How much resistance?

kuksul08

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Can anyone tell me how many ohms these resistors are?

IMG_1675LargeMedium.jpg


Side note... would it be worthwhile to upgrade the LED's in this flashlight? I think they are luxeon III's now. Maybe I could add a driver or something for 700mA. It's a 3D flashlight. Does the heatsinking look sufficient? It's basically just a chunk of aluminum.

IMG_1676LargeMedium.jpg
 

kuksul08

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I think it's green, orange, black, silver, gold. But I just wanted to check. That would mean each LED gets 146mA which isn't that much.

.3 ohm seems quite low to me
 

VegasF6

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Yah, but one of the resistors is soldered on backwards, so they actually cancel each other out.....

:crackup:

Well, I make myself laugh at least.
 

Mr Happy

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I don't think the green band is part of the resistance value. I agree with EasySt, it looks like orange-black-silver-gold, or 0.3 ohms +/-5%.
 

kuksul08

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VegasF6

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How does that thing work? Is it a simple on/off switch with both leds and both resistors in parallel?
 

Mr Happy

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4.5 volts for the power source (3 D batteries).
.3 Ohms series resistor

V=IR
I=4.5/.3
I=15amps
That calculation would work IF the resistor was the only resistance in the circuit, and IF the power source had no internal resistance of its own. But of course, you and I know that is not true, don't we? ;)

Here's a puzzle: what would be the current through the LEDs if we replaced the resistors with a plain piece of wire having no resistance at all?
 

kuksul08

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How does that thing work? Is it a simple on/off switch with both leds and both resistors in parallel?

It's an on-off switch. The positive end of the batteries go into the switch, then into the parallel resistors and then into each LED. The negative terminal of each LED has a copper connector which grounds onto the body of the flashlight which is connected to the negative terminal of the batteries. Really simple

That calculation would work IF the resistor was the only resistance in the circuit, and IF the power source had no internal resistance of its own. But of course, you and I know that is not true, don't we? ;)

Here's a puzzle: what would be the current through the LEDs if we replaced the resistors with a plain piece of wire having no resistance at all?

Well in my experience, the LEDs have very minimal resistance compared to an actual resistor. I forget what I measured, but it was a small percentage. The batteries don't have a huge internal resistance either.

If the resistors were just plain wires, too much current would flow and likely burn up the LEDs right? Once they burn out idk if they short circuit or open... am I missing something?
 

kuksul08

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Actually this makes me wonder... with an ideal battery with a piece of wire creating a short circuit, how much current flows through the wire? V=IR doesn't work, hmm. I've only taken one EE class :D
 

VegasF6

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The length and diameter of the wire of course make a difference. And even the shape somewhat. If you coil it up it seems to lessen the resistance. Well I aint no EE but I wouldn't worry about 2 LUX's in parallel with a decent heatsink if you are using alkaline batteries. I will be darned if I can figure the resistor values. I tried looking it up here but I am too dumb to understand it :)
Apparently it was some attempt to balance the current between the 2.

Personally I would check the voltage at each led (why I don't know) and then put a clamp meter between the resistor and the led to measure current. Lacking that, I would put a dc current meter inline. Yes, it does induce some more resistance, but it's better than nothing.

You didn't mention it, but you normally use alkaline D cells with it?
 

Mr Happy

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am I missing something?
Yes, you are missing the resistance inside the battery, and you are missing the resistance inside the LED.

A D size alkaline battery has an internal resistance of about 0.15 ohms, and an LED has a forward voltage drop somewhere in the range of 3 to 4 volts depending on the current. (LED emitters don't have resistance in the classical sense because the graph of voltage against current is not a straight line as it is with a true resistor.)

Actually this makes me wonder... with an ideal battery with a piece of wire creating a short circuit, how much current flows through the wire? V=IR doesn't work, hmm. I've only taken one EE class :D
Yes, V=IR works, as long as you take all resistances into account. For instance, if a D size alkaline battery has a voltage of 1.5 V and an internal resistance of 0.15 ohms, then the short circuit current will be about 10 A.

The length and diameter of the wire of course make a difference. And even the shape somewhat. If you coil it up it seems to lessen the resistance.
Actually, coiling the wire should increase the resistance. This is because resistance increases with increasing temperature, and a coiled wire gets hotter than a straight wire. (Which is why bulb filaments are coiled.)
 

Turak

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Can anyone tell me how many ohms these resistors are?

IMG_1675LargeMedium.jpg


Well, using the Standard EIA Color Code table, you would end up with the following.

Reading the bands as;

orange
black
silver
gold
blue (since there is no green band for 'temp coefficient' band)

Also reading the bands in the other direction does not work, since there is no black band for the 'tolerance' band.

That would be....

Resistance - .3 Ohm
Tolerance - plus or minus 5%
Temp. Coefficient - 25 ppm/'C


Here is a good web site resister claculator thant handles 5 color band resistors; http://www.okaphone.nl/calc/resistor.shtml
 

kuksul08

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Okay then. So what type of resistance does that voltage drop correspond to? If there is .3ohms from the resistor, .45 ohms resistance from the batteries, 4.5 total volts... that still means 6 amps.
 

2xTrinity

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Well in my experience, the LEDs have very minimal resistance compared to an actual resistor. I forget what I measured, but it was a small percentage. The batteries don't have a huge internal resistance either.
LED's don't have a fixed resistance, they are nonlinear devices. Current through them increases exponentially with the voltage applied to them. This means that their resistance changes dramatically depending on the circumstances. If you apply a low- or reverse- voltage to them, their resistance will be effectively infinite (zero current). If you apply a high voltage, their resistance will be effectively zero (extremely large current).

Don't think of LEDs as resistors at all. A more useful model is to assume that the LED will drop a certain fixed voltage, usually referred to as the Vf in your LED data sheet.

Let's say the Vf of your LED is 3.5V (a fairly typical value). Let's say battery internal resistance is 0.5V Expected current would then be:

I = V / R

V = Vbatt - Vdiode = 4.5 - 3.5 = 1
R = R + Rbatt = 0.3 + 0.5 = 0.7

I = 1/0.5 = 1.4A

This may sound high, but in all likelyhood, your alkalines voltage will drop and internal resistnace will rise as they drain, so this could easily settle to around ~0.7A for most of the runtime. It is also possible that the LEDs being used in this case are ones with higher Vf (say 3.8V), and that there is even more resistance elsewhere in the system (such as in the LED leads, wire traces etc), such that the total system resistance is more like 1 ohm:

I = (4.5-3.8) / (1) = 0.7 amp peak

To find out what the Vf of your LED is, simply measure the voltage drop across it while the LED is on sometime When I build direct-drive flaslhights, what I will actually do is hook up my LEDs to my bench power supply, then use the current limiter to set the current to my goal current for the project -- that way i can instantly see what the Vf of my LED is. Then I will know whether my choice of battery and resistor (or regulator) is suitable before installing anything.

If the resistors were just plain wires, too much current would flow and likely burn up the LEDs right? Once they burn out idk if they short circuit or open... am I missing something?
If running on Alkaline batteries, you might be able to still survive even with no extra resistance. Those have very high internal resistance. However, If you were to throw something like NiMH cells in your light (which have almost no internal resistance) you might cook the LEDs. Or you might be fine. Depends on the Vf of the LEDs:

Let's say your LED Vf is 3.8, and you drop in a few fresh NiMH Cells (1.4V hot off the charger):

I = (4.2 - 3.8) / 0.3 Ohms = 1.33 Amps. Not great for longevity, but the LEDs may live.

Now let's assume your LED Vf is actually 3.3:

I = ( 4.2 - 3.3 ) / 0.3 Ohms = 3 Amps. Cooked.
 
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VegasF6

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Actually, coiling the wire should increase the resistance. This is because resistance increases with increasing temperature, and a coiled wire gets hotter than a straight wire. (Which is why bulb filaments are coiled.)[/quote]

Thanks Mr. Happy, I knew it was one or the other :) And now I know why.
 

kuksul08

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LED's don't have a fixed resistance, they are nonlinear devices. Current through them increases exponentially with the voltage applied to them. This means that their resistance changes dramatically depending on the circumstances. If you apply a low- or reverse- voltage to them, their resistance will be effectively infinite (zero current). If you apply a high voltage, their resistance will be effectively zero (extremely large current).

Don't think of LEDs as resistors at all. A more useful model is to assume that the LED will drop a certain fixed voltage, usually referred to as the Vf in your LED data sheet.

Let's say the Vf of your LED is 3.5V (a fairly typical value). Let's say battery internal resistance is 0.5V Expected current would then be:

I = V / R

V = Vbatt - Vdiode = 4.5 - 3.5 = 1
R = R + Rbatt = 0.3 + 0.5 = 0.7

I = 1/0.5 = 1.4A

This may sound high, but in all likelyhood, your alkalines voltage will drop and internal resistnace will rise as they drain, so this could easily settle to around ~0.7A for most of the runtime. It is also possible that the LEDs being used in this case are ones with higher Vf (say 3.8V), and that there is even more resistance elsewhere in the system (such as in the LED leads, wire traces etc), such that the total system resistance is more like 1 ohm:

I = (4.5-3.8) / (1) = 0.7 amp peak

To find out what the Vf of your LED is, simply measure the voltage drop across it while the LED is on sometime When I build direct-drive flaslhights, what I will actually do is hook up my LEDs to my bench power supply, then use the current limiter to set the current to my goal current for the project -- that way i can instantly see what the Vf of my LED is. Then I will know whether my choice of battery and resistor (or regulator) is suitable before installing anything.

If running on Alkaline batteries, you might be able to still survive even with no extra resistance. Those have very high internal resistance. However, If you were to throw something like NiMH cells in your light (which have almost no internal resistance) you might cook the LEDs. Or you might be fine. Depends on the Vf of the LEDs:

Let's say your LED Vf is 3.8, and you drop in a few fresh NiMH Cells (1.4V hot off the charger):

I = (4.2 - 3.8) / 0.3 Ohms = 1.33 Amps. Not great for longevity, but the LEDs may live.

Now let's assume your LED Vf is actually 3.3:

I = ( 4.2 - 3.3 ) / 0.3 Ohms = 3 Amps. Cooked.

:twothumbs

I measured the voltage of each new battery as 1.621V, and voltage drop across the LEDs as 3.425V.


Is it possible to measure the internal resistance of a battery? I put the multimeter in current mode and touched the terminals of a battery and it shows 7.45A (not consistent...slowly dropping). That would imply an internal resistance of .22 Ohms?

Factoring all this into the equations:
(4.863-3.425)/(.3 + 3*.22) = 1.49A

Still doesnt really make sense, but I don't really know where I'm going with this anyway. Just more knowledge I guess. Thanks for the responses
 
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