FreeTrack Forum

Zaphod @ 13/05/2012 - 09h41 a dit:

dewey1, the need for resistors in parallel circuit is well established, the original question is about series circuit.

Sinister @ 07/05/2012 - 23h47 a dit:

For Example, 3 LEDs rated at 1.2v wired in series would have a max of 3.6v. I don't see why running this off of a 3v battery would be a problem.

This is the original question;

Would it be possible to connect 3 LEDs to a 3V battery without resistors? Would such a battery be enough to power the LEDs?
Read the title of the posting.

Steph @ 12/05/2012 - 21h29 a dit:

I took LEDs out of the computation because, as already said, in a DC environment, if the source voltage is equal to the LED forward voltage drop, no resistor is needed.

Call me Simson, but I don't understand this argumentation.  
Sorry, perhaps I missed something.

Since what we want to achieve is to match the forward voltage drop of the LED to the voltage of the source, what we have to take into account is only the behaviour of the circuit, minus the LED itself, regardless of its behaviour.

We would compute also the non-ohmic LED behaviour just in case V or I of the source wouldn't match the nominal values of the LED.

Regarding the current-voltage curve of the sfh485 1.5V let more than 100mA current flow.
But anyway, this is theory.

Following the graph, yes, but then we should only complain to Osram for its discrepancies.

Put a sfh485 on a 1.5 Alkaline and you never will measure more than 80mA in circuit. The internal resistance of battery prevents current rising.
So I agree with you, you can drive a LED on battery without a resistor.

Isn't an average AA Alkaline capable of generating around 700 mA? Moreover, internal resistance of a battery is inversely proportional to size, so a bigger 1.5 V Alkaline could provide a larger amount of current.

dewey1 @ 13/05/2012 - 14h19 a dit:

This is the original question;

Sartoris @ 07/05/2012 - 18h37 a dit:

Would it be possible to connect 3 LEDs to a 3V battery without resistors? Would such a battery be enough to power the LEDs?

Near-infrared LEDs have a forward voltage drop below 2.1 V and below 1.63 V the infrared ones. For instance, I own infrared LEDs with a nominal forward voltage drop of 1.2 V.

Since his source is of 3 V and he doesn't want or can't employ resistors, either he should use 3 V LEDs in parallel, which would have a wavelength unsuitable for FreeTrack, or put suitable LEDs in series, taking into account their cut-off voltage.

Therefore, we have been considering the series case the whole time. To avoid further complications and misunderstandings, we could now consider the behaviour of a hypothetical (I'm sorry if you hate this so much) circuit with just one LED and with the voltage of the source equal to the forward voltage drop of said LED.

Reached an agreement on this scenario, we could then proceed to our real (now you can breathe again) case.

Anyway, I beg your pardon if my own neuronal quantity does not match yours; I do what I can.

Gian92;

Do not take my reply the wrong way, it was not meant to be offensive.
There are just to many people that think they know Electronics in the DIY world.

Whether you realize it or not, what you are referring to is a constant current source. This is what Steph was referring to.
Please read this link and you will see that there are not any resistors for the LEDs in a series circuit.
The output voltage varies to maintain the desired current.
This allows for series connection of LEDs with different Vf that exist in a real world environment.

The only other method is to match the Vf of LEDs so that they are all the same and have a voltage source to match the combined total of the individual Vf.
Very impractical because now you need a variable voltage power source.

http://forum.free-track.net/index.php?showtopic=2666&page=1#14745

The internal resistance of cells is more dependent on the chemistry construction of the cell than its size. There is a slight difference in cell internal resistance as the physical size increases.

Good information here:

http://www.duracell.com/en-US/Global-Technical-Content-Library/Product-Data-Sheets.jspx?icn=AdLob/ProductDataSheets&cc=AdLob

dewey1 @ 13/05/2012 - 18h24 a dit:

The only other method is to match the Vf of LEDs so that they are all the same and have a voltage source to match the combined total of the individual Vf.
Very impractical because now you need a variable voltage power source.

Normally for FreeTrack you buy the same LEDs, so all the specifications will be the same.
Why would you need a variable voltage power source? For instance, I have rechargeable 1.2 V Nickel-Metal Hydride AA batteries which I use to power 1.2 V LEDs directly without any issues so far.

Thanks for the links.

This topic is a hoot. All I know is that I was able to solder 3 IR LEDs in series to  a battery pack with 3 1.2V NiMH batteries without resistors.

It works, they don't get hot, and I've been using them for hours at a time.

If this shortens the life of the LEDs (the ones that I ground down with the sanding disk of my dremel tool) to some appreciable value below their rated 50,000 hour life expectancy, then I shall un-tape them from my baseball cap and spend another $6 dollars to replace them.  LOL

Zaphod @ 13/05/2012 - 09h41 a dit:

dewey1, the need for resistors in parallel circuit is well established, the original question is about series circuit.

The diagram is not showing them in parallel as you think but rather what affect Vf of different voltages has on current.

Attached is an updated drawing with series circuits added. The point is to show that the LEDs have a Vf range of about 3 Volts. This is what is known as a tolerance.
Keep in mind that all manufactured things need a tolerance.

Gian92 wrote:
Normally for FreeTrack you buy the same LEDs, so all the specifications will be the same.

The spec sheet may be same but actual Vf will vary from one LED to another.
Do you not understand tolerances and the fact Vf of the SFH485P is 1.5 to <1.8 at 100mA?

In the diagram I show different Vf that could occur, for example with 3 LEDs with 1.50, 1.60 and 1.70 in series without a resistor to a 4.90 Volt supply.
Using the same supply with 3 "identical" LEDs with Vf of 1.50 and note what happens.

What I am explaining is that you can use LEDs without a resistor if the Vf of the LEDs is measured for each then add these up to get the desired supply voltage.
One hell of alot work for something as trivial as just adding a cheap resistor.
I am also trying to empasize worst case scenarios because that is how you should design/build circuits.

For those that are not using any resistor, it is just pure luck that the Vf are very close to each other or the power source is current limiting/voltage dropping off preventing any noticable damage.

Example:I use a CR2032 3 Volt battery to check my SFH485P LEDs. The only reason it does not damage the LED is because the battery can not supply enough current due to its internal resistance. If I had used a 2.4 V NiCd it would have damaged it.

Zaphod @ 13/05/2012 - 03h41 a dit:

dewey1, the need for resistors in parallel circuit is well established, the original question is about series circuit.

Sinister @ 07/05/2012 - 23h47 a dit:

For Example, 3 LEDs rated at 1.2v wired in series would have a max of 3.6v. I don't see why running this off of a 3v battery would be a problem.

Newbie here, but a lifetime of electronics. And there are different constants in a serial and parallel circuit.

In serial circuits, the current/amperage/I is constant through all components. If there are, say, 4 resistors or different values (5, 10, 15 and 20 ohms), and a power source of 6 VDC, Ohm's Law (E=IR) says that I=E/R. Resistance is 50 ohms total, and the current (I) is 120ma (.12A). This current is the same through all 4 resistors, and the voltage is divided unequally by each resistor.

In a parallel circuit, the voltage (E) is constant in each leg, and the current drop differs through each leg. With the same 4 resistors, each would be pulling the same 6 VDC, but the current would differ with the resistance (I=E/R), i.e. the 5 ohm would pull 1.2A, the 10 ohm would pull 600ma, etc.

So, if you drive 4 50 ohm LEDs (200 ohms total) from a 6 VDC supply (4x1.5 AA), then the current across every LED is 30ma. If the forward current in each LED is 100ma, the LEDs are being driven at ~33% of their rating. It would take the failure of 3 of the LEDs to over-drive the last one. So I cannot see the value of adding a current-limiting resistor to such a low-power circuit. Now, if I decided to drive my LEDs with a 9VDC battery, that might be different. Or a power supply that could fail.

gpbarth @ 24/12/2012 - 22h19 a dit:

So I cannot see the value of adding a current-limiting resistor to such a low-power circuit.

That's exactly the point. Moreover, at those levels of power the internal resistance of the battery plays a not negligible role.

Resistance of the solder joints can also be significant if the wires aren't in contact properly or at all and only the solder is the electrical conductor: a typical PbSn solder (63% lead, 37% tin) has ten times the resistivity of the copper, and it affects the overall resistance of the circuit as well.

In any case, my resistor-less tracking system is still working flawlessly.