter New Member

Dec 23, 2012
12
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in other words if i fix/limit voltage and everything else is calced correctly shouldnt measured current be a constant fixed value based on ohms law? Im guessing im missunderstanding the concept here or maybe the heat build up has a exponential affect thats greater than i imagine..

5. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
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LEDs should not be powered in series with a Constant Voltage source. When LEDs warm up, the junction voltage will decrease, causing current to increase. This turns into "Thermal Runaway" ending with the LED failing.

There are some good single transistor Constant Current drivers in This older thread.

Higher power LEDs benefit from more efficient (and complex) current regulation circuits, such as a switching regulator set up for current mode . This method is what most good 1W+ Flashlights use to be very bright and not waste a lot of heat in current limiting resistors.

6. ### aguaman99 Thread Starter New Member

Dec 23, 2012
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ahhh that might explain why when i set the power supply to constant current mode the voltage dropped slowly in contrast to the original issue of current rise.

I targeted 3.05 V @ 700 mA and the data sheet states this is within the limits of the leds.

8 * 3.05 = 24.4V. I set the bench power supply to 25V giving me a voltage overhead of .6V. .6V/.7A = .86 ohms. I used a 1 ohm resistor and expected .6V/1ohm = 600mA.

am i thinking about it wrong? sounds like maybe i need a constant current circuit as you suggested. I assumed it might be a temp thing things did get hot quick.

7. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
Depends on the application and how easily you can dissipate heat in the environment.

The other problem with constant voltage is if you set the voltage for the running current, it may be too low for initial brightness. A current limiting circuit will be a requirement for your application.

Post a schematic of your LEDs and power setup you plan for the end use, and there may be off the shelf LED Drivers you can use as well. The drivers are essentially a switching regulator with a constant current regulation, rather than constant voltage regulation.

Example is the "Buck Puck", which takes a wide range of input voltages and produces fixed output current for LEDs. If running off a battery, you may want to separate your LEDs into 2 strings, each powered by it's own current source. This would allow a "dim" mode in addition to full brightness, and reduce the required initial voltage by ½.

aguaman99 likes this.
8. ### aguaman99 Thread Starter New Member

Dec 23, 2012
12
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well i wish i were that organized or experienced enough to have a completed plan and know how to draw and post it . As i said im new to this and home schooling myself in electronics. to start out I was just experimenting with the 8 led sting and a constant voltage supply to make sure i had that part understood correctly before designing the rest of the circuit. I wanted to be able to build the driver circuit myself. I liked the other thread you sent me to, looks like a good example of a constant current design i can make myself. already have a few lm317's in the shop so maybe with some transistors and tinkering i can get something going without going the off the shelf direction. thanks for you time and help. not sure i totally understand why but ill take your word for it and abandon the constant voltage for a constant current circuit. its a lot of fun trying to figure it all out, had no idea electronics could be this fun. ive learned a lot but obviously have a long long way to go.

9. ### aguaman99 Thread Starter New Member

Dec 23, 2012
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Following your advice Ive created a simple constant current transistor circuit out of parts I salvaged here in my shop. Things are working much better, thank you. At this point I'm trying to tweak things to get the best efficiency and lowest power disipation and I'm having trouble with the math or my understanding of how to bias the transistor. When I set the base voltage via the voltage devider resistors I can get the voltage I want (think i need) to limit the current to the emmiter there by setting the constant current through collector and emmitter. I'm able to make it work however; I keep having to replace R2 with a pot and adjust the voltage at the base. In short, i do the math and figure out what R2 should be to get 2V to the base and i get that when i test the leed to the base before i connect it to the base pin. Once i connect to the transistor base pin and retest the voltage i get at least a 1V drop. do i have something connected incorrectly or is this normal and i am just missing something about how the bjt works....

10. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
The Base-Emitter junction behaves like a diode, this means that the transistor will show a 0.6V or slightly higher (depending on type and current) when operating.

Bipolar Junction Transistors are current controlled devices, rather than voltage controlled. See "Vol III, Semiconductors" in the bar at the top of the page for the e-Book, which explains this much better than I can in a single post.

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13. ### aguaman99 Thread Starter New Member

Dec 23, 2012
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attached is a spice drawing of my circuit. I expect to get 5.4 mA and 5.08V to the base. Before connecting to the transistor, I do get the 5.4V(very close 5.04). Once I connect to the base pin of the transistor in the circuit I only get .68V at the base pin and my emitter voltage is .71? Is'nt making sense to me why I dont still have the 5.04 volts at the base pin node and then .7V less or 4.34V at the emitter pin???

Sorry if im not as clear as i should be here, as i said im new to electronics and trying to learn on my own.

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14. ### aguaman99 Thread Starter New Member

Dec 23, 2012
12
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Thanks I'll read through that. I just tried to upload my schematic (hope it worked) and put in a few values im getting. If you wouldnt mind looking at that and telling me if im headed down the right path i would greatly appreciate it. Im still learning and loving it.

15. ### takao21203 Distinguished Member

Apr 28, 2012
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Yes the current will rise.

But it depends very much on temperature, initial current, and LED type.

-You have to cool the LEDs
-You have to choose a lower initial current, then measure when the LED is hot.

I have a blue LEDs lamp, where current is rising from 0.7 Amps initially to more than 1.5 Amps. Not less than 4 dropping diodes are required.

I measured after 12 hours: Exactly 0.98 Amps.

Another lamp made from red LEDs: Initial current 0.7 Amps, 0.8 Amps when hot (cooling fan not spinning for some minutes).

You really have to preset a current of about 0.7 Amps for a 1 A LED. And it depends on the LED color + cooling.

Last edited: Jan 5, 2013
16. ### Markd77 Senior Member

Sep 7, 2009
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I'm not sure how to analyse that circuit, you need to find the minimum possible Hfe for the transistor you are using from the datasheet.
If you divide the LED current (0.7A) by the minimum Hfe (can be around 30 for higher powered transistors). That would be the minimum current that needs to flow into the base of the transistor. With the 3930 ohm resistor the most current that could flow into the base (ignoring R2 for now) is around 6ma so the transistor Hfe would have to be over 100. You also need some current to flow through through R2 otherwise the voltage at the midpoint of R1 and R2 will not be right. If you used a Zener diode instead of R2 you wouldn't need to worry about that.
I'd suggest as a first try dividing the values of both resistors by 5 and see how much that improves things.

17. ### Markd77 Senior Member

Sep 7, 2009
2,803
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I just re-read this and with a 1 ohm resistor the voltage at the base would be 0.7V for the resistor (at 700mA) plus about 0.6V for the base emitter junction of the transistor, so 1.3V.
I don't think Zeners go that low but if you replaced R2 with two normal silicon diodes (cathodes towards ground) that would give you around 1.3V. If you calculate R1 as above then the current through the diodes should be low enough to use pretty much any diodes.