# Beginner getting into electonics - help please

Discussion in 'General Electronics Discussion' started by dbov22, Aug 27, 2012.

1. ### CocaCola

3,635
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Apr 7, 2012
I see your math but I disagree and believe there is an error with it, or best a missing variable...

1.25/9=0.139 approximate impedance of 9 LEDs
0.139 + 0.642 = 0.781 add in battery impedance
1.1 / 0.781 = 1.41 Amps

Can you verify if said flashlight is drawing 1.41 Amps? My guess is closer to 800mA...

To verify my guess I just tested one of my cheap LED flashlights (mine had 6 LEDs) so by your math...

1.25/6= 0.208 approximate impedance of 6 LEDs
0.208 + 0.642 = 0.85 add in battery impedance
1.1 / 0.85 = 1.294 Amps

BUT, when I measure my 6 LED flashlight (fresh batteries and all) it's only drawing 550mA, that is 43% of what your calculation said it would be... That is a HUGE discrepancy and IMO shows a clear fault in the calculation... It also indicates that 'something' else is limiting the current... As I have been stating that is the LED itself... These flashlights do not depend on the battery to limit the current, they simply exploit the fixed voltage and what the LED will pass at that voltage to limit current... It's still a poor way to manage current but it will work, in a calculated shortened life product...

BTW, yes I verified no resistors in the flash light...

My measurements are pretty much fully inline with what I would expect assuming that the LED is actually doing the limiting, not the battery... Look at this chart below of a typical super bright white LED... I measured the 3 AAA batteries in the flashlight they measure at 4.56 volts, I highlighted this point on the chart...

Based on this chart, I would expect each LED in my 6 LEDs flashlight to max out at about 85mA, or 510mA total... Since I actually measured 550mA (only a 7% error) I shows the LEDs to be the true source of the current limiting, in said flashlight...

And just for kicks I tested 1 single LED and 2 LEDs in parallel based on your math... They are 14000mcd bright white, no solid datasheet information...

1.25 + 0.642 = 1.892
1.1 / 1.892 = 581mA (1 LED)

1.25/2 + 0.642 = 1.267
1.1 / 1.267 = 868mA (2 LED)

Those values above will likely instantly pop the LEDs...

Real measurements, LEDs that were quite warm to the touch nearly instantly, so they were being pushed hard...

1 LED = 101mA ~ 17% of what was expected by your math
2 LED = 196mA ~ 23% "

Now as the number of LEDs increase X-fold there will certainly be a point where the batteries internal resistance comes into play and will supersede the existing limiting being done by the LEDs, but that is also the point at which you are drawing so much current the battery is in a rapid terminal overheat self destruct mode due to the high current drain, so there is no stability in such a circuit and it's actually a potential hazard...

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Last edited: Aug 29, 2012
2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
My calculations indicate the "resistance" of the LED at this voltage is around 10 ohms.

I'm not sure where the resistance stuff came from that KJ6EAD was quoting.

The figure from the graph indicates that around this point the current changes by 10mA for each change in voltage of 0.1V.

The small signal resistance if a diode decreases as the voltage across it increases. I don't recall a formula for it, but I expect it's significantly different for a LED vs normal silicon diodes, but I suggest the significant difference in Vf would make the results very different at any particular voltage.

Last edited: Aug 30, 2012

1,114
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Aug 13, 2011
Don't blame me. I didn't do it. It was John Monks or CocaCola in their debate about the number of electrons that can dance on a diode junction. The OP doesn't know the difference between parallel and series circuits and has started a new thread on this topic over at AAC.

4. ### rodrilyx

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Aug 29, 2012
You would put a resistors!

5W at 12V = 416,6 mA.

A red led need it 1,83V and something like 20mA so,

40 red leds = 20mA * 40 = 800mA

5. ### CocaCola

3,635
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Apr 7, 2012
I believe we see a case of the OP not getting the answer he/she wants to hear... I gave suggestions in post #17 for them to do this, easily and straight forward, but I suspect it's not a case of how to do this 'properly' it' more how can I make this work... The same types of suggestions and issues that are posed here are also posed in his/her new thread..

Last edited: Aug 29, 2012
6. ### john monks

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Mar 9, 2012
Steve, where did you get the datasheet?

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,401
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Jan 21, 2010
Oops, I wrote datasheet when I should have written graph.

I was referring to CocaCola's graph

8. ### dbov22

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Aug 27, 2012
Guys, the power transformer pictured in original post - I found my multimeter and providing it s on the correct setting (V with two horizontal lines, top line solid and bottom line is 3 dashes like this - - -)

That reads at 13.64 so does that mean its kicking out more than 12 volts?

In my parallel circuit, at the point where the wires start joining onto the LED circuit it was reading at 3.2

I disconnected HALF of the LEDs and all was fine.

I went into Maplin (electronic shop) equivalent of radio shack and the 7 members of staff didn't have a clue, they didn't know the difference between series and parallel but all absolutely insisted that the supply was AC and not DC...due to the symbol on it and the fact it was reading more than 12v

9. ### john monks

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Mar 9, 2012
CocaCola, let me explain something to you.
The resistance I used was derived from the short circuit current (7A) I got from you. Even though the resistance I got seemed low I used it just to make a point. Last night I took a brand new AAA cell out of the original wrapper and loaded it with 131mA. The cell went from 1.56 volts to 1.51 volts giving me an impedance of 0.382 ohms. Therefore three AAA cells gives me 1.15 ohms. The textbook formula (25mV/I) is good only around the knee of
the IV curve, not around the extremes, close to zero current or up into the straight high current part of the curve. This is only an approximation, not an exact value. So based on this information I recalculated the forward current per LED like this. One AAA cell has .382 ohms. Therefore three cells has 1.145 ohms. An LED impedance will be approximately (25mV/20mA) or 1.25 ohms. All in parallel will give around 0.03125 ohms. Now you subtract the 3.4 volt per LED (manufacturers specification) from the 4.5 volt battery voltage and divide by the total impedance 1.1763 ohms and you get 935mA. Divide this by the 40 LEDs and you get 23.4mA per LED.
Mistake #1 using a textbook approximation (25mV/I) without carefully explaining it.
Mistake #2 using my cheap \$2.00 9-LED flashlight as an example without characterizing the LEDs.
Mistake #3 not explaining that the battery resistance is not the only protection for the LEDs.

These were my mistakes not yours. Mistake #1 led to some great absurdities with your flashlight example.

Last edited: Aug 30, 2012
10. ### BobK

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Jan 5, 2010
If you got a reading of 13.6V on the DC Voltage setting of the multimeter it is a DC supply. When unloaded, an unregulated supply will typically read higher than the rated voltage just like yours did. When drawing the rated current it will read closer to 12V. The fact that you read 3.2V when your LEDs are connected just reinforces what I said in my initial reply:

I am surprised the power supply has lasted this long. You are undoubtedly damaging it by overloading it the way you are. If you would reconfigure your LEDs to be 3 in series with a current limiting resistor in each string, your power supply will be much happier and will read something close to 12V when it is running.

Bob

11. ### CocaCola

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Apr 7, 2012

It's one of the better resolution graphs I was able to source quickly, as many of the datasheets only contain the graph at the recommended working voltages, or no graph at all...

Here is another datasheet nearly (if not) identical

http://www.surplustronics.co.nz/library/Mega-White-LED.pdf

And here is how I found most of the datasheets with the voltage cut off at just above 3.5V, thus didn't apply in this situation...

http://www1.futureelectronics.com/doc/EVERLIGHT /334-15__T1C1-4WYA.pdf

This was my biggest objection, the batteries limiting in the circuit is in the end insignificant and offer little protection, even with a 'perfect' power source at 4.5V the LEDs will survive on their own for awhile due to their own limiting...

Not a shock, but sad altogether...

Probably not, you just extended the life of the transformer a little bit... As you have been told repeatably, you are doing this wrong and the only reason the LEDs are not turning into popcorn is because the transformer is puking it's guts out...

12. ### john monks

693
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Mar 9, 2012
Not so insignificant CocaCola.
Using your short circuit figure of 7A for the batteries and zero impedance for the LEDs you get 43mA per LED.

Last edited: Aug 31, 2012
13. ### arg733

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Dec 14, 2010
I know from experience that using LEDs without resistors will permanently damage them.
I used some amber and white LEDs connected in parallel and with no resistors, the white ones didn't even light up and after a few seconds some of the amber ones became black. When i replaced those that had burned out and added resistors I noticed that the amber ones that had survived were emitting less light than the new ones besides the fact that the LEDs were completely identical.

It is possible that if all your LEDs are identical you MAY avoid catastrophic overload BUT WHY TAKE THE RISK AND NOT JUST ADD A FEW RESISTORS? I also don't like the idea of drawing the last ounce of power from the power adapter, always try to leave a 5-10% unused power to avoid overheating and to prolong the life of the adapter / transformer / power supply

ALWAYS REMEMBER THAT YOU MUST PUT RESISTORS WHENEVER YOU USE LEDs

Last edited: Sep 3, 2012
14. ### john monks

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Mar 9, 2012
I'm sure that I and CocaCola can agree that LED's are connected in parallel, with no resistors, in at least some flashlights. Now, as one chemistry professor recently told me, "when a person says 'whenever' be suspicious." you only have to come up with one example to the contrary to prove the statement wrong"

As I remember an amber LED typically runs at about 2.7 volts and a white LED runs at about 3.4 volts. A white LED is typically a blue LED with a phosphor that make white light. I suspect that a white LED requires more energy to cause an electron to fall into a hole to release the higher energy photon than an amber LED and that's why more voltage is required. Dbov22 gave me the impression that all his LED's had a forward voltage spec of 3.4 volts @ 20 ma. And he already soldered them together. So if I were him I wouldn't waste the time changing it now.

Last edited: Sep 3, 2012
15. ### CocaCola

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Apr 7, 2012
nb

Step back there, I stated that the batteries internal resistance is nil overall in limiting the circuit, it contributes little and I stand by that... You must keep all other factors the same if you are going to argue one factors effect on the circuit... You can't magically make the LEDs impedance zero when it's NOT! You intentionally twisted and falsified the data and facts to fit your argument ... My real world test and numbers posted previous clearly show that the LED impedance @ 4.5V is FAR from zero, and more in the 10-12 Ohm (give or take) range... You can't change this fact to fit your argument... No matter what 4.5V source you use the LEDs impedance will remain steady at that voltage, you can't just eliminate it to zero to fit an argument... The drive difference between LEDs with a perfect 4.5V power source and a 4.5V battery with low internal resistance is nil, just as I stated, the batteries internal resistance contributes little to the overall limiting of the circuit...

This is simply false, if the voltage is held at a constant level where the LEDs is passing a safe level of current, it will do no more damage than driving the LED by another limiting means at the same drive level...

Sounds like you put too much voltage into them, if you held the voltage at a 'safe level' for those amber LEDs no damage would have occurred to them... And yes the whites would have likely not lit up at that voltage level as they require more...

Because sometimes it's cost effective for a design and will still function within expected perimeters (see the led flashlights discussed in this thread) and/or it's necessary due to other design restraints or no resistors are needed due to the design limiting current/voltage by another means...

A good practice in laymen's terms, but blatantly false in real practice... There are many instances when you don't need resistors... In fact almost all LED driver circuits and ICs are constant current driven and thus require no resistors, once you set the current to a safe level for the LEDs in the design...

16. ### john monks

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Mar 9, 2012
CocaCola. Let me take one more crack at this.
Extreme#1 Taking your short circuit current of 7a you obtain a total battery impedance of 0.643 ohms. The LEDs running at 20ma you obtain a total LED current of 0.8A. So the battery voltage drops to 3.99 volts (4.5-0.8*0.643 ohms).
The LED impedance will be 0.732 ohms (3.99-3.40)/0.8
Extreme#2 You have a battery impedance of 0.643 ohms and assuming you have an LED impedance of zero. Your current LED current will be 42.8ma (4.5-3.4)/(0.643*40).

So at one extreme you have a battery impedance of 0.643ohms vs Led impedance of 0.732 ohms. Not so insignificant
On the other extreme you have a zero LED impedance.

The true impedance must lie somewhere in between.

The point is that the battery to LED impedance ratio is at a minimum of 0.878 and at most much higher.

Whenever I don't get my point across I always blame myself, not the other person.
Maybe I need to explain what extremes are in terms of mathematics. Extremes relate to the maximum and minimum possible values. This term is ubiguitous in calculus. Maybe I should have explained that I was using extremes to make my point. Maybe I don't explain my logic well enough. Maybe I assumed the other person understands more than they do. But usually I am willing to continue even for weeks if that is what it takes. And what is so great about the forum is that I get a chance to explain these thing and improve my articulation. And for that I say thanks.

Last edited: Sep 4, 2012
17. ### BobK

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Jan 5, 2010
Coca Cola an John Monks,

I think one of you is analyzing a circuit with 1 LED and the other one with 40 LEDs in parallel.

Bob

18. ### CocaCola

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Apr 7, 2012
I'm analyzing it with any number of LEDs, as should be clearly evidenced in my actual testing of several different LED numbers, I also clearly stated earlier in post #21

Where did you derive the 20mA current of the LEDs?

19. ### john monks

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Mar 9, 2012
The 20mA came from dbov22. This was one extreme where the significance of the battery impedance becomes clear. The other extreme is when considering the LED impedance at zero ohms.

Sorry it wasn't clear. We do not have the datasheet for this LED so I was calculating both extremes (the extrema). We only have the numbers dbov22 gave us (3.4V @ 20mA). And at one point even used a textbook formula (25mV / amperes) to estimate the LED impedance.

Let me know if there is further confusion. I will be happy to clear it up.

20. ### CocaCola

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Apr 7, 2012
20mA is not an extreme, it's just a single value on the curve, not an extreme at all in fact it's a suggest median value... Your math is only as good as the numbers you plug in, and I keep seeing you specifically pick numbers that work in your favor...

Try your math with 1.5mA @ 3.0V for example... Certainly a little closer to an 'extreme'...